## Alternative proof of unique factorization for ideals in a Dedekind ring

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I'm writing some commutative algebra notes, but I'm facing a difficulty in organizing the order of the topics. I'd like to have the topics about factorization before speaking of integral closure. This is fine, as long as I talk of UFD and primary decomposition.

The problem is that a topic worth mentioning is the factorization theory for ideals in a Dedekind ring. Now, there are a few ways to define a Dedekind ring, but I guess one of the most natural is a Noetherian domain, integrally closed, of dimension 1.

At this point I haven't yet introduced the concept of dimension, nor integral closure. It is easy not to speak of dimension, and just say that every prime ideal is maximal. I'm also fine in writing out explicitly what integrally closed means.

The real problem is to get a proof of unique factorization for ideals without using anything about integral closure, apart from direct arguments. For instance, I'd be fine in saying: "...so this element satisfies this monic equation, hence it is in A." Less so in saying "...so this element lies in a ring which is finitely generated as an A-module, hence it is integral. Since it is in the field of fractions of A, is must belong to A."

The only missing step in proving that ideals in a Dedekind ring satisfy unique factorization is the fact that primary ideals are prime powers.

Is there a direct proof of this fact which does not rely on anything about integrally closed domains, apart from the definition?

I should make clear that other standard techniques are available at this point: localization, Noetherian and Artinian stuff, primary decomposition, symbolic powers and so on.

I should also say that changing the order of the topics woule a major headache. I have thought out for long the order, and this is the only point where I get things in the wrong order. If possible, I would like to leave it as it is.

Edit (added in response to KConrad comment).

The steps which are easy are the following. Since $A$ is Noetherian, a primary decomposition exists. Since every prime is maximal, there are no embedded primes, so all primary components are unique. Finally, using again that every prime is maximal, all primary components are coprime, so intersections become products. So the only step where one uses integrality is the proof that the primary ideals are actually prime powers.

For the definition, there is no need to speak about integral closure, let alone proving that the integral closure is a ring. The integral domain $A$ is said to be integrally closed if every element $x$ of the quotient field of $A$, which satisfies a monic equation $x^n + a_{n-1} x^{n-1} + \cdots + a_0$ with coefficients in $A$, is itself in $A$.

As for the examples, the compromise for now is to list some number rings, with the promise that it will be shown in a later section that these are actually Dedekind rings. Of course I'm not happy with this solution. But I'm also not happy with putting an aside on integral closure in the middle of a section about factorization and primary decomposition; even less so because there IS a later section on integral closure.

I cannot even reverse the two, because in the section of integral closure I want to be able to speak about the integral closures of $\mathbb{Z}$, so I need the factorization theory for Dedekind rings.

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Could you tell us what steps you do have, and not just the missing step in your argument? You write that you are okay with defining some ring A as an integral closure (even if you don't use that term), but you don't want to bring in theorems that tell you something is integral. How then do you know that the integral closure is a ring in general? Can you tell us which examples you will be using to illustrate Dedekind rings? (I mean of course examples that are not UFDs.) – KConrad Mar 9 2010 at 17:30
I really like the local-global approach. A Dedekind domain is just a noetherian ring whose localizations are DVR. The theory of DVR is really simple and the factorisation property there is more or less trivial. Then you globalize. – YBL Mar 9 2010 at 18:48
Yes, but how do I show that localizations of a Dedekind ring are DVR? I have the same problem with using the property of being integrally closed. Moreover, I already have a global theory of factorization, which is primary decomposition, and I would like to be able to use it. – Andrea Ferretti Mar 9 2010 at 18:52
@Andrea - This is part of Theorem 3.16 in Janusz' "Algebraic Number Fields". – Ben Linowitz Mar 9 2010 at 21:05
@Ben: it was a rethorical question. :-) More explicitly, I just wanted to say that proving this fact requires using nontrivial properties of the integral closure. – Andrea Ferretti Mar 9 2010 at 22:32

## 5 Answers

I believe that the proof in Marcus's "Number Fields" contains a proof which does not rely on integral closure except to say that if an element satisfies a monic polynomial, it is in the domain. I'll summarize the lemmas he uses:

1) For any ideal $I$, there is an ideal $J$ so that $IJ$ is principle.

2) For any proper ideal $I$, there is an element $x$ in the field of fractions and not in the Dedekind domain so that $xI$ is still in the Dedekind domain.

To prove the second lemma he uses integrally closed, but only to show that an element of the field of fractions satisfies a monic polynomial, and is thus in the Dedekind domain.

3) The ideal classes form a group. This is a quick consequence of the previous lemmas.

4) Some group results about the ideals. (the google books view which I am using is missing the last page).

I think after that there is no more use of integrally closed, but as I said I'm missing the last page of the proof. Hope this helps.

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The proof is on pages 56-60. – Ben Weiss Mar 9 2010 at 1:31
Sadly, in the proof of Lemma 2 the monic equation is obtained by the determinant trick, which is the same that allows you to prove the assertion "an element which lies in a ring which is finitely generated as an A-module is integral over A". So it seems that Marcus is essentially avoiding the general theory, but he actually uses the more general argument in a specific case. – Andrea Ferretti Mar 9 2010 at 12:31

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Ireland-Rosen give a similar proof, based on the finiteness of the class number a la Kronecker. In my opinion, however, the dependence on integral closure should be spelled out as clearly as possible. Kummer, who was not aware of the concept of integral closure, gave two proofs that factorization into ideal prime numbers is unique, and both contained gaps that could not be closed without using integral closure. – Franz Lemmermeyer Mar 9 2010 at 19:13
Andrea, is the determinant trick so terrible? It is self-contained. And students will benefit from seeing it. – Ravi Vakil Mar 9 2010 at 22:13
Of course it is not so terrible. Indeed up to now this is the best solution. But the determinant trick will appear later anyway. Moreover, It would be nicer to deduce unique factorization from primary decomposition, which is already done. This is why I'm more interested in a proof of the fact that primary ideals in a Dedekind ring are powers of primes than in changing completely approach. That is, if it is possible. So my problem with this solution is both having to use the determinant trick now and having to rework from scratch the theory of factorization. – Andrea Ferretti Mar 9 2010 at 22:25
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Can you show, from whatever you have available to you in the course at this point, that maximal ideals have inverses (as $A$-modules)? It would then follow that if a max. ideal $\mathfrak m$ contains an ideal $\mathfrak a$ then $\mathfrak m$ is a factor of $\mathfrak a$. Then maybe by a Noetherian inductive process show any primary ideal is a prime power by starting with a primary ideal, picking a maximal ideal containing it (we know secretly there's only 1 choice), write your primary ideal as a product of that maximal ideal and another ideal, show that other ideal is primary, and repeat. At the end you could read off that all the maximal ideal factors must be the same. I'm not saying I have worked out the details on that, but it's still not completely clear to me what is known and not known to the students, so this is just a suggestion.

I looked at my own lecture notes to see how I showed a maximal ideal in a Dedekind domain has an inverse, and I used a variant on the determinant trick together with the ring being integrally closed. I vote for using the determinant trick. Then you'd use it now once and you have already told us that you will use it again later. After you use it twice, it becomes a method and not a trick. :) Let's think about it this way: you are willing to use the raw definition of being integrally closed, and you certainly need to use integral closedness somewhere, but if you want to produce a monic polynomial at some point so you can use the fact that your ring $A$ is integrally closed, where in the world are you going to get such polynomials from? The determinant trick is one way. What other way is there in this proof?

I am not sure that any solution you eventually find will be satisfying to the students, to whom this is being seen for the first time. The next time you teach the course, consider covering the material in a different order so you don't get caught in the same way.

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I may go for the determinant trick way. Anyway, I should point out that I'm not currently teaching the course, but only writing some notes. Sorry for the misunderstanding. – Andrea Ferretti Mar 9 2010 at 23:48
As for the maths, your idea seems very good. It is fine for me to use the determinant trick, as long as it allows me to reduce to primary factorization (as opposed to the proposal of Ben, which would also require reworking everything from scratch). The fact that a primary ideal is contained in only one maximal ideal follows easy from a localization argument. – Andrea Ferretti Mar 9 2010 at 23:53

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Oh, then of course you should reorder of the notes, regardless of the pain, if nobody else is relying on them (yet). Unless you're trying to satisfy yourself that you can really do something in a very specific logical order. – KConrad Mar 9 2010 at 23:55

In the specific case at stake, the determinant trick is quite easy to explain. Let $R$ be the Dedekind ring, $K$ its field of fractions, $I$ a nonzero ideal of $R$ and $w\in K$ such that $wI\subset I$. One wants to prove that $w\in R$.

Pick up a finite generating family of $I$, $(w_1,\dots,w_n)$. By assumption, $ww_i\in I$ for any $i$, so there exists a $n\times n$-matrix with coefficients in $R$, say $A=(a_{i,j})$, such that $ww_i=\sum a_{i,j}w_j$ for all $i$. Now this means that the vector $(w_1,\dots,w_n)$ is an eigenvector of $A$, with eigenvalue $w$. So $w$ is a root of the characteristic polynomial $P$ of $A$. Since $A$ has its coefficients in $R$, $P\in R[X]$. By definition of a Dedekind ring (the integrally closed property), $w\in R$.

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You could define a Dedekind ring as a noetherian domain s.t. the localization at any nonzero prime ideal is a discrete valuation ring (see the beginning of Serre's Local fields). From there it is easy to show unique factorization. It is of course a cheat since the equivalence between the two definitions relies on the "integrally closed" property.

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Here is a route I took when teaching this material to bright high school students at PROMYS. (Of course, not using this language.) Let $R$ be a subring of $\mathbb{C}$, of finite rank over $\mathbb{Z}$. I was actually only doing the particular case of $\mathbb{Z}[\sqrt{-D}]$, for some positive integer $D$, but you could presumably be more general with your audience.

It is easy to show that ideal classes form a semi-group, and that this semi-group is finite (using Minkowski's theorem). Moreover, the proof is constructive; they can compute the class semi-group in practice without difficulty. It is also easy to show that, if the class semi-group is a group, then unique factorization into prime ideals holds.

I then had them compute lots of examples, and see that the class semigroup often was a group. You can then discuss those examples without mentioning integral closure at all. When you do get to integral closure, you can have them check their list of examples and see that the class semi-group is a group precisely when the ring is integrally closed. Hopefully, this will make the notion seem better motivated. I never actually got to proving that "all ideals invertible" is equivalent to "integrally closed", but I don't see why I couldn't have if I had more time.

In your setting of general commutative algebra, my proposal is to define the class semi-group; show that one dimensional, Noetherian and class semi-group is a group implies unique factorization into ideals; and compute class semi-groups, using Minkowski's theorem, for the number fields which you wish to exhibit.

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Imaginary quadratic rings of integers are also nicer because in that case the inverse of an ideal class is the same as its complex conjugate. There's a simple proof of that fact here: math.umass.edu/~weston/oldpapers/cnf.pdf (theorem 2.13 on page 30); this is also the approach used in chapter 11 of Artin's /Algebra/. I don't know if this argument can be generalized to higher degree number fields; it seems like there would be difficulties extending this technique to non-Galois extensions of Q. – Alison Miller May 27 2010 at 21:34
That's a really cute trick! I did not know that. Now I'm thinking about whether I can generalize it. – David Speyer May 27 2010 at 22:13
I'm not actually teaching now; only writing the notes. Thank you for your answer, though! – Andrea Ferretti May 27 2010 at 22:32
Alison and David: the correct involution to use on ideal classes which generalizes the inverse formula to higher degree is dual lattices. If K is a number field and L is a Z-lattice in K with dual lattice L' (I mean dual w.r.t. the trace-pairing K x K ---> Q, as used in defining the different ideal for instance) then the "master formula" is LL' = R(L)', where R(L) = {x in K : xL \subset L} is the order associated to L. Now in the special case that L = Z[a], R(L) = (1/f'(a))Z[a] for f = min. poly. of a over Q. Passing to Z[a]-ideal classes, the eqn. LL' = R(L)' becomes [L][L'] = [1]. – KConrad May 28 2010 at 5:22
What's very special about the quadratic setting is that in a quadratic field all orders have the form Z[a] for some a, hence all Z-lattices L in the field are invertible fractional ideals relative to their natural associated order R(L), which is the only order w.r.t which the lattice could be an invertible fractional ideal at all. To emphasize that this is very special to the quadratic case, one can show that in every number field of degree greater than 2 there are infinitely many Z-lattices L that are not invertible as fractional R(L) ideals. :( – KConrad May 28 2010 at 5:25
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