# Commutative Algebra/Generators and chain conditions
## Generators

Definition 6.1 (generators of modules):
Let $M$ be a module over the ring $R$. A generating set of $M$ is a subset $\{m_{j}\}_{j\in J}\subseteq M$ such that
    $\forall n\in M:\exists j_{1},\ldots ,j_{k}\in J,r_{1},\ldots ,r_{k}\in R:n=\sum _{l=1}^{k}r_{l}m_{j_{l}}$.
Example 6.2:
For every module $M$, the whole module itself is a generating set.
Definition 6.3:
Let $M$ be a module. $M$ is called finitely generated if there exists a generating set of $M$ which has a finite cardinality.
Example 6.4: Every ring $R$ is a finitely generated $R$-module over itself, and a generating set is given by $\{1_{R}\}$.
Definition 6.5 (generated submodules):

### Exercises
## Noetherian and Artinian modules

Definition 6.6 (Noetherian modules):
Let $M$ be a module over the ring $R$. $M$ is called a Noetherian module iff for every ascending chain of submodules
    $N_{1}\subseteq N_{2}\subseteq N_{3}\subseteq \cdots \subseteq N_{k}\subseteq \cdots$
of $M$, there exists an $l\in \mathbb {N}$ such that
    $\forall k\geq l:N_{k}=N_{l}$.
We also say that ascending chains of submodules eventually become stationary.
Definition 6.7 (Artinian modules):
A module $M$ over a ring $R$ is called Artinian module iff for every descending chain of submodules
    $N_{1}\supseteq N_{2}\supseteq N_{3}\supseteq \cdots \supseteq N_{k}\supseteq \cdots$
of $M$, there exists an $l\in \mathbb {N}$ such that
    $\forall k\geq l:N_{k}=N_{l}$.
We also say that descending chains of submodules eventually become stationary.
We see that those definitions are similar, although they define a bit different objects.
Using the axiom of choice, we have the following characterisation of Noetherian modules:
Theorem 6.8:
Let $M$ be a module over $R$. The following are equivalent:
* $M$ is Noetherian.
* All the submodules of $M$ are finitely generated.
* Every nonempty set of submodules of $M$ has a maximal element.
Proof 1:
We prove 1. $\Rightarrow$ 2. $\Rightarrow$ 3. $\Rightarrow$ 1.
1. $\Rightarrow$ 2.: Assume there is a submodule $N$ of $M$ which is not finitely generated. Using the axiom of dependent choice, we choose a sequence $(n_{k})_{k\in \mathbb {N} }$ in $N$ such that
    $\forall k\in \mathbb {N} :\langle n_{1},\ldots ,n_{k}\rangle \subsetneq \langle n_{1},\ldots ,n_{k+1}\rangle$;
it is possible to find such a sequence since we may just always choose $n_{k+1}\in N\setminus \langle n_{1},\ldots ,n_{k}\rangle$, since $N$ is not finitely generated. Thus we have an ascending sequence of submodules
    $\langle n_{1}\rangle \subsetneq \langle n_{1},n_{2}\rangle \subsetneq \cdots \subsetneq \langle n_{1},\ldots ,n_{k}\rangle \subsetneq \langle n_{1},\ldots ,n_{k+1}\rangle \subsetneq \cdots$
which does not stabilize.
2. $\Rightarrow$ 3.: Let ${\mathcal {M}}$ be a nonempty set of submodules of $M$. Due to Zorn's lemma, it suffices to prove that every chain within ${\mathcal {N}}$ has an upper bound (of course, our partial order is set inclusion, i.e. $N_{1}\leq N_{2}:\Leftrightarrow N_{1}\subseteq N_{2}$). Hence, let ${\mathcal {N}}$ be a chain within ${\mathcal {M}}$. We write
    ${\mathcal {N}}=\left(N_{1}\subseteq N_{2}\subseteq \cdots \right)=\left(\langle n_{1},\ldots ,n_{k_{1}}\rangle \subseteq \langle n_{1},\ldots ,n_{k_{1}},n_{k_{1}+1},\ldots ,n_{k_{2}}\rangle \subseteq \cdots \right)$.
Since every submodule is finitely generated, so is
    $\langle n_{1},n_{2},\ldots ,n_{k},n_{k+1},\ldots \rangle =\langle m_{1},\ldots ,m_{l}\rangle$.
We write $m_{j}=\sum _{u\in \mathbb {N} }r_{u}n_{u}$, where only finitely many of the $r_{u}$ are nonzero. Hence, we have
    $\langle n_{1},n_{2},\ldots ,n_{k},n_{k+1},\ldots \rangle =\langle n_{u_{1}},\ldots ,n_{u_{r}}\rangle$
for suitably chosen $u_{1},\ldots ,u_{r}$. Now each $u_{i}$ is eventually contained in some $N_{j}$. Since the $N_{j}$ are an ascending sequence with respect to inclusion, we may just choose $j$ large enough such that all $u_{i}$ are contained within $N_{j}$. Hence, $N_{j}$ is the desired upper bound.
3. $\Rightarrow$ 1.: Let
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
be an ascending chain of submodules of $M$. The set $\{N_{j}|j\in \mathbb {N} \}$ has a maximal element $N_{l}$ and thus this ascending chain becomes stationary at $l$.$\Box$
Proof 2:
We prove 1. $\Rightarrow$ 3. $\Rightarrow$ 2. $\Rightarrow$ 1.
1. $\Rightarrow$ 3.: Let ${\mathcal {N}}$ be a set of submodules of $M$ which does not have a maximal element. Then by the axiom of dependent choice, for each $N\in {\mathcal {N}}$ we may choose $N'\in {\mathcal {N}}$ such that $N\subsetneq N'$ (as otherwise, $N$ would be maximal). Hence, using the axiom of dependent choice and starting with a completely arbitrary $N_{1}\in {\mathcal {N}}$, we find an ascending sequence
    $N_{1}\subsetneq N_{2}\subsetneq \cdots \subsetneq N_{k}\subsetneq N_{k+1}\subsetneq \cdots$
which does not stabilize.
3. $\Rightarrow$ 2.: Let $N\leq M$ be not finitely generated. Using the axiom of dependent choice, we choose first an arbitrary $x_{1}\in N$ and given $x_{1},\ldots ,x_{k}$ we choose $x_{k+1}$ in $N\setminus \langle x_{1},\ldots ,x_{k}\rangle$. Then the set of submodules
    $\{\langle x_{1},\ldots ,x_{k}\rangle {\big |}k\in \mathbb {N} \}$
does not have a maximal element, although it is nonempty.
2. $\Rightarrow$ 1.: Let
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
be an ascending chain of submodules of $M$. Since these are finitely generated, we have
    $\left(N_{1}\subseteq N_{2}\subseteq \cdots \right)=\left(\langle n_{1},\ldots ,n_{k_{1}}\rangle \subseteq \langle n_{1},\ldots ,n_{k_{1}},n_{k_{1}+1},\ldots ,n_{k_{2}}\rangle \subseteq \cdots \right)$
for suitable $(k_{j})_{j\in \mathbb {N} }$ and $(n_{j})_{j\in \mathbb {N} }$. Since every submodule is finitely generated, so is
    $\langle n_{1},n_{2},\ldots ,n_{k},n_{k+1},\ldots \rangle =\langle m_{1},\ldots ,m_{l}\rangle$.
We write $m_{j}=\sum _{u\in \mathbb {N} }r_{u}n_{u}$, where only finitely many of the $r_{u}$ are nonzero. Hence, we have
    $\langle n_{1},n_{2},\ldots ,n_{k},n_{k+1},\ldots \rangle =\langle n_{u_{1}},\ldots ,n_{u_{r}}\rangle$
for suitably chosen $u_{1},\ldots ,u_{r}$. Now each $u_{i}$ is eventually contained in some $N_{j}$. Hence, the chain stabilizes at $l$, if $l$ is chosen as the maximum of those $j$.$\Box$
The second proof might be advantageous since it does not use Zorn's lemma, which needs the full axiom of choice.
We can characterize Noetherian and Artinian modules in the following way:
Theorem 6.9:
Let $M$ be a module over a ring $R$, and let $N\leq M$. Then the following are equivalent:
* $M$ is Noetherian.
* $N$ and $M/N$ are Noetherian.
Proof 1:
We prove the theorem directly.
1. $\Rightarrow$ 2.: $N$ is Noetherian since any ascending sequence of submodules of $N$
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
is also a sequence of submodules of $M$ (check the submodule properties), and hence eventually becomes stationary.
$M/N$ is Noetherian, since if
    $M_{1}\subseteq M_{2}\subseteq \cdots \subseteq M_{k}\subseteq M_{k+1}\subseteq \cdots$
is a sequence of submodules of $M/N$, we may write
    $M_{k}=N_{k}/N$,
where $N_{k}:=\{m+n|m+N\in M_{k},n\in N\}$. Indeed, "$\subseteq$" follows from $m+N\in M_{k}\Rightarrow m+0+N\in N_{k}/N$ and "$\supseteq$" follows from
    $l+N\in N_{k}/N\Rightarrow \exists m+N\in M_{k},n,n'\in N:l=m+n+n'\Rightarrow l+N=m+N\in M_{k}$.
Furthermore, $N_{k}$ is a submodule of $M$ as follows:
* $l,l'\in N_{k}\Rightarrow \exists m+N,m'+N\in M_{k},n,n'\in N:l=m+n,l'=m'+n'\Rightarrow (m+m')+(n+n')=l+l'\in N_{k}$ since $m+m'+N\in M_{k}$ and $n+n'\in N$,
* $l\in N_{k}\Rightarrow \exists m+N\in M_{k},n\in N:l=m+n\Rightarrow al\in N_{k}$ since $a(m+N)\in M_{k}$ and $an\in N$.
Now further for each $k\in \mathbb {N}$ $N_{k}\subseteq N_{k+1}$, as can be read from the definition of the $N_{k}$ by observing that $m+N\in M_{k},n\in N\Rightarrow m+N\in M_{k+1},n\in N$. Thus the sequence
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
becomes stationary at some $j\in \mathbb {N}$. But If $N_{k}=N_{k+1}$, then also $M_{k}=M_{k+1}$, since
    $m+N\in M_{k+1}\Rightarrow m\in N_{k+1}\Rightarrow m\in N_{k}\Rightarrow m=m'+n,m'\in M_{k},n\in N\Rightarrow m+N=m'+N\in M_{k}$.
Hence,
    $M_{1}\subseteq M_{2}\subseteq \cdots \subseteq M_{k}\subseteq M_{k+1}\subseteq \cdots$
becomes stationary as well.
2. $\Rightarrow$ 1.: Let
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
be an ascending sequence of submodules of $M$. Then
    $N\cap N_{1}\subseteq N\cap N_{2}\subseteq \cdots \subseteq N\cap N_{k}\subseteq N\cap N_{k+1}\subseteq \cdots$
is an ascending sequence of submodules of $N$, and since $N$ is Noetherian, this sequence stabilizes at an $l\in \mathbb {N}$. Furthermore, the sequence
    $N_{1}/N\subseteq N_{2}/N\subseteq \cdots \subseteq N_{k}/N\subseteq N_{k+1}/N\subseteq \cdots$
is an ascending sequence of submodules of $M/N$, which also stabilizes (at $j\in \mathbb {N}$, say). Set $N:=\max\{l,j\}$, and let $k\geq N$. Let $n\in N_{k+1}$. Then $n+N\in N_{k+1}/N$ and thus $n+N\in N_{k}/N$, that is $n=m+n'$ for an $m\in N_{k}$ and an $n'\in N$. Now $n'=n-m\in N_{k+1}$, hence $n'\in N_{k+1}\cap N=N_{k}\cap N$. Hence $n\in N_{k}$. Thus,
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
is stable after $N$.$\Box$
Proof 2:
We prove the statement using the projection morphism to the factor module.
1. $\Rightarrow$ 2.: $N$ is Noetherian as in the first proof. Let
    $M_{1}\subseteq M_{2}\subseteq \cdots \subseteq M_{k}\subseteq M_{k+1}\subseteq \cdots$
be a sequence of submodules of $M/N$. If $\pi :M\to M/N$ is the projection morphism, then
    $N_{k}:=\pi ^{-1}(M_{k})$
defines an ascending sequence of submodules of $M$, as $\pi ^{-1}$ preserves inclusion (since $\pi$ is a function). Now since $M$ is Noetherian, this sequence stabilizes. Hence, since also $\pi$ preserves inclusion, the sequence
    $M_{1}\subseteq M_{2}\subseteq \cdots =\pi (\pi ^{-1}(M_{1}))\subseteq \pi (\pi ^{-1}(M_{2}))\subseteq \cdots =\pi (N_{1})\subseteq \pi (N_{2})\subseteq \cdots$
also stabilizes ($\pi (\pi ^{-1}(M_{k}))=M_{k}$ since $\pi$ is surjective).
2. $\Rightarrow$1.: Let
    $N_{1}\subseteq N_{2}\subseteq \cdots \subseteq N_{k}\subseteq N_{k+1}\subseteq \cdots$
be an ascending sequence of submodules of $M$. Then the sequences
    $\pi (N_{1})\subseteq \pi (N_{2})\subseteq \cdots$ and $N\cap N_{1}\subseteq N\cap N_{2}\subseteq \cdots$
both stabilize, since $M/N$ and $N$ are Noetherian. Now $\pi ^{-1}(\pi (N_{k}))=N_{k}+N$, since $\pi (m)\in \pi (N_{k})\Leftrightarrow m=n'+n,n'\in N_{k},n\in N$. Thus,
    $N_{1}+N\subseteq N_{2}+N\subseteq \cdots \subseteq N_{k}+N\subseteq N_{k+1}+N\subseteq \cdots$
stabilizes. But since $N_{k}=N_{k+1}\Leftrightarrow N_{k}\cap N=N_{k+1}\cap N\wedge N_{k}+N=N_{k+1}+N$, the theorem follows.$\Box$
Proof 3:
We use the characterisation of Noetherian modules as those with finitely generated submodules.
1. $\Rightarrow$ 2.: Let $K\leq N$. Then $K\leq M$ and hence $K$ is finitely generated. Let $J\leq M/N$. Then the module $\pi _{N}^{-1}(J)$ is finitely generated, with generators $g_{1},\ldots ,g_{n}$, say. Then the set $\pi _{N}(g_{1}),\ldots ,\pi _{N}(g_{n})$ generates $J$ since $\pi _{N}$ is surjective and linear.
2. $\Rightarrow$ 1.: Let now $K\leq M$. Then $J:=K\cap N$ is finitely generated, since it is also a submodule of $N$. Furthermore,
    $L:=\{k+N|k\in K\}$
is finitely generated, since it is a submodule of $M/N$. Let $\{k_{1}+N,\ldots ,k_{n}+N\}$ be a generating set of $L$. Let further $S$ be a finite generating set of $J$, and set $S':=\{k_{1},\ldots ,k_{n}\}$. Let $k\in K$ be arbitrary. Then $k+N\in L$, hence $k+N=\sum _{j=1}^{n}r_{j}k_{j}+N$ (with suitable $r_{j}\in R$) and thus $k=\sum _{j=1}^{n}r_{j}k_{j}+n$, where $n\in N$; we even have $n\in J$ due to $n=k-\sum _{j=1}^{n}r_{j}k_{j}\in K$, which is why we may write it as a linear combination of elements of $S$.$\Box$
Proof 4:
We use the characterisation of Noetherian modules as those with maximal elements for sets of submodules.
1. $\Rightarrow$ 2.: If $\{K_{\alpha }\}_{\alpha \in A}$ is a family of submodules of $N$, it is also a family of submodules of $M$ and hence contains a maximal element.
If $\{J_{\alpha }\}_{\alpha \in A}$ is a family of submodules of $M/N$, then $\{\pi _{N}^{-1}(J_{\alpha })\}_{\alpha \in A}$ is a family of submodules of $M$, which has a maximal element $\pi _{N}^{-1}(J_{\beta })$. Since $\pi _{N}$ is inclusion-preserving and $\pi _{N}(\pi _{N}^{-1}(J))$ for all $J\leq M/N$, $J_{\beta }$ is maximal among $\{J_{\alpha }\}_{\alpha \in A}$.
2. $\Rightarrow$ 1.: Let $\{K_{\alpha }\}_{\alpha \in A}$ be a nonempty family of submodules of $M$. According to the hypothesis, the family $\{K_{\alpha }\cap N\}_{\alpha \in B}$, where $B$ is defined such that the corresponding $K_{\alpha }\cap N,\alpha \in B$ are maximal elements of the family $\{K_{\alpha }\cap N\}_{\alpha \in A}$, is nonempty. Hence, the family $\{L_{\alpha }\}_{\alpha \in B}$, where
    $L_{\alpha }:=\{k+N|k\in K_{\alpha }\}$,
has a maximal element $L_{\gamma }$. We claim that $K_{\gamma }$ is maximal among $\{K_{\alpha }\}_{\alpha \in A}$. Indeed, let $K_{\delta }\supseteq K_{\gamma }$. Then $K_{\delta }\cap N=K_{\gamma }\cap N$ since $\gamma \in B$. Hence, $\delta \in B$. Furthermore, let $k\in K_{\gamma }$. Then $k+N\in L_{\delta }\Rightarrow k+N\in L_{\gamma }$, since $\delta \in B$. Thus $k+n\in K_{\delta }$ for a suitable $n\in N$, which must be contained within $K_{\gamma }$ and thus also in $K_{\delta }$.
We also could have first maximized the $L_{\alpha }$ and then the $K_{\alpha }\cap N$.$\Box$
These proofs show that if the axiom of choice turns out to be contradictory to evident principles, then the different types of Noetherian modules still have some properties in common.
The analogous statement also holds for Artinian modules:
Theorem 6.10:
Let $M$ be a module over a ring $R$, and let $N\leq M$. Then the following are equivalent:
* $M$ is Artinian.
* $N$ and $M/N$ are Artinian.
That statement is proven as in proofs 1 or 2 of the previous theorem.
Lemma 6.11:
Let $M,N$ be modules, and let $\varphi :M\to N$ be a module isomorphism. Then
    $M{\text{ Noetherian}}\Leftrightarrow N{\text{ Noetherian}}$.
Proof:
Since $\varphi ^{-1}$ is also a module isomorphism, $\Rightarrow$ suffices.
Let $M$ be Noetherian. Using that $\varphi$ is an inclusion-preserving bijection of submodules which maps generating sets to generating sets (due to linearity), we can use either characterisation of Noetherian modules to prove that $\varphi (M)=N$ is Noetherian.$\Box$
Theorem 6.12:
Let $M,N$ be modules and let $\varphi :M\to N$ be a surjective module homomorphism. If $M$ is Noetherian, then so is $N$.
Proof:
Let $K\leq N$ be a submodule of $N$. By the first isomorphism theorem, we have $N\cong M/\ker \varphi$. By theorem 6.9, $M/\ker \varphi$ is Noetherian. Hence, by lemma 6.11, $N$ is Noetherian.$\Box$

### Exercises

* Exercise 6.2.1: Is every Noetherian module $M$ finitely generated?
* Exercise 6.2.2: We define the ring $R$ as the real polynomials in infinitely many variables, i.e. $$. Prove that $R$ is a finitely generated $R$-module over itself which is not Noetherian.