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Postmodern Algebra

Sam Macdonald

Section 8.4 Noetherian Rings

“It is worth ascending unexiting heights if for nothing else than to see the big ones from nearer their own level.”
―Nan Shepherd

Subsection Noetherian Rings

Most rings that commutative algebraists naturally want to study are noetherian. Noetherian rings are named after Emmy Noether, who is in many ways the mother of modern commutative algebra.

Definition 8.38. Noetherian Ring.

Suppose \(R\) is a commutative ring. Then \(R\) is called a noetherian ring if \(R\) satisfies the ascending chain condition on ideals - i.e., for every chain of ideals of \(R\)
\begin{equation*} I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots \end{equation*}
there exists a positive integer \(n\) such that \(I_n = I_{n+1} = I_{n+2} = \cdots=I_{n+k}\) for all \(k\geq 0\text{.}\)

Proof.

(1)\(\Rightarrow\)(2): Suppose there is a nonempty family of ideals with no maximal element. This means that we can keep inductively choosing larger ideals from this family to obtain an infinite properly ascending chain, so \(R\) is not noetherian.
(2)\(\Rightarrow\)(1): An ascending chain of ideals is a family of ideals, and the maximal ideal in the family indicates where our chain stabilizes.
(1)\(\Rightarrow\)(3): Contained within the definition of a noetherian ring.
(3)\(\Rightarrow\)(4): Suppose that there is an ideal \(I\) and a generating set \(S\) for \(I\) such that no finite subset of \(S\) generates \(I\text{.}\) So for any finite \(S'\subseteq S\) we have \((S') \subsetneq (S) =I\text{,}\) so there is some \(s \in S \setminus (S').\) Thus, \((S') \subsetneq (S' \cup \{s\})\text{.}\) Inductively, we can continue this process to obtain an infinite proper chain of finitely generated ideals, so (c) does not hold.
(4)\(\Rightarrow\)(5): This is the definition of being finitely generated.
(5)\(\Rightarrow\)(1): Given an ascending chain of ideals
\begin{equation*} I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots \end{equation*}
let \(I=\bigcup_{n\in \N} I_n\text{.}\) In general, the union of two ideals might fail to be an ideal, but the union of a chain of ideals is an ideal (exercise). By assumption, the ideal \(I\) is finitely generated, say \(I=(a_1,\dots,a_t)\text{.}\) Now since each \(a_i\) is in \(I\text{,}\) it must be in some \(I_{n_i}\text{,}\) by definition. Thus for any \(N \geqslant \max n_i\text{,}\) we have \(a_1, \ldots, a_t \in I_N\text{.}\) But then \(I_N=I\text{,}\) and thus \(I_n = I_{n+1}\) for all \(n \geqslant N\text{.}\) Thus the original chain stabilizes, and \(R\) is noetherian.

Remark 8.40.

When we say that every non-empty family of ideals has a maximal element, that maximal element does not have to be unique in any way. An ideal \(I\) is maximal in the family \(\mathcal{F}\) if \(I \subseteq J\) for some \(J \in \mathcal{F}\) implies \(I = J\text{.}\) However, we might have many incomparable maximal elements in \(\mathcal{F}\text{.}\) For example, every element in the family of ideals in \(\mathbb{Z}\) given by
\begin{equation*} \mathcal{F} = \lbrace (p) \mid p \textrm{ is a prime integer} \rbrace \end{equation*}
is maximal.

Remark 8.41.

If \(R\) is a noetherian ring and \(S\) is a non-empty set of ideals in \(R\text{,}\) not only does \(S\) have a maximal element, but every element in \(S\) must be contained in a maximal element of \(S\text{.}\) Given an element \(I \in S\text{,}\) the subset \(T\) of \(S\) of ideals in \(S\) that contain \(I\) is nonempty, and must then contain a maximal element \(J\) by . If \(J \subseteq L\) for some \(L \in S\text{,}\) then \(I \subseteq L\text{,}\) so \(L \in T\text{,}\) and thus by maximality of \(J\) in \(T\text{,}\) we must \(J = L\text{.}\) This proves that \(J\) is in fact a maximal element in \(S\text{,}\) and by construction it contains \(I\text{.}\)

Example 8.42. Noetherian Rings.

  1. If \(R=k\) is a field, the only ideals in \(k\) are \((0)\) and \((1)=k\text{,}\) so \(k\) is a noetherian ring.
  2. \(\mathbb{Z}\) is a noetherian ring, since all ideals are principal. More generally, if \(R\) is a PID, then \(R\) is noetherian. Indeed, every ideal is finitely generated!
  3. As a special case of the previous example, consider the ring of germs of complex analytic functions near \(0\text{,}\)
    \begin{equation*} \C\{z\} := \{ f(z) \in \CC[[z]] \ | \ f \text{ is analytic on a neighborhood of }z=0\}. \end{equation*}
    This ring is a PID: every ideal is of the form \((z^n)\text{,}\) since any \(f\in \CC\{z\}\) can be written as \(z^n g(z)\) for some \(g(z)\neq 0\text{,}\) and any such \(g(z)\) is a unit in \(\C\{z\}\text{.}\)
  4. A ring that is not noetherian is a polynomial ring in infinitely many variables over a field \(k\text{,}\) \(R = k[x_1, x_2, \ldots]\text{:}\) the ascending chain of ideals
    \begin{equation*} (x_1)\subseteq (x_1,x_2) \subseteq (x_1,x_2,x_3) \subseteq \cdots \end{equation*}
    does not stabilize.
  5. The ring \(R=k[x,x^{1/2},x^{1/3},x^{1/4},x^{1/5},\dots]\) is not noetherian. A nice ascending chain of ideals is
    \begin{equation*} (x) \subsetneq (x^{1/2}) \subsetneq (x^{1/3})\subsetneq (x^{1/4}) \subsetneq \cdots. \end{equation*}
  6. The ring of continuous real-valued functions \(\mathcal{C}(\R,\R)\) is not noetherian. The chain of ideals
    \begin{equation*} I_{n}=\{ f(x) \in \mathcal{C}(\R,\R) \ | \ f|_{[-1/n,1/n]}\equiv 0 \} \end{equation*}
    is increasing and proper. The same construction shows that the ring of infinitely differentiable real functions \(\mathcal{C}^{\infty}(\R,\R)\) is not noetherian: properness of the chain follows from, e.g., Urysohn’s lemma (though it’s not too hard to find functions distinguishing the ideals in the chain). Note that if we asked for analytic functions instead of infinitely-differentiable functions, every element of the chain would be the zero ideal!

Proof.

There is an order preserving bijection
\begin{equation*} \{ \text{ideals of } R \text{ that contain } I \} \longleftrightarrow \{\text{ideals of }R/I\} \end{equation*}
that sends the ideal \(J \supseteq I\) to \(J/I\text{;}\) its inverse is the map that sends each ideal in \(R/I\) to its preimage. Given this bijection, chains of ideals in \(R/I\) come from chains of ideals in \(R\) that contain \(J\text{.}\) This implies that if \(R\) is noetherian, then \(R/I\) is noetherian as well.
This gives us many more examples of noetherian rings, by simply taking quotients of the examples above. We will soon show that any polynomial ring over a noetherian ring is also noetherian; as a consequence, we obtain that any quotient of a polynomial ring over a field is noetherian. This is the content of Hilbert’s Basis Theorem.
But first, we need to talk about noetherian modules.

Subsection Noetherian Modules

Definition 8.44. Noetherian Module.

An \(R\)-module \(M\) is noetherian if every ascending chain of submodules of \(M\) eventually stabilizes.

Example 8.46. Noetherian Ring iff Noetherian Module over Itself.

A ring \(R\) is a noetherian ring if and only if \(R\) is noetherian as a module over itself.

Example 8.47. Noeth Ring Not Noeth Mod over Subring.

A noetherian ring need not be a noetherian module over a subring.
For example, consider \(\Z\subseteq \Q\text{.}\) These are both noetherian rings but \(\Q\) is not a noetherian \(\Z\)-module; for example, the following is an ascending chain of submodules which does not stabilize:
\begin{equation*} 0 \subsetneq \frac{1}{2} \Z \subsetneq \frac{1}{2} \Z + \frac{1}{3} \Z \subsetneq \frac{1}{2} \Z + \frac{1}{3} \Z + \frac{1}{5} \Z \subsetneq \cdots . \end{equation*}

Proof.

Assume \(B\) is noetherian. Since \(A\) is a submodule of \(B\text{,}\) and its submodules are also submodules of \(B\text{,}\) \(A\) is noetherian. Moreover, any submodule of \(B/A\) is of the form \(D/A\) for some submodule \(D \supseteq A\) of \(B\text{.}\) Since every submodule of \(B\) is finitely generated, every submodule of \(C\) is also finitely generated. Therefore, \(C\) is noetherian.
Conversely, assume that \(A\) and \(C\) are noetherian, and let
\begin{equation*} M_1 \subseteq M_2 \subseteq M_3 \subseteq \cdots \end{equation*}
be a chain of submodules of \(B\text{.}\) First, note that
\begin{equation*} M_1 \cap A \subseteq M_2 \cap A \subseteq \cdots \end{equation*}
is an ascending chain of submodules of \(A\text{,}\) and thus it stabilizes. Moreover,
\begin{equation*} g(M_1) \subseteq g(M_2) \subseteq g(M_3) \subseteq \cdots \end{equation*}
is a chain of submodules of \(C\text{,}\) and thus it also stabilizes. Pick a large enough index \(n\) such that both of these chains stabilize. We claim that \(M_n = M_{n+1}\text{,}\) so that the original chain stabilizes as well. To show that, take \(x \in M_{n+1}\text{.}\) Then
\begin{equation*} g(x) \in g(M_{n+1}) = g(M_n) \end{equation*}
so we can choose some \(y \in M_n\) such that \(g(x) = g(y)\text{.}\) Then \(x-y \in \ker g = \im f = A\text{.}\) Now note that \(y \in M_n \subseteq M_{n+1}\text{,}\) so \(x-y \in M_{n+1}\text{,}\) and thus
\begin{equation*} x-y \in M_{n+1} \cap A = M_n \cap A. \end{equation*}
Then \(x-y \in M_n\text{,}\) and since \(y \in M_n\text{,}\) we must have \(x \in M_n\) as well.

Proof.

Apply Lemma 8.48 to the short exact sequence
\begin{equation*} \begin{CD} 0@>>>A@>>>A\oplus B@>>>B@>>>0 \end{CD} \end{equation*}

Proof.

We will do induction on \(n\text{.}\) The case \(n=1\) is a tautology. For \(n>1\text{,}\) consider the short exact sequence
\begin{equation*} \begin{CD} 0@>>>M^{n-1}@>>>M^{n}@>g>>M@>>>0 \end{CD} \end{equation*}
Lemma 8.48 and the inductive hypothesis give the desired conclusion.

Proof.

If \(M\) is noetherian, \(M\) is finitely generated, and so are all of its submodules.
Now let \(R\) be noetherian and \(M\) be a finitely generated \(R\)-module. Then \(M\) is isomorphic to a quotient of \(R^n\) for some \(n\text{,}\) which is noetherian by Corollary 8.50 and Lemma 8.43.

Example 8.52. Noetherian Required for Noetherian F.G.

If \(R\) is a non-noetherian ring and \(M\) is a finitely generated \(R\)-module, \(M\) might not be noetherian.
Solution.
For a dramatic example, note that \(R\) itself is a finitely generated \(R\)-module, but not noetherian.

Subsection Hilbert’s Basis Theorem

Now we are ready to prove Hilbert’s Basis Theorem. David Hilbert was a big influence in the early years of commutative algebra, in many different ways. Emmy Noether’s early work in algebra was in part inspired by some of his work, and he later invited her to join the Göttingen Math Department — many of her amazing contributions to algebra happened during her time in Göttingen. Unfortunately, some of the faculty opposed a woman joining the department, and for her first two years in Göttingen, Noether did not have an official position nor was she paid. Hilbert’s contributions also include three of the most fundamental results in commutative algebra — Hilbert’s Basis Theorem, the Hilbert Syzygy Theorem, and Hilbert’s Nullstellensatz.

Proof.

We will give the proof for polynomial rings, and at the end we will indicate what the difference is in the argument for the power series ring case. First, note that by induction on \(d\text{,}\) we can reduce to the case \(d=1\text{.}\)
Given an ideal \(I\subseteq R[x]\text{,}\) consider the set of leading coefficients of all polynomials in \(I\text{,}\)
\begin{equation*} J := \{ a \in R \ \mid \ \textrm{there is some } a x^n + \text{lower order terms (with respect to }x) \in I\}. \end{equation*}
By the [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Exercise - Hilbert Basis Lemma|Hilbert Basis Lemma]], we see that this is an ideal of \(R\text{.}\) Since \(R\) is noetherian, \(J\) is finitely generated, so let \(J = (a_1,\dots,a_t)\text{.}\) Pick \(f_1,\dots,f_t\in R[x]\) such that the leading coefficient of \(f_i\) is \(a_i\text{,}\) and set \(N=\displaystyle\max_i \{\deg{f_i} \}\text{.}\)
Let \(f \in I\text{.}\) The leading coefficient of \(f\) is an \(R\)-linear combination of \(a_1, \ldots, a_t\text{.}\) If \(f\) has degree greater than \(N\text{,}\) then we can cancel off the leading term of \(f\) by subtracting a suitable combination of the \(f_i\text{.}\) Therefore, any \(f \in I\) can be written as \(f = g+ h\) for some \(h \in (f_1, \ldots, f_t)\) and \(g \in I\) with degree at most \(N\text{.}\) In particular, note that
\begin{equation*} g \in I \cap \frac{R + Rx + \cdots + R x^N}. \end{equation*}
Since \(I \cap (R + Rx + \cdots + R x^N)\) is a submodule of the finitely generated free \(R\)-module \(R + Rx + \cdots + R x^N\text{,}\) it must also be finitely generated as an \(R\)-module. Given such a generating set, say \(I \cap (R + Rx + \cdots + Rx^N) = (f_{t+1}, \ldots, f_s)\text{,}\) we can write any element \(f \in I\) as an \(R[x]\)-linear combination of these generators \(f_{t+1}, \ldots, f_s\) and the original \(f_{1}, \ldots, f_t\text{.}\) Therefore, \(I = (f_1, \ldots, f_t, f_{t+1}, \ldots, f_s)\) is finitely generated as an ideal in \(R[x]\text{,}\) and \(R[x]\) is a noetherian ring.
In the power series case, take \(J\) to be the set of coefficients of lowest degree terms.

Remark 8.54.

We can rephrase Hilbert’s Basis Theorem in a way that can be better understood by those with high school algebra (as opposed to abstract algebra) knowledge:
Any system of polynomial equations in finitely many variables can be written in terms of finitely many equations.
Finally, note that a standard corollary of Hilbert’s Basis Theorem is that finitely generated algebras over noetherian rings are also noetherian.

Proof.

Any finitely generated \(R\)-algebra is isomorphic to a quotient of a polynomial ring over \(R\) in finitely many variables; polynomial rings over noetherian rings are noetherian by the Hilbert’s Basis Theorem, and quotients of noetherian rings are noetherian.

Remark 8.56.

The converse to this statement is false: there are lots of noetherian rings that are not finitely generated algebras over a field. For example, \(\C\{z\}\) is not algebra-finite over \(\C\text{.}\) We will see more examples of these when we talk about local rings.
Finally, we can now prove a technical sounding result that puts together all our finiteness conditions in a useful way.

Proof.

Let \(C=A[f_1,\dots,f_r]\) and \(C=B g_1 + \cdots + B g_s\text{.}\) Then,
\begin{equation*} f_i = \sum_j b_{ij} g_j \quad \text{and} \quad g_i g_j = \sum_k b_{ijk} g_k \end{equation*}
for some \(b_{ij}, b_{ijk}\in B\text{.}\) Let \(B_0 = A[\{b_{ij}, b_{ijk}\}] \subseteq B\text{.}\) This is a finitely generated \(A\)-algebra; by this Corollary 8.55, since \(A\) is noetherian, so is \(B_0\text{.}\)
We claim that \(C=B_0 \, g_1 + \cdots + B_0 g_s\text{.}\) Given an element \(c\in C\text{,}\) write \(c\) as a polynomial expression in \(f_1, \ldots, f_r\text{.}\) Since the \(f_i\) are linear combinations of the \(g_i\) with coefficients in the \(b_{ij},\) we have \(c\in A[\{b_{ij}\}][g_1,\dots,g_s]\text{.}\) Then using the equations for \(g_i g_j\) repeatedly, we can rewrite \(c\) as a linear combination of the \(g_i\) with coefficients in \(B_0\text{.}\)
Since \(B_0\) is noetherian and \(C\) is a finitely generated \(B_0\)-module, \(C\) is a noetherian \(B_0\)-module, by Proposition 8.51. Since \(B\subseteq C\text{,}\) then \(B\) is also a finitely generated \(B_0\)-module. In particular, \(B_0 \subseteq B\) is algebra-finite. Since \(A \subseteq B_0\) is algebra-finite, we conclude that \(A \subseteq B\) is algebra-finite, as required.
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