Yan Sheng's site A math blog

Commutative Algebra (VII): Noetherian Rings

17 Feb 2019

Even though the definitions of the Noetherian and Artinian properties are dual to each other, it turns out that the Noetherian condition is more important. For instance, we have already seen that every Artinian ring is Noetherian. In this post, we will prove more properties of Noetherian modules and rings.

Noetherian modules

We can characterise Noetherian modules by yet another finiteness property.

Proposition: An $A$-module $M$ is Noetherian if and only if every submodule of $M$ is finite (ie. finitely generated).

Proof: ($\Rightarrow$) For any submodule $M’\subseteq M$, let $S$ be the collection of all finite submodules of $M’$. Note that $S$ is nonempty since $\{0\}\in S$; hence $S$ has a maximal element $N$.

If there exists $x\in M’\backslash N$, then $N+Ax$ is a finite submodule of $M’$ that strictly contains $N$, contradicting maximality of $N$. Hence $N=M’$, so $M’$ is finite.

($\Leftarrow$) Let $N_1\subsetneq N_2\subsetneq\cdots$ be an ascending chain of submodules of $M$. Then $N\coloneqq\bigcup_kN_k$ is also a submodule of $M$, so $N$ is finite by assumption, say

Now each $n_i$ is contained in some $N_{k_i}$. Taking $K=\max_i(k_i)$, we see that $N_K$ contains all the $n_i$, so $N_K=N$, and the ascending chain must stop at that point. $\qed$

Corollary: A ring $A$ is Noetherian if and only if every ideal of $A$ is finitely generated. $\qed$

Hilbert basis theorem

In this section, we show that if $A$ is a Noetherian ring, then so are $A[X]$ and $A⟦X⟧$, even though they are not finitely generated as $A$-modules.

Hilbert Basis Theorem: Let $A$ be a Noetherian ring. Then the polynomial ring $A[X]$ is also Noetherian.

Proof: We will show that any ideal $I\subseteq A[X]$ is finitely generated. Consider the set of all leading coefficients of polynomials in $I$; we check that these form an ideal $J\subseteq A$. Since $A$ is Noetherian, $J$ is finitely generated, say $J=Aa_1+\cdots+Aa_n$. For each $i$, choose $f_i\in A[X]$ with leading coefficient $a_i$, and let $r$ be the maximal degree among all $f_i$.

Write $I’=A[X]f_1+\cdots+A[X]f_n$. Now for any $f=aX^m+\cdots\in I$ with $m\geq r$, note that $a\in J$ implies that $a$ is an $A$-linear combination of the $a_i$. Hence by multiplying each $f_i$ by a suitable power of $X$, we see that there is a polynomial $aX^m+\cdots\in I’$. Thus we can repeatedly subtract $f$ by elements of $I’$ to obtain a polynomial of smaller degree, so we can write

Writing $M=A+AX+\cdots+AX^{m-1}$, the above statement implies

But $M$ is a finitely generated, hence Noetherian, $A$-module; thus $I\cap M$ is a finitely generated $A$-module. Hence the union of $\{f_1,\ldots,f_n\}$ and a finite generating set for $I\cap M$ forms a finite generating set for $I$, and we are done. $\qed$

Corollary: If $A$ is a Noetherian ring, then so is $A[X_1,\ldots,X_n]$.

Proof: Induction on $n$ using $A[X_1,\ldots,X_n]\cong A[X_1,\ldots,X_{n-1}][X_n]$. $\qed$

Corollary: If $A$ is a Noetherian ring and $B\supseteq A$ is a finitely generated $A$-algebra, then $B$ is also Noetherian.

Proof: Note that $B$ is isomorphic to a quotient of some polynomial ring $A[X_1,\ldots,X_n]$. $\qed$

In particular, every finitely generated ring (ie. finitely generated $\bb Z$-algebra) is Noetherian; so is every finitely generated algebra over a field.

Theorem: Let $A$ be a Noetherian ring. Then the ring of formal power series $A⟦X⟧$ is also Noetherian.

Proof: The proof is analogous to that for the Hilbert basis theorem, but with “degree” replaced with “minimal exponent of $X$ among terms with nonzero coefficient,” and “leading coefficient” replaced by “coefficient of the term with lowest degree.” $\qed$

Eakin-Nagata theorem

In this section, we look at when the existence of a sufficiently “nice” module implies that the base ring is Noetherian; in particular, when a subring of a Noetherian ring is Noetherian.

We know that a finite module over a Noetherian ring is a Noetherian module. The following result is a partial converse: a ring with a faithful Noetherian module is a Noetherian ring.

Proposition: Let $A$ be a ring, and let $M$ be a Noetherian $A$-module. Then $A/\ann(M)$ is a Noetherian ring.

Proof: Write $\overline A=A/\ann(M)$. Then the $A$-submodules of $M$ are exactly the $\overline A$-submodules of $M$, so $M$ is also a Noetherian $\overline A$-module.

In particular, $M$ is a finite $\overline A$-module, say $M=\overline A\omega_1+\cdots+\overline A\omega_n$. Now the map

is a $\overline A$-module map from $\overline A$ to $M^n$, which is injective since $\ann_{\overline A}(M)=0$; hence $\overline A$ is isomorphic to an $\overline A$-submodule of $M^n$. But $M$ is a Noetherian $\overline A$-module, hence so is $M^n$, thus so is $\overline A$. $\qed$

Theorem: (Formanek) Let $A$ be a ring, and let $B$ be a finite $A$-module which is faithful (ie. $\ann(B)=0$). Suppose that the collection of submodules of $B$ given by

satisfies the a.c.c.; then $A$ is a Noetherian ring.

Proof: By the previous result, it suffices to show that $B$ is a Noetherian $A$-module.

Assume otherwise for sake of contradiction; consider the collection

This collection is nonempty (since it contains $\{0\}$), so it has some maximal element $I_0B$. Replacing $B\to B/I_0B$ and $A\to A/\ann(B/I_0B)$, we may assume that $B$ is not Noetherian as an $A$-module, but $B/IB$ is Noetherian for any nonzero ideal $I\subseteq A$.

Next, consider the family

Assume for sake of contradiction that for some chain $(N_i)_ {i\in I}$ in $\Gamma$, the union $N=\bigcup_{i\in I}N_i$ is not in $\Gamma$. Now $B$ is a finite $A$-module, say $B=Ab_1+\cdots+Ab_n$. Then $N\not\in\Gamma$ implies

But then $ab_k\in N_{i_k}$ for some $i_k\in I$; hence if $i=\max_ki_k$, then $ab_k\in N_i$ for all $k$, so $N_i\not\in\Gamma$, contradiction.

Thus every chain in $\Gamma$ has an upper bound in $\Gamma$, namely its union. Hence by Zorn’s lemma, $\Gamma$ has a maximal element $N_0$. Now $B/N_0$ is a finite faithful $A$-module, so if it is also Noetherian then $A$ is a Noetherian ring, so $B$ is a Noetherian $A$-module, contradiction. Thus by replacing $B\to B/N_0$, we get a module $B$ with the following properties:

  1. $B$ is not Noetherian as an $A$-module;
  2. For any nonzero ideal $I\subseteq A$, we have $B/IB$ is Noetherian as an $A$-module; and
  3. For any nonzero submodule $N\subseteq B$, we have $B/N$ is not faithful as an $A$-module.

Now take any nonzero submodule $N\subseteq B$. By (3), there is a nonzero element $a\in A$ with $a(B/N)=0$, ie. $aB\subseteq N$. By (2), $B/aB$ is a Noetherian $A$-module, so $N/aB$ is finitely generated. But $B$ is finitely generated, hence so is $aB$; thus $N$ is finitely generated. Hence $B$ is a Noetherian $A$-module, which is our desired contradiction. $\qed$

Eakin-Nagata Theorem: Let $B$ be a Noetherian ring, and let $A\subseteq B$ be a subring such that $B$ is finite over $A$. Then $A$ is also a Noetherian ring.

Proof: If $a\in\ann(B)$ then $a1_B=0$ implies $a=0$. Hence $B$ is a finite faithful $A$-module, and $B$ Noetherian implies that the $A$-submodules

satisfies the a.c.c.. Hence $A$ is a Noetherian ring by the previous theorem. $\qed$

Related Posts

Comments (1)

  • fattypiggy123 07 Mar 2019, 09:58 UTC
    Eakin-Nagata's theorem reminds me of Artin-Tate's lemma: Let $A$ be a Noetherian ring, $B\subset C$ algebras over $A$. If $C$ is a finitely generated as an $A$-algebra, and $C$ is a finite $B$-module, then $B$ is also a finitely generated $A$-algebra.$\\$ So over here, the Noetherian property is replaced by finitely generated which is also a measure of "smallness". The proof can be found on Wikipedia and is pretty short but I think rather tricky. The lemma was invented to prove Hilbert's Nullstellensatz (according to Wikipedia, and through Zariski's lemma I believe) but it has a neat application to invariant theory.$\\$ Let $K$ be a field, $R = K[x_1,x_2,\ldots,x_n]$ the polynomial ring over $K$ in $n$ variables and let $G$ be a group of $K$-automorphisms on $R$. Then a classical problem in invariant theory is to determine the ring $$R^G = \{ r \mid g \cdot r = r \text{ for all } g \in G \},$$ also known as the ring of invariants under $G$. More precisely, one can ask whether $R^G$ is finitely generated as a $K$-algebra, and if so, what is a possible set of generators. $\\$ When $G$ is of finite size, we can apply Artin-Tate lemma by letting $A = K, B = R^G$ and $C = R$. Then all one has to show is that $R$ is finitely generated as a $R^G$-module, which it then suffices to show that $R$ is an integral extension of $R^G$ (actually equivalent by Nakayama). Now we borrow a construction from Galois theory: given $r \in R$, we consider the polynomial $$P(t) = \prod_{g \in G} (t - g \cdot r)$$ which expands out to polynomial with coefficients in $R^G$ that $r$ satisfies, since $1_G \cdot r = r$. The conclusion then follows.$\\$ Using Easkin-Nagata, one can also immediately conclude that $R^G$ is Noetherian since $R$ is Noetherian by Hilbert basis and under some additional natural conditions such as elements of $G$ send homogeneous polynomials to homogeneous polynomials, is enough to conclude that $R^G$ is finitely generated. I would like to mention another proof that $R^G$ is Noetherian, which is closer to Hilbert's original proof for $G = SL_n$.$\\$ Just like Formanek's theorem, given a chain of ideals $\{I_j\}$ in $R^G$ we extend it to $R$ via $\{I_j R\}$. Then as $R$ is Noetherian, this chain will stabilize and hopefully we can conclude that the $\{ I_j \}$ will stabilize too. The issue arises when one could have distinct $I_j$'s such that $I_j R$ are equal to each other. Our aim is to show that this won't happen. In fact, we will show something more general.$\\$ Let $R \subseteq S$ be rings. Call $S$ a direct summand of $R$ if it is a direct summand as a $R$-module, that is $S = R \oplus R'$ as $R$-modules. Then it is easy to see that if this is the case, then for any ideal $I$ of $R$, one has $IS \cap R = I$. Indeed, if $r = \sum_{i=1}^{n} i_k s_k$, then writing $s_k = r_k + r'_k$, we have $$r - \sum_{i=1}^{n} i_k r_k = \sum_{i=1}^{n} i_k r'_k$$ and since $R \cap R' = \{ 0 \}$, the above expression must also be $ = 0$ and so $r = \sum_{i=1}^{n} i_k r_k \in I$ as desired. In particular, if $R$ was a direct summand of $R^G$ then the stabilization of $\{ I_j R \}$ immediately implies that of $\{ I_j \}$, giving us the Noetherianess of $R$. $\\$ To show that $R$ is a direct summand of $R^G$, it suffices to give a $R$-module projection map $\pi: R \to R^G$ such that $\pi \circ i$ is the identity on $R^G$ where $i: R^G \to R$ is the inclusion map. We now borrow an idea from representation theory: define $\pi : R \to R^G$ by $$\pi(r) = \frac{1}{|G|} \sum_{g \in G} g \cdot r.$$ This only works if $|G|$ is coprime to the characteristic of $K$, but assuming that, it is easy to verify $\pi$ satisfies the required properties and thus proving that $R$ is indeed a direct summand of $R^G$ and hence $R^G$ is Noetherian as $R$ is.$\\$ I would like to point out that while this proof only works for certain characteristic, the one via Artin-Tate lemma works for all of them. A key thing that was considered is the construction $P(t) = \prod_{g \in G} (t - g \cdot r)$ and I have found that this is a very useful thing to consider when dealing with arbitrary characteristic. For instance, if we had a non-zero function $f$ on some vector space $V$ carrying a $G$-action, then if we wanted to symmetrize $f$ to be invariant under $G$, a possible consideration is $\sum_{g \in G} g \cdot f$. But if $V$ had positive characteristic then this could be possibly just the zero function, especially if the stabilizer of each $v \in V$ had cardinality $p$. You can try other possible symmetrizations but there will always be a possibility that it ends up at $0$. On the other hand, if we take $$0 = (f - g \cdot f) = f^n + a_1 \cdot f^(n-1) + \cdots + a_n$$ where $a_i$ are functions invariant under $G$, then there must exist at least one $a_i$ where $a_i$ is non-zero as otherwise, $f$ must be the zero function.

Write a comment…