1 This request for a preliminary ruling concerns the interpretation of the concept of a 'work', as referred to in Directive 2001/29/EC of the European Parliament and of the Council of 22 May 2001 on the harmonisation of certain aspects of copyright and related rights in the information society (OJ 2001 L 167, p. 10).
2 The request has been made in proceedings between Levola Hengelo BV ('Levola') and Smilde Foods BV ('Smilde') concerning an alleged infringement, by Smilde, of Levola's intellectual property rights relating to the taste of a food product.
'(1) The expression "literary and artistic works" shall include every production in the literary, scientific and artistic domain, whatever may be the mode or form of its expression, such as books, pamphlets and other writings; lectures, addresses, sermons and other works of the same nature; dramatic or dramatico-musical works; choreographic works and entertainments in dumb show; musical compositions with or without words; cinematographic works to which are assimilated works expressed by a process analogous to cinematography; works of drawing, painting, architecture, sculpture, engraving and lithography; photographic works to which are assimilated works expressed by a process analogous to photography; works of applied art; illustrations, maps, plans, sketches and three-dimensional works relative to geography, topography, architecture or science.
5 Under Article 9(1) of the Berne Convention, authors of literary and artistic works protected by that convention are to have the exclusive right of authorising the reproduction of these works, in any manner or form.
10 Article 5 of Directive 2001/29 sets out a series of exceptions to, and limitations on, the exclusive rights which Articles 2 to 4 of the directive confer on authors over their works.
14 'Heksenkaas' or 'Heks'nkaas'('Heksenkaas') is a spreadable dip containing cream cheese and fresh herbs, which was created by a Dutch retailer of vegetables and fresh produce in 2007. By an agreement concluded in 2011 and in return for remuneration linked to the turnover to be achieved by sales of Heksenkaas, its creator transferred his intellectual property rights over that product to Levola.
15 A patent for the method of manufacturing Heksenkaas was granted on 10 July 2012.
16 Since January 2014 Smilde has been manufacturing a product called 'Witte Wievenkaas' for a supermarket chain in the Netherlands.
17 Levola took the view that the production and sale of 'Witte Wievenkaas' infringed its copyright in the 'taste' of Heksenkaas and brought proceedings against Smilde before the Rechtbank Gelderland (Gelderland District Court, Netherlands).
18 After stating that, from its point of view, copyright in a taste refers to the 'overall impression on the sense of taste caused by the consumption of a food product, including the sensation in the mouth perceived through the sense of touch', Levola asked the Rechtbank Gelderland (Gelderland District Court) to rule (i) that the taste of Heksenkaas is its manufacturer's own intellectual creation and is therefore eligible for copyright protection as a work, within the meaning of Article 1 of the Copyright Law, and (ii) that the taste of the product manufactured by Smilde is a reproduction of that work. It also asked that court to issue a cease and desist order against Smilde in relation to all infringements of its copyright and, in particular, in relation to the production, purchase, sale, supply or other trade in the product known as 'Witte Wievenkaas'.
19 By judgment of 10 June 2015, the Rechtbank Gelderland (Gelderland District Court) held that it was not necessary to rule on whether the taste of Heksenkaas was protectable under copyright law, given that Levola's claims had, in any event, to be rejected since it had not indicated which elements, or combination of elements, of the taste of Heksenkaas gave it its unique, original character and personal stamp.
20 Levola appealed against that judgment before the referring court.
21 The latter considers that the key issue in the case before it is whether the taste of a food product may be eligible for copyright protection. It adds that the parties to the main proceedings have adopted diametrically opposed positions on this issue.
22 According to Levola, the taste of a food product may be classified as a work of literature, science or art that is eligible for copyright protection. Levola relies by analogy, inter alia, on the judgment of 16 June 2006 of the Hoge Raad der Nederlanden (Supreme Court of the Netherlands), Lancôme (NL:HR:2006:AU8940), in which that court accepted in principle the possibility of recognising copyright in the scent of a perfume.
23 Conversely, Smilde submits that the protection of tastes is not consistent with the copyright system, as the latter is intended purely for visual and auditory creations. Moreover, the instability of a food product and the subjective nature of the taste experience preclude the taste of a food product qualifying for copyright protection as a work. Smilde further submits that the exclusive rights of the author of a work of intellectual property and the restrictions to which those rights are subject are, in practical terms, inapplicable in the case of tastes.
24 The referring court notes that the Cour de cassation (Court of Cassation, France) has categorically rejected the possibility of granting copyright protection to a scent, in particular in its judgment of 10 December 2013 (FR:CCASS:2013:CO01205). There is therefore divergence in the case-law of the national supreme courts of the European Union when it comes to the question –– which is similar to that raised in the case in the main proceedings –– as to whether a scent may be protected by copyright.
(b) Is copyright protection precluded by the fact that the expression "literary and artistic works" in Article 2(1) of the Berne Convention, which is binding on all the Member States of the European Union, includes "every production in the literary, scientific and artistic domain, whatever may be the mode or form of its expression", but that the examples cited in that provision relate only to creations which can be perceived by sight and/or by hearing?
(c) Does the (possible) instability of a food product and/or the subjective nature of the taste experience preclude the taste of a food product being eligible for copyright protection?
(d) Does the system of exclusive rights and limitations, as governed by Articles 2 to 5 of Directive [2001/29], preclude the copyright protection of the taste of a food product?
(a) What are the requirements for the copyright protection of the taste of a food product?
(b) Is the copyright protection of a taste based solely on the taste as such or (also) on the recipe of the food product?
(c) What evidence should a party who, in infringement proceedings, claims to have created a copyright-protected taste of a food product, put forward? Is it sufficient for that party to present the food product involved in the proceedings to the court so that the court, by tasting and smelling, can form its own opinion as to whether the taste of the food product meets the requirements for copyright protection? Or should the applicant (also) provide a description of the creative choices involved in the taste composition and/or the recipe on the basis of which the taste can be considered to be the author's own intellectual creation?
26 Smilde claims that the present request for a preliminary ruling is inadmissible because the action in the main proceedings should, in any event, be dismissed. It argues that Levola has not identified the elements of Heksenkaas which allegedly make it its author's own intellectual creation.
27 It must be recalled in that regard that it is solely for the national court before which the dispute has been brought, and which must assume responsibility for the subsequent judicial decision, to determine in the light of the particular circumstances of the case both the need for a preliminary ruling in order to enable it to deliver judgment and the relevance of the questions which it submits to the Court. Consequently, where the questions submitted concern the interpretation of EU law, the Court is, in principle, bound to give a ruling (judgments of 10 March 2009, Hartlauer, C‑169/07, EU:C:2009:141, paragraph 24, and of 1 July 2010, Sbarigia, C‑393/08, EU:C:2010:388, paragraph 19).
28 Indeed, it is settled case-law that questions on the interpretation of EU law referred by a national court in the factual and legislative context which that court is responsible for defining, and the accuracy of which is not a matter for the Court to determine, enjoy a presumption of relevance. The Court may refuse to rule on a question referred by a national court only where it is quite obvious that the interpretation of EU law that is sought bears no relation to the actual facts of the main action or its purpose, where the problem is hypothetical, or where the Court does not have before it the factual or legal material necessary to give a useful answer to the questions submitted to it (judgments of 24 June 2008, Commune de Mesquer, C‑188/07, EU:C:2008:359, paragraph 30 and the case-law cited, and of 21 May 2015, Verder LabTec, C‑657/13, EU:C:2015:331, paragraph 29).
29 In the light of the information provided by the referring court, it cannot be held that the questions raised bear no relation to the actual facts of the main action or its purpose or concern a hypothetical problem. A different conclusion cannot be reached merely because the lower court, whose decision has been challenged before the referring court, took the view –– unlike the referring court –– that it was in a position to rule on the dispute before it without deciding upon the preliminary issue of whether the taste of a food product may be eligible for copyright protection.
30 Moreover, the referring court has, in accordance with Article 94 of the Rules of Procedure, provided the Court with the factual and legal material necessary to enable it to answer the questions raised.
31 Accordingly, the questions referred are admissible.
32 By its first question, the referring court asks, in essence, whether Directive 2001/29 must be interpreted as precluding (i) the taste of a food product from being protected by copyright under that directive and (ii) national legislation from being interpreted in such a way that it grants copyright protection to such a taste.
33 In that regard, Articles 2 to 4 of Directive 2001/29 state that the Member States are to provide for a set of exclusive rights relating, in the case of authors, to their 'works', while Article 5 sets out a series of exceptions and limitations to those rights. The directive makes no express reference to the laws of the Member States for the purpose of determining the meaning and scope of the concept of a 'work'. Accordingly, in view of the need for a uniform application of EU law and the principle of equality, that concept must normally be given an autonomous and uniform interpretation throughout the European Union (see, to that effect, judgments of 16 July 2009, Infopaq International, C‑5/08, EU:C:2009:465, paragraphs 27 and 28, and of 3 September 2014, Deckmyn and Vrijheidsfonds, C‑201/13, EU:C:2014:2132, paragraphs14 and 15).
34 It follows that the taste of a food product can be protected by copyright under Directive 2001/29 only if such a taste can be classified as a 'work' within the meaning of the directive (see, by analogy, judgment of 16 July 2009, Infopaq International, C‑5/08, EU:C:2009:465, paragraph 29 and the case-law cited).
35 In that regard, two cumulative conditions must be satisfied for subject matter to be classified as a 'work' within the meaning of Directive 2001/29.
36 First, the subject matter concerned must be original in the sense that it is the author's own intellectual creation (judgment of 4 October 2011, Football Association Premier League and Others, C‑403/08 and C‑429/08, EU:C:2011:631, paragraph 97 and the case-law cited).
37 Secondly, only something which is the expression of the author's own intellectual creation may be classified as a 'work' within the meaning of Directive 2001/29 (see, to that effect, judgments of 16 July 2009, Infopaq International, C‑5/08, EU:C:2009:465, paragraph 39, and of 4 October 2011, Football Association Premier League and Others, C‑403/08 and C‑429/08, EU:C:2011:631, paragraph 159).
38 It should be recalled in that regard that although the European Union is not a party to the Berne Convention, it is nevertheless obliged, under Article 1(4) of the WIPO Copyright Treaty, to which it is a party and which Directive 2001/29 is intended to implement, to comply with Articles 1 to 21 of the Berne Convention (see, to that effect, judgments of 9 February 2012, Luksan, C‑277/10, EU:C:2012:65, paragraph 59 and the case-law cited, and of 26 April 2012, DR and TV2 Danmark, C‑510/10, EU:C:2012:244, paragraph 29).
39 Under Article 2(1) of the Berne Convention, literary and artistic works include every production in the literary, scientific and artistic domain, whatever the mode or form of its expression may be. Moreover, in accordance with Article 2 of the WIPO Copyright Treaty and Article 9(2) of the Agreement on Trade-Related Aspects of Intellectual Property Rights, which is mentioned in paragraph 6 of this judgment and which also forms part of the EU legal order (see, to that effect, judgment of 15 March 2012, SCF, C‑135/10, EU:C:2012:140, paragraphs 39 and 40), copyright protection may be granted to expressions, but not to ideas, procedures, methods of operation or mathematical concepts as such (see, to that effect, judgment of 2 May 2012, SAS Institute, C‑406/10, EU:C:2012:259, paragraph 33).
40 Accordingly, for there to be a 'work' as referred to in Directive 2001/29, the subject matter protected by copyright must be expressed in a manner which makes it identifiable with sufficient precision and objectivity, even though that expression is not necessarily in permanent form.
41 That is because, first, the authorities responsible for ensuring that the exclusive rights inherent in copyright are protected must be able to identify, clearly and precisely, the subject matter so protected. The same is true for individuals, in particular economic operators, who must be able to identify, clearly and precisely, what is the subject matter of protection which third parties, especially competitors, enjoy. Secondly, the need to ensure that there is no element of subjectivity –– given that it is detrimental to legal certainty –– in the process of identifying the protected subject matter means that the latter must be capable of being expressed in a precise and objective manner.
42 The taste of a food product cannot, however, be pinned down with precision and objectivity. Unlike, for example, a literary, pictorial, cinematographic or musical work, which is a precise and objective form of expression, the taste of a food product will be identified essentially on the basis of taste sensations and experiences, which are subjective and variable since they depend, inter alia, on factors particular to the person tasting the product concerned, such as age, food preferences and consumption habits, as well as on the environment or context in which the product is consumed.
43 Moreover, it is not possible in the current state of scientific development to achieve by technical means a precise and objective identification of the taste of a food product which enables it to be distinguished from the taste of other products of the same kind.
44 It must therefore be concluded, on the basis of all of the foregoing considerations, that the taste of a food product cannot be classified as a 'work' within the meaning of Directive 2001/29.
45 In view of the requirement, referred to in paragraph 33 of this judgment, for a uniform interpretation of the concept of a 'work' throughout the European Union, it must also be concluded that Directive 2001/29 prevents national legislation from being interpreted in such a way that it grants copyright protection to the taste of a food product.
46 Accordingly, the answer to the first question is that Directive 2001/29 must be interpreted as precluding (i) the taste of a food product from being protected by copyright under that directive and (ii) national legislation from being interpreted in such a way that it grants copyright protection to such a taste.
47 In the light of the answer to the first question, there is no need to reply to the second question.


The Tobacco Cessation Health Coach provides assessment, counseling, education and behavioral intervention to tobacco users. This position directly assists tobacco users by using motivational interviewing techniques in deciding whether or not to make a quit attempt, obtaining and using cessation medicine and staying abstinent from tobacco. Counseling is conducted over the phone and/or through the internet with participants.
Educates and coaches a diverse population of participants in quitting tobacco by following motivational interviewing techniques. Explains Health Initiatives programs to participants, builds rapport, and engages callers following guidelines, protocols and scripting.
Completes an intake questionnaire with participants to verify program eligibility. Documents participant data in compliance with HIPAA requirements, departmental standards and specific client guidelines.
Maintains friendly, confident, open, and effective communication with participants to identify program eligibility and inspire continued participation. Assists participants to help determine participant's goals, readiness to change, and supports the behavior change.
Engages in a process of continuous training, learning, and professional growth including QA scores and feedback to ensure delivery of the latest evidence-based practices, content knowledge and skills.
Maintains call center metrics and individual performance standards.
Bachelor's Degree required. A degree in Psychology, Social Work or other Human Services field preferred.
A minimum of two (2) years of experience as a health coach preferred.


RideOnTrack and Zenitel agreed on a partnership co-operation.
RideOnTrack successfully integrated the Zenitel intercom solution with the RideOnTrack Operational Telecommunication product line. Customers that use the RideOnTrack Sip Call Flow Manager, e.g. for dispatching and call recording, can now also seemlessly connect the Zenitel intercom products. The dispatcher will be able to connect to the Zenitel intercom devices, and likewise, the Zenitel intercom devices can be reached by the dispatchers. This integration will further facilitate the communication between all operational communication systems in a customer network and there will be no need for an extra intercom switch (IpPbx) or separate voice recording. It is RideOnTrack's aim to improve the operational communication capabilities of their customers while reducing complexity and cost. Both companies will work together in the regional express train project, a railway development in Dakar, Senegal, for the interconnection between Dakar and the AIBD airport.
For some of the (hardware) product developments, RideOnTrack selected 'strategical' partners. For our gateway products we work closely together with the Canadian company, Sangoma Technology Inc.. Both R&D departments cooperate intensively to 'tune' the products to t the typical operational telephony requirements. We use several of the Sangoma PCIe plug in boards for our server based solutions.


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Chance The Rapper: "I feel like... some people want me to feel ashamed."
In a Twitter thread on Monday, Chance opened up about his feelings a week on from the release of The Big Day.
Chance The Rapper speaks onstage at WE Day California at The Forum on April 25, 2019 in Inglewood, California. Emma McIntyre/Getty
Chance The Rapper has posted a lengthy Twitter thread in which he appears to argue back against his belief that people want him to feel "ashamed."
Chance opened the thread by claiming that he's "getting this crazy feeling that people want me to kill myself," adding: "That's not my vibe at all, but i feel the push." He later went on to say that "some people" were trying to shame him, though he didn't specify who or in what way.
Read Next: Listen to Chance The Rapper's holiday-themed "The Return"
I'm getting this crazy feeling that people want me to kill myself 😢
— The Big Day out now (@chancetherapper) August 5, 2019
And I feel like even more than that, some people want me to feel ashamed.
Shame is heavy yo. And feeling shame for something that you were once prideful about is super heavy.
A running theme of the discourse around The Big Day has been Chance's repeated declarations of love for his wife that crop up regularly across an album inspired by their wedding day. In his final Tweets, Chance said this feeling remains more true than ever.
That I LOVE my wife
I just want to reiterate that I don't want to kill myself, nor am I ashamed of loving my wife. I think I just wanted to say out loud that I see the vibes 👀👀
The Big Day narrowly missed out on the No.1 spot on the Billboard 200. The album was outsold by Christian rapper NF's latest record, Perception. He will tour the album across North America later this year.
chance the rapper, Hip-Hop

redpajama_set_name: RedPajamaCommonCrawl


I'm a regular research conference goer. For the most part the shows I've attended are pretty vanilla and go off as planned. A good speaker or two, some networking, a forced promotion here and there, sponsors on the hunt for business cards (guilty as charged), and mediocre lunch buffets. I learn something useful, meet and re-connect with some great people, and head off to the next one.
I was told before the Qualtrics Insight Summit that this event was different. Skeptical, I packed my bags (and skis) for Salt Lake. A soon as things kicked off, I realized this was not my father's stodgy research conference. And it wasn't the great swag that made me a believer. This conference had energy and passion. The data and insights talk track can at times be a sleepy one. But our profession, what we do, the value we have, and the hurdles we face came alive at the show through a mixture of unconventional presentation formats, entertainment, and engaging speakers.
Whenever myself or one of my colleagues gets back from a conference, two questions always ensue: How was it? What did you learn? In a word, it was refreshing. And, as you will see, the learnings were many and varied.
1) We all have our work issues, but perspective is a powerful asset. Having a bad day at the office took on a whole new meaning as I listened to Captain Phillips' inspiring story of survival and triumph as he battled Somali Pirates. His tale of adversity came on the heels of a moving day-one presentation, interview session, and campaign launch centered around eradicating Cancer. It's nice to be reminded that research problems as well as issues with co-workers and clients aren't nearly as big when viewed through the right lens. And, if Captain Phillips can beat back pirates and individuals can fight Cancer so valiantly, don't we all have what it takes to overcome our own sources of adversity?
2) Saying what you really think can make all the difference. Author Kim Scott introduced attendees to the idea of Radical Candor and how it can make or break the boss/employee relationship. Saying exactly what you think is not however as simple as blurting out the truth in any way, shape, or form. Radical Candor needs to be accompanied by giving a damn and guiding team members, not giving them orders. Without those ideas working in unison, you risk being just an a-hole. But, when you care and challenge people directly, magic can happen.
3) Pregnant turkeys can teach the value of getting behind the data. Stephen Dubner, author of Freakonomics, Superfeakonomics, and Think Like A Freak wore many hats at the show, including game show host. However, his talk on getting behind data and approaching problems differently stuck with me. In particular, his example which centered around this fun fact: most turkeys in the US are bred through artificial insemination. That data point piqued Dubner's interest as he pondered what makes Turkeys different from similar farm animals that procreate the old fashioned way.
So he did some digging. As it happens, most US consumers prefer turkey breast meat. The fact that turkeys now have been bred to have large breasts renders them completely unable to procreate on their own (a bump into each other issue) – hence, the need for artificial insemination. Without the story behind that fact, it's not nearly as sticky, relevant or clear. We can't take things for face value, and should always keep digging until we know why.
4) Effecting change takes the Rider & the Elephant. Chip Heath, co-author of Switch, Made to Stick, and Decisive discussed how to create change – not an easy proposition. In his talk and in Switch, he cites University of Virginia psychologist Jonathan Haidt who likens our emotional side to an Elephant and our rational side to its Rider. The Rider holds the reins and seems to be the leader. But the Rider is so small relative to the Elephant that control is not a given. Thus appealing to both Rider and Elephant is key to making an impact and effecting change. In his talk, Chip gave some sterling examples of how that dual approach has played out and the unbelievable outcomes that result.
5) Great talent can beat back father time. Attendees were treated to a private concert featuring Steven Tyler and The Loving Mary Band. Tyler, of Aerosmith fame, is 67 years old. And whether you love his music, hate it or are just plain indifferent, what's indisputable is that he hasn't lost his touch. He can still sing and perform with the best of them. Inspiring and entertaining even if that's not your thing.
6) The Insight Suite is pretty sweet. Move over research suite, the insight suite is here. Yes, Qualtrics released its new platform and it's got all types of new features and functions to aide people in their market research, customer experience, and employee insight crusades. The functionality that really struck a chord with me is the ability to launch surveys right within an email, send SMS survey invites, and employ new coding features that help classify text as it comes in. Overall, a really nice next step for them and their customers.
7) It's Barnaby's world, we're all just living in it. Barnaby, the office dog is, and will be for the foreseeable future, the star of the show. Apologies to Steven Tyler, but Barnaby was the biggest rock star in the hotel.
See you next year in Salt Lake.


text:
Guignardia citricarpa är en svampart som beskrevs av Kiely 1948. Guignardia citricarpa ingår i släktet Guignardia och familjen Botryosphaeriaceae.   Inga underarter finns listade i Catalogue of Life.

Källor

Bildgalleri

Sporsäcksvampar
citricarpa

redpajama_set_name: RedPajamaWikipedia


text:
"Belated Fantasy Accolade Encounter"
After the Oscars, during which Birdman was selected "Best Screenplay", "Best Director" and "Best Picture" and Boyhood captured a single Oscar for "Best Supporting Actress", I imagined making my way over to Boyhood's writer-director Richard Linklater during some post-Oscars "after-party", tapping him gently on the shoulder, and saying,
"Birdman is a winner. But Boyhood is a classic."
I meant every word of that fake interaction. I am nothing if not sincere in my fabricated illusions.
(Note: You know the interaction is fake because I would never be invited to a post-Oscars "after-party", nor would I ever tap a complete stranger on the shoulder, either gently or otherwise.)
I had seen Birdman in the theater. It made me, as I wrote earlier, extremely uncomfortable – a clichéd story of show business redemption gussied up with impressive camerawork and hyperventilated acting. (That's a little facile, but so, to a substantial degree, is the movie, so it fits. Which, thinking it over, is furtherly facile. I just think I'll move on.)
I watched Boyhood in my bedroom. (Owing to my consummate mastery of the DVD-playing apparatus. I never tire of bragging about my technological advancements. iPhone-5 – you're next! Even though while amassing the courage to tackle it, I have already fallen one iPhone behind.)
One evening, because Dr. M was hosting a psychoanalytic event in our living room, I was summarily exiled upstairs. Boyhood would be required to bear the brunt of my grumpy disposition, as I did not want to be exiled upstairs. Who wants to be exiled anywhere? It makes you feel sorry for Napoleon. Poor little Emperor got exiled twice!
Also, Boyhood was reputedly two hours and forty-five minutes long. That was my evening's agenda – being penned up in my bedroom, watching an overlong movie I may possibly not enjoy.
And I didn't enjoy it at first. As you are probably aware, Boyhood, the story of a family highlighting the younger, male sibling, was filmed piecemeal over a twelve-year period, allowing the actors to age naturally while continuing to play the same roles. (So there was no six year-old "Mason Evans Jr." played by one actor and an eighteen year-old "Mason Evans Jr." played by a different actor because who would believe a six year-old playing an eighteen year-old? Or vice versa. This way, it was actor Ellar Coltrane playing "Mason Evans Jr." the whole time. And all the other actors playing the same characters the whole time as well. Because you can't do just one.)
I don't know why I didn't enjoy Boyhood at first. Maybe it was because I had been forced into watching it and I am a vindictive old coot. Maybe it was my unfamiliarity with the family, who lived in Texas and none of them was Jewish.
No matter. It short time, the movie grew on me, and by an hour or so into it, I was hooked. My favorite moment? – talk about unfamiliarity – Mason Jr., having turned fifteen receives two birthday presents from his divorced Dad's new wife's parents – a personalized Bible and a vintage shotgun.
The thing is, these items were presented with such generosity and kindness that a non-shotgun-shooting Jewish man cooped up in his bedroom was viscerally affected by the gesture. Who'd have thought that a scene bestowing a gun and a Bible on a adolescent boy who had little enthusiasm for either would be so unexpectedly moving?
Why was it moving? There was an identifiable humanity shimmering right through it. They may have been misguided, perhaps, but these people didn't have to give that kid anything. And instead, they delivered from their hearts.
The entire movie – Boyhood often reminding me of The Graduate for its ability to accurately encapsulate a cultural moment – sparkled with meaningful interactions and reverberating surprises, like the disappointed-in-life patriarch turning out to be an excellent father.
Conditioned to expecting cinematic hackery, I kept anticipating, "Oh, here's where she announces to the family that she has cancer" or "Here's where the unsophisticated country boy succumbs to hard drugs", I was instead relieved – nay, delighted – to discover that Boyhood eschews hackneyed pyrotechnics in favor of chronicling the mundane realities, a choice which for me is unceasingly rewarding.
Opting for the "every day moment" turns out to be a deliberate strategy. Allow me to excerpt from the recent Writers' Guild "Written By" magazine, in which writer Lisa Rosen profiles Linklater, and his idiosyncratic storytelling process:
"The drama feels completely lived in. Things don't escalate the way you expect in a film because the usual plot twists don't apply. Because there's no plot. 'Somewhere along the way it hit me, I have dumped plot completely in favor of just character and story,' he {Linklater} says. 'So many movies have a structure that's built around satisfying plot and leave no room, or very little, for these real moments. You hear all the time, What are the stakes, what's the payoff? That's all artificial.
'When you're going for the rhythms of real life,' he asks, 'are you sitting right now thinking, What are the stakes in this thing I'm trying to do? Naw, it's something I'm compelled to do. I thought maybe I could do a whole movie without a plot. That doesn't mean it doesn't make sense. It's plot that's fake. What's real is the way life unfolds, the way time moves, the way things don't always pay off in big ways. Because they don't. Your life doesn't have a plot. It has character and story.'"
No formal plotline, yet it remains compelling to the end. Not all movies have to be like that. But I am delighted that some of them are. When courageous filmmakers like Richard Linklater can abandon formula and turn people's ordinary existence into memorable entertainment…
For me, at least,
Those movies are classics.
"Two Historians Tackling A Significant Issue"
Hitler has refused to retreat from Poland and an international conflict is about to begin. Not saying this was the most important issue of the day, but recently released documents reveal that two eminent historians were instructed to meet in order to determine what to name it.
We are fortunate to have the "minutes" of that meeting. And here, for your education and enlightenment, they are.
Two superannuated historians, Cosgrove and Bumbershoot, assemble to discuss an issue of utmost urgency and significance.
Cosgrove: I cannot for the life of me see why they selected us for this assignment. Can you?
Bumbershoot: The important historian said "No."
C: "You're probably right. Sparing their reputations due to faulty decision-making."
B: "Our decision today will be preserved for the ages, you know. Muck it up and we shall be laughingstocks for eternity."
"I never quite thought of it that way. 'Laughingstocks for eternity.' Ah well. At least we'll be remembered."
"It's difficult to believe, isn't it? It seems like just yesterday they were naming the last war."
"They did a damn fine job of it. 'The Great War'."
"It suits it to a 'T." Because it was in every regard…"
"A great war."
"The greatest. Howitzers. Dogfights in the air. Mustard gas. 'Waterloo, by comparison?' – a Sunday School picnic! " And yet, here we are, about to undo their magnificent labeling."
"We are at the dawn of a devastating conflict. Better technology. A more insidious opponent."
"Worse than the Kaiser?"
"Significantly worse. The previous war's labeling is about to become obsolete. I mean, what are the chances of the upcoming encounter being less great than the last one?"
"I see what you mean. We are unlikely to go backwards. (AFTER A BEAT) We could call it 'The Greater War'."
"And breed competition between the respective participants? (PROVOCATIVELY SING-SONGY) 'Our war was greater than your war.' How insensitively unfair!"
"I was just trying to preserve the previous accomplishment. But I suppose we shall have to dig deeper. You know, Cosgrove…"
"You're Cosgrove."
"Am I? Oh yes, I suppose I am – sorry. You know, Bumbershoot, historically this upcoming war is to a great degree a consequence of the Treaty of Versailles which brought an end to the last war, making this war arguably a continuation of the last war. Why then not label the upcoming war… "B."
"B?"
"Do you follow my reasoning? If w label the original war 'A', the continuation thereof would then inevitably be called 'B'."
"You want to call a war 'B'?"
"Signifying a continuation of 'A.'"
"It doesn't sound like a continuation. It sounds, first, ridiculously inconsequential, considering the death toll. Plus, as with your earlier suggestion…
"Which you most sensibly shot down…"
"Thank you… it, once again, makes the original war seem comparatively less important. 'What war did you fight in?' 'B.' 'I fought in "A." But I'm sure "B" was fine too."
"I see what you mean. Historically correct, perhaps. But a psychological faux pas."
"Let's get away from the comparatives. I believe that's what's catching us up."
"All right then. What if we renamed the previous war "Yellow", and label the war we are about to embark upon "Red"?
"You want to name wars after colours?"
"Yellow is comparatively no more important than red. So we cannot be called on the carpet for prioritizing."
"Well why then restrict the possibilities to colours? Why not call one war 'Teacup' and the other war 'Chest of Drawers.' Or one war 'Omelet' and the other war, 'Suet Pudding.' What about actual names! No name is superior to any other name. We can name the formerly-labeled 'Great War' 'Harry' and the impending conflict 'Bartholomew.'"
"My name's Bartholomew. I don't a war named after me."
"Nobody does. A 'war name' must be epic. Not colours. Not furniture. After, mistakenly it turns out, naming the last war 'The Great War', we now need to come up with something gigantic!"
"How about 'Gigantic'?"
"'The Gigantic War'?"
"It's not a direct comparative. The 'Great' in 'The Great War' could refer to its 'moral significance' while 'gigantic' for this one could refer to its sheer magnitude. I like it, don't you?"
"Keep thinking."
THE ROOM FALLS INTO THUNDEROUS SILENCE. FOR THE NEXT SIX HOURS. YOU CAN HEAR A SUSPENDED WALL CLOCK TICKING, BUT THAT'S ALL.
IT IS NOW MIDNIGHT. A MAINTENANCE MAN ENTERS, SURPRISED THAT THE HISTORIANS ARE STILL AT WORK.
"Sorry, gents. I could come back later."
"No, no. You are interrupting nothing… unfortunately. Do what you must."
"I'll just empty these ashtrays and be on my way. (AS HE PROCEEDS) Terrible times, these is. I mean, who would have thought that in the span of just a single generation, we could have one big war, and then Number Two."
TWO LIGHTBULBS SIMULTANEOUSLY COME TO LIFE.
"'War One' and 'War Two'!"
"Labeled exclusively in order of chronology!"
"I think we've got something, Pendrake."
"You're Pendrake."
"Come to think of it, I believe neither of us is Pendrake."
"I shall leave you two to your business. And apologies again for the inconvenience."
THE MAINTENANCE MAN EXITS.
"'War One' and 'War Two.' (AFTER A BEAT) Something's missing."
(DESPERATELY WANTING TO GO HOME) "Really?"
"It's just 'war counting.' One – two. I don't know, it needs some… pizzazz. What in the world could we add?"
Ultimately hearing themselves, they appended the word "World." And from then on, the respective conflicts were known forever as World War One and World War Two, the superannuated not only successfully completing their assignment, but creating the template for all the international wars yet to come.
We, of course, hope that there aren't any. But if there are, at least there will not be a meeting to name them.
Another Time, Maybe: The Explosive Brouhaha Over Naming The Old And New Testament.
"The Anointed Ones - A Follow-Up"
As recently promised…
The top athletes, entertainment participants at the highest level, there is something remarkable about them, not to overstate as I shall provide further examples to diminish their exclusivity.
These people, though not special, if they are truthful, feel privately special to themselves. (And I suppose to their fans, though I know very little about that. I did, however, meet Jimmy Stewart once and felt detectably weak-kneed, and I followed Henry Fonda into a Baskin-Robbins just to hear him say, "I want Cheeocolate.")
You had a dream, and you made it. You have beaten enormous odds, gaining entry into "The Club", becoming members of a select and elite minority – not the kind that is unable get a cab, the kind where a nice car (and a driver in a quality suit) seem always conveniently to be waiting for you.
But it's not just the perks that feel wonderful, though you, hopefully gratefully, take comfortable advantage of them all. It is the irrefutable feeling of accomplishment.
"Look where I am!"
the achievers rarely reticent about mythologizing their destinations. Baseball honors my profession by nicknaming the Major League level "The Show." And I am not aware of chartered accountancy ever calling itself "The Business." (Though they might, and keep it quietly to themselves.)
Personal Experience Of Exaltedness: I was standing in the lobby at the Emmys and a recognized comedian aware of my "elevated status" is trying desperately to make me laugh. I felt uncomfortable, but also, you know, "Look what he's doing!", imagining, "I must really be somebody!" To my immeasurable shock and surprise – and also secret satisfaction – I was the high school "princess" the boys were eagerly trying to impress.
You have arrived at your destination and you are enjoying the benefits (here's where I throw in paralleling examples in other arenas) – the CEO's corner office, the principal's office (as the principal, not a student miscreant sent in to see the principal, though if you're a gang leader, you are in a similar situation), the United States Senate (or any other political office without term limits), a prestigious law firm's founder, the chief surgeon in the hospital, the head coach or manager of any organized team.
… and a list of other positions, too troublesome to accumulate.
You are "Top Dog" in your respective profession. (Or an accredited includee. "Top Dog" merely accentuates the condition.) You have a "team jacket" to prove it.
The next step – and it is never easy –
Is to stay there.
You do everything it takes. You work hard, upgrading your performance, boasting a consistently winning record.
And they allow you to stick around. Patting you on the back. Regularly sweetening your compensation package.
But at some point, and it's surprisingly soon…
You hear the proverbial "footsteps." Somebody is gaining on you, hungry to displace you, knock you permanently off of your perch. The transition is breathtaking. In the seeming blinking of an eye, you go from "Talented Rookie" to "It's time for a change."
You are not ready to go.
And it goes without saying why.
You step down, hang 'em up, or are intentionally moved aside…
And you are no longer an "Anointed One."
With all that that banishment entails. (Including being relieved of your jacket.)
In baseball, they say, "They're going to have to tear off my uniform." I can imagine a corporate counterpart releasing with agonizing reluctance the key to the Executive Fitness Center.
Due to its tangible and symbolic meaning to you, you are unwilling to give it all up, doing whatever it takes to tenaciously stay in the game.
I have witnessed this phenomenon. I have experienced it myself.
Personal Observation: I have watched a loftily regarded television writer stooping to easy laughs and transparently suggestive "low road" pandering, struggling desperately to hold on. (It didn't work.)
Personal Experience: Sparing both you and myself the embarrassing details, I have seen myself subverting my integrity to prolong my career. (And it did work. Temporarily.)
During a recent mealtime conversation, I was asked, concerning the activity I had aspired to, had participated in and was required to relinquish, "Do you miss it?" I turned to a retired physician in the group and I said, "You know." His look of longing and regret gave indication that he did.
I am unsure I can articulate that feeling. And if you have been there, I don't have to.
There is little better than accomplishing your dream.
And little worse than surrendering it.
"A Simple Point Made With Music And Prose"
I have been suffering from a cold for the past few weeks. Is it a virus, or a bacterial infection?
I was invited to a hockey game by a friend who's a financial adviser, and as we drove to the Staples Center, the man explained to me how bonds work, and how their value fluctuates with interest rates. Or is it inflation?
"Do those paragraphs fit together?"
Indeed they do, blue-highlighted, gender-unspecified "Italics Person." And now, this:
"They sing of Yancey Derringer
On every danger trail
On riverboat, in manor house
And now and then in jail.
They say that Yancey Derringer had ruffles at his wrists
Brocade and silver buckles
And iron in his fists."
"Okay, now, I'm really confused."
Me too. But in an entirely different direction.
The preceding assembled factoids land on opposite sides – the first two on one side, the third example on the other – of the Earl Pomerantz unilaterally uncrossable "Retention Line."
How many times during interludes of bronchial discomfort has a friend generously distinguished on my behalf a virus from a bacterial infection and I still cannot for the life of me remember the difference?
How many times over the twenty-plus years that I have known him has my friend/slash/financial adviser patiently explained to me how bonds work and I am still unable to recall if the determining factor is interest rates or inflation?
And yet:
"Ringo, Johnny Ringo,
His fears were never shown
The fastest gun in all the West
The quickest ever known."
(An equally obscure second example, to avoid boredom through repetition.)
The answer is easy, you say.
"You are interested in westerns so you remember their theme songs. The other stuff does not interest you, leaving you unmotivated to assimilate the information.
Oh, really? It's that simple?
I respectfully – though without scientific verification beyond my own Guinea Piggal participation – beg to differ.
A GENEROUS FRIEND: "Would you like me to explain the difference between a virus and a bacterial infection?"
MY FRIEND/SLASH/FINANCIAL ADVISER: "Do you want me to tell you (again) how bonds work?"
That is not the way it happened. This is:
EARLO: "Can you explain to me the difference between a virus and a bacterial infection?"
EARLO: "Tell me (again) how bonds work?"
Both of those conversations were initiated by me. Out of disinterest? Of course not! I sincerely wanted to know. Why? Because of, respectively, an ongoing medical condition, and a substantial investment in bonds. This was important and meaningful information.
And yet, it went into my brain and passed immediately right through.
Every single time.
I am interested. I listen. But it does not seem to stick.
On the other hand, certain seemingly meaningless information that went into my brain appears to remain archived in there forever.
Like,
The father of a cabin-mate at Camp Ogama in the early 1960's named Robbie Krangle was the owner of the Cheerio Yoyo and Bo-lo Bat Company.
Why would I remember that?
I was going to write "Why do I choose to remember that?" But therein lies my message.
What I seem to remember – and, by extension, me not being some kind of biological anomaly, what people seem to remember – does not appear to be a choice, the product of neither preference, nor interest nor intention.
It is not that I don't want to retain something. Over the years, I have attended four separate classes in "Philosophy" at UCLA Extension. And I do not remember any of it. I recall, with enjoyment in fact, the ambient philosophical music, but not the tiniest sliver of the content. And believe me, I was really trying!
Unofficial Conclusion: People's proclivities for retention are surprisingly various. What one person's brain holds on to, another person's immediately forgets.
Is that truly possible? Two people of equal intelligence, or even unequal intelligence – one's brain can internalize one category of information and the other's brain, as hard as they try and as interested as they may be, cannot?
Hm. So who then is "The Smart One"?
The person who can rattle off the entire Periodic Table of Elements?
Or the person who can sing:
"He cleaned up the country, the old Wild West country
He made law and order prevail.
And none can deny it, the legend of Wyatt
Forever will live on the trail."
My vote's with the latter.
But I may possibly be prejudiced.
"The Smart One", it turns out, based on the category in question,
May be everyone.
Note: I forgot that I promised yesterday I would do a followup today. Instead I wrote something completely different instead. Did I say "instead" twice in that sentence? Uh-oh, I may be losing it. As promised however - but this time I mean it, I will follow up on yesterday's post. Yesterday, it was "The Anointed Ones." Tomorrow, or perhaps another day in the future if I forget tomorrow again, why the Anointed Ones can't leave. Explained substantially by the fact that if they left, they would not be Anointed Ones anymore, but there is a little more to it than that, though I am not at the moment entirely certain what that is.
"The Anointed Ones"
"Lorne Michaels' Surprising Secret To Success"
"On Returning Home (From That Fitness Place That W...
"What Exactly Am I Looking At?"
"Seriously Misunderstood"
"The Measure OF A Man"
"Messing With The Prototype"
"I Once Met Oprah Winfrey At A Fitness Spa In Mexico"
"An Unusual Coincidence""
'What's Entertainment? - The Follow-Up"
"What's Entertainment?"
"Earl Pomerantz Is Seventy"
"Birthday Eve"
"The Under-Appreciated Essence Of 'Fiddler on the ...


text:
CHARLES STEELE, CONSERVATIONIST AND HORSE OWNER, DIES
by NSA / December 19, 2019 / Published in News
Charles N. Steele (Douglas Lees photo)
Charles Nevett Steele of Markham, Va., a longtime public-interest lawyer, a conservationist, and a member of Noble Stables with his wife, Helen, died at home on Dec. 4 from complications following surgery. He was 80.
He was known for his combination of striking looks, sharp intellect, dry wit, and modest demeanor. He was self-confident and hardworking, yet low-keyed and welcoming with colleagues and friends, with whom he shared a lifetime of loyalty and very good times.
He was born June 21, 1939, in New York City to J. Murray Steele and Sylvia Ward Steele. He graduated from Harvard College in 1960 and Harvard Law School in 1965.
After law school Charlie began his career in public service. He was a staff attorney with the Appellate Court Branch of the National Labor Relations Board from 1965-75. He then served at the Federal Elections Commission as associate general counsel for enforcement and litigation from 1976-79, and as general counsel between 1979-87. He argued cases before the U.S. Supreme Court, including ones concerning the constitutionality of laws limiting campaign spending by various types of organizations.
He left the FEC to become general counsel of Conservation International, and in 1990 came to Fauquier County, Va., as county attorney. In 1991 he went to work in the administration of newly elected Gov. William F. Weld of Massachusetts, serving there and in Washington until 1997.
An avid outdoorsman, he traveled widely, consulted on conservation projects globally, and carefully managed the woodlands at his Markham home. He was a birdwatcher, fisherman, tennis player, and golfer.
In addition to his wife, he is survived by two granddaughters, Emma Bruce and Helen Livingston (Livvie) Steele-Smith. He also leaves his son-in-law, Gerald (Jerry) Smith of Oakland, and sister, Lucy Anne Steele, of Orient, N.Y. His daughter, Helen Livingston (Haley) Steele, and his brother, John, predeceased him.
A memorial service will be held at 11 a.m. on Saturday, Jan. 4, at Grace Episcopal Church in The Plains, Va. In lieu of flowers, donations in his name may be made to the American Bird Conservancy in The Plains, Va. Moser Funeral Home handled arrangements.


text:
The State of Britain
UK Stats
NW Stats
NE Stats
EM Stats
EE Stats
WM Stats
LDN Stats
SE Stats
SW Stats
YH Stats
SCOT Stats
WALES Stats
NI Stats
IOM | GIB | CI
IOM | GIB | CI Stats
The YH economy grew by 1.2% in 2018, the Barnsley, Doncaster and Rotherham economy the fastest growing in the region but North East Lincolnshire declines, the growth in YH house prices a UK outlier
December 2019 January 2020 / Leave a Comment
Following its first publication of quarterly GDP estimates for the regions in September, the ONS has now published its 2018 full year estimate of economic activity by UK country, region and local area using gross domestic product.
The figures showed the YH economy grew by 1.2% in 2018, down from the 2017 growth rate of 2.2%. This placed YH sixth (2017 ranking fifth) out of the twelve UK 'regions.'
The UK and England growth rate in 2018 was 1.4%. Growth in Wales was 1.3%, Scotland grew by 0.9% and the Northern Ireland economy shrank by 0.5%.
London topped the 2018 table with growth of 2.0% whilst Northern Ireland was at the bottom.
Within the region, the Barnsley, Doncaster and Rotherham economy grew the fastest at 4.6%, followed by Leeds at 4.5% and York at 3.9%. Across the UK, the highest annual growth of sub national areas was in Falkirk at 10.5%.
Five areas of the region saw their GDP decline in 2018. The worst performer was North and North East Lincolnshire at -3.2% followed by Bradford at -2.2% and East Yorkshire and Northern Lincolnshire at -0.6%. In UK terms, the lowest annual growth of sub national areas was in Mid and East Antrim at -10.1%.
GDP per head growth of 3.9% to £36,500 was seen in Leeds. GDP per head fell by 3.5% in North and North East Lincolnshire to £30,320 but despite growing by 4.0%, Barnsley, Doncaster and Rotherham posted the lowest GDP per head in the region at £20,326.
In terms of UK extremes, GDP per head was £395,309 in Camden and the City of London and £15,034 in Ards and North Down. These figures are a guide and are influenced by commuter flows.
In 2018, key drivers of the YH economy were information/communication and arts/entertainment both at 6% and water supply and services at 4%. Those areas that did not perform well were agriculture down by 6%, public administration/defence fell by 4% and construction also declined by 1%. Overall the services sector grew by 1.7% and production by 0.3%.
The 2018 performance of the region's enterprise partnerships was also highlighted by the ONS. Of the UK's 45 development bodies, Sheffield City Region was ranked 4th in the UK (2017 ranking 18th) with growth of 2.3%, with Leeds City Region moving up the rankings most from 22nd to 5th with growth of 2.2%. The region's three other LEPs all slipped down the rankings. York, North Yorkshire and East Riding LEP slipped from 5th in 2017 to 19th in 2018, Greater Lincolnshire fell from 15th to 37th, but the worst performer was the Humber which dropped from 12th to 38th as the economy declined by 0.4% compared with growth of 3.2% in 2017.
More data from the ONS showed unemployment in YH fell by 6,000 to 110,000 between August and October 2019; the decrease of 0.2% took the overall rate to 4.1%. Northern Ireland had the lowest rate at 2.3% with the UK rate at 3.8%. The highest rate was 6.1% which was recorded in the North East.
The South West had the highest employment rate at 80.8% which compared with 73.8% in YH. UK employment was estimated at 76.2%.
The only part of England where average property prices grew was in YH, by 0.9% during October 2019 to £166,904, which took annual growth to an England best of 3.2%. In comparison, UK prices fell by 0.7% to £232,944, an annual growth rate of 0.7%.
Website @ CSU Web Design


text:
FRA ex-employees cannot work in authority-supervised bodies for 1 year - Daily News Egypt
Business FRA ex-employees cannot work in authority-supervised bodies for 1 year
FRA ex-employees cannot work in authority-supervised bodies for 1 year
FRA Chairperson Mohammed Omran said the ban extends to former employees whose service ended with a clear or implied resignation who may look for employment at any of the entities subject to its supervision and control.
Daily News Egypt July 6, 2020 Comments Off on FRA ex-employees cannot work in authority-supervised bodies for 1 year
The Financial Regulatory Authority's (FRA) Board of Directors approved, at a virtual meeting on Sunday, an amendment to the authority's bylaws.
It stipulates that former FRA employees must avoid involvement in bodies engaged in activities under the authority's supervision for one year, to prevent conflict of interests.
This would be valid for a period of one year from the former employee's end of service. The year-long ban may be reduced to three months subject to a decision by the FRA chairperson.
Omran stressed that, given the importance of the FRA's role in this regard, employees have significant duties and responsibilities due to the nature of the field in which they work.
As a result, employees are expected to adhere to the FRA's Human Resources Regulations that ensure their neutrality and that their interests do not conflict with business interests.
The FRA has addressed all units subject to its supervision with the contents of the amendment and the importance of implementing it.
Topics: ex-employees FRA Mohammed Omran
Careem, iCheck, and Cargas partner to convert cars to natural gas with payment facilities
https://www.dailynewsegypt.com/2020/07/06/fra-ex-employees-cannot-work-in-authority-supervised-bodies-for-1-year/
FRA launches guide for "Empowering Women" application
El-Dabaa, Amer shortlisted for MCSD chairperson position
Egypt, Eritrea vow to ensure Horn of Africa stability
Opinion| How socially-led innovation can bring positive change in unchartered times
Unidentified jets strike Al-Watiya airbase in western Libya
July 6, 2020 Breaking News


Place uncooked chicken in greased crock-pot. Add jalapeno and mushrooms.
Mix together peanut butter and tomatoes will make a chunky sauce. Pour over chicken; make sure all pieces are covered in sauce.
Place lid and cook all-night on low setting. Serve over wild rice or noodles.


text:
\section{Introduction}
Let~$R$ be a Dedekind domain. A fundamental and well known property of Dedekind domains is that every ideal $\mathfrak{a}\vartriangleleft R$ has a unique factorization into a product of powers of prime ideals. There are cases when this factorization is algorithmically computable. For instance, if $R = \mathbb{Z}_K$ is the ring of algebraic integers (i.e. the integral closure of~$\mathbb{Z}$) in some algebraic number field $K = \mathbb{Q}(\vartheta)$, then a suitable algorithm can be found e.g. in \cite[Algorithm 2.3.22]{Cohen00} or \cite[\S2.2]{GMN13}. The algorithms can be adapted also to global function fields. They depend however on knowing an embedding of the ring of integers (or polynomials) into~$R$. In this paper we discuss the problem of performing the computations intrinsically in the monoid of $R$-ideals without relaying on these embeddings. The procedure of factoring ideals, that we propose, resembles a method of factoring polynomials over finite fields. We show how to generalize known algorithms for polynomial factorization to make them work with ideals in maximal orders of global function fields. The ideal to be factored passes through a three-stage process: radical decomposition, distinct degree factorization and equal degree factorization.

The algorithms presented in \cite{Cohen00, GMN13} are quite efficient, hence the aim of developing intrinsic methods is not so much to reduce the computation time but rather to construct algorithm that do not dependent on the particular structure of global fields and so have potential to be generalized to other rings. In particular the first step of the process, namely the radical decomposition, can be performed in \emph{any} Dedekind domain in which three elementary operation on ideals are computable. This class of rings include coordinate rings of  smooth, algebraically irreducible curves over a computable, perfect field (see Proposition~\ref{prop_coord_is_computable}). Some early experiments of the authors suggest that the algorithms presented here can be generalized to compute primary decomposition of ideals in affine algebras. This subject need further investigation, though.

The paper is organized as follows. In Section~\ref{sec_radical} we discuss the radical decomposition of ideals, which is an analog of a square-free factorization of polynomials. Given an ideal $\mathfrak{a}\vartriangleleft R$, this procedure produces a list of radical ideals $\gg_1, \dotsc, \gg_m$ such that~$\mathfrak{a}$ is a product of their respective powers. Next, in Section~\ref{sec_distinct} we show how to factor a radical ideal (i.e. any of the ideals $\gg_1, \dotsc, \gg_m$) into a product of (radical) ideals such that each one of these new ideals is a product of primes of the same residual degree. Finally in Section~\ref{sec_equal} we present a variant of Cantor--Zassenhaus algorithm (Algorithm~\ref{alg_ed_factor}) capable of factoring radical ideals with prime divisors of a fixed degree. The algorithms discussed in this paper were implemented by the authors in a computer algebra system Magma \cite{magma}. In the closing section we presented two examples obtained with our implementation.

In the whole paper the letter~$R$ always denotes a (fixed) Dedekind domain with a field of fractions~$K$. For readers convenience our notation follows the one used in \cite{AM16}, in particular fraktur letters are used to denote ideals. All the ideals in this paper are integral ideals.

\section{Radical decomposition of ideals}\label{sec_radical}
Let~$R$ be a Dedekind domain and $\mathfrak{a}\vartriangleleft R$ be an ideal in~$R$. Assume that~$\mathfrak{a}$ factors into primes as
\begin{equation}\label{eq_factor_a}
\mathfrak{a} = \mathfrak{p}_1^{k_1}\dotsm \mathfrak{p}_s^{k_s},
\end{equation}
where $\mathfrak{p}_1, \dotsc, \mathfrak{p}_s$ are distinct (and unknown) prime ideals and $k_1, \dotsc, k_s > 0$ their multiplicities. Collate the factors of equal multiplicities. For any $j\leq m := \max\{ k_1, \dotsc, k_s \}$ denote
\[
\gg_j := \bigcap_{\substack{1\leq i\leq s\\ k_i = j}} \mathfrak{p}_i.
\]
This way we may write~$\mathfrak{a}$ as a product analogous to a square-free factorization of a polynomial:
\begin{equation}\label{eq_radical_decomp}
\mathfrak{a} = \gg_1\cdot \gg_2^2\dotsm \gg_m^m.
\end{equation}
We shall call~\eqref{eq_radical_decomp} the \term{radical decomposition} of the ideal~$\mathfrak{a}$. The name is justified by the following observation.

\begin{obs}
Ideals $\gg_1, \dotsc, \gg_m$ are radical.
\end{obs}

Indeed, radicals are preserved by intersection (see e.g. \cite[Ch.~1]{AM16}), hence
\[
\rad(\gg_j)
= \rad\Bigl( \bigcap_{k_i = j}\mathfrak{p}_i \Bigr)
= \bigcap_{k_i = j} \rad(\mathfrak{p}_i)
= \bigcap_{k_i = j} \mathfrak{p}_i
= \gg_j.
\]
In our settings, the ideals $\gg_1, \dotsc, \gg_m$ play roles analogous to square-free factors of a polynomial in case of the square-free factorization, so that we shall call them \emph{radical factors} of~$\mathfrak{a}$.

The following operations are the basic building blocks for our first algorithm:
\begin{itemize}
\item given an ideal~$\mathfrak{a}$ compute its radical $\rad\mathfrak{a}$,
\item given two ideals~$\mathfrak{a}$ and~$\mathfrak{b}$ compute their sum $\mathfrak{a}+\mathfrak{b}$ and the colon ideal $(\mathfrak{a}:\mathfrak{b}) = \{ x\;|\  x\mathfrak{b}\subseteq \mathfrak{a}\}$.
\end{itemize}
We shall say that~$R$ is a ring with \term{computable ideal arithmetic} if all the three operations are computable for ideals of~$R$.

\begin{prop}\label{prop_coord_is_computable}
Let~$\Bbbk$ be a perfect, computable field and $C := \{ F = 0\}$ be a smooth, geometrically irreducible algebraic curve over~$\Bbbk$, defined by a bivariate polynomial $F\in \kk[X,Y]$. Then the coordinate ring $R = \Bbbk[C] = \sfrac{\kk[X,Y]}{\Ideal{F}}$ admits computable ideal arithmetic.
\end{prop}

The proof of the proposition needs to be preceded by a lemma. Let $\kappa: \kk[X,Y]\twoheadrightarrow R$ be the canonical epimorphism. By superscripts $\cdot^c$, $\cdot^e$ we shall denote respectively the ideal contraction and extension with respect to~$\kappa$.

\begin{lem}
Keep the assumptions of the proposition. If $\mathfrak{a}, \mathfrak{b}\vartriangleleft R$ are two ideals, then
\[
\mathfrak{a}^{ce} = \mathfrak{a},\qquad
\rad(\mathfrak{a}) = \bigl( \rad(\mathfrak{a}^c)\bigr)^e,\qquad
(\mathfrak{a} : \mathfrak{b}) = ( \mathfrak{a}^c : \mathfrak{b}^c)^e.
\]
\end{lem}

\begin{proof}
The inclusion $\mathfrak{a}^{ce}\subseteq \mathfrak{a}$ holds always (see e.g. \cite[Proposition 1.17]{AM16}). The other inclusion follows from the fact that~$\kappa$ is an epimorphism. Consequently we have
\[
\rad(\mathfrak{a})
= \bigl( \rad(\mathfrak{a}) \bigr)^{ce}
= \bigl( \rad(\mathfrak{a}^c) \bigr)^e,
\]
where the last equality follows from \cite[Exercise~1.18]{AM16}. Likewise we may write
\[
( \mathfrak{a}^c : \mathfrak{b}^c )^e
\subseteq ( \mathfrak{a}^{ce} : \mathfrak{b}^{ce} )
= ( \mathfrak{a} : \mathfrak{b} )
= ( \mathfrak{a} : \mathfrak{b} )^{ce}
\subseteq ( \mathfrak{a}^c : \mathfrak{b}^c )^e.
\]
This concludes the proof.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop_coord_is_computable}]
If we do not insist on obtaining the $2$-generators representation of the result, the computation of the sum $\mathfrak{a}+ \mathfrak{b}$ of two ideals can be as simple as a concatenation of their lists of generators. Next, an algorithm for computing a quotient of two ideal in a multivariate polynomial ring is well known and so it follows from the above lemma that one may compute the quotient of ideals in~$R$. Finally, being a Dedekind domain, the ring~$R$ has dimension one. Consequently every nontrivial ideal $\mathfrak{a}\vartriangleleft R$ lifts to a zero-dimensional ideal $A\vartriangleleft \kk[X,Y]$. The radical of a zero-dimensional ideal in a multivariate polynomial ring over a perfect field is computable using Seidenberg's formula (see \cite{Seidenberg74}). Thus, the radical of~$\mathfrak{a}$ is computable, as well by the previous lemma.
\end{proof}

We are now ready to present an algorithm for the radical decomposition. The reader may wish to observe that it is a generalization of Musser's algorithm \cite{Musser71} for the square-free factorization of polynomials over a field of characteristic zero.

\begin{alg}[Radical decomposition of an ideal]\label{alg_radical_decomp}
\mbox{}\\
\textbf{Input:} an ideal~$\mathfrak{a}$ in a Dedekind domain~$R$ with computable ideal arithmetic.\\
\textbf{Output:} radical factors $\gg_1, \dotsc, \gg_m$ of~$\mathfrak{a}$.
\begin{algorithmic}[1]
\LineComment{Initialization}
\State $\mathfrak{a}_0 \gets \mathfrak{a}$
\State $i     \gets 1$
\State $\mathfrak{b}_1 \gets \rad(\mathfrak{a})$
\State $\mathfrak{a}_1 \gets (\mathfrak{a}_0:\mathfrak{b}_1)$
\LineComment{Main loop}
\While{$\mathfrak{b}_i\neq R$}
	\State $\mathfrak{b}_{i+1} \gets \mathfrak{a}_i + \mathfrak{b}_i$
	\State $\mathfrak{a}_{i+1} \gets (\mathfrak{a}_i : \mathfrak{b}_{i+1})$
	\State $\gg_i     \gets (\mathfrak{b}_i : \mathfrak{b}_{i+1})$
	\State $i \gets i + 1$
\EndWhile
\State \Return $\gg_1, \dotsc, \gg_i$
\end{algorithmic}
\end{alg}

Before we show the correctness of the algorithm, we present a slightly technical lemma that gives an explicit description of ideals~$\mathfrak{a}_i$ and~$\mathfrak{b}_i$ constructed during the execution of the algorithm.

\begin{lem}
Keep the notation as in Algorithm~\ref{alg_radical_decomp}. The ideals~$\mathfrak{a}_i$ and~$\mathfrak{b}_i$ satisfy:
\[
\mathfrak{a}_i = \gg_{i+1}\cdot \gg_{i+2}^2\dotsm \gg_m^{m-i}
\qquad\text{and}\qquad
\mathfrak{b}_i = \gg_i\cdot \gg_{i+1}\dotsm \gg_m.
\]
\end{lem}

\begin{proof}
We proceed by induction. The assertion is trivially true for~$\mathfrak{a}_0$ and~$\mathfrak{b}_1$. Assume that the two formulas hold for ideals~$\mathfrak{a}_{i-1}$ and~$\mathfrak{b}_i$. Take any $x\in \gg_{i+1}\cdot \gg_{i+2}^2\dotsm \gg_m^{m-i}$ and $y\in \mathfrak{b}_i = \gg_i\cdot \gg_{i+1}\dotsm \gg_m$. Then their product~$xy$ lies in $\gg_i\cdot \gg_{i+1}^2\dotsm \gg_m^{m-i+1} = \mathfrak{a}_{i-1}$. Hence $x\in (\mathfrak{a}_{i-1} : \mathfrak{b}_i) = \mathfrak{a}_i$ proving an inclusion $\mathfrak{a}_i \supseteq \gg_{i+1}\cdot \gg_{i+2}^2\dotsm \gg_m^{m-i}$.

Conversely, take $x\in \mathfrak{a}_i = (\mathfrak{a}_{i-1} : \mathfrak{b}_i)$. Fix any prime ideal~$\mathfrak{p}$ dividing~$\mathfrak{a}_{i-1}$ and let $k := \ord_\mathfrak{p}(\mathfrak{a})$ be the multiplicity of~$\mathfrak{p}$ in the factorization~\eqref{eq_factor_a} of~$\mathfrak{a}$. By the inductive hypothesis, the ideal~$\mathfrak{p}$ divides~$\mathfrak{b}_i$ and $k - i + 1$ is the multiplicity of~$\mathfrak{p}$ in the factorization of~$\mathfrak{a}_{i-1}$. By the strong approximation theorem (see e.g. \cite[Corollary~10.5.11]{Cohn03}) there exists an element $y\in R$ such that
\[
\ord_\mathfrak{p} y = 1\qquad\text{and}\qquad \ord_\mathfrak{q} y \geq 1\quad\text{for all }\mathfrak{q}\mid \mathfrak{b}_i.
\]
In particular, $y$ is an element of~$\mathfrak{b}_i$ and $y\notin \mathfrak{p}^2$. By the definition of the colon ideal, $xy\in x\cdot \mathfrak{b}_i\subseteq \mathfrak{a}_{i-1}\subseteq \mathfrak{p}^{k - i + 1}$. It follows that the $\mathfrak{p}$-adic valuation of the product~$xy$ is at least $k - i + 1$. Therefore, we have
\[
k - i + 1
\leq \ord_\mathfrak{p}(xy)
= \ord_\mathfrak{p} x + 1.
\]
Consequently $\ord_\mathfrak{p} x\geq k - i$, which means that $x\in \mathfrak{p}^{k - i}$. As this holds for every prime~$\mathfrak{p}$ dividing~$\mathfrak{a}_{i-1}$, we see that
\[
x\in \bigcap_{\substack{\mathfrak{p}\mid \mathfrak{a}\\ \mathpalette\mathclapinternal{\ord_\mathfrak{p}(\mathfrak{a})\geq i}}} \mathfrak{p}^{\ord_\mathfrak{p}(\mathfrak{a}) - i}
= \prod_{\substack{\mathfrak{p}\mid \mathfrak{a}\\ \mathpalette\mathclapinternal{\ord_\mathfrak{p}(\mathfrak{a})\geq i}}} \mathfrak{p}^{\ord_\mathfrak{p}(\mathfrak{a}) - i}
= \prod_{k\geq i}\biggl( \prod_{\substack{\mathfrak{p}\mid \mathfrak{a}\\ \mathpalette\mathclapinternal{\ord_\mathfrak{p}(\mathfrak{a})=k}}} \mathfrak{p}\biggr)^{k - i}
\hspace{-1em}= R\cdot \gg_{i+1}\cdot \gg_{i+2}^2\dotsm \gg_m^{m-i}.
\]
This shows that $\mathfrak{a}_i \subseteq \gg_{i+1}\cdot \gg_{i+2}^2\dotsm \gg_m^{m-i}$.

We now prove the equality $\mathfrak{b}_{i+1} = \gg_{i+1}\dotsm \gg_m$. One inclusion is immediate.
\begin{align*}
\mathfrak{b}_{i+1}
&= \mathfrak{a}_i + \mathfrak{b}_i = \Ideal{ \gg_{i+1}\cdot \gg_{i+1}^2\dotsm \gg_m^{m-i}\cup \gg_i\cdot \gg_{i+1}\dotsm \gg_m }\\
\intertext{The radical ideals~$\gg_i$ are pairwise coprime, hence}
&= \Ideal{ \bigcap_{j\geq i+1} \gg_j^{j-i}\cup \Bigl(\gg_i\cap \bigcap_{j\geq i+1}\gg_j\Bigr) }\\
&= \Ideal{ \Bigl( \bigcap_{j\geq i+1} \gg_j^{j-i}\cup \gg_i\Bigr) \cap \Bigl(\bigcap_{j\geq i+1}\gg_j^{j-i}\cup \bigcap_{j\geq i+1}\gg_j\Bigr) }\\
&\subseteq \Ideal{ \Bigl(\bigcap_{j\geq i+1}\gg_j\cup \gg_i\Bigr) \cap  \bigcap_{j\geq i+1}\gg_j}\\
&= \Ideal{ \bigcap_{j\geq i+1}\gg_j} = \gg_{i+1}\cdot \gg_{i+1}\dotsm \gg_m.
\end{align*}
In order to show the other inclusion fix an element $x\in \gg_{i+1}\dotsm \gg_m$. Ideals~$\gg_i$ and $\gg_{i+1}\gg_{i+2}^2\dotsm \gg_m^{m-i} = \mathfrak{a}_i$ are relatively prime, hence there exist elements $y\in \gg_i$ and $z\in \mathfrak{a}_i$ such that $x = y + z$. Therefore, for any $j\geq i+1$ we have
\[
y = x - z \in \gg_j + \mathfrak{a}_i\subseteq \gg_j + \gg_j = \gg_j.
\]
It follows that $y\in \gg_{i+1}\cap\dotsc \cap\; \gg_m = \mathfrak{b}_i$. Consequently, $x = y + z\in \mathfrak{b}_i + \mathfrak{a}_i = \mathfrak{b}_{i+1}$.
\end{proof}

We are now ready to show the correctness of the algorithm.

\begin{proof}[Proof of correctness of Algorithm~\ref{alg_radical_decomp}]
It follows immediately from the preceding lemma that the algorithm terminates. All we need to show is that for every index~$i$ the colon ideal $(\mathfrak{b}_i:\mathfrak{b}_{i+1})$ equals the sought radical ideal~$\gg_i$. One inclusion is immediate. By the lemma we have
\[
\gg_i\cdot \mathfrak{b}_{i+1} = \gg_i\cdot \bigl( \gg_{i+1}\dotsm \gg_m\bigr) = \mathfrak{b}_i
\]
and so $\gg_i\subseteq (\mathfrak{b}_i : \mathfrak{b}_{i+1})$. We need to prove the other inclusion. To this end take any $x\in (\mathfrak{b}_i : \mathfrak{b}_{i+1})$ and fix a prime divisor~$\mathfrak{p}$ of~$\gg_i$. The multiplicity of~$\mathfrak{p}$ in the factorization of~$\mathfrak{a}$ is thus $\ord_\mathfrak{p}(\mathfrak{a}) = i$. By the strong approximation theorem there is an element $y\in R$ such that $y\in \mathfrak{b}_{i+1}\setminus \mathfrak{p}$. Now, $xy \in x\cdot \mathfrak{b}_{i+1}\subseteq \mathfrak{b}_i\subseteq \mathfrak{p}$ but $y\notin \mathfrak{p}$, it follows that $x\in \mathfrak{p}$. This shows that~$x$ belongs to every prime divisor~$\mathfrak{p}$ of~$\mathfrak{a}$ of multiplicity $\ord_\mathfrak{p}(\mathfrak{a}) = i$. Therefore
\[
x\in \bigcap_{\substack{\mathfrak{p}\mid a\\ \mathpalette\mathclapinternal{\ord_\mathfrak{p}(\mathfrak{a}) = i}}}\mathfrak{p} = \gg_i.
\]
This proves the correctness of the algorithm.
\end{proof}

\section{Distinct degree factorization}\label{sec_distinct}
In this and the next section we restrict our attention to maximal orders in global function fields. Thus, let~$\Bbbk$ be a fixed finite field and let $R = \Bbbk[C] = \sfrac{\kk[X,Y]}{\Ideal{F}}$ be a Dedekind domain which is a coordinate ring of a smooth and geometrically irreducible curve~$C$. In particular~$R$ is a maximal order in a field~$\Bbbk(C)$ of rational functions on~$C$ (i.e. in a global function field). Given a radical ideal $\mathfrak{a}\vartriangleleft R$, consider its factorization into primes
\[
\mathfrak{a} = \mathfrak{p}_1\dotsm \mathfrak{p}_s.
\]
Collate the primes with respect to their residual degrees setting
\[
\mathfrak{h}_j := \prod_{\substack{\mathfrak{p}\mid \mathfrak{a}\\ \deg \mathfrak{p} = j}} \mathfrak{p}.
\]
Consequently the ideal~$\mathfrak{a}$ may be expressed as a product
\begin{equation}\label{eq_dd_factor}
\mathfrak{a} = \mathfrak{h}_1\dotsm \mathfrak{h}_m,\qquad\text{where}\quad m := \max\{\deg\mathfrak{p}\;|\  \mathfrak{p}\text{ divides }\mathfrak{a}\}.
\end{equation}
By analogy to the polynomial case, we shall call~\eqref{eq_dd_factor} the \term{distinct degree factorization} of~$\mathfrak{a}$.

We will compute the distinct degree factorization of a given ideal~$\mathfrak{a}$ by constructing successive greatest common divisors (in the lattice of $R$-ideals) of~$\mathfrak{a}$ and~$\mathfrak{u}_k$, where~$\mathfrak{u}_k$ is the intersection of all primes of residual degrees dividing~$k$:
\[
\mathfrak{u}_k := \prod_{\substack{\mathfrak{p}\text{ prime}\\\deg\mathfrak{p} \mid k}} \mathfrak{p}.
\]
Before we continue, recall that with every prime ideal~$\mathfrak{p}$ one can associate a unique point $(x_\mathfrak{p}, y_\mathfrak{p})$ on the curve~$C$, with coordinates in the algebraic closure~$\algcl{\Bbbk}$ of~$\Bbbk$. To this end treat elements of $R = \Bbbk[C]$ as polynomial functions on~$C$ and set $(x_\mathfrak{p}, y_\mathfrak{p})$ to be a unique point where all elements of~$\mathfrak{p}$ vanish simultaneously. In this section~$\Bbbk$ is a finite field, say $\Bbbk = \mathbb{F}_q$ for some prime power~$q = p^l$. The degree of~$\mathfrak{p}$ divides~$k$ if and only if $x_\mathfrak{p}, y_\mathfrak{p}$ lie in~$\mathbb{F}_{q^k}$. It is well known that $\mathbb{F}_{q^k}$ consists of elements satisfying $a^{q^k}-a = 0$. Apply this fact to both coordinates.

\begin{lem}\label{lem_universal_ideal}
For every $k\geq 1$, the ideal $\mathfrak{u}_k$ is generated by $x^{q^k}-x$ and $y^{q^k}-y$, where $x,y$ are images in~$R$ of $X,Y\in \kk[X,Y]$.
\end{lem}

\begin{proof}
Fix $k\geq 1$ and denote $\mathfrak{v}_k := \tIdeal{ x^{q^k}-x, y^{q^k}-y}\vartriangleleft R$. We shall prove first that the ideal~$\mathfrak{v}_k$ is contained in~$\mathfrak{u}_k$. It suffices to show that both its generators belong to every prime ideal~$\mathfrak{p}$ of~$R$ whose residual degree divides~$k$. Take any such prime~$\mathfrak{p}$. The coordinates $x_\mathfrak{p}, y_\mathfrak{p}$ of the associated point belong to~$\mathbb{F}_{q^k}$, hence $x_\mathfrak{p}^{q^k}-x_\mathfrak{p} = 0 = y_\mathfrak{p}^{q^k} - y_\mathfrak{p}$. Thus the generators of~$\mathfrak{v}_k$ vanish on $(x_\mathfrak{p}, y_\mathfrak{p})$ and so $\mathfrak{v}_k\subseteq \mathfrak{p}$.

Next, we show that~$\mathfrak{v}_k$ is not contained in any prime ideal~$\mathfrak{p}$ whose degree does not divide~$k$. Suppose that $\mathfrak{v}_k\subset \mathfrak{p}$ for some prime ideal~$\mathfrak{p}$. In particular the generators $x^{q^k}-x, y^{q^k}-y$ belong to~$\mathfrak{p}$. Therefore, they vanish on the associated point $(x_\mathfrak{p}, y_\mathfrak{p})$, which means that $x_\mathfrak{p}, y_\mathfrak{p}\in \mathbb{F}_{q^k}$ and so $\deg \mathfrak{p}$ divides~$k$.

From what we have proved so far it follows that~$\mathfrak{u}_k$ is the radical of~$\mathfrak{v}_k$. In order to conclude the proof, it suffices to show that~$\mathfrak{v}_k$ is a radical ideal itself. To this end we show that for every prime ideal~$\mathfrak{p}$, $\deg \mathfrak{p}\mid k$ the valuation of at least one of the generators of~$\mathfrak{v}_k$ equals~$1$. Consider two (reducible) algebraic curves $C_1 := \bigl\{x^{q^k}= x\bigr\}$ and $C_2 := \bigl\{ y^{q^k}=y\bigr\}$. They both consist of parallel lines but they are not parallel to each other. Suppose that $\ord_\mathfrak{p} (x^{q^k}-x)> 1$ for some~$\mathfrak{p}$. This means that~$\mathfrak{p}$ is a ramified extension of an ideal $p\cdot \mathbb{F}_q[x]$ for some irreducible polynomial~$p$ in~$x$. We may identify the valuation $\ord_\mathfrak{p}(x^{q^k}-x)$ with the intersection index $I\bigl( (x_\mathfrak{p},y_\mathfrak{p}), C\cap C_1\bigr)$. If
\[
\ord_\mathfrak{p}\bigl( x^{q^k}-x ) = I\bigl( (x_\mathfrak{p},y_\mathfrak{p}), C\cap C_1\bigr) > 1,
\]
then~$C$ is tangent to~$C_1$ at $(x_\mathfrak{p}, y_\mathfrak{p})$. Consequently it cannot be tangent to~$C_2$ at $(x_\mathfrak{p}, y_\mathfrak{p})$ as~$C$ is non-singular. Therefore
\[
\ord_\mathfrak{p}\bigl( y^{q^k}-y ) = I\bigl( (x_\mathfrak{p},y_\mathfrak{p}), C\cap C_2\bigr) = 1.
\]
This shows that for every prime ideal~$\mathfrak{p}$, whose residual degree divides~$k$, either $x^{q^k}-x\in \mathfrak{p}\setminus\mathfrak{p}^2$ or $y^{q^k}-y\in \mathfrak{p}\setminus \mathfrak{p}^2$. This implies that~$\mathfrak{v}_k$ is radical.
\end{proof}

We may now present an algorithm for distinct degree factorization.
\begin{alg}[Distinct degree factorization]\label{alg_dd_factor}
\mbox{}\\
\textbf{Input:} a radical ideal~$\mathfrak{a}\vartriangleleft R$.\\
\textbf{Output:} distinct degree factors $\mathfrak{h}_1, \dotsc, \mathfrak{h}_m$ of~$\mathfrak{a}$.
\begin{algorithmic}[1]
\LineComment{Initialization}
\State $k     \gets 1$
\State $\mathfrak{a}_1 \gets \mathfrak{a}$
\LineComment{Main loop}
\While{$\mathfrak{a}_k\neq R$}
	\State $\mathfrak{u}_k \gets \tIdeal{ x^{q^k}-x, y^{q^k}-y }$
	\State $\mathfrak{h}_k \gets \mathfrak{u}_k + \mathfrak{a}_k$
	\State $\mathfrak{a}_{k+1} \gets (\mathfrak{a}_k : \mathfrak{h}_k)$
	\State $k \gets k + 1$
\EndWhile
\State \Return $\mathfrak{h}_1, \dotsc, \mathfrak{h}_k$
\end{algorithmic}
\end{alg}

\begin{poc}
We proceed by an induction on~$k$. Assume that $\mathfrak{h}_{k-1}$ is the $(k-1)$-th distinct degree factor of $\mathfrak{a}$ and~$\mathfrak{a}_k$ is the product of the prime divisors of~$\mathfrak{a}$ with residual degrees at least~$k$. This is trivially true for $\mathfrak{a}_1 = \mathfrak{a}$ and $\mathfrak{h}_0 := R$. Lemma~\ref{lem_universal_ideal} asserts that $\tIdeal{ x^{q^k}-x, y^{q^k}-y } = \mathfrak{u}_k$. Compute
\begin{multline*}
\mathfrak{u}_k + \mathfrak{a}_k
= \Ideal{ \mathfrak{u}_k\cup \mathfrak{a}_k}
= \Ideal{ \bigcap_{\deg\mathfrak{p}\mid k}\mathfrak{p}\ \cup \bigcap_{\substack{\mathfrak{q}\mid\mathfrak{a}\\ \deg\mathfrak{q}\geq k}}\mathfrak{q} }\\
= \Ideal{ \bigcap_{\substack{\deg \mathfrak{p} \mid k\\ \mathfrak{q}\mid \mathfrak{a}\\ \deg \mathfrak{q}\geq k}} (\mathfrak{p}\cup \mathfrak{q}) }
= \Ideal{ \bigcap_{\substack{\deg \mathfrak{p} \mid k\\ \mathfrak{q}\mid \mathfrak{a}\\ \deg \mathfrak{q}\geq k}} (\mathfrak{p} + \mathfrak{q}) }.
\end{multline*}
Now prime ideals $\mathfrak{p},\mathfrak{q}$ are either equal or relatively prime. Hence $\mathfrak{p}+ \mathfrak{q} = \mathfrak{p}$ when $\mathfrak{p} = \mathfrak{q}$ and $\mathfrak{p}+\mathfrak{q} = R$ if $\mathfrak{p}\neq \mathfrak{q}$. Consequently the above formula simplifies to
\[
\mathfrak{u}_k + \mathfrak{a}_k = \bigcap_{\substack{\mathfrak{p}\mid \mathfrak{a}\\ \deg \mathfrak{p} = k}} \mathfrak{p} = \mathfrak{h}_k.
\]
It follows that $\mathfrak{a}_{k+1} = (\mathfrak{a}_k : \mathfrak{h}_k)$ is the product off all those prime divisors of~$\mathfrak{a}$ that have degrees strictly greater than $k$.
\end{poc}

\section{Equal-degree factorization}\label{sec_equal}
After performing a radical decomposition and distinct degree factorization, we are left with a list of radical ideals such that each one is a product of primes all having the same (known) residual degree. We can deal with such ideals using a generalization of a classical Cantor--Zassenhaus algorithm. We shall first note the following fact.

\begin{lem}
If $\mathfrak{a}\vartriangleleft R$ is a nonzero radical ideal, then the number of elements of the residue ring $\sfrac{R}{\mathfrak{a}}$ is algorithmically computable.
\end{lem}

\begin{proof}
As in the proof of Proposition~\ref{prop_coord_is_computable} we use the ideal contraction with respect to the canonical epimorphism $\kappa: \kk[X,Y]\twoheadrightarrow R$. The ring~$R$ is a Dedekind domain and $\mathfrak{a}\neq \{0\}$, hence $\sfrac{R}{\mathfrak{a}}$ is a finite ring isomorphic to $\sfrac{\kk[X,Y]}{\mathfrak{a}^c}$. The number of elements of the latter ring is computable using a well known trick of counting monomials not in $\lm(\mathfrak{a}^c)$, where  $\lm(\mathfrak{a}^c)$ is an ideal spanned by leading monomials of~$\mathfrak{a}^c$ with respect any monomial order in $\kk[X,Y]$.
\end{proof}

From now on we assume that $\mathfrak{a}\vartriangleleft R$ is a radical ideal with some (unknown) factorization
\[
\mathfrak{a} = \mathfrak{p}_1\dotsm \mathfrak{p}_m
\]
and the residual degrees of $\mathfrak{p}_1, \dotsc, \mathfrak{p}_m$ are all the same and a priori known. Denote this common degree by~$d$.

\begin{lem}
Let~$b$ be an element of~$R$ not in~$\mathfrak{a}$. Denote $\overline{b} := b + \mathfrak{a}$ the class of~$b$ in $\sfrac{R}{\mathfrak{a}}$ and $e := q^d-1$. The following conditions are equivalent:
\begin{enumerate}
\item\label{it_proper_divisor} the ideal $\mathfrak{b} := \Ideal{b} + \mathfrak{a}$ is a proper divisor of~$\mathfrak{a}$;
\item\label{it_zero_divisor} the element $\overline{b}$ is a zero-divisor in $\sfrac{R}{\mathfrak{a}}$;
\item\label{it_exponent} $\overline{b}^{e}\neq 1$.
\end{enumerate}
\end{lem}

\begin{proof}
Assume that~$\mathfrak{b}$ is a proper divisor of~$\mathfrak{a}$. This means that $\mathfrak{a}\varsubsetneq \mathfrak{b}\varsubsetneq R$. In particular~$b$ cannot lie in~$\mathfrak{a}$ and so $\overline{b}\neq 0$. The ring $\sfrac{R}{\mathfrak{a}}$ is finite, hence it suffices to show that~$\overline{b}$ is not invertible. Suppose a contrario that there is an element $c\in R$ such that $\overline{c}\cdot \overline{b} = 1$. But then $1\in \mathfrak{b}$ and this contradicts the assumption that $\mathfrak{b}\neq R$.

The implication $\eqref{it_zero_divisor}\Longrightarrow\eqref{it_exponent}$ is trivial. In order to prove the remaining implication $\eqref{it_exponent}\Longrightarrow\eqref{it_proper_divisor}$, assume that $\overline{b}^e\neq 1$. By the Chinese reminder theorem there is an isomorphism
\[
\varphi : \sfrac{R}{\mathfrak{a}} \xrightarrow{\;\sim\;} \sfrac{R}{\mathfrak{p}_1}\times \dotsb \times \sfrac{R}{\mathfrak{p}_m},
\]
where each quotient ring $\sfrac{R}{\mathfrak{p}_i}$ is in turn isomorphic to $\mathbb{F}_{q^d}$. Let $\pi_i : \sfrac{R}{\mathfrak{p}_1}\times \dotsb \times \sfrac{R}{\mathfrak{p}_m}\twoheadrightarrow \sfrac{R}{\mathfrak{p}_i}$ be the projection onto the $i$-th coordinate. For every $i\leq m$, the image $(\pi_i\circ \varphi)(\overline{b}^e)$ is either~$1$ if $b\notin \mathfrak{p}_i$, or~$0$ if $b\in \mathfrak{p}_i$. Not all coordinates can be equal~$1$, because $\overline{b}^e\neq 1$. Neither all the coordinates are equal zero, since $b\notin \mathfrak{a}$. Denote
\[
I := \bigl\{ i\leq m : (\pi\circ \varphi)(b^e) = 0\bigr\} = \bigl\{ i\leq m: b\in \mathfrak{p}_i\},
\]
we then have
\[
\mathfrak{b} = \prod_{i\in I} \mathfrak{p}_i
\]
and it is clear that $\mathfrak{a}\varsubsetneq \mathfrak{b}\varsubsetneq R$.
\end{proof}

We may now present a randomized recursive algorithm, in a spirit of Cantor--Zassenhaus, for factoring radical ideals of constant residual degree.
\begin{alg}[Equal degree factorization]\label{alg_ed_factor}
\mbox{}\\
\textbf{Input:} a radical ideal~$\mathfrak{a}\vartriangleleft R$ and an integer~$d$ such that the residual degree of every prime factor of~$\mathfrak{a}$ equals~$d$.\\
\textbf{Output:} prime factors $\mathfrak{p}_1, \dotsc, \mathfrak{p}_m$ of~$\mathfrak{a}$.
\algrenewcommand\algorithmicuntil{\textbf{end repeat}}
\begin{algorithmic}[1]
\LineComment{Recursion termination}
\If{$\card{\sfrac{R}{\mathfrak{a}}} = q^d$}
	\State \Return $\mathfrak{a}$
\EndIf
\LineComment{Main loop}
\Repeat
	\State $b\gets\mbox{}$ random element of $R\setminus\mathfrak{a}$
	\State $\overline{b} \gets b + \mathfrak{a}\in \sfrac{R}{a}$
	\If{$\overline{b}^{q^d-1}\neq 1$}
		\State $\mathfrak{b} \gets \Ideal{b} + \mathfrak{a}$
		\State $\mathfrak{c} \gets ( \mathfrak{a} : \mathfrak{b} )$
		\Statex {\hspace{\algorithmicindent}\hspace{\algorithmicindent}// Recursion}
		\State $r_1 \gets \mbox{}$ Equal degree factorization of~$\mathfrak{b}$
		\State $r_2 \gets \mbox{}$ Equal degree factorization of~$\mathfrak{c}$
		\State \Return $r_1\cup r_2$
	\EndIf
\Until{}
\end{algorithmic}
\end{alg}

The correctness of the algorithm follows immediately from the lemma preceding it. For the sake of completeness we present an algorithm for the complete factorization of an ideal, that summarizes the whole discussion.
\begin{alg}[complete factorization]\label{alg_factorization}
\mbox{}\\
\textbf{Input:} an ideal~$\mathfrak{a}$ in~$R$.\\
\textbf{Output:} the list of pairs $(\mathfrak{p}_i, k_i)$ of prime divisors and multiplicities, see Eq.~\eqref{eq_factor_a}.
\begin{algorithmic}[1]
\State $\mathop{Factors} \gets []$
\State $G\gets\mbox{}$ radical decomposition of~$\mathfrak{a}$ \textup(Algorithm~\ref{alg_radical_decomp}\textup)
\For{$j\leq \card{G}$}
	\State $\gg_j\gets G[j]$
	\State $H \gets\mbox{}$ distinct degree factorization of~$\gg_j$ \textup(Algorithm~\ref{alg_dd_factor}\textup)
	\For{$d\leq \card{H}$}
		\State $\mathfrak{h}_d \gets H[d]$
		\State $P \gets \mbox{}$ equal degree factorization of $\mathfrak{h}_d$ \textup(Algorithm~\ref{alg_ed_factor}\textup)
		\State $\mathop{Factors} \gets \mathop{Factors}\cup \bigl[ (\mathfrak{p}, j) : \mathfrak{p} \in P \bigr]$
	\EndFor
\EndFor
\State \Return $\mathop{Factors}$
\end{algorithmic}
\end{alg}

\section{Examples}\label{sec_examples}
The authors implemented algorithms described in this paper in a computer algebra system Magma~\cite{magma}. Below we preset two examples computed using our implementation.

\subsubsection*{Example} Let $K = \mathbb{F}_{13}(x, y)$ be a hyperelliptic function field  given by a generating polynomial
\[
F = y^2 - (x^5 - x)(x^4 + 2)
\]
and let $R:=\sfrac{\mathbb{F}_{13}[x,y]}{\Ideal{F}}$. Consider the ideal $\mathfrak{a} \vartriangleleft R$
\begin{multline*}
\mathfrak{a} =  \langle
x^9 + 8x^7 + 5x^6 + 10x^5 + 6x^4 + 4x^3 + 9x^2 + 6x + 4,\\
\qquad 11x^8 + 8x^7 + 2x^6 + 10x^5 + 6x^4 + x^3y + x^3 + 4x^2y + 7x^2 + 4xy + 9y + 7 \rangle
\end{multline*}
Use Algorithm~\ref{alg_radical_decomp} to compute the radical decomposition $\mathfrak{a} = \gg_1\cdot \gg_2^2$, where
\begin{align*}
\gg_1 &= \langle x^6 + 9x^5 + 7x^4 + 10x^3 + 4x^2 + 4x + 12,\\
&\qquad y + 12x^5 + x^4 + 11x^3 + 10x^2 + 3x + 8 \rangle, \\
\gg_2 &= \Ideal{ x^3 + 4x^2 + 4x + 9, y + 7x^2 + 9x + 12 }.\\
\end{align*}
Next, using Algorithm~\ref{alg_dd_factor}, we compute the distinct degree factorization for each element of the radical decomposition. For $\gg_1$ it returns two trivial factors $\mathfrak{h}_{11} = \mathfrak{h}_{12} = R$ and one nontrivial, degree~$3$ factor
\begin{multline*}
\mathfrak{h}_{13} = \langle 8x^5y + 5x^4y + 9x^3y + xy + 5y + 1,\\
x^6y + 9x^5y + 7x^4y + 10x^3y + 4x^2y + 4xy + 12y \rangle.
\end{multline*}
For $\gg_2$ the situation is fully analogous: $\mathfrak{h}_{21}= \mathfrak{h}_{22} = R$ and
\[
\mathfrak{h}_{23} = \Ideal{ 5x^2y + 5xy + 6y + 1, x^3y + 4x^2y + 4xy + 9y }.
\]
Finally we compute the equal degree factorization for each of the above factors using Algorithm~\ref{alg_ed_factor}. For $\mathfrak{h}_{13}$ we obtain the following primes
\begin{align*}
\mathfrak{p}_1 &= \Ideal{ x^3 + 4x^2 + 4x + 9, y + 6x^2 + 4x + 1 }\\
\mathfrak{p}_2 &= \Ideal{  x^3 + 5x^2 + 9x + 10, y + 3x^2 + 7x + 4 }\\
\intertext{and for $\mathfrak{h}_{23}$ we get}
\mathfrak{p}_3 &= \Ideal{ x^3 + 4x^2 + 4x + 9, y + 7x^2 + 9x + 12 }.
\end{align*}
Hence the complete factorization of~$\mathfrak{a}$ is $\mathfrak{p}_1\cdot \mathfrak{p}_2\cdot \mathfrak{p}_3^2$.

\subsubsection*{Example}
In this example, we consider an elliptic function field $K = \mathbb{F}_{19}(x, y)$ with full constant field $\mathbb{F}_{19}$, where
\[
y^2 + y = x^3 - 2x^2 + 1
\]
Take a and ideal
\begin{align*}
\mathfrak{a}
&= \langle x^{21} + 14x^{20} + 9x^{19} + 4x^{18} + 5x^{17} + 12x^{16} + 9x^{15} + 7x^{14} + 12x^{13} + 8x^{12}\\
&\quad + 3x^{11} + 8x^{10} + 14x^9 + 7x^8 + 12x^7 + x^6 + 9x^5 + 13x^4 + 9x^3 + 4x^2 + 18x + 4, \\
& x^3y + 6x^2y + 3xy + 17y + 7x^{18} + 7x^{17} + 11x^{16} + x^{15} + 18x^{13} + 8x^{12} + 9x^{11}\\
&\quad + 15x^{10} + 13x^9 + 18x^8 + 12x^7 + x^6 + 14x^5 + 10x^4 + 7x^3 + 15x^2 + 9x + 5\rangle.
\end{align*}
We again use Algorithm~\ref{alg_radical_decomp} to factor~$I$ into a product of radical ideals. It returns one trivial factor $\gg_{3} = R$ and three nontrivial factors $\gg_{1}, \gg_{2}$ and $\gg_{4}$ where
\begin{align*}
\gg_1 &= \langle x^3 + 6x^2 + 3x + 17, x^3y + 6x^2y + 3xy + 17y \rangle,\\
\gg_2 &= \Ideal{ x^3 + 4x + 17, y + 8x^2 + 2x + 9 }, \\
\gg_4 &= \Ideal{ x^3 + 2x^2 + 10x + 4, y + 8x^2 + 3x }.
\end{align*}
Now we compute the distinct degree factors of the above ideals. For $\gg_1$ we have two trivial factors $\mathfrak{h}_{11} = \mathfrak{h}_{13} = R$ and two nontrivial one, degrees~$2$ and~$4$, respectively:
\begin{align*}
\mathfrak{h}_{12} &= \Ideal{ x + 1 }.\\
\mathfrak{h}_{14} &= \Ideal{ x^2 + 5x + 17 }
\intertext{For $\gg_2$ it returns two trivial factors $\mathfrak{h}_{21} = \mathfrak{h}_{22} = R$ and one nontrivial, degree~$3$ factor}
\mathfrak{h}_{23} &= \Ideal{x^3 + 4x + 17, y + 8x^2 + 2*x + 9}.
\intertext{Similarly for~$\gg_4$ we have $\mathfrak{h}_{41} = \mathfrak{h}_{42} = R$ and }
\mathfrak{h}_{43} &= \Ideal{x^3 + 2x^2 + 10x + 4, y + 8x^2 + 3x}.
\end{align*}
Finally we use Algorithm~\ref{alg_ed_factor} to compute the equal degree factorization. It turns out that all four ideals $\mathfrak{h}_{12}, \mathfrak{h}_{14}, \mathfrak{h}_{23}$ and~$\mathfrak{h}_{43}$ are in fact prime. Denoting
\[
\mathfrak{p}_1 := \mathfrak{h}_{12},\quad
\mathfrak{p}_2 := \mathfrak{h}_{14},\quad
\mathfrak{p}_3 := \mathfrak{h}_{23},\quad
\mathfrak{p}_4 := \mathfrak{h}_{43},
\]
we obtain the complete factorization of~$\mathfrak{a}$, namely $\mathfrak{a} = \mathfrak{p}_1\cdot \mathfrak{p}_2\cdot \mathfrak{p}_3^2\cdot \mathfrak{p}_4^3$.

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redpajama_set_name: RedPajamaArXiv


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