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

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

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

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

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

$\displaystyle 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.

William Stein 2004-05-06