text:
\section{Introduction}

Around 1990, Hopkins \cite{Hopkins} and Neeman \cite{Neeman} gave a classification theorem of the thick subcategories of the derived category of perfect complexes (i.e. finite complexes of finitely generated projective modules) over a commutative noetherian ring in terms of the ring spectrum.
After that, Thomason \cite{Thomason} generalized this classification theorem to quasi-compact and quasi-separated schemes, in particular, to arbitrary commutative rings.
Let $\D_\mathrm{perf}(R)$ denote the derived category of perfect complexes over a commutative ring $R$.
The classification theorem (for commutative rings) can be stated as follows.

\begin{thm*}[Hopkins-Neeman-Thomason]
Let $R$ be a commutative ring.
Then there is an isomorphism
\begin{align*}
& \big\{\,\text{thick subcategories of }\D_\mathrm{perf}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{complements of intersections of quasi-compact open subsets of }\mathrm{Spec}\,R\,\big\}
\end{align*}
of lattices.
\end{thm*}

Here we recall the definitions of several subcategories of an abelian category.
A coherent subcategory is defined to be a full subcategory which is closed under kernels, cokernels and extensions.
A Serre subcategory is defined to be a coherent subcategory which is closed under subobjects.
A torsion class is defined to be a Serre subcategory which is closed under arbitrary direct sums.
Let $R$ be a commutative ring.
We denote by $\mathrm{Mod}\,R$ the category of $R$-modules and by $\mathrm{mod}\,R$ the full subcategory of finitely presented $R$-modules.
If $R$ is noetherian, then the lattice of Serre subcategories of $\mathrm{mod}\,R$, the lattice of torsion classes of $\mathrm{Mod}\,R$, and the lattice of subsets of $\mathrm{Spec}\,R$ which are closed under specialization are isomorphic to each other.
Taking advantage of the Hopkins-Neeman-Thomason theorem, Garkusha and Prest \cite{GP,GP2} gave the following result very recently.
(A torsion class ${\mathcal X}$ of ${\mathcal M}=\mathrm{Mod}\,R$ is said to be of finite type if the inclusion functor ${\mathcal M}/{\mathcal X}\to{\mathcal M}$, where ${\mathcal M}/{\mathcal X}$ denotes the quotient category, preserves arbitrary direct sums.)

\begin{thm*}[Garkusha-Prest]
Let $R$ be a commutative ring.
Then the following hold.
\begin{enumerate}
\item[{\rm (1)}]
One has lattice isomorphisms
\begin{align*}
& \big\{\,\text{thick subcategories of }\D_\mathrm{perf}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{torsion classes of finite type of }\mathrm{Mod}\,R\,\big\}\\
\,\cong\, & \big\{\,\text{complements of intersections of quasi-compact open subsets of }\mathrm{Spec}\,R\,\big\}.
\end{align*}
\item[{\rm (2)}]
Suppose that $R$ is coherent.
Then one has lattice isomorphisms
\begin{align*}
& \big\{\,\text{thick subcategories of }\D_\mathrm{perf}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{Serre subcategories of }\mathrm{mod}\,R\,\big\}\\
\,\cong\, & \big\{\,\text{complements of intersections of quasi-compact open subsets of }\mathrm{Spec}\,R\,\big\}.
\end{align*}
\end{enumerate}
\end{thm*}

Also by using the Hopkins-Neeman-Thomason theorem, Hovey \cite{Hovey} proved the following classification theorem of coherent subcategories.

\begin{thm*}[Hovey]
Let $R$ be a quotient ring of a commutative coherent regular ring by a finitely generated ideal.
Then the following hold.
\begin{enumerate}
\item[{\rm (1)}]
One has a lattice isomorphism
\begin{align*}
& \big\{\,\text{thick subcategories of }\D_\mathrm{perf}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{coherent subcategories of }\mathrm{mod}\,R\,\big\}.
\end{align*}
\item[{\rm (2)}]
Every coherent subcategory of $\mathrm{mod}\,R$ is a Serre subcategory.
\end{enumerate}
\end{thm*}

As Hovey pointed it out as an interesting fact, there was no direct proof of the second assertion of the above theorem; it could not be proved without resorting to the rather difficult classification of thick subcategories of the derived category, namely the Hopkins-Neeman-Thomason theorem.
Hovey conjectures that the isomorphism stated in the above theorem always holds for commutative coherent rings.
(Recall that a commutative ring is called coherent if every finitely generated ideal is finitely presented.)

\begin{conj*}[Hovey]
Let $R$ be a commutative coherent ring.
Then one has a lattice isomorphism
\begin{align*}
& \big\{\,\text{thick subcategories of }\D_\mathrm{perf}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{coherent subcategories of }\mathrm{mod}\,R\,\big\}.
\end{align*}
\end{conj*}

One of the main purposes of this paper is to prove that this conjecture is true if $R$ is noetherian.
First of all, we will directly prove that every coherent subcategory of $\mathrm{mod}\,R$ is Serre; in the proof, we will not apply the Hopkins-Neeman-Thomason theorem.
Actually, we will not use the notion of a derived category in the proof.
Furthermore, the proof we will give is much simpler than Hovey's.
After that, we will prove that if $R$ is noetherian, then all the lattices appearing in the above part are isomorphic to each other.
Our first main theorem is the following.

\begin{thma}
Let $R$ be a commutative noetherian ring.
Then every coherent subcategory of $\mathrm{mod}\,R$ is a Serre subcategory, and one has the following isomorphisms of lattices:
\begin{align*}
& \big\{\,\text{thick subcategories of }\D_\mathrm{perf}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{coherent subcategories of }\mathrm{mod}\,R\,\big\}\\
\,=\, & \big\{\,\text{Serre subcategories of }\mathrm{mod}\,R\,\big\}\\
\,\cong\, & \big\{\,\text{torsion classes of }\mathrm{Mod}\,R\,\big\}\\
\,\cong\, & \big\{\,\text{subsets of }\mathrm{Spec}\,R\text{ closed under specialization}\,\big\}\\
\,=\, & \big\{\,\text{complements of intersections of quasi-compact open subsets of }\mathrm{Spec}\,R\,\big\}.
\end{align*}
\end{thma}

On the other hand, Neeman \cite{Neeman} showed the following theorem.

\begin{thm*}[Neeman]
Let $R$ be a commutative noetherian ring.
Then one has an isomorphism
\begin{align*}
& \big\{\,\text{localizing subcategories of }{\mathcal D}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{subsets of }\mathrm{Spec}\,R\,\big\}\\
\intertext{of lattices.
Moreover, this induces an isomorphism}
& \big\{\,\text{smashing subcategories of }{\mathcal D}(R)\,\big\}\\
\,\cong\, & \big\{\,\text{subsets of }\mathrm{Spec}\,R\text{ closed under specialization}\,\big\}
\end{align*}
of lattices.
\end{thm*}

Here, ${\mathcal D}(R)$ denotes the derived category of $\mathrm{Mod}\,R$.
A localizing subcategory is defined as a full triangulated subcategory which is closed under arbitrary direct sums, and a smashing subcategory is defined as a localizing subcategory such that Bousfield localization commutes with arbitrary direct sums.
The second main purpose of this paper is to construct a module version of the above Neeman's theorem:

\begin{thmb}
Let $R$ be a commutative noetherian ring.
Then one has an isomorphism
\begin{align*}
& \big\{\,\text{full subcategories of }\mathrm{mod}\,R\text{ closed under submodules and extensions}\,\big\}\\
\,\cong\, & \big\{\,\text{subsets of }\mathrm{Spec}\,R\,\big\}\\
\intertext{of lattices.
Moreover, this induces the isomorphism}
& \big\{\,\text{Serre subcategories of }\mathrm{mod}\,R\,\big\}\\
\,\cong\, & \big\{\,\text{subsets of }\mathrm{Spec}\,R\text{ closed under specialization}\,\big\}
\end{align*}
of lattices given in Theorem A.
\end{thmb}

In Section 2, we will give the precise definitions of subcategories which are stated above, and study several basic properties of them.
In Sections 3 and 4, we shall give proofs of Theorems A and B, respectively.

\begin{rem*}
It is known that Hovey's paper \cite{Hovey} contains an error, but it is not relevant to the results and arguments in this paper.
The error has recently been corrected by Krause \cite{Krause}.
\end{rem*}

\section{Basic properties}

In this section, we will give some definitions and several basic results most of which are necessary to state and prove the main results of this paper.

Let ${\mathcal A}$ be an additive category.
A full subcategory ${\mathcal X}$ of ${\mathcal A}$ is said to be {\em closed under isomorphisms} (or {\em replete}) provided that if $X$ is an object of ${\mathcal X}$ and $Y$ is an object of ${\mathcal A}$ which is isomorphic to $X$, then $Y$ is also an object of ${\mathcal X}$.
In this paper, by a {\em subcategory} we always mean a nonempty full subcategory which is closed under isomorphisms.

First of all, we recall the definitions of various types of closedness of a subcategory of an additive category.

\begin{defn}
Let ${\mathcal A}$ be an additive category, and let ${\mathcal X}$ be a full subcategory of ${\mathcal A}$.
We say that
\begin{enumerate}
\item[(1)]
${\mathcal X}$ is {\em closed under subobjects} (resp. {\em closed under quotient objects}) provided that if $X$ is an object of ${\mathcal X}$ and $Y\in{\mathcal A}$ is a subobject (resp. a quotient object) of $X$, then $Y$ is also an object of ${\mathcal X}$.
\item[(2)]
${\mathcal X}$ is {\em closed under direct summands} (or {\em closed under retracts}) provided that if $X$ is an object of ${\mathcal X}$ and $Y\in{\mathcal A}$ is a direct summand of $X$, then $Y$ is also an object of ${\mathcal X}$.
\item[(3)]
${\mathcal X}$ is {\em closed under finite direct sums} (resp. {\em closed under arbitrary direct sums}) if all finite direct sums (resp. arbitrary direct sums) of objects of ${\mathcal X}$ are objects of ${\mathcal X}$.
\item[(4)]
${\mathcal X}$ is {\em closed under extensions} provided that for any exact sequence $0 \to A \to B \to C \to 0$ in ${\mathcal A}$, if both $A$ and $C$ are objects of ${\mathcal X}$, then so is $B$.
\item[(5)]
${\mathcal X}$ is {\em closed under kernels} (resp. {\em closed under images}, {\em closed under cokernels}) if the kernel (resp. the image, the cokernel) of every morphism of objects of ${\mathcal X}$ is also an object of ${\mathcal X}$.
\item[(6)]
${\mathcal X}$ is {\em closed under homologies} if the homologies of every chain complex of objects of ${\mathcal X}$ are objects of ${\mathcal X}$.
\item[(7)]
${\mathcal X}$ is {\em closed under direct limits} provided that if $\{ X_\lambda\}_{\lambda\in\Lambda}$ is a direct system of objects of ${\mathcal X}$ then the direct limit $\varinjlim _{\lambda\in\Lambda}X_\lambda$ is an object of ${\mathcal X}$.
\end{enumerate}
\end{defn}

Here we study the relationships among the closed properties of a subcategory which are defined above.

\begin{prop}\label{bbasic}
Let ${\mathcal A}$ be an abelian category and ${\mathcal X}$ a subcategory of ${\mathcal A}$.
Then the following hold.
\begin{enumerate}
\item[{\rm (1)}]
If ${\mathcal X}$ is closed under kernels or cokernels, then ${\mathcal X}$ is closed under direct summands and contains the zero object of ${\mathcal A}$.
\item[{\rm (2)}]
If ${\mathcal X}$ is closed under extensions, then ${\mathcal X}$ is closed under finite direct sums.
\item[{\rm (3)}]
If ${\mathcal X}$ is closed under kernels and cokernels, then ${\mathcal X}$ is closed under images and homologies.
\item[{\rm (4)}]
If ${\mathcal X}$ is closed under subobjects and cokernels, then ${\mathcal X}$ is closed under quotient objects.
\item[{\rm (5)}]
If ${\mathcal X}$ is closed under arbitrary direct sums and quotient objects, then ${\mathcal X}$ is closed under direct limits.
\end{enumerate}
\end{prop}

\begin{pf}
(1) Let $X$ be an object of ${\mathcal X}$ and $Y$ an object of ${\mathcal A}$ which is a direct summand of $X$.
Then we can write $X=Y\oplus Z$ for some object $Z$ of ${\mathcal A}$.
Considering the morphism $f: X \to X$ given by $(y,z) \mapsto (0,z)$ for $y\in Y$ and $z\in Z$, we see that both the kernel and the cokernel of $f$ are isomorphic to $Y$.
By the assumption that ${\mathcal X}$ is closed under kernels or cokernels, the object $Y$ is in ${\mathcal X}$.
Therefore ${\mathcal X}$ is closed under direct summands.

On the other hand, since ${\mathcal X}$ is nonempty, there exists an object $W\in{\mathcal X}$.
Let $g:W\to W$ be the identity morphism.
Then both the kernel and the cokernel of $g$ are the zero object.
Since ${\mathcal X}$ is closed under kernels or cokernels, ${\mathcal X}$ contains the zero object.

(2) Let $X$ and $Y$ be two objects of ${\mathcal X}$.
Then there exists a natural split exact sequnece
$$
0 \to X \to X\oplus Y \to Y \to 0
$$
in ${\mathcal A}$.
Since ${\mathcal X}$ is closed under extensions, we see from this exact sequence that the direct sum $X\oplus Y$ is an object of ${\mathcal X}$.
This argument shows that ${\mathcal X}$ is closed under finite direct sums.

(3) Let $f:X \to Y$ be a morphism of objects of ${\mathcal X}$.
Then there is an exact sequence
$$
0 \to \mathrm{Im}\,f \to Y \overset{\pi}{\to} \mathrm{Coker}\,f \to 0
$$
in ${\mathcal A}$.
Since ${\mathcal X}$ is closed under cokernels, the object $\mathrm{Coker}\,f$ is in ${\mathcal X}$.
Noting that the object $\mathrm{Im}\,f$ is the kernel of $\pi$ and ${\mathcal X}$ is closed under kernels, we see that $\mathrm{Im}\,f$ is in ${\mathcal X}$.
Therefore ${\mathcal X}$ is closed under images.
That ${\mathcal X}$ is closed under kernels and images implies that ${\mathcal X}$ is closed under homologies.

(4) Let $X$ be an object of ${\mathcal X}$ and $Y\in{\mathcal A}$ a quotient object of $X$.
Then there exists a subobject $Z\in{\mathcal A}$ of $X$ such that $Y=X/Z$.
Since ${\mathcal X}$ is closed under subobjects, $Z$ is an object of ${\mathcal X}$.
Let $i:Z\to X$ be the natural inclusion.
Since $Y$ coincides with the cokernel of $i$ and ${\mathcal X}$ is closed under cokernels, we have $Y\in{\mathcal X}$.
This says that ${\mathcal X}$ is closed under quotient objects.

(5) Let $\{ X_\lambda\}_{\lambda\in\Lambda}$ be a direct system of objects of ${\mathcal X}$, and let $X=\varinjlim _{\lambda\in\Lambda}X_\lambda$ be the direct limit.
Then, by definition, $X$ is a quotient object of $Y:=\bigoplus _{\lambda\in\Lambda}X_\lambda$.
Since ${\mathcal X}$ is closed under arbitrary direct sums, $Y$ is an object of ${\mathcal X}$.
Since ${\mathcal X}$ is closed under quotient objects, $X$ is an object of ${\mathcal X}$.
Consequently, ${\mathcal X}$ is closed under direct limits.
\qed
\end{pf}

Next, we recall the definitions of a coherent subcategory, a Serre subcategory and a torsion class, which will play important roles throughout this paper.

\begin{defn}
Let ${\mathcal A}$ be an abelian category, and let ${\mathcal X}$ be a subcategory of ${\mathcal A}$.
Then
\begin{enumerate}
\item[(1)]
${\mathcal X}$ is called a {\em coherent subcategory} of ${\mathcal A}$ if it is closed under kernels, cokernels and extensions.
\item[(2)]
${\mathcal X}$ is called a {\em Serre subcategory} of ${\mathcal A}$ if it is a coherent subcategory which is closed under subobjects.
\item[(3)]
${\mathcal X}$ is called a {\em (hereditary) torsion class} of ${\mathcal A}$ if it is a Serre subcategory which is closed under arbitrary direct sums.
\end{enumerate}
\end{defn}

\begin{rem}
The original definition of a coherent subcategory is as follows: let ${\mathcal A}$ be an abelian category, and let ${\mathcal X}$ be a subcategory of ${\mathcal A}$.
It is said that ${\mathcal X}$ is coherent provided that for any exact sequence
$$
A \to B \to C \to D \to E
$$
in ${\mathcal A}$, if $A$, $B$, $D$ and $E$ are in ${\mathcal A}$, then so is $C$.
One can easily check that this definition is equivalent to our definition.
\end{rem}

A coherent subcategory, a Serre subcategory and a torsion class have the following properties, which immediately follow from Proposition \ref{bbasic}.

\begin{cor}\label{basic}
Let ${\mathcal A}$ be an abelian category.
\begin{enumerate}
\item[(1)]
Let ${\mathcal X}$ be a coherent subcategory of ${\mathcal A}$.
Then ${\mathcal X}$ contains the zero object of ${\mathcal A}$, and ${\mathcal X}$ is closed under finite direct sums, direct summands, images and homologies.
\item[(2)]
Let ${\mathcal X}$ be a Serre subcategory of ${\mathcal A}$.
Then ${\mathcal X}$ is closed under quotient objects.
\item[(3)]
Let ${\mathcal X}$ be a torsion class of ${\mathcal A}$.
Then ${\mathcal X}$ is closed under direct limits.
\end{enumerate}
\end{cor}

Throughout the rest of this paper, let $R$ be a commutative ring.
We denote by $\mathrm{Mod}\,R$ the category of $R$-modules, and by $\mathrm{mod}\,R$ the full subcategory of $\mathrm{Mod}\,R$ consisting of finitely presented $R$-modules.
Let us recall the definitions of a lattice and a homomorphism of lattices.

\begin{defn}
\begin{enumerate}
\item[(1)]
Let ${\mathcal L}$ be an ordered set.
\begin{enumerate}
\item[(i)]
Let $x,y\in{\mathcal L}$ be elements.
If the supremum (resp. infimum) of the set $\{ x,y\}$ exists, then it is called the {\em join} (resp. {\em meet}) of $x$ and $y$, and denoted by $x\vee y$ (resp. $x\wedge y$).
\item[(ii)]
It is said that ${\mathcal L}$ is a {\em lattice} if any two elements of ${\mathcal L}$ have both the join and the meet.
\end{enumerate}
\item[(2)]
A map $f:{\mathcal L} \to {\mathcal L}'$ of lattices is called a {\em (lattice) homomorphism} if $f(x\vee y)=f(x)\vee f(y)$ and $f(x\wedge y)=f(x)\wedge f(y)$ for all $x,y\in{\mathcal L}$.
A bijective lattice homomorphism is called a {\em (lattice) isomorphism}.
\end{enumerate}
\end{defn}

This paper will deal with the following lattices of subcategories of modules.

\begin{defn}
\begin{enumerate}
\item[(1)]
Let $\mathrm{coh}(R)$ be the coherent subcategory of $\mathrm{Mod}\,R$ generated by $R$.
We denote by ${\mathcal L}_{\sf coh}(\mathrm{coh}(R))$ the lattice of all coherent subcategories of $\mathrm{coh}(R)$.
\item[(2)]
Let $\mathrm{Serre}(R)$ be the Serre subcategory of $\mathrm{Mod}\,R$ generated by $R$.
We denote by ${\mathcal L}_{\sf Serre}(\mathrm{Serre}(R))$ the lattice of all Serre subcategories of $\mathrm{Serre}(R)$.
\item[(3)]
We denote by ${\mathcal L}_{\sf tors}(\mathrm{Mod}\,R)$ the lattice of all torsion classes of $\mathrm{Mod}\,R$.
\end{enumerate}
\end{defn}

\begin{rem}\label{infact}
If $R$ is noetherian, then one has $\mathrm{coh}(R) = \mathrm{Serre}(R) = \mathrm{mod}\,R$; see the first sentence and the latter half of Page 3185 of \cite{Hovey}.
Hence, whenever $R$ is a noetherian ring, one has ${\mathcal L}_{\sf coh}(\mathrm{coh}(R))={\mathcal L}_{\sf coh}(\mathrm{mod}\,R)$ and ${\mathcal L}_{\sf Serre}(\mathrm{Serre}(R))={\mathcal L}_{\sf Serre}(\mathrm{mod}\,R)$.
\end{rem}

A {\it perfect} $R$-complex $P_\bullet$ is defined to be an $R$-complex of the form
$$
P_\bullet=(0 \to P_s \to P_{s-1} \to \cdots \to P_{t+1} \to P_t \to 0),
$$
where each $P_i$ is a finitely generated projective $R$-module.
We denote by ${\mathcal D} (R)$ the derived category of the category $\mathrm{Mod}\,R$, and by $\D_\mathrm{perf}(R)$ the full subcategory of ${\mathcal D}(R)$ consisting of $R$-complexes which are isomorphic to perfect $R$-complexes.

\begin{rem}
Recall that an object $X$ of an additive category ${\mathcal A}$ is called {\em small} (or {\em compact}) if the functor $\mathrm{Hom} _{\mathcal A}(X,-)$ preserves arbitrary direct sums.
It is well-known that a complex of $R$-modules is quasi-isomorphic to a perfect complex if and only if it is a small object of ${\mathcal D}(R)$ (cf. \cite[3.7]{DGI}).
\end{rem}

\begin{defn}
\begin{enumerate}
\item[(1)]
Let ${\mathcal T}$ be a triangulated category and ${\mathcal X}$ a subcategory of ${\mathcal T}$.
Then we say that ${\mathcal X}$ is a {\it thick subcategory} if it satisfies the following two conditions.
\begin{enumerate}
\item[(a)]
${\mathcal X}$ is closed under direct summands.
\item[(b)]
For any exact triangle $A \to B \to C \to \Sigma A$ in ${\mathcal T}$, if two of the objects $A,B,C$ are in ${\mathcal X}$, then so is the third.
\end{enumerate}
\item[(2)]
We denote by ${\mathcal L}_{\sf thick} (\D_\mathrm{perf}(R))$ the lattice of all of the thick subcategories of ${\mathcal D} (R)$ whose objects are small, namely, the lattice of all thick subcategories of $\D_\mathrm{perf}(R)$.
\end{enumerate}
\end{defn}

Recall that a subset $S$ of $\mathrm{Spec}\,R$ is said to be {\em closed under specialization} provided that if $\mathfrak p$ is a prime ideal in $S$ and $\mathfrak q$ is a prime ideal containing $\mathfrak p$ then $\mathfrak q$ is also in $S$.
Dually, $S$ is said to be {\em closed under generalization} provided that if $\mathfrak p$ is a prime ideal in $S$ and $\mathfrak q$ is a prime ideal contained in $\mathfrak p$ then $\mathfrak q$ is also in $S$.
Note that every union of closed subsets of $\mathrm{Spec}\,R$ is closed under specialization.
Similarly, every intersection of open subsets of $\mathrm{Spec}\,R$ is closed under generalization.

\begin{defn}
We denote by ${\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$ the lattice of all subsets of $\mathrm{Spec}\,R$ that are closed under specialization, and by ${\mathcal L}_{\sf spcl}^0(\mathrm{Spec}\,R)$ the sublattice of ${\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$ consisting of all complements of arbitrary intersections of quasi-compact open subsets of $\mathrm{Spec}\,R$.
\end{defn}

\begin{rem}\label{compact}
An open subset $U$ of $\mathrm{Spec}\,R$ is quasi-compact if and only if $U=D(I):=\{\mathfrak p\in\mathrm{Spec}\,R\mid I\nsubseteq\mathfrak p\}$ for some finitely generated ideal $I$ of $R$; see the argument on the top of \cite[Page 72]{Hartshorne}.
Therefore, if $R$ is noetherian, then every open subset of $\mathrm{Spec}\,R$ is quasi-compact, and ${\mathcal L}_{\sf spcl}^0(\mathrm{Spec}\,R)$ consists of all unions of closed subsets of $\mathrm{Spec}\,R$.
\end{rem}

Hovey \cite{Hovey} constructs the following order-preserving maps among the lattices which we defined above:
$$
\begin{CD}
{\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R) \\
\begin{smallmatrix}
@A{\tau}AA & @VV{\sigma}V
\end{smallmatrix}\\
{\mathcal L}_{\sf tors}(\mathrm{Mod}\,R) \\
\begin{smallmatrix}
@A{\nu}AA & @VV{\mu}V
\end{smallmatrix}\\
{\mathcal L}_{\sf Serre}(\mathrm{Serre}(R)) \\
\begin{smallmatrix}
@A{\beta}AA & @VV{\alpha}V
\end{smallmatrix}\\
{\mathcal L}_{\sf coh}(\mathrm{coh}(R)) \\
\begin{smallmatrix}
@A{g}AA & @VV{f}V
\end{smallmatrix}\\
{\mathcal L}_{\sf thick}(\D_\mathrm{perf}(R))
\end{CD}
$$
The above maps are defined as follows.
\begin{enumerate}
\item[(1)]
For $S\in {\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$, let $\sigma(S)$ be the full subcategory of $\mathrm{Mod}\,R$ consisting of all $R$-modules $M$ with $\mathrm{Supp}\,M\subseteq S$.
For ${\mathcal X}\in{\mathcal L}_{\sf tors} (\mathrm{Mod}\,R)$, let $\tau({\mathcal X})$ be the union $\bigcup_{M\in{\mathcal X}}\mathrm{Supp}\,M$.
\item[(2)]
For ${\mathcal X}\in{\mathcal L}_{\sf tors} (\mathrm{Mod}\,R)$, let $\mu({\mathcal X})$ be the full subcategory of $\mathrm{Mod}\,R$ consisting of all $R$-modules $M$ in the intersection of ${\mathcal X}$ and $\mathrm{Serre}(R)$.
For ${\mathcal Y}\in{\mathcal L}_{\sf Serre} (\mathrm{Serre}(R))$, let $\nu({\mathcal Y})$ be the torsion class of $\mathrm{Mod}\,R$ generated by ${\mathcal Y}$.
\item[(3)]
For ${\mathcal Y}\in{\mathcal L}_{\sf Serre} (\mathrm{Serre}(R))$, let $\alpha ({\mathcal Y})$ be the full subcategory of $\mathrm{Mod}\,R$ consisting of all $R$-modules $M$ in the intersection of ${\mathcal Y}$ and $\mathrm{coh}(R)$.
For ${\mathcal Z}\in{\mathcal L}_{\sf coh} (\mathrm{coh}(R))$, let $\beta ({\mathcal Z})$ be the Serre subcategory of $\mathrm{Mod}\,R$ generated by ${\mathcal Z}$.
\item[(4)]
For ${\mathcal Z}\in{\mathcal L}_{\sf coh} (\mathrm{coh}(R))$, let $f({\mathcal Z})$ be the full subcategory of ${\mathcal D} (R)$ consisting of all complexes $X_\bullet\in\D_\mathrm{perf}(R)$ such that the $i$th homology $H_i(X_\bullet)$ is in ${\mathcal Z}$ for any $i\in\Bbb Z$.
For ${\mathcal W}\in{\mathcal L}_{\sf thick} (\D_\mathrm{perf}(R))$, let $g({\mathcal W})$ be the coherent subcategory of $\mathrm{Mod}\,R$ generated by $\{\, H_i(X_\bullet)\,|\,X_\bullet\in{\mathcal W},i\in\Bbb Z\,\}$.
\end{enumerate}

We recall here the definition of an adjoint pair of order-preserving maps.
Let $\phi: A \to B$ and $\psi: B \to A$ be two order-preserving maps between ordered sets.
Then the pair $(\phi,\psi)$ is said to be an {\em adjoint pair} provided that $\phi(a)\le b$ if and only if $a\le \psi(b)$ for any $a\in A$ and $b\in B$.
Concerning the above order-preserving maps among lattices, the following proposition holds.

\begin{prop}\label{morphisms}
\begin{enumerate}
\item[{\rm (1)}]
\begin{enumerate}
\item[{\rm (a)}]
The pair $(\tau,\sigma)$ is an adjoint pair.
\item[{\rm (b)}]
The composite map $\tau\sigma$ is the identity map.
\item[{\rm (c)}]
If $R$ is noetherian, then $\sigma$ is a lattice isomorphism and $\tau$ is the inverse homomorphism.
\end{enumerate}
\item[{\rm (2)}]
\begin{enumerate}
\item[{\rm (a)}]
The pair $(\nu,\mu)$ is an adjoint pair.
\item[{\rm (b)}]
The composite map $\nu\mu$ is the identity map.
\item[{\rm (c)}]
If $R$ is noetherian, then $\mu$ is a lattice isomorphism and $\nu$ is the inverse homomorphism.
\end{enumerate}
\item[{\rm (3)}]
\begin{enumerate}
\item[{\rm (a)}]
The pair $(\beta,\alpha)$ is an adjoint pair.
\item[{\rm (b)}]
If $R$ is noetherian, then the composite map $\beta\alpha$ is the identity map.
\end{enumerate}
\item[{\rm (4)}]
\begin{enumerate}
\item[{\rm (a)}]
The pair $(g,f)$ is an adjoint pair.
\item[{\rm (b)}]
The composite map $fg$ is the identity map.
\end{enumerate}
\end{enumerate}
\end{prop}

\begin{pf}
It is easy to check that every order-preserving bijective map between two lattices is a lattice isomorphism.
Hence we see from the arguments in Page 3185 of \cite{Hovey} that the assertions (1), (2) and (3) hold.
(Here, note that derived categories do not appear in those arguments.)

The assertion (a) and (b) in (4) are shown in \cite[Proposition 1.4]{Hovey} and \cite[Corollary 2.2]{Hovey}, respectively.
\qed
\end{pf}

The above proposition especially says that one has the following relationships between two modules whose supports have inclusion relation.

\begin{cor}\label{generated}
Let $R$ be a noetherian ring, and let $M,N$ be $R$-modules with $\mathrm{Supp}\,M\subseteq\mathrm{Supp}\,N$.
Then $M$ belongs to the torsion class of $\mathrm{Mod}\,R$ generated by $N$.
If $M$ and $N$ are finitely generated, then $M$ belongs to the Serre subcategory of $\mathrm{mod}\,R$ generated by $N$.
\end{cor}

\begin{pf}
Proposition \ref{morphisms} says that the maps $\sigma,\mu$ are isomorphisms whose inverse maps are $\tau,\nu$ respectively.
Let ${\mathcal T}$ be the torsion class of $\mathrm{Mod}\,R$ generated by $N$.
We have ${\mathcal T}=\sigma\tau({\mathcal T})$, which is the full subcategory of $\mathrm{Mod}\,R$ consisting of all $R$-modules $K$ with $\mathrm{Supp}\,K\subseteq\bigcup_{L\in{\mathcal T}}\mathrm{Supp}\,L$.
Since $N$ is in ${\mathcal T}$, we get $\mathrm{Supp}\,M\subseteq\mathrm{Supp}\,N\subseteq\bigcup_{L\in{\mathcal T}}\mathrm{Supp}\,L$.
Therefore $M$ is in ${\mathcal T}$.

Suppose that both $M$ and $N$ are finitely generated.
Let ${\mathcal S}$ be the Serre subcategory of $\mathrm{mod}\,R$ generated by $N$.
Then we have ${\mathcal S}=\mu\sigma\tau\nu({\mathcal S})$, which consists of all finitely generated $R$-modules $K$ with $\mathrm{Supp}\,K\subseteq\bigcup_{L\in{\mathcal T}}\mathrm{Supp}\,L(=\bigcup_{L\in{\mathcal S}}\mathrm{Supp}\,L)$.
Hence $M$ is in ${\mathcal S}$.
\qed
\end{pf}

\section{Coherent subcategories are Serre}

Throughout this section, let $R$ be a commutative ring.
We begin with proving the following theorem.

\begin{thm}\label{main1}
Let $R$ be a noetherian ring.
Let ${\mathcal X}$ be a full subcategory of $\mathrm{mod}\,R$ which is closed under finite direct sums, kernels and cokernels.
Then ${\mathcal X}$ is closed under submodules and quotient modules.
\end{thm}

\begin{pf}
According to Proposition \ref{bbasic}(4), it is enough to prove that ${\mathcal X}$ is closed under submodules.
Assume that ${\mathcal X}$ is not closed under submodules.
Then there exist an $R$-module $X$ in ${\mathcal X}$ and an $R$-submodule $M$ of $X$ such that $M$ does not belong to ${\mathcal X}$.
Since $R$ is noetherian and $X$ is a finitely generated $R$-module, $X$ is a noetherian $R$-module.
Hence we can choose $M$ to be a maximal element, with respect to the inclusion relation, of the set of $R$-submodules $M'$ of $X$ such that $M'$ does not belong to ${\mathcal X}$.
Since $M$ does not coincide with $X$, there is an element $x\in X-M$.
Set $Y=M+Rx$.
Note that $Y$ is an $R$-submodule of $X$ strictly containing $M$.
By the maximality of $M$, the module $Y$ is in ${\mathcal X}$.
Put $I=(M:x):=\{ a\in R\mid ax\in M \}$.
This is an ideal of $R$, and we easily see that the quotient $R$-module $Y/M$ is isomorphic to $R/I$.
There is an exact sequence
$$
0 \to M \to Y \overset{\pi}{\to} R/I \to 0
$$
of $R$-modules.
Since $M\notin{\mathcal X}$ and $Y\in{\mathcal X}$ and ${\mathcal X}$ is closed under kernels, we see from this exact sequence that $R/I$ must not be in ${\mathcal X}$.

On the other hand, the map $\pi$ in the exact sequence induces a surjective homomorphism
$$
\overline\pi :Y/IY \to R/I
$$
of $R/I$-modules, which sends the residue class of $y\in Y$ in $Y/IY$ to $\pi (y)$.
Of course $R/I$ is a projective $R/I$-module, so $\overline\pi$ is a split epimorphism.
Therefore $R/I$ is isomorphic to a direct summand of $Y/IY$.
The noetherian property of $R$ implies that the ideal $I$ is finitely generated; write $I=(a_1,a_2,\dots,a_n)R$ for some elements $a_1,a_2,\dots,a_n\in R$.
There is an exact sequence
$$
\begin{CD}
R^{\oplus n} @>(a_1,\dots,a_n)>> R @>>> R/I @>>> 0
\end{CD}
$$
of $R$-modules.
Tensoring the $R$-module $Y$ with this exact sequence yields another exact sequence of $R$-modules:
$$
\begin{CD}
Y^{\oplus n} @>(a_1,\dots,a_n)>> Y @>>> Y/IY @>>> 0.
\end{CD}
$$
The assumption of the theorem says that ${\mathcal X}$ is closed under finite direct sums, cokernels, and direct summands; see Proposition \ref{bbasic}(1).
Hence the direct sum $Y^{\oplus n}$ belongs to ${\mathcal X}$, and so does the module $Y/IY$, and therfore so does $R/I$.
This is a contradiction, which says that ${\mathcal X}$ is closed under submodules.
Thus the proof of the theorem is completed.
\qed
\end{pf}

Corollary \ref{basic}(1) says that any coherent subcategory of $\mathrm{mod}\,R$ is closed under finite direct sums, kernels and cokernels.
Thus, according to Theorem \ref{main1}, we obtain the following result, which is the former half part of Theorem A in the first section of this paper.

\begin{cor}\label{wideserre}
Let $R$ be a noetherian ring.
Then every coherent subcategory of $\mathrm{mod}\,R$ is a Serre subcategory of $\mathrm{mod}\,R$.
\end{cor}

Now, let us check that the subset ${\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$ coincides with ${\mathcal L}_{\sf spcl}^0(\mathrm{Spec}\,R)$ if $R$ is a noetherian ring.

\begin{prop}\label{J}
\begin{enumerate}
\item[{\rm (1)}]
Let $Z$ be a subset of $\mathrm{Spec}\,R$ which is closed under specialization.
Then one has
$$
Z=\bigcup_{\mathfrak p\in Z}V(\mathfrak p).
$$
\item[{\rm (2)}]
Let $R$ be a noetherian ring.
Then
$$
{\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)={\mathcal L}_{\sf spcl}^0(\mathrm{Spec}\,R).
$$
\end{enumerate}
\end{prop}

\begin{pf}
(1) Let $\mathfrak q$ be a prime ideal in $Z$.
Then $\mathfrak q$ is in $V(\mathfrak q)$, which is contained in $\bigcup_{\mathfrak p\in Z}V(\mathfrak p)$.
As to the opposite inclusion relation, take a prime ideal $\mathfrak q$ in $\bigcup_{\mathfrak p\in Z}V(\mathfrak p)$.
Then $\mathfrak q$ is in $V(\mathfrak p)$ for some $\mathfrak p\in Z$.
Since $Z$ is closed under specialization, we get $\mathfrak q\in Z$, as required.

(2) This immediately follows from Remark \ref{compact} and the assertion (1).
\qed
\end{pf}

To prove our next result, we prepare here the following two lemmas.

\begin{lem}\label{gf}
Let $R$ be a noetherian ring and ${\mathcal M}$ a subcategory of $\mathrm{mod}\,R$ which is closed under finite direct sums and cokernels.
Let $M$ be a cyclic $R$-module in ${\mathcal M}$.
Then there exists a perfect $R$-complex $X_\bullet$ such that $H_0(X_\bullet)$ is isomorphic to $M$ and that $H_j(X_\bullet)$ belongs to ${\mathcal M}$ for any $j\in\Bbb Z$.
\end{lem}

\begin{pf}
Since $M$ is cyclic, there exists an ideal $I$ of $R$ such that $M$ is isomorphic to $R/I$.
The noetherian property of $R$ implies that the ideal $I$ is finitely generated; let $\text{\boldmath $x$} = x_1,x_2,\dots,x_r$ be a system of generators of $I$.
Consider the Koszul complex $K_\bullet := K_\bullet (\text{\boldmath $x$} , R)$ of the sequence $\text{\boldmath $x$}$.
The complex $K_\bullet$ is a perfect $R$-complex, and the zeroth homology $H_0(K_\bullet)$ is equal to $R/I$, which is isomorphic to $M$.
Thus, to show the lemma, it suffices to check that the homology $H_j(K_\bullet)$ belongs to ${\mathcal M}$ for each $j\in\Bbb Z$.
Note from \cite[Proposition 1.6.5(b)]{BH} that the $R$-module $H_j(K_\bullet)=H_j(\text{\boldmath $x$},R)$ is annihilated by the ideal $I=\text{\boldmath $x$} R$.
Hence $H_j(K_\bullet)$ can be regarded as an $R/I$-module.
Since $R$ is noetherian, $H_j(K_\bullet)$ is finitely generated, hence finitely presented as an $R/I$-module.
It follows that there is an exact sequence
$$
(R/I)^{\oplus n} \to (R/I)^{\oplus m} \to H_j(K_\bullet) \to 0
$$
of $R/I$-modules.
Since ${\mathcal M}$ is closed under finite direct sums, the sum $(R/I)^{\oplus i}\cong M^{\oplus i}$ is an object of ${\mathcal M}$ for any $i\ge 0$.
Since ${\mathcal M}$ is closed under cokernels, the module $H_j(K_\bullet)$ is in ${\mathcal M}$, as desired.
\qed
\end{pf}

\begin{lem}\label{filt}
Let $M$ be a finitely generated $R$-module.
Then there exist exact sequences
$$
\begin{cases}
0 \to R/I_1 \to M_0 \to M_1 \to 0,\\
0 \to R/I_2 \to M_1 \to M_2 \to 0,\\
\qquad \vdots \\
0 \to R/I_{n-1} \to M_{n-2} \to M_{n-1} \to 0,\\
0 \to R/I_n \to M_{n-1} \to M_n \to 0
\end{cases}
$$
of $R$-modules such that $I_1,I_2, \dots , I_{n-1}, I_n$ are ideals of $R$, and $M_0=M$ and $M_n=0$.
\end{lem}

\begin{pf}
Let $x_1,x_2,\dots,x_n$ be a system of generators of $M$.
Set $M_0=M$.
We have an isomorphism $Rx_1\cong R/I_1$, where $I_1=\mathrm{Ann}(x_1)$.
Putting $M_1=M_0/Rx_1$, we have an exact sequence
$$
0 \to R/I_1 \to M_0 \to M_1 \to 0
$$
of $R$-modules.
Note that the $R$-module $M_1$ is generated by $n-1$ elements.
By induction on $n$, we can obtain such a system of exact sequences as in the lemma.
\qed
\end{pf}

Now we are in a position to prove the following theorem, which is the latter half part of Theorem A in the first section.
In the proof, we should note that all the isomorphisms in the theorem except ${\mathcal L}_{\sf coh}(\mathrm{coh}(R))\cong{\mathcal L}_{\sf thick}(\D_\mathrm{perf}(R))$ are obtained without using derived categories.

\begin{thm}
Let $R$ be a noetherian ring.
Then the homomorphisms $\sigma,\mu,\alpha,f$ (defined in the previous section) are lattice isomorphisms, and $\tau,\nu,\beta,g$ are their inverse homomorphisms, respectively.
Consequently, one has
\begin{align*}
{\mathcal L}_{\sf spcl}^0(\mathrm{Spec}\,R) & = {\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R) \\
& \cong {\mathcal L}_{\sf tors}(\mathrm{Mod}\,R) \\
& \cong {\mathcal L}_{\sf Serre}(\mathrm{Serre}(R)) \\
& = {\mathcal L}_{\sf coh}(\mathrm{coh}(R)) \\
& \cong {\mathcal L}_{\sf thick}(\D_\mathrm{perf}(R)).
\end{align*}
\end{thm}

\begin{pf}
The equality ${\mathcal L}_{\sf spcl}^0(\mathrm{Spec}\,R) = {\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$ is already shown in Proposition \ref{J}.
Proposition \ref{morphisms} says that the homomorphisms $\sigma,\mu$ are lattice isomorphisms and $\tau,\nu$ are the inverse homomorphisms of $\sigma,\mu$ respectively.
Hence we have ${\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)\cong {\mathcal L}_{\sf tors}(\mathrm{Mod}\,R)\cong {\mathcal L}_{\sf Serre}(\mathrm{Serre}(R))$.
It is seen from Corollary \ref{wideserre} and Remark \ref{infact} that both of the homomorphisms $\alpha$ and $\beta$ are the identity maps, and we have ${\mathcal L}_{\sf Serre}(\mathrm{Serre}(R))= {\mathcal L}_{\sf coh}(\mathrm{coh}(R))$.

It remains to prove that $f$ is an isomorphism with the inverse homomorphism $g$.
The composite map $fg$ is the identity homomorphism by Proposition \ref{morphisms}(4)(b), and the subcategory $gf({\mathcal X})$ is contained in ${\mathcal X}$ for every ${\mathcal X}\in{\mathcal L}_{\sf coh} (\mathrm{coh}(R))$ by Proposition \ref{morphisms}(4)(a).
Let us show that ${\mathcal X}$ is contained in $gf({\mathcal X})$.
Let $M$ be an $R$-module in ${\mathcal X}$.
Since Remark \ref{infact} guarantees that ${\mathcal X}$ is a subcategory of $\mathrm{mod}\,R$, $M$ is a finitely generated $R$-module, and we have a system of exact sequences as in Lemma \ref{filt}.
(In the following, we use the same notation.)
Since ${\mathcal X}$ is a coherent subcategory of $\mathrm{mod}\,R$ by Remark \ref{infact}, we see from Corollary \ref{wideserre} that ${\mathcal X}$ is a Serre subcategory of $\mathrm{mod}\,R$.
Hence ${\mathcal X}$ is closed under submodules and quotient modules in $\mathrm{mod}\,R$ by Corollary \ref{basic}(2).
From the above exact sequences we easily see that the cyclic $R$-module $R/I_i$ belongs to ${\mathcal X}$ for every $1\le i\le n$.
Note by Corollary \ref{basic}(1) that ${\mathcal X}$ is closed under finite direct sums.
Hence Lemma \ref{gf} shows that for each integer $1\le i\le n$ there exists a perfect $R$-complex $X_\bullet^{(i)}$ such that $R/I_i$ is isomorphic to $H_0(X_\bullet^{(i)})$ and that $H_j(X_\bullet^{(i)})$ belongs to ${\mathcal X}$ for any $j\in\Bbb Z$.
It easily follows from the definitions of the homomorphisms $f,g$ that the $R$-module $R/I_i$ belongs to $gf({\mathcal X})$.
Since $gf({\mathcal X})$ is a coherent subcategory, it is closed under extensions.
Hence from the system of exact sequences, we see that $M$ belongs to $gf({\mathcal X})$.
Therefore ${\mathcal X} $ is contained in $gf({\mathcal X} )$, and thus $gf$ is the identity homomorphism.
This completes the proof of our theorem.
\qed
\end{pf}

\section{In relation to Neeman's classification}

In this section, we shall give a module version of Neeman's classification theorem of localizing categories and smashing categories, which we stated in the first section of this paper.
Throughout this section, let $R$ be a commutative noetherian ring.

We denote by ${\mathcal L}_{\sf subext}(\mathrm{mod}\,R)$ the lattice of all subcategories of $\mathrm{mod}\,R$ which are closed under submodules and extensions, and by ${\mathcal L}(\mathrm{Spec}\,R)$ the lattice of all subsets of $\mathrm{Spec}\,R$.
We define maps
$$
\begin{cases}
\Phi:{\mathcal L}(\mathrm{Spec}\,R) \to {\mathcal L}_{\sf subext}(\mathrm{mod}\,R) \\
\Psi:{\mathcal L}_{\sf subext}(\mathrm{mod}\,R) \to {\mathcal L}(\mathrm{Spec}\,R)
\end{cases}
$$
by $\Phi(S)=\{\,M\in\mathrm{mod}\,R\mid\mathrm{Ass}\,M\subseteq S\,\}$ and $\Psi({\mathcal M})=\bigcup_{M\in{\mathcal M}}\mathrm{Ass}\,M$.
It is easy to check that these maps are lattice homomorphisms.
Let $\phi:{\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R) \to {\mathcal L}_{\sf Serre}(\mathrm{mod}\,R)$ and $\psi:{\mathcal L}_{\sf Serre}(\mathrm{mod}\,R) \to {\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$ be the composite maps $\mu\sigma$ and $\tau\nu$, where $\sigma$, $\tau$, $\mu$, $\nu$ are the homomorphisms defined in Section 2.
Note that the maps $\phi,\psi$ are given by $\phi(S)=\{\,M\in\mathrm{mod}\,R\mid\mathrm{Supp}\,M\subseteq S\,\}$ and $\psi({\mathcal M})=\bigcup_{M\in{\mathcal M}}\mathrm{Supp}\,M$ for $S\in{\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R)$ and ${\mathcal M}\in{\mathcal L}_{\sf Serre}(\mathrm{mod}\,R)$.

Recall that $\phi$ is an isomorphism and $\psi$ is its inverse homomorphism since $R$ is noetherian.
This section is mainly devoted to proving the following theorem.

\begin{thm}\label{main}
Let $R$ be a commutative noetherian ring.
Then the homomorphism $\Phi$ is an isomorphism and $\Psi$ is its inverse homomorphism.
Moreover, $\Phi$ and $\Psi$ induce the isomorphisms $\phi$ and $\psi$, respectively.
Thus one has the following commutative diagram.
$$
\begin{CD}
{\mathcal L}_{\sf subext}(\mathrm{mod}\,R) @>{\Psi}>{\cong}> {\mathcal L}(\mathrm{Spec}\,R) @>{\Phi}>{\cong}> {\mathcal L}_{\sf subext}(\mathrm{mod}\,R) \\
@AA{\subseteq}A @AA{\subseteq}A @AA{\subseteq}A \\
{\mathcal L}_{\sf Serre}(\mathrm{mod}\,R) @>{\psi}>{\cong}> {\mathcal L}_{\sf spcl}(\mathrm{Spec}\,R) @>{\phi}>{\cong}> {\mathcal L}_{\sf Serre}(\mathrm{mod}\,R)
\end{CD}
$$
\end{thm}

This theorem will be proved after showing the following two lemmas.

\begin{lem}\label{coprimary}
Let ${\mathcal M}$ be a subcategory of $\mathrm{mod}\,R$ which is closed under submodules and extensions, and let $M$ be a finitely generated $R$-module.
Suppose that $M$ has a unique associated prime $\mathfrak p$.
If $R/\mathfrak p$ is in ${\mathcal M}$, then so is $M$.
\end{lem}

\begin{pf}
Assume that $M$ is not in ${\mathcal M}$.
Set $M_0=M$, and let $f_{0,1},\dots,f_{0,s_0}$ be a system of generators of the $R$-module $\mathrm{Hom}_R(M_0,R/\mathfrak p)$.
There is an exact sequence
$$
\begin{CD}
0 @>>> M_1 @>>> M_0 @>{\left(
\begin{smallmatrix}
f_{0,1}\\
\vdots\\
\\
f_{0,s_0}
\end{smallmatrix}
\right)}>> (R/\mathfrak p)^{\oplus s_0}
\end{CD}
$$
of $R$-modules.
Since $M_0=M$ is not in ${\mathcal M}$ and $(R/\mathfrak p)^{\oplus s_0}$ is in ${\mathcal M}$ and ${\mathcal M}$ is closed under submodules and extensions, it is easily seen that $M_1$ must not be in ${\mathcal M}$.
In particular, $M_1\ne 0$ and hence $\mathfrak p$ is the unique associated prime of $M_1$.
Letting $f_{1,1},\dots,f_{1,s_1}$ be a system of generators of the $R$-module $\mathrm{Hom}_R(M_1,R/\mathfrak p)$, we have an exact sequence
$$
\begin{CD}
0 @>>> M_2 @>>> M_1 @>{\left(
\begin{smallmatrix}
f_{1,1}\\
\vdots\\
\\
f_{1,s_1}
\end{smallmatrix}
\right)}>> (R/\mathfrak p)^{\oplus s_1}.
\end{CD}
$$
Since $M_1$ is not in ${\mathcal M}$ and $(R/\mathfrak p)^{\oplus s_1}$ is in ${\mathcal M}$, we see that $M_2$ is not in ${\mathcal M}$, and that $\mathfrak p$ is the unique associated prime of $M_2$.
Iterating this procedure, for each integer $i\ge 0$ we obtain an exact sequence
$$
\begin{CD}
0 @>>> M_{i+1} @>>> M_i @>{\left(
\begin{smallmatrix}
f_{i,1}\\
\vdots\\
\\
f_{i,s_i}
\end{smallmatrix}
\right)}>> (R/\mathfrak p)^{\oplus s_i},
\end{CD}
$$
where $f_{i,1},\dots,f_{i,s_i}$ is a system of generators of the $R$-module $\mathrm{Hom}_R(M_i,R/\mathfrak p)$ and $\mathfrak p$ is the unique associated prime of $M_i$.
Localizing the descending chain $M=M_0\supseteq M_1\supseteq M_2\supseteq\cdots$ at $\mathfrak p$ yields a descending chain
$$
M_\mathfrak p=(M_0)_\mathfrak p\supseteq (M_1)_\mathfrak p\supseteq (M_2)_\mathfrak p\supseteq\cdots
$$
of $R_\mathfrak p$-modules.
Since the $R_\mathfrak p$-module $(M_i)_\mathfrak p$ has finite length for every $i$, there exists an integer $t$ such that $(M_t)_\mathfrak p=(M_{t+1})_\mathfrak p=(M_{t+2})_\mathfrak p=\cdots$.
The exact sequence
$$
\begin{CD}
0 @>>> (M_{t+1})_\mathfrak p @>{=}>> (M_t)_\mathfrak p @>{\left(
\begin{smallmatrix}
(f_{t,1})_\mathfrak p\\
\vdots\\
\\
(f_{t,s_t})_\mathfrak p
\end{smallmatrix}
\right)}>> \kappa(\mathfrak p)^{\oplus s_t},
\end{CD}
$$
shows that $\mathrm{Hom}_{R_\mathfrak p}((M_t)_\mathfrak p,\kappa(\mathfrak p))=R_\mathfrak p(f_{t,1})_\mathfrak p +\cdots +R_\mathfrak p(f_{t,s_t})_\mathfrak p =0$.
Therefore $(M_t)_\mathfrak p =0$.
This is a contradiction since $\mathfrak p\in\mathrm{Ass}\,M_t\subseteq\mathrm{Supp}\,M_t$.
Thus we conclude that $M$ is in ${\mathcal M}$.
\qed
\end{pf}

\begin{lem}\label{seisitu}
Let ${\mathcal M}$ be a subcategory of $\mathrm{mod}\,R$ which is closed under submodules and extensions.
Let $M$ be a finitely generated $R$-module.
Suppose that $R/\mathfrak p$ belongs to ${\mathcal M}$ for every $\mathfrak p\in\mathrm{Ass}\,M$.
Then $M$ also belongs to ${\mathcal M}$.
\end{lem}

\begin{pf}
Let $\mathfrak p_1,\dots,\mathfrak p_s$ be the associated primes of $M$, and let
$$
0 = N_1\cap\cdots\cap N_s
$$
be an irredundant primary decomposition of the zero submodule $0$ of $M$, where $N_i$ is a $\mathfrak p_i$-primary submodule of $M$ for $1\le i\le s$.
Then the natural homomorphism
$$
M=M/N_1\cap\cdots\cap N_s\to M/N_1\oplus\cdots\oplus M/N_s
$$
is injective.
Since $\mathfrak p_i$ is the unique associated prime of the $R$-module $M/N_i$, Lemma \ref{coprimary} implies that $M/N_i$ belongs to ${\mathcal M}$ for $1\le i\le s$.
Hence $M/N_1\oplus\cdots\oplus M/N_s$ belongs to ${\mathcal M}$, and so does $M$.
\qed
\end{pf}

Now we can achieve the main purpose of this section.

\begin{tpf}
Let $S$ be a subset of $\mathrm{Spec}\,R$.
The set $\Psi\Phi(S)$ is the union of $\mathrm{Ass}\,M$ where $M$ runs through finitely generated $R$-modules all of whose associated primes are in $S$.
It is trivial that this set is contained in $S$.
For a prime ideal $\mathfrak p\in S$, we have $\mathrm{Ass}_R\,R/\mathfrak p =\{\mathfrak p\}\subseteq S$.
Hence $\mathfrak p$ belongs to $\Psi\Phi(S)$, and therefore $\Psi\Phi(S)=S$.
Let ${\mathcal M}$ be a subcategory of $\mathrm{mod}\,R$ which is closed under submodules and extensions.
We have that $\Phi\Psi({\mathcal M})$ is the subcategory of $\mathrm{mod}\,R$ consisting of all finitely generated $R$-modules $N$ with $\mathrm{Ass}\,N\subseteq\bigcup_{M\in{\mathcal M}}\mathrm{Ass}\,M$, and it is obvious that $\Phi\Psi({\mathcal M})$ contains ${\mathcal M}$.
Let $N$ be a finitely generated $R$-module with $\mathrm{Ass}\,N\subseteq\bigcup_{M\in{\mathcal M}}\mathrm{Ass}\,M$.
Fix a prime ideal $\mathfrak p\in\mathrm{Ass}\,N$.
Then there exists an $R$-module $M\in{\mathcal M}$ with $\mathfrak p\in\mathrm{Ass}\,M$.
There is an injective homomorphism $R/\mathfrak p\to M$, and $R/\mathfrak p$ belongs to ${\mathcal M}$ since ${\mathcal M}$ is closed under submodules.
It follows from Lemma \ref{seisitu} that $N$ is in ${\mathcal M}$.
Hence $\Phi\Psi({\mathcal M})={\mathcal M}$.
Thus we conclude that $\Phi$ is an isomorphism whose inverse homomorphism is $\Psi$.

On the other hand, let $S$ be a subset of $\mathrm{Spec}\,R$ which is closed under specialization.
Let $M$ be a finitely generated $R$-module such that $\mathrm{Ass}\,M$ is contained in $S$, and take $\mathfrak p\in\mathrm{Supp}\,M$.
Then there is a prime ideal $\mathfrak q\in\mathrm{Min}\,M\subseteq\mathrm{Ass}\,M$ that is contained in $\mathfrak p$.
Since $\mathfrak q$ is in $S$ and $S$ is closed under specialization, $\mathfrak p$ is also in $S$.
Thus $\Phi(S)=\{\,M\in\mathrm{mod}\,R\mid\mathrm{Ass}\,M\subseteq S\,\}$ coincides with $\phi(S)=\{\,M\in\mathrm{mod}\,R\mid\mathrm{Supp}\,M\subseteq S\,\}$.
Let ${\mathcal M}$ be a Serre subcategory of $\mathrm{mod}\,R$.
Let $N\in{\mathcal M}$ and $\mathfrak p\in\mathrm{Supp}\,N$.
Choose a prime ideal $\mathfrak q\in\mathrm{Min}\,N$ which is contained in $\mathfrak p$.
Then $\mathfrak q$ is an associated prime of $N$, so there is an injective homomorphism $R/\mathfrak q\to N$.
Since ${\mathcal M}$ is closed under submodules, the module $R/\mathfrak q$ is in ${\mathcal M}$.
Noting that there is a surjective homomorphism $R/\mathfrak q \to R/\mathfrak p$ and that ${\mathcal M}$ is closed under quotient modules, $R/\mathfrak p$ is also in ${\mathcal M}$.
Hence we get $\mathfrak p\in\mathrm{Ass}\,R/\mathfrak p\subseteq\bigcup_{M\in{\mathcal M}}\mathrm{Ass}\,M$.
Therefore the set $\psi({\mathcal M})=\bigcup_{M\in{\mathcal M}}\mathrm{Supp}\,M$ is contained in $\Psi({\mathcal M})=\bigcup_{M\in{\mathcal M}}\mathrm{Ass}\,M$, and we see that $\Psi({\mathcal M})=\psi({\mathcal M})$.
It follows that $\phi$ and $\psi$ are induced from $\Phi$ and $\Psi$, respectively.
\qed
\end{tpf}

Here, let us check that an analogous result to Corollary \ref{generated} holds.
This actually follows from Lemma \ref{seisitu}.

\begin{cor}
Let $R$ be a noetherian ring.
Let $M$ and $N$ be finitely generated $R$-modules with $\mathrm{Ass}\,M\subseteq\mathrm{Ass}\,N$.
Then $M$ is in the full subcategory of $\mathrm{mod}\,R$ closed under submodules and extensions which is generated by $N$.
\end{cor}

\begin{pf}
Let ${\mathcal E}$ be the full subcategory of $\mathrm{mod}\,R$ closed under submodules and extensions which is generated by $N$.
According to Lemma \ref{seisitu}, we have only to show that the $R$-module $R/\mathfrak p$ is in ${\mathcal E}$ for every $\mathfrak p\in\mathrm{Ass}\,M$.
Let $\mathfrak p$ be a prime ideal in $\mathrm{Ass}\,M$.
The assumption says that $\mathfrak p$ is in $\mathrm{Ass}\,N$.
Hence there exists an injective homomorphism $R/\mathfrak p\to N$ of $R$-modules.
Since $N$ is in ${\mathcal E}$ and ${\mathcal E}$ is closed under submodules, $R/\mathfrak p$ is also in ${\mathcal E}$, as required.
\qed
\end{pf}

In the following example, we will give several correspondences between subcategories of $\mathrm{mod}\,R$ which are closed under submodules and extensions and subsets of $\mathrm{Spec}\,R$, which are made by the isomorphisms $\Phi$ and $\Psi$.
Before that, we need to prepare some notation.
Let $I$ be an ideal of $R$, and let $M,N$ be $R$-modules.
We denote by $\Gamma_I(M)$ the {\it $I$-torsion submodule} of $M$, namely, the set of elements of $M$ which are annihilated by some power of $I$.
Recall that an $R$-module $M$ is called {\it $I$-torsion} if $\Gamma_I(M)=M$, and that $M$ is called {\it $I$-torsionfree} if $\Gamma_I(M)=0$.
It is well-known and easy to see that $M$ is $I$-torsion if and only if $\mathrm{Ass}\,M\subseteq V(I)$, and that $M$ is $I$-torsionfree if and only if $\mathrm{Ass}\,M\cap V(I)=\emptyset$.
We set $\mathrm{grade}(N,M)=\inf\{\,i\,|\,\mathrm{Ext}_R^i(N,M)\ne 0\,\}$, $\mathrm{grade}(I,M)=\mathrm{grade}(R/I,M)$, $\mathrm{grade}\,I=\mathrm{grade}(I,R)$ and $\mathrm{grade}\,M=\mathrm{grade}(\mathrm{Ann}\,M,R)$.

\begin{ex}
The isomorphisms $\Phi$ and $\Psi$ make the following correspondences.
Let $n$ be a nonnegative integer, $I$ an ideal of $R$ and $X$ a finitely generated $R$-module.
\begin{enumerate}
\item[{\rm (1)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ M\text{ is }I\text{-torsion}\,\big\}\,\leftrightarrow\,V(I)$.\\
\item[{\rm (2)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{grade}(X,M)>0\,\big\}\,\leftrightarrow\,\mathrm{Spec}\,R\setminus\mathrm{Supp}\,X$.\\
\item[{\rm (3)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ M\text{ is }I\text{-torsionfree}\,\big\}\\
\,=\,\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{grade}(I,M)>0\,\big\}\,\leftrightarrow\,D(I)$.\\
\item[{\rm (4)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{grade}(M,X)\ge n\,\big\}\,\leftrightarrow\,\big\{\,\mathfrak p\in\mathrm{Spec}\,R\ \big|\ \mathrm{grade}(\mathfrak p,X)\ge n\,\big\}$.\\
\item[{\rm (5)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{rank}\,M=0\,\big\}\,=\,\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{grade}\,M>0\,\big\}\\
\,\leftrightarrow\,\big\{\,\mathfrak p\in\mathrm{Spec}\,R\ \big|\ \mathrm{grade}\,\mathfrak p>0\,\big\}$.\\
\item[{\rm (6)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \text{every }X\text{-regular element is }M\text{-regular}\,\big\}\\
\,\leftrightarrow\,\big\{\,\mathfrak p\in\mathrm{Spec}\,R\ \big|\ \mathrm{grade}(\mathfrak p,X)=0\,\big\}$.\\
\item[{\rm (7)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ M\text{ is torsionfree}\,\big\}\,\leftrightarrow\,\big\{\,\mathfrak p\in\mathrm{Spec}\,R\ \big|\ \mathrm{grade}\,\mathfrak p=0\,\big\}$.\\
\item[{\rm (8)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{ht}\,\mathrm{Ann}\,M\ge n\,\big\}\,\leftrightarrow\,\big\{\,\mathfrak p\in\mathrm{Spec}\,R\ \big|\ \mathrm{ht}\,\mathfrak p\ge n\,\big\}$.\\
\item[{\rm (9)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \mathrm{dim}\,M\le n\,\big\}\,\leftrightarrow\,\big\{\,\mathfrak p\in\mathrm{Spec}\,R\ \big|\ \mathrm{dim}\,R/\mathfrak p\le n\,\big\}$.\\
\item[{\rm (10)}]
$\big\{\,M\in\mathrm{mod}\,R\ \big|\ \ell(M)<\infty\,\big\}\,\leftrightarrow\,\mathrm{Max}\,R$.
\end{enumerate}
\end{ex}

\begin{pf}
In each correspondence, we denote by ${\mathcal M}$ the left-hand subcategory of $\mathrm{mod}\,R$, and by $S$ the right-hand subset of $\mathrm{Spec}\,R$.
Note that it is enough to check either that $\Psi({\mathcal M})=S$ or that $\Phi(S)={\mathcal M}$ since $\Phi$ is an isomorphism with the inverse homomorphism $\Psi$.

(1) The subcategory ${\mathcal M}$ consists of all finitely generated $R$-modules $M$ with $\mathrm{Ass}\,M\subseteq V(I)$, which coincides with $\Phi(V(I))$.
Hence $\Phi(S)={\mathcal M}$.

(2) Let $M$ be a finitely generated $R$-module.
Note that $\mathrm{Ass}\,M\subseteq\mathrm{Spec}\,R\setminus\mathrm{Supp}\,X$ if and only if $\mathrm{Ass}\,M\cap\mathrm{Supp}\,X=\emptyset$, if and only if $\mathrm{Ass}\,\mathrm{Hom}(X,M)=\emptyset$ (cf. \cite[Exercise 1.2.27]{BH}), if and only if $\mathrm{Hom}(X,M)=0$.
Thus we have $\Phi(S)={\mathcal M}$.

(3) The equality is well-known.
Putting $X=R/I$ in (2), we obtain the correspondence.

(4) Let $M$ be a finitely generated $R$-module with $\mathrm{grade}(M,X)\ge n$, and let $\mathfrak p\in\mathrm{Ass}\,M$.
Then there is an $X$-regular sequence $\text{\boldmath $a$}=a_1,\dots,a_n$ in $\mathrm{Ann}\,M$, and $\mathfrak p\in\mathrm{Supp}\,M$.
Hence $\text{\boldmath $a$}$ is an $X$-regular sequence in $\mathfrak p$, and we have $\mathrm{grade}(\mathfrak p,X)\ge n$.
Therefore $\Psi({\mathcal M})$ is contained in $S$.
Conversely, if $\mathfrak p$ is a prime ideal with $\mathrm{grade}(\mathfrak p,X)\ge n$, then $R/\mathfrak p\in{\mathcal M}$ and $\mathfrak p\in\mathrm{Ass}\,R/\mathfrak p$.
Hence $\mathfrak p$ is in $\Psi({\mathcal M})$.
Therefore $S$ is contained in $\Psi({\mathcal M})$, and thus we get $\Psi({\mathcal M})=S$.

(5) Let $M$ be an $R$-module.
We have that $\mathrm{rank}\,M=0$ if and only if $M_\mathfrak p=0$ for every $\mathfrak p\in\mathrm{Ass}\,R$, if and only if $\mathrm{Supp}\,M\cap\mathrm{Ass}\,R=\emptyset$, if and only if $\mathrm{Ass}\,\mathrm{Hom}(M,R)=\emptyset$, if and only if $\mathrm{Hom}(M,R)=0$, namely $\mathrm{grade}\,M>0$.
Thus the equality holds.
For the correspondence, put $X=R$ and $n=1$ in (4).

(6) Let $\mathfrak p$ be a prime ideal in $\Psi({\mathcal M})$.
Then there exists an $R$-module $M\in{\mathcal M}$ of which $\mathfrak p$ is an associated prime.
Assume that $\mathrm{grade}(\mathfrak p,X)>0$.
Then there is an $X$-regular element $a\in\mathfrak p$, and this is also $M$-regular.
This is a contradiction since $\mathfrak p\in\mathrm{Ass}\,M$.
Thus $\mathrm{grade}(\mathfrak p,X)=0$, namely $\mathfrak p$ belongs to $S$.
On the contrary, let $\mathfrak p$ be a prime ideal with $\mathrm{grade}(\mathfrak p,X)=0$.
Then there exists an associated prime $\mathfrak q$ of $X$ which contains $\mathfrak p$.
Let $a$ be an $X$-regular element.
Then $a$ is not in $\mathfrak q$, so is not in $\mathfrak p$.
Hence $a$ is an $R/\mathfrak p$-regular element, and $R/\mathfrak p$ belongs to ${\mathcal M}$.
Since $\mathfrak p\in\mathrm{Ass}\,R/\mathfrak p$, the prime ideal $\mathfrak p$ is in $\Psi({\mathcal M})$, and it holds that $\Psi({\mathcal M})=S$.

(7) Put $X=R$ in (6), and we get this correspondence.

(8) If $M$ is a finitely generated $R$-module with $\mathrm{ht}\,\mathrm{Ann}\,M\ge n$, then $\mathrm{ht}\,\mathfrak p\ge n$ for all $\mathfrak p\in\mathrm{Supp}\,M$, hence for all $\mathfrak p\in\mathrm{Ass}\,M$.
Therefore $\Psi({\mathcal M})$ is contained in $S$.
If $\mathfrak p$ is a prime ideal of height at least $n$, then the ideal $\mathrm{Ann}\,R/\mathfrak p$ of $R$ also has height at least $n$ and $\mathfrak p\in\mathrm{Ass}\,R/\mathfrak p$.
Hence $\Psi({\mathcal M})$ contains $S$, and thus $\Psi({\mathcal M})=S$.

(9) Let $M$ be a finitely generated $R$-module.
Then $\mathrm{Ass}\,M$ is contained in $S$ if and only if $\mathrm{dim}\,R/\mathfrak p\le n$ for every $\mathfrak p\in\mathrm{Ass}\,M$, if and only if $\mathrm{dim}\,M\le n$.
Therefore $\Phi(S)={\mathcal M}$.

(10) Putting $n=0$ in (9) yields this correspondence.
\qed
\end{pf}

Note that in the correspondences (1), (4), (5), (8), (9) and (10) in the above example, the left-hand subcategories of $\mathrm{mod}\,R$ are Serre subcategories and the right-hand subsets of $\mathrm{Spec}\,R$ are closed under specializations, hence those correspondences are in fact obtained by the isomorphisms $\phi$ and $\psi$.

\begin{ac}
The author would like to give his deep gratitude to Luchezar L. Avramov and Srikanth Iyengar for valuable discussions and useful suggestions, and thanks Henning Krause and Osamu Iyama very much for helpful comments.
Most of the results stated in this paper were obtained while the author visited the University of Nebraska-Lincoln in the summer of 2006.
The author is also greatly indebted to the University of Nebraska-Lincoln for the hospitality and support.
The author also thanks an anonymous referee for important comments.
\end{ac}

redpajama_set_name: RedPajamaArXiv


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Daily Goal: I have hopes of being able to take a walk this afternoon. It's pretty cold, but I have a coat.
Daily Goal: Should be a good day with few temptations. Stay the course. Tai Chi this afternoon.


text:
Erdogan Says Rights of Turkish Minority in Greece Must Be Respected
By Staff Writer	 On Dec 8, 2017
Turkish President Erdogan visits Western Thrace on Friday.
In a visit marred by tense exchanges and rebukes, President Recep Tayyip Erdogan called for the rights of Turkish-speaking Muslim minority in Western Thrace to be respected.
The Turkish leader's trip to northeastern Greece, home to a Turkish population, aroused a spirited fervor among the region's Turks.
"We (Turkey) continue to work hard to improve the quality of life of our Greek origin citizens, like our citizens. And we believe it is our right to expect the same approach from Greece," Mr. Erdogan said during a speech in the town of Komotini.
Nearly 1,000 members of the minority group — which numbers between 120,000 and 150,000 in total — warmly welcomed the Turkish president to the region which borders Turkey.
Mr. Erdogan's official visit to Greece on Thursday was the first by a Turkish president in 65 years.
He denounced the "discrimination" of the Muslim minority during talks with Greek Prime Minister Alexis Tsipras, citing terms in the 1923 Lausanne Treaty which he said had not been "respected."
The treaty, which defined the borders of the Turkish state at the end of the Ottoman Empire, includes a Greco-Turkish chapter devoted to minority rights.
Contrary to its terms, Athens appoints religious jurists known as muftis instead of allowing the local community to do so, the Turkish president said in Athens.
"You are the bridge between Turkey and Greece, we see you like this," Mr. Erdogan told the crowd as he called for "unity" and "solidarity."
"Right now, in the Greek parliament, four MPs represent you. These lawmakers must work hard."
Mr. Tsipras on Thursday countered President Erdogan's claims by urging him to speed up social reforms in Turkey, where many are still imprisoned after a crackdown by his government that followed the attempted army coup last year.
President Erdogan's remarks for revising borders between the two countries sent a chilling echo, aired unexpected feeling of shock among the hosts during the press conference broadcasted live on TVs.
The Turkish leader's call for revisiting Lausanne Treaty stumbled into a swift rebuke from his Greek counterpart who reminded that the treaty could not be updated without the consent of all international powers involved in peace talks in 1923, including Japan.
For Greek leaders, the treaty has been a sensitive issue and any irredentist move to revisit it could upset the delicate balance in the region and in relations between two countries.
AFP with Staff Writer
Zarrab: Erdogan's Son-In-Law Demanded Revival of Iran Gold Trade
In Major Media Trial, Turkish Court Releases Only Three Journalists
Turkey Reports Death of Fmr. Army Commander from Coronavirus, Pledges to Step Up…
Turkey Calls for Fresh Arrest of Rights Defender Kavala After Acquittal

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Gridview doesnt update after the delete command Basically, I use objectdatasource with gridview.. Everything works, but the gridview, doesnt update as soon as I click the delete button. This forces me to use Server.Transfer on the gridviews events:
protected void GridView1_RowDeleted(object sender, GridViewDeletedEventArgs e)
{
   //Server.Transfer(~/SamePageAsTheGridview);
}

Is there a way for the data to be updated without full post back?

A: After deleting the selected item from your objectsource,
Just rebind the data to the gridview after delete.

A: protected void GridView1_RowDeleted(object sender, GridViewDeletedEventArgs e)
{
   GridView1.DataBind();
}

redpajama_set_name: RedPajamaStackExchange


text:
Q: Flask-Admin Cannot find reverse relation I already have an application using my SQLAlchemy model, and I'm now trying to add a flask admin website on top to manage some of the data in the DB. This model is already setup and working throughout the existing application so I know the relationships are configured correctly.
In my DB the data_package table has a FK to the supplier table, and this has been added as a relationship. However, when I create a flask model view to allow editing of the data_package I'm receiving the error:

Exception: Cannot find reverse relation for model

I've added a subset of the code below, there are a lot more columns and logic but this is the bare minimum I have to try and test a fix to the issue.The issue is happening because I am adding inline_models = (Supplier,) which I need as I want to have a select to allow the supplier to be changed for a data package.
The line of code throwing the exception is https://github.com/flask-admin/flask-admin/blob/master/flask_admin/contrib/sqla/form.py#L558. The reason for this is the relationship direction is defined as ONETOMANY, and there is a check to ensure the direction is MANYTOONE or MANYTOMANY
Base = declarative_base()

class Supplier(Base):
    __tablename__ = 'supplier'
    id = Column(Integer, primary_key=True)
    name = Column(String(100), nullable=False)

class DataPackage(Base):
    __tablename__ = 'data_package'
    id = Column(Integer, primary_key=True)
    name = Column(String(100), nullable=False)

    supplier_id = Column(Integer, ForeignKey('supplier.id'), index=True, nullable=False)
    supplier = relationship("Supplier", backref='packages')

class DataPackageAdminView(ModelViewOnly):
    form_columns = ['name', 'supplier']
    inline_models = (Supplier,)

def create_app():
    settings.configure_orm()

    app = Flask(__name__)
    app.config.from_pyfile('config.py')

    @app.route('/')
    def index():
        return 'Click me to get to Admin!'

    admin = Admin(app, name='Test Editor', template_mode='bootstrap3')
    admin.add_view(DataPackageAdminView(DataPackage, settings.Session, category='Data Sources'))

    return app

if __name__ == '__main__':
    app = create_app()
    app.run(debug=True)


The Red Orange of Sicily has enjoyed PGI status since 1996. The Consorzio di Tutela Arancia Rossa di Sicilia IGP is an association of 600 producers and 65 packaging firms.
A limited production: the maximum yield per hectare is 25 000 kg/ha (Tarocco), 35 000 kg/ha (Moro), 32 000 kg/ha (Sanguinello).
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The University of Maryland (commonly referred to as UMD, or just Maryland) is one of the most frequently searched universities in the United States, according to a recent internet keyword search of American colleges.
UMD is a public research university whose main campus is located in College Park, approximately four miles from Washington DC.
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UMD offers nearly 100 undergraduate majors, as well as just as many more graduate degrees and professional certificate programs.
The undergraduate acceptance rate for Maryland is about 45%. The average student faculty ratio is about 17 to 1.
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The main campus of UMD is located in College Park, Maryland, although UMD has additional locations in Baltimore County.
Above: The Eastern Shore campus of UMD.
2. What is the organizational structure of the University of Maryland?
3. What undergraduate studies are offered at the University of Maryland?
Undergraduate course offerings at UMD cover a wide range of topics.
Above: Students bask near a peaceful waterfall in front of the McKeldin Library.
4. What graduate programs are available at the University of Maryland?
Click here to see a Complete List of PhD Programs at UMD.
Above: The UMD campus in Baltimore.
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$35,216 per year for enrollments from out of state.
The above cost is for tuition only, and does not include room & board, books & supplies, personal expenses, and mandatory fees.
$1,548 per credit hour for enrollments from out of state.
The above costs do not include room and board, books and supplies, personal expenses, or mandatory fees.
Above: Fans gather to watch the Terrapins play at Maryland Stadium.
Above: The UMD School of Nursing.
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The main campus of UMD spans about 1,250 acres in College Park, with a 400-acre "urban forest" right next to the campus. College Park is conveniently located just about ten miles northeast of Washington DC.
The central lawn of the College Park campus (known as McKeldin Mall) is surrounded by stunning, white-columned buildings of red brick Georgian architecture. The campus has so many trees (7,500+), that the American Public Gardens Association has officially designated the College Park campus of UMD as an "arboretum and botanical garden."
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Just east of the campus is Baltimore Avenue, featuring many conveniences and affordable eateries. The campus is bordered on the west by University Boulevard.
Student housing is available at UMD, both on and off campus.
The Baltimore campus of UMD consists of 58 buildings, and is located very near Baltimore's famous Lexington Market.
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The school colors are red, gold, white, and black. The varsity athletes at UMD are called the Terrapins. They play for the NCAA Division 1 - Big Ten Conference.
UMD also features an award winning dance team.
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Aside form athletics, UMD also offers numerous on-campus student organizations, in a broad range of interests. Student activity groups include an African Languages association, a campus magazine, a campus ministry, a cooking club, a karate club, a theater group, a film society, and an ROTC club, to name only a few.
Above: The dance team from UMD at the 2014 National Championship, where they took 1st place.
7. What are the distance learning options at the University of Maryland?
8. What is the contact information for the University of Maryland?
To reach the main campus of UMD, or for general information, call 301-405-1000.
To reach the office of Undergraduate Admissions at UMD, call 301-314-8377.
For the Graduate Admissions office of UMD, call 301-405-3644.
To reach UMD toll free, call 800-422-5867.
To contact the Eastern Shore campus of UMD, call 410-651-2200.
To reach the Baltimore County facility of UMD, call 410-455-1000.


text:
title: Python Lesson 15
layout: post
author: samuel.brown
permalink: /python-lesson-15/
source-id: 1IHux56BAQDFqkWuf87tPkU6MB51yV3ZEpF8zxPE7bK8
published: true
---

    Title:
    Python Lesson 15
    Date:
    17/07/17

    Starting point:
    38%

    Target for this lesson?
    45%

    Did I reach my target?
(add details to "Lesson Review")
    Yes I reached 52%

    What was your favourite part of IT in year 8?

    My favourite part of IT in year 8, was when my percentage went up a bit. It was a nice feeling to know that I was getting somewhere. Also, there was the competitive side which involved me trying to get past my friends. Lucky for me, seeing as they seemed to get stuck on one of the questions for a while today, I managed to speed ahead and finish with 52%.

    What have I found hard this year?

    Seeing as I joined year 8 late, one of my struggles was to catch up with everyone else. Another thing I found hard, is the actual website. At the beginning of the year, I struggled with making and personalising my github website but by the second term I was fine and ready to carry on.

    What am I looking forward to next year?

    Next year I look forward to taking my github website a step further and look forward to the successes of my work

redpajama_set_name: RedPajamaGithub


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Hi everyone! I've received a few enquiries recently about Order of Service cards and the best way to format them. In my opinion, the style you go for will very much depend on the amount of detail and wording you would like to show.
I've explained each option below, which I hope will give you a better idea on which style to order!
This card is perfect if you would only like to show a few details. One side can show your names, a Thank You message and/or the names of your wedding party. The other side can show the title of your ceremony songs and readings, but there won't be room to show the lyrics or words from the readings.
With a cover, you can spread out the wording a little more. You can show the titles of your songs/readings, but there wouldn't be enough room for the lyrics.
A cover also gives you the option of printing your own inserts at home, which is useful if you're on a budget. Keep in mind that you may need to hole punch these in and then tie together with ribbon.
Double-fold cards are really popular, as you can show a lot of wording, and also have a really sleek design! They're small enough to fit into handbags and guests love to take them away as little keepsakes.
On these cards, you can show a thank you message, the names of your wedding party, your schedule, and service details. They're ideal if you only need to show the lyrics of 1 or 2 songs.
An Order of Service booklet is perfect if you have over two songs and readings in your ceremony and would like to show the lyrics and words for each of them. You can have one, two, three or four pages in your booklet, and most people wouldn't need more than four.
When putting the design together, I always ensure that all the content is carefully laid out. It's important to me that you have a beautiful and clean design which your guests will find easy to read and follow.
If you have any questions about Order of Service, or would like to enquire about a bespoke design, please don't hesitate to get in touch!


Thank you for your ongoing support to connect Albertans with disabilities to jobs!
You're invited to provide us with comments, questions, or concerns.
Every time you donate your gently-used items to Goodwill you are helping Albertans with disabilities get jobs. Find out donating DOs and DON'Ts.
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Arctic Paper Mochenwangen GmbH is situated in a beautiful village located in the state of Baden-Württemberg in Southern Germany and has about 4200 inhabitants. The closest city, Ravensburg, is approx. 15 kilometres away. Just outside the mill the river Schussen flows by, through Ravensburg and into beautiful Lake Constance (Bodensee).
The mill was founded in 1868 by the Müller family and was bought by the Arctic Paper group in 2008. The mill has a long tradition in producing high quality wood-containing bulky book paper and also has a high production of offset paper.
As the Management Board announced in Current Report No. 10/2015 of 28 July 2015 a decision was taken to cease production at the Arctic Paper Mochenwangen mill as by December 2015.


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latest tailored shirt designs for men 2013 1.details 1) It is one of our many excellent products 2) Your logo and other patterns can printed or embroidered on your shirts 3) Any material, style, color can made for you4) Our price is very competitive 2.The steps to cooperation 1) Make the sample for you 2) Confirm the quality when you receive the samples 3) Place the firm order 3.about us: We are a OEM garment manufacturer in China for 10 years, have very rich experience in manufacturing and have our firm materials supply channels, it is believed will not let you down. 4.advantage of us: 1) Have factory in shenzhen city which is the most businesslike city in China2) High quality material and apparel control team3) Professional creative garment designers to make your idea to be come ture4) High efficiency customers service team 5) OEM orders are welcome6) size :as customers' requirments 5.new design and OEM services: 1.Fabric: polyester fabric but can be converted to CVC, TC, Cotton , Bamboo, Silk, etc 2.Our Company is Specialized in producing Men's & boys Shirts and neckties 3.Supply Capability: 50,000 PCS/month 4.Quality: 10 years experience of shirt manufacturing and processing 5.Certification: ISO9001:2000 6.Style: The latest Casual,fashionable style 7.Packing: According to Customers Requirements 8.Workmanship: According to Customer's Requirements 9.Size: S,M,L,XL,XXL,etc. 14.Guarantee: Good after Service 6.we also sell other style WE BELIEVE: Honest attitude + professional + sincerely service + competitive price + high quality product == your satisfaction == long-term relationship of cooperation.
samples need 7 working days to finish once detail confirmed. 6. Lead time: samples need 7-9 working days to finish once detail confirmed. Believe that we can give you our best price and best service.
If you don't have design,just let us know your ideas,we can design for you,it's free. If you have your own sample or design , pls send to us and we can copy it for you. We have been in this line for 13 years, and had supplied various countries with high quality items.
The designs, style ,size, can be customized as you need. The embroidery style can be customized. The beading & others special decoration can be customized.
founded in 1983, nearly 30 years of rich experience in garment (man suit) production, The company developed the man suits, shirts, t-shirts, ties, leather goods, shoes, caps, etc. The company has a group of high stability of the technical level of the garment, and invested heavily introduce the modern advanced level of production lines, equipped with clothing CAD design system, and have a suit of advanced production lines, automatic plate making, cutting, typesetting system. Our b randed products in the large and medium cities of inland have sales , and also exported to the United States, Japan, southeast Asia, Europe, Africa, Hong Kong, Taiwan, M acau and other countries and regions.
(branded quality shirts, dresses is our strongest line. our price is competitive among the suppliers which supply the same quality clothes.
How To Do Business With us 1.Send a enquiry to us to tell me which one you want to buy ,the quantity and the sizes . 3.We do sample for your reference and you should pay USD 20 for the cost of the sample ,then pay for the express . 4.We send the sample to you ,after you confirm the sample ,we can start our business .
COTTON/SPANDEX MENS ITALIAN SHIRT WITH DOUBLE COLLAR 1.Color: according to our Color provided by us or by customers 2.Size: Europe Size, Asian Size or other size as customers requirement. 3.Your logo and design are available 4.OEM and ODM are welcomed 5. Competitive price,high quality and well service If you want know more about our products,don't hesitate to contact us. Or visit our website: to get more imformation.
Shipping and handing Before ordering , please confirm the shipment. Please contact us , we will do as your requirement. All the design can be followed customer's design, colors, Material and quantity.
2.Can you ship to my country9 We can ship to all the countries around the world via Express,Sea,Air. 3.How much is the shipping charge to my country9 The shipping charge depends on your located country and the weight of the package. 5.How can I track the whole shipping process9 We will send you a BL tracking number.
Fashion Tiger head 3D printing design mens shirts Size Chart Packaging & Shipping 1pc/polybag, 50-100pcs/ctn or according to client's request About 15 days after order confirmation(depend on quantity) Sea shipment / Air shipment / via Express ( DHL,UPS, FedEx, TNT ) (Quotations not include shipping cost) Destination Shipping Company Processing & Delivery USA UPS, DHL 2~3 days UK UPS, DHL 3~4 days Canada UPS, DHL 3~4 days Australia, New Zealand UPS, DHL 3~4 days Middle East UPS, DHL 3~4 days Southeast Asia UPS, DHL 3~4 days West European, East European UPS, DHL 3~4 days Middle America, South America UPS, DHL 5~6 days Africa country UPS, DHL 5~6 days We have in stock, also accept custom. Your design are Europe size, UK size, USA size, UN size are available,also can customizable.
welcome to take ordersand we will try our best to meet your requirement . our products are sold well around the world and in good reputation.
and if the samples are very ungrent, we can push it. and we will charge samples fee, if we sign the contract, we will refound the samples fee. 3. We will make sample for you to check quality before the bulk order.
Some new custom fabrics will be more than 500 pcs per color and design . if you have strict request on fabric color . And we bet you can get good quality products with good service and low price.
12 YRS Wuhan Lianli Fashion Co., Ltd.
all orders entrusted to us will be given our careful and prompt attention.
if there are the same fabric,we can make if for you. Emails are typically responded to within 24hrs, though on weekends & holidays it may take slightly longer. If you have any questions or concerns, please don`t hesitate to contact us.
7 YRS Nanchang Dimiya Plastic Industry Co., Ltd.
Custom Samples: The price is based on your artwork, print or embroidery designs, colors, and fabric materials. 5.What is your payment terms9 We accept 30% T/T in advance,70% in the period of shipment. 6.What is your company name and address9 Company Name: Nanchang Dimiya Garment Manufacturing Co.
Give us your need on our goods or send us your design, we will make sample for you. 4,30%Deposit in advance and balance should be paid after you get the copy of B/L based on we nominate the forwarder. 2.we can follow your oder carefully and show the photo information and problems solution in time.
3. Size and color9 --- Size chart(including chest width, body length, sleeve length ,etc. 7. Payment terms9 --- 30% T/T in advance,70% in the period of shipment. 8. Delievery time9 --- 30-45 days after recieving the bank deposit. 9 .
Alibaba.com offers 719 100% cotton men shirts 2013 products. About 56% of these are men's t-shirts, 38% are men's shirts, and 20% are plus size shirts & blouses.
A wide variety of 100% cotton men shirts 2013 options are available to you, such as eco-friendly, breathable, and anti-wrinkle. You can also choose from free samples, paid samples.
There are 719 100% cotton men shirts 2013 suppliers, mainly located in Asia. The top supplying country is China (Mainland), which supply 100% of 100% cotton men shirts 2013 respectively.
00% cotton men shirts 2013 products are most popular in North America, Western Europe, and South America.
You can ensure product safety by selecting from certified suppliers, including 103 with Other, 46 with ISO9001, and 25 with BSCI certification.


text:
Q: Moving checked values to another block and appending them within
* lists I want to move the selected list items from one block (block 1) another block (block 2). After moving to block 2, if I delete the item from block 2, it should append back to block 1.
Here is my code:
The HTML:

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>

<input type="checkbox" value="Option 1" onclick="addval(this)" />Option 1

<input type="checkbox" value="Option 2" onclick="addval(this)" />Option 2

The script:
function addval(x){
     values1 = $(x).val();
     $('#right ul').append("
"+ values1 +"
");
     $(x).parent().remove();

 }
 function removeval(y,val){
     values2 = val;
     $('#left ul').append("
<input type='checkbox' value="+ values2 +" name="+ values2 +" onclick='addval(this)' />"+ values2 +"
");
     $(y).parent().remove();

 }

The CSS:
div { border:1px #ccc solid; height:100px; overflow:auto; width:200px; background:#f1f1f1; }
ul { margin:0; padding:0; }
ul li { list-style:none; }

Working demo of the same is available online at : http://jsfiddle.net/prajan55/maLqV/
Kindly help.. thanks in advance.

A: You're just missing a few quotes.
Look here: http://jsfiddle.net/maLqV/7/
Fixed code:
function addval(x){
     values1 = $(x).val();
     $('#right ul').append("
"+ values1 +"
");
     $(x).parent().remove();

 }
 function removeval(y,val){
     values2 = val;
     $('#left ul').append("
<input type='checkbox' value='"+ values2 +"' name='"+ values2 +"' onclick='addval(this)' />"+ values2 +"
");
     $(y).parent().remove();

 }

A: May I suggest a slightly different implementation:
var checkedList = $('#right ul');
var unCheckedList = $('#left  ul');

$('input:checkbox').click(
    function(){
        if (this.checked){
            $(this).closest('li').appendTo(checkedList);
        }
        else if (!this.checked){
            $(this).closest('li').appendTo(unCheckedList);
        }
    });

JS Fiddle demo.
This preserves the details of the checkbox and text, whereas your implementation seemed to remove the label, or the number from the label (which seems to have been resolved by @sje397's answer), but also this version allows for the click handler to be removed from the onclick attribute of the elements. Which may, or may not, be a bonus.


The Marine Corps plans to transform its future force of unmanned aerial vehicles into more versatile weapon systems with cyber, electronic warfare and, perhaps, lethal capabilities, in addition to their current role as tactically valuable intelligence, reconnaissance and surveillance (ISR) platforms.
The transformation, detailed in the 2015 Marine Aviation Plan and in recent congressional testimony, is focused on the Corps' newest unmanned aerial system, the MQ-21A Blackjack, a Group 3 unmanned air system (UAS) that is being fielded.
But the plan includes development of a larger and more capable Group 4/5 vehicle, currently designated the MQ-X, which would be "network-enabled, digitally interoperable, and built to execute responsive, persistent, lethal and adaptive full-spectrum operations." Originally designated the RQ- 21, the Blackjack has been redesignated MQ "since it will have more payload than simple reconnaissance," said Maj Paul Greenberg, aviation programs spokesman at Marine Corps headquarters.
and change detection radar and moving target indicators to assist warfighters in battlespace awareness and force application," Davis said.
President Barack Obama's fiscal 2016 defense budget requests $6.4 million in research and testing for the Marines' MQ-21A and $84.9 million to buy four MQ- 21A systems, which includes 20 air vehicles, "to address Marine Corps ISR capability requirements currently supported by service contracts," Davis said in his testimony.
"This Group 3 UAS will provide persistent ship- and land-based ISR support for expeditionary tactical-level maneuver decisions and unit-level force defense and force protection missions," he said.
The aviation plan says the Blackjack also could carry hyperspectral payloads capable of detecting explosives, synthetic aperture radar "capable of detecting targets through clouds and tree cover," and a laser designator to guide precision munitions.
Produced by the Boeing subsidiary Insitu, the MQ-21A will replace the Marines' current Group 3 UAS, the RQ-7B Shadow, produced by Textron Unmanned Systems. It also will allow the Corps to end its contract ISR service being provided by the Aerosonde UAS, another Textron product.
The Marine Corps plan for its future unmanned air systems force is focused on the MQ-21A Blackjack small tactical unmanned aircraft system (STUAS).
; The Blackjack also could carry hyperspectral payloads capable of detecting explosives, synthetic aperture radar and a laser designator to guide precision munitions.
; The Corps' aviation plan includes development of a larger UAS capable of full-spectrum operations.


(September 7, 2018, Barberton, Ohio) – The Barberton Community Foundation Board of Directors is preparing for a transition in the role of President and CEO. The precise time frame for the transition is still being determined, but the Board is eager to have new leadership in place by its annual meeting in January 2019.
This transition comes on the heels of other recent staff transitions, including Debby Rolland who is moving out of state with her family, and Amber Genet who returned to the for-profit world with an area business.
Rolland and Genet have both remained engaged with the Foundation and are providing help during the transition period.
To facilitate this transition in leadership, and to help with the implementation of the strategic plan adopted in August 2018, the Barberton Community Foundation has engaged Finding Leaders, an organization with an excellent reputation for identifying strong leaders.
For questions about the transition, please contact Barberton Community Foundation Board Chair Michael Chisnell at 330.745.5995.
The Barberton Community Foundation Board of Directors is seeking two individuals to fill available seats for the 2019 Board. Successful candidates will strengthen the Foundation Board by adding knowledge, diversity, experience, vision and commitment to improve the quality of life of Barberton residents, now and forever. Please click the link below to review the qualifications. Deadline to apply is Friday, September 28, 2018.
The Barberton Community Foundation (BCF) is seeking a candidate to fill an available seat on our Board of Directors (BOD). The successful candidate will strengthen the Foundation Board by adding knowledge, diversity, experience, vision and commitment to improve the quality of life of Barberton residents, now and forever.
Interested applicants can email Jim Stonkus at JStonkus@barbertoncf.org or by calling the BCF offices at 330-745-5995 by Friday, September 29th.
View full description by clicking here.
The Barberton Community Foundation Board of Directors is seeking a candidate to fill an available seat. The successful candidate will strengthen the Foundation Board by adding knowledge, diversity, experience, vision and commitment to improve the quality of life of Barberton residents, now and forever. Please click here to review the qualifications. Deadline to apply is Friday, April 1, 2016.