Noetherian Rings and Modules

Let $R$ be a commutative ring with unit element. We will frequently work with $R$-modules, which are like vector spaces but over a ring. More precisely, recall that an is an additive abelian group $M$ equipped with a map $R\times M \to M$ such that for all  $r, r'\in R$ and all $m, m'\in M$ we have $(r r')m = r(r' m )$, $(r + r')m = rm + r' m$, $r(m+m') = rm + rm'$, and $1m=m$. A is a subgroup of $M$ that is preserved by the action of $R$.

Example 4.1.1   The set of abelian groups are in natural bijection with $\mathbf{Z}$-modules.

A of $R$-modules $\varphi :M\to N$ is a abelian group homomorphism such that for any $r\in R$ and $m\in M$ we have $\varphi (rm) = r\varphi (m)$. A of $R$-modules

$$0 \to L \xrightarrow{f} M \xrightarrow{g} N \to 0$$

is a specific choice of injective homomorphism $f:L\to M$ and a surjective homomorphism $g:M\to N$ such that $\im (f) = \ker(g)$.

Definition 4.1.2 (Noetherian)   An $R$-module $M$ is if every submodule of $M$ is finitely generated. A ring $R$ is if $R$ is Noetherian as a module over itself, i.e., if every ideal of $R$ is finitely generated.

Notice that any submodule $M'$ of $M$ is Noetherian, because if every submodule of $M$ is finitely generated then so is every submodule of $M'$, since submodules of $M'$ are also submodules of $M$.

Definition 4.1.3 (Ascending chain condition)   An $R$-module $M$ satisfies the if every sequences $M_1\subset M_2 \subset M_3 \subset \cdots$ of submodules of $M$ eventually stabilizes, i.e., there is some $n$ such that $M_n=M_{n+1}=M_{n+2}=\cdots$.

Proposition 4.1.4   If $M$ is an $R$-module, then the following are equivalent:

1. $M$ is Noetherian,
2. $M$ satisfies the ascending chain condition, and
3. Every nonempty set of submodules of $M$ contains at least one maximal element.

Proof. $1\implies 2$: Suppose $M_1\subset M_2\subset \cdots$ is a sequence of submodules of $M$. Then $M_\infty=\cup _{n=1}^{\infty} M_n$ is a submodule of $M$. Since $M$ is Noetherian, there is a finite set $a_1,\ldots, a_m$ of generators for $M$. Each $a_i$ must be contained in some $M_j$, so there is an $n$ such that $a_1,\ldots, a_m\in M_n$. But then $M_{k}=M_n$ for all $k\geq n$, which proves that the ascending chain condition holds for $M$.

$2\implies 3$: Suppose 3 were false, so there exists a nonempty set $S$ of submodules of $M$ that does not contain a maximal element. We will use $S$ to construct an infinite ascending chain of submodules of $M$ that does not stabilize. Note that $S$ is infinite, otherwise it would contain a maximal element. Let $M_1$ be any element of $S$. Then there is an $M_2$ in $S$ that contains $M_1$, otherwise $S$ would contain the maximal element $M_1$. Continuing inductively in this way we find an $M_3$ in $S$ that properly contains $M_2$, etc., and we produce an infinite ascending chain of submodules of $M$, which contradicts the ascending chain condition.

$3\implies 1$: Suppose 1 is false, so there is a submodule $M'$ of $M$ that is not finitely generated. We will show that the set $S$ of all finitely generated submodules of $M'$ does not have a maximal element, which will be a contradiction. Suppose $S$ does have a maximal element $L$. Since $L$ is finitely generated and $L\subset M'$, and $M'$ is not finitely generated, there is an $a\in M'$ such that $a\not\in L$. Then $L'=L+Ra$ is an element of $S$ that strictly contains the presumed maximal element $L$, a contradiction. $\qedsymbol$

Lemma 4.1.5   If

$$0 \to L \xrightarrow{f} M \xrightarrow{g} N \to 0$$

is a short exact sequence of $R$-modules, then $M$ is Noetherian if and only if both $L$ and $N$ are Noetherian.

Proof. First suppose that $M$ is Noetherian. Then $L$ is a submodule of $M$, so $L$ is Noetherian. If $N'$ is a submodule of $N$, then the inverse image of $N'$ in $M$ is a submodule of $M$, so it is finitely generated, hence its image $N'$ is finitely generated. Thus $N$ is Noetherian as well.

Next assume nothing about $M$, but suppose that both $L$ and $N$ are Noetherian. If $M'$ is a submodule of $M$, then $M_0=\varphi (L)\cap M'$ is isomorphic to a submodule of the Noetherian module $L$, so $M_0$ is generated by finitely many elements $a_1,\ldots, a_n$. The quotient $M'/M_0$ is isomorphic (via $g$) to a submodule of the Noetherian module $N$, so $M'/M_0$ is generated by finitely many elements $b_1,\ldots, b_m$. For each $i\leq m$, let $c_i$ be a lift of $b_i$ to $M'$, modulo $M_0$. Then the elements $a_1,\ldots, a_n, c_1,\ldots, c_m$ generate $M'$, for if $x\in M'$, then there is some element $y\in M_0$ such that $x-y$ is an $R$-linear combination of the $c_i$, and $y$ is an $R$-linear combination of the $a_i$. $\qedsymbol$

Proposition 4.1.6   Suppose $R$ is a Noetherian ring. Then an $R$-module $M$ is Noetherian if and only if it is finitely generated.

Proof. If $M$ is Noetherian then every submodule of $M$ is finitely generated so $M$ is finitely generated. Conversely, suppose $M$ is finitely generated, say by elements $a_1,\ldots, a_n$. Then there is a surjective homomorphism from $R^n=R\oplus \cdots \oplus R$ to $M$ that sends $(0,\ldots,0,1,0,\ldots,0)$ ($1$ in $i$th factor) to $a_i$. Using Lemma 4.1.5 and exact sequences of $R$-modules such as $0\to R\to R\oplus R\to R\to 0$, we see inductively that $R^n$ is Noetherian. Again by Lemma 4.1.5, homomorphic images of Noetherian modules are Noetherian, so $M$ is Noetherian. $\qedsymbol$

Lemma 4.1.7   Suppose $\varphi :R\to S$ is a surjective homomorphism of rings and $R$ is Noetherian. Then $S$ is Noetherian.

Proof. The kernel of $\varphi$ is an ideal $I$ in $R$, and we have an exact sequence

$$0 \to I \to R \to S \to 0$$

with $R$ Noetherian. By Lemma 4.1.5, it follows that $S$ is a Noetherian $R$-modules. Suppose $J$ is an ideal of $S$. Since $J$ is an $R$-submodule of $S$, if we view $J$ as an $R$-module, then $J$ is finitely generated. Since $R$ acts on $J$ through $S$, the $R$-generators of $J$ are also $S$-generators of $J$, so $J$ is finitely generated as an ideal. Thus $S$ is Noetherian. $\qedsymbol$

Theorem 4.1.8 (Hilbert Basis Theorem)   If $R$ is a Noetherian ring and $S$ is finitely generated as a ring over $R$, then $S$ is Noetherian. In particular, for any $n$ the polynomial ring $R[x_1,\ldots, x_n]$ and any of its quotients are Noetherian.

Proof. Assume first that we have already shown that for any $n$ the polynomial ring $R[x_1,\ldots, x_n]$ is Noetherian. Suppose $S$ is finitely generated as a ring over $R$, so there are generators $s_1,\ldots, s_n$ for $S$. Then the map $x_i\mapsto s_i$ extends uniquely to a surjective homomorphism $\pi: R[x_1,\ldots, x_n] \to S$, and Lemma 4.1.7 implies that $S$ is Noetherian.

The rings $R[x_1,\ldots, x_n]$ and $(R[x_1,\ldots,x_{n-1}])[x_n]$ are isomorphic, so it suffices to prove that if $R$ is Noetherian then $R[x]$ is also Noetherian. (Our proof follows [Art91, §12.5].) Thus suppose $I$ is an ideal of $R[x]$ and that $R$ is Noetherian. We will show that $I$ is finitely generated.

Let $A$ be the set of leading coefficients of polynomials in $I$ along with 0. If $a,b\in A$ are nonzero with $a+b\neq 0$, then there are polynomials $f$ and $g$ in $I$ with leading coefficients $a$ and $b$. If $\deg(f)\leq \deg(g)$, then $a+b$ is the leading coefficient of $x^{\deg(g)-\deg(f)}f + g$, so $a+b\in A$. If $r\in R$ and $a\in A$ with $ra\neq 0$, then $ra$ is the leading coefficient of $rf$, so $ra\in A$. Thus $A$ is an ideal in $R$, so since $R$ is Noetherian there exists $a_1,\ldots, a_n$ that generate $A$ as an ideal. Since $A$ is the set of leading coefficients of elements of $I$, and the $a_j$ are in $I$, we can choose for each $j\leq n$ an element $f_j\in I$ with leading coefficient $a_j$. By multipying the $f_j$ by some power of $x$, we may assume that the $f_j$ all have the same degree $d$.

Let $S_{<d}$ be the set of elements of $I$ that have degree strictly less than $d$. This set is closed under addition and under multiplication by elements of $R$, so $S_{<d}$ is a module over $R$. The module $S_{<d}$ is submodule of the $R$-module of polynomials of degree less than $n$, which is Noetherian because it is generated by $1,x,\ldots, x^{n-1}$. Thus $S_{<d}$ is finitely generated, and we may choose generators $h_1,\ldots, h_m$ for $S_{<d}$.

Suppose $g\in I$ is an arbitrary element. We will show by induction on the degree of $g$ that $g$ is an $R[x]$-linear combination of $f_1,\ldots, f_n, h_1,\ldots h_m$. Thus suppose this statement is true for all elements of $I$ of degree less than the degree of $g$. If the degree of $g$ is less than $d$, then $g\in S_{<d}$, so $g$ is in the $R[x]$-ideal generated by $h_1,\ldots, h_m$. Next suppose that $g$ has degree $e\geq d$. Then the leading coefficient $b$ of $g$ lies in the ideal $A$ of leading coefficients of $g$, so there exist $r_i\in R$ such that $b=r_1 a_1 + \cdots + r_n a_n$. Since $f_i$ has leading coefficient $a_i$, the difference $g- x^{e-d} r_i f_i$ has degree less than the degree $e$ of $g$. By induction $g- x^{e-d} r_i f_i$ is an $R[x]$ linear combination of $f_1,\ldots, f_n, h_1,\ldots h_m$, so $g$ is also an $R[x]$ linear combination of $f_1,\ldots, f_n, h_1,\ldots h_m$. Since each $f_i$ and $h_j$ lies in $I$, it follows that $I$ is generated by $f_1,\ldots, f_n, h_1,\ldots h_m$, so $I$ is finitely generated, as required. $\qedsymbol$

Properties of Noetherian rings and modules will be crucial in the rest of this course. We have proved above that Noetherian rings have many desirable properties.