Commutative Algebra/Generators and chain conditions

Generators

Example 6.2:

For every module $M$, the whole module itself is a generating set.

Example 6.4: Every ring $R$ is a finitely generated $R$-module over itself, and a generating set is given by $\{1_R\}$.

Noetherian and Artinian modules

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:

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

$\mathrm{\forall }k\in \mathbb{N}:⟨n_1,\dots ,n_k⟩⊊⟨n_1,\dots ,n_{k+1}⟩$;

it is possible to find such a sequence since we may just always choose $n_{k+1}\in N\setminus ⟨n_1,\dots ,n_k⟩$, since $N$ is not finitely generated. Thus we have an ascending sequence of submodules

$⟨n_1⟩⊊⟨n_1,n_2⟩⊊\cdots ⊊⟨n_1,\dots ,n_k⟩⊊⟨n_1,\dots ,n_{k+1}⟩⊊\cdots$

which does not stabilize.

2. $\Rightarrow$ 3.: Let $ℳ$ be a nonempty set of submodules of $M$. Due to Zorn's lemma, it suffices to prove that every chain within $𝒩$ has an upper bound (of course, our partial order is set inclusion, i.e. $N_1\le N_2:\iff N_1\subseteq N_2$). Hence, let $𝒩$ be a chain within $ℳ$. We write

$𝒩=(N_1\subseteq N_2\subseteq \cdots )=(⟨n_1,\dots ,n_{k_1}⟩\subseteq ⟨n_1,\dots ,n_{k_1},n_{k_1+1},\dots ,n_{k_2}⟩\subseteq \cdots )$.

Since every submodule is finitely generated, so is

$⟨n_1,n_2,\dots ,n_k,n_{k+1},\dots ⟩=⟨m_1,\dots ,m_l⟩$.

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

$⟨n_1,n_2,\dots ,n_k,n_{k+1},\dots ⟩=⟨n_{u_1},\dots ,n_{u_r}⟩$

for suitably chosen $u_1,\dots ,u_r$. 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 $l$.$◻$

Proof 2:

We prove 1. $\Rightarrow$ 3. $\Rightarrow$ 2. $\Rightarrow$ 1.

1. $\Rightarrow$ 3.: Let $𝒩$ be a set of submodules of $M$ which does not have a maximal element. Then by the axiom of dependent choice, for each $N\in 𝒩$ we may choose $N^{\prime }\in 𝒩$ such that $N⊊N^{\prime }$ (as otherwise, $N$ would be maximal). Hence, using the axiom of dependent choice and starting with a completely arbitrary $N_1\in 𝒩$, we find an ascending sequence

$N_1⊊N_2⊊\cdots ⊊N_k⊊N_{k+1}⊊\cdots$

which does not stabilize.

3. $\Rightarrow$ 2.: Let $N\le M$ be not finitely generated. Using the axiom of dependent choice, we choose first an arbitrary $x_1\in N$ and given $x_1,\dots ,x_k$ we choose $x_{k+1}$ in $N\setminus ⟨x_1,\dots ,x_k⟩$. Then the set of submodules

$\{⟨x_1,\dots ,x_k⟩|k\in \mathbb{N}\}$

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

$(N_1\subseteq N_2\subseteq \cdots )=(⟨n_1,\dots ,n_{k_1}⟩\subseteq ⟨n_1,\dots ,n_{k_1},n_{k_1+1},\dots ,n_{k_2}⟩\subseteq \cdots )$

for suitable $(k_j{)}_{j\in \mathbb{N}}$ and $(n_j{)}_{j\in \mathbb{N}}$. Since every submodule is finitely generated, so is

$⟨n_1,n_2,\dots ,n_k,n_{k+1},\dots ⟩=⟨m_1,\dots ,m_l⟩$.

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

$⟨n_1,n_2,\dots ,n_k,n_{k+1},\dots ⟩=⟨n_{u_1},\dots ,n_{u_r}⟩$

for suitably chosen $u_1,\dots ,u_r$. 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 $j$.$◻$

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:

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

$l+N\in N_k/N\Rightarrow \mathrm{\exists }m+N\in M_k,n,n^{\prime }\in N:l=m+n+n^{\prime }\Rightarrow l+N=m+N\in M_k$.

Furthermore, $N_k$ is a submodule of $M$ as follows:

• $l,l^{\prime }\in N_k\Rightarrow \mathrm{\exists }m+N,m^{\prime }+N\in M_k,n,n^{\prime }\in N:l=m+n,l^{\prime }=m^{\prime }+n^{\prime }\Rightarrow (m+m^{\prime })+(n+n^{\prime })=l+l^{\prime }\in N_k$ since $m+m^{\prime }+N\in M_k$ and $n+n^{\prime }\in N$,
• $l\in N_k\Rightarrow \mathrm{\exists }m+N\in M_k,n\in N:l=m+n\Rightarrow al\in N_k$ 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

$m+N\in M_{k+1}\Rightarrow m\in N_{k+1}\Rightarrow m\in N_k\Rightarrow m=m^{\prime }+n,m^{\prime }\in M_k,n\in N\Rightarrow m+N=m^{\prime }+N\in M_k$.

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 $k\ge N$. Let $n\in N_{k+1}$. Then $n+N\in N_{k+1}/N$ and thus $n+N\in N_k/N$, that is $n=m+n^{\prime }$ for an $m\in N_k$ and an $n^{\prime }\in N$. Now $n^{\prime }=n-m\in N_{k+1}$, hence $n^{\prime }\in N_{k+1}\cap N=N_k\cap N$. 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 $N$.$◻$

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 $\pi (m)\in \pi (N_k)\iff m=n^{\prime }+n,n^{\prime }\in N_k,n\in N$. Thus,

$N_1+N\subseteq N_2+N\subseteq \cdots \subseteq N_k+N\subseteq N_{k+1}+N\subseteq \cdots$

stabilizes. But since $N_k=N_{k+1}\iff N_k\cap N=N_{k+1}\cap N\wedge N_k+N=N_{k+1}+N$, the theorem follows.$◻$

Proof 3:

We use the characterisation of Noetherian modules as those with finitely generated submodules.

1. $\Rightarrow$ 2.: Let $K\le N$. Then $K\le M$ and hence $K$ is finitely generated. Let $J\le M/N$. Then the module ${\pi }_N^{-1}(J)$ is finitely generated, with generators $g_1,\dots ,g_n$, say. Then the set ${\pi }_N(g_1),\dots ,{\pi }_N(g_n)$ generates $J$ since ${\pi }_N$ is surjective and linear.

2. $\Rightarrow$ 1.: Let now $K\le M$. 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 $\{k_1+N,\dots ,k_n+N\}$ be a generating set of $L$. Let further $S$ be a finite generating set of $J$, and set $S^{\prime }:=\{k_1,\dots ,k_n\}$. 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 $S$.$◻$

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 $J\le M/N$, $J_{\beta }$ 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 $K_{\alpha }\cap N$.$◻$

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:

That statement is proven as in proofs 1 or 2 of the previous theorem.

Lemma 6.11:

Let $M,N$ be modules, and let $\phi :M\to N$ be a module isomorphism. Then

$M\text{ Noetherian}\iff N\text{ Noetherian}$.

Proof:

Since ${\phi }^{-1}$ is also a module isomorphism, $\Rightarrow$ suffices.

Let $M$ be Noetherian. Using that $\phi$ 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 $\phi (M)=N$ is Noetherian.$◻$

Proof:

Let $K\le N$ be a submodule of $N$. By the first isomorphism theorem, we have $N\cong M/\ker \phi$. By theorem 6.9, $M/\ker \phi$ is Noetherian. Hence, by lemma 6.11, $N$ is Noetherian.$◻$

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. ${\displaystyle }$. Prove that $R$ is a finitely generated $R$-module over itself which is not Noetherian.