Basic Notions

In this category, every ring that appears is a commutative ring. Also, any \(A\)-algebra is always understood to be a commutative associative unital \(A\)-algebra. In particular, since we saw after [Algebraic Structures] §Algebras, ⁋Definition 1 that an associative unital \(A\)-algebra \(E\) and a ring homomorphism \(A\rightarrow Z(E)\) are the same thing, it suffices for our subsequent discussion to think of an \(A\)-algebra as a ring homomorphism \(A\rightarrow E\).

Basic Definitions

In this category, we consider a commutative ring \(A\) and a module \(M\) defined over it. Since for the ring \(A\), any ideal \(\mathfrak{a}\) can always be regarded as an \(A\)-module, in many cases we develop the theory of \(A\)-modules. Previously, in the posts of the [Algebraic Structures] category, to avoid confusion we wrote elements of an \(A\)-module \(M\) as \(x,y,\ldots\) and elements of \(A\) as \(\alpha,\beta,\ldots\); however, if we regard \(\mathfrak{a}\) also as an \(A\)-module, distinguishing the notation in this way rather causes more confusion, so in this category we do not make such a distinction.

Definition 1 For any \(A\)-module \(M\), we define the annihilator of \(M\), \(\ann(M)\), by the formula

\[\ann(M)=\{a\in A\mid aM=0\}\]

On the other hand, in a ring \(A\), for two ideals \(\mathfrak{a},\mathfrak{b}\) we define the ideal quotient \((\mathfrak{a}:\mathfrak{b})\) by the formula

\[(\mathfrak{a}:\mathfrak{b})=\{a\in A\mid a\mathfrak{b}\subseteq \mathfrak{a}\}\]

and similarly, in an \(A\)-module \(M\), for two submodules \(N_1,N_2\) we define

\[(N_1:N_2)=\{a\in A\mid aN_2\subseteq N_1\}\]

The ideal quotient \((\mathfrak{a}:\mathfrak{b})\) can roughly be thought of as something like \(\mathfrak{a}/\mathfrak{b}\), and for any \(A\)-module \(M\) we have \(\ann(M)=(0:M)\).

Meanwhile, in [Multilinear Algebra] §Exact Sequences, ⁋Proposition 7 we examined two useful short exact sequences; it is worth adding the following short exact sequence to them and keeping it in mind:

\[0 \longrightarrow A/(\mathfrak{a}:(a)) \overset{a}{\longrightarrow} A/\mathfrak{a}\longrightarrow A/(\mathfrak{a}+(a)) \longrightarrow 0\]

The first map \(A/(\mathfrak{a}:(a)) \rightarrow A/\mathfrak{a}\) is given by the formula

\[x+(\mathfrak{a}:(a))\mapsto ax+\mathfrak{a}\]

and that this is well defined is obvious from

\[y\in (\mathfrak{a}:(a))\iff ay\in \mathfrak{a}\]

Now the second map \(A/\mathfrak{a} \rightarrow A/(\mathfrak{a}+(a))\) is defined by the formula

\[x+\mathfrak{a}\mapsto x+(\mathfrak{a}+(a))\]

and one can check that this is surjective and that its kernel is exactly the submodule generated by \(a+\mathfrak{a}\) in \(A/\mathfrak{a}\).

Finiteness condition

In many cases we assume some kind of finiteness. For example, in the posts of [Multilinear Algebra] we assumed that a given module is a finitely generated \(A\)-module, and by choosing a basis we could reduce many computations to matrix computations. In a similar vein we define the notions of finiteness that we will frequently use.

Definition 2 We say that an \(A\)-module \(M\) satisfies the ascending chain condition if, whenever an increasing sequence

\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]

of submodules of \(M\) is given, there exists a suitable \(k\) such that \(M_k=M_{k+1}=\cdots\). Similarly, we say that \(M\) satisfies the descending chain condition if, whenever a decreasing sequence

\[M_0\supseteq M_1\supseteq M_2\supseteq\cdots\]

of submodules of \(M\) is given, there exists a suitable \(k\) such that \(M_k=M_{k+1}=\cdots\). An \(A\)-module \(M\) satisfying the ascending chain condition is called Noetherian. An \(A\)-module \(M\) satisfying the descending chain condition is called Artinian. A ring \(A\) being Noetherian or Artinian means that \(A\) is Noetherian or Artinian as a module over itself.

Then the following holds.

Theorem 3 For any \(A\)-module \(M\), the following are all equivalent.

1. \(M\) is Noetherian.
2. Every submodule of \(M\) is finitely generated.
3. Any nonempty collection of submodules of \(M\) always has a maximal element with respect to inclusion.

Proof

First assume condition 1 and show condition 2. Suppose for contradiction that \(M\) has a submodule \(N\) that is not finitely generated. Then from \(N\) we can choose an arbitrary element \(x_1\neq 0\), and from the fact that \(N\) is not finitely generated, since \(N\neq \langle x_1\rangle\), we can choose \(x_2\in N\setminus \langle x_1\rangle\). Repeating this, we obtain an increasing sequence of submodules of \(N\)

\[\langle x_1\rangle\subsetneq \langle x_1,x_2\rangle\subsetneq\cdots\]

and since these are also submodules of \(M\), this contradicts the assumption that \(M\) is Noetherian.

Now assume condition 2 and show condition 1. Given an ascending chain of submodules of \(M\)

\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]

let \(M'=\bigcup M_k\); then since \(M'\) is finitely generated, write \(M'=\langle x_1,\ldots, x_n\rangle\). Now for each \(i\), we can choose \(k_i\) such that \(x_i\in M_{k_i}\) holds, and now letting the largest among these \(k_i\) be \(k\), \(M_k\) must be equal to \(M'\).

Now we show that conditions 1 and 3 are equivalent. First, if condition 1 is satisfied, then whenever any collection of submodules of \(M\) is given, the premise of [Set Theory] §Axiom of Choice, ⁋Theorem 4 (Zorn’s lemma) is satisfied by the ACC, so it is obvious that condition 3 holds. Conversely, if condition 3 is satisfied, then given an ascending chain of submodules of \(M\)

\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]

there must exist a maximal element of this collection, so condition 1 holds.

Thus it is obvious that any submodule of a Noetherian module is also Noetherian. Moreover, the following holds.

Proposition 4 For a Noetherian \(A\)-module \(M\), any quotient \(M/N\) is also Noetherian.

Proof

Any submodule of \(M/N\) is, for a suitable submodule of \(M\), say \(L\), of the form \(L/N\), and now since \(L\) is finitely generated and the generators of \(L\) generate \(L/N\) under the canonical surjection, this is obvious.

Proposition 5 For any \(A\)-module \(M\) and any submodule \(N\), \(M\) being Noetherian is equivalent to \(N,M/N\) both being Noetherian.

Proof

One direction has already been proved. Thus it suffices to assume that \(N, M/N\) are Noetherian and show that \(M\) is Noetherian. In \(M\), fix an arbitrary submodule \(L\). Then the image of \(L\) in \(M/N\), \((L+N)/N\), is finitely generated, and since \(L\cap N\) is also a submodule of \(N\), it is finitely generated. Now suppose that the images of \(x_1,\ldots, x_m\in L\) in \((L+N)/N\) are generators of \((L+N)/N\), and let \(y_1,\ldots, y_n\in L\cap N\) be generators of \(L\cap N\). Then for any \(x\in L\), we have

\[x\equiv \alpha_1x_1+\cdots+\alpha_m x_m\pmod{N}\]

for some \(\alpha_i\in A\). Therefore

\[x-\sum \alpha_i x_i\in L\cap N\]

and rewriting this using the generators of \(L\cap N\) gives the desired result.

Hence the following holds.

Corollary 6 For a ring \(A\) and two Noetherian \(A\)-modules \(M,N\), \(M\oplus N\) is a Noetherian \(A\)-module.

Proof

It suffices to apply Proposition 5 to \(M\oplus N\) and its submodule \(M\oplus 0\cong M\).

The condition for a finitely generated \(A\)-module that we examined in the [Multilinear Algebra] category is that there exists the following exact sequence

\[A^{\oplus n} \rightarrow M \rightarrow 0\]

and in this case, for a basis of \(A^{\oplus n}\), the images \(x_1,\ldots, x_n\) generate \(M\). However, in general, when we write \(M\) as

\[M=\langle x_1,\ldots, x_n\mid \text{relations on $x_i$}\rangle\]

there may be infinitely many relations on the \(x_i\). Since these relations are determined by the kernel of the surjection \(A^{\oplus n} \rightarrow M\), we define the following.

Definition 7 An \(A\)-module \(M\) is called finitely presented if there exist suitable \(m,n\) such that the following exact sequence exists:

\[A^{\oplus m} \rightarrow A^{\oplus n} \rightarrow M \rightarrow 0\]

In general, a finitely presented module is finitely generated, but the converse does not hold. However, for any Noetherian ring \(A\), the two notions coincide. This is because if \(M\) is a finitely generated \(A\)-module, we obtain the following exact sequence

\[0\longrightarrow\ker u \longrightarrow A^{\oplus n} \overset{u}{\longrightarrow} M \longrightarrow 0\]

and on the other hand \(A^{\oplus n}\) is Noetherian by Corollary 6, so its submodule \(\ker u\) is finitely generated. Now, considering

\[A^{\oplus m} \rightarrow \ker u \rightarrow 0\]

and using the composition \(A^{\oplus m} \rightarrow \ker u \rightarrow A^{\oplus n}\), we obtain the following exact sequence

\[A^{\oplus m} \rightarrow A^{\oplus n} \rightarrow M \rightarrow 0\]

On the other hand, we define the following.

Definition 8 An \(A\)-module \(M\) is called a coherent module if \(M\) is finitely generated and, whenever any \(A\)-linear map \(A^{\oplus n} \rightarrow M\) is given, the kernel of this linear map is finitely generated.

Then the following proposition is obvious.

Proposition 9 For a Noetherian ring \(A\) and an \(A\)-module \(M\), the following are all equivalent.

1. \(M\) is finitely generated.
2. \(M\) is finitely presented.
3. \(M\) is coherent.

Proof

We have already seen that conditions 1 and 2 are equivalent. Also, by definition, a coherent \(A\)-module is always finitely generated. Thus it suffices to assume that \(M\) is finitely generated and show that \(M\) is coherent. This follows from the fact that, given any \(A\)-linear map \(A^{\oplus n}\rightarrow M\), the kernel of this linear map is a submodule of \(A^{\oplus n}\), and since \(A^{\oplus n}\) is Noetherian by Corollary 6, it is finitely generated by Theorem 3.

Prime Ideals

Finally, we need the notion of prime ideal defined in [Algebraic Structures] §Field of Fractions, ⁋Proposition 9.

Definition 10 For a ring \(A\), an ideal \(\mathfrak{p}\subsetneq A\) is called a prime ideal if, whenever \(ab\in \mathfrak{p}\), then necessarily \(a\in \mathfrak{p}\) or \(b\in \mathfrak{p}\) holds.

Then we can further refine the fourth result of [Algebraic Structures] §Quotient Rings and Ring Isomorphisms, ⁋Theorem 3 to show the following.

Proposition 11 For a ring \(A\) and any ideal \(\mathfrak{a}\), there is a one-to-one correspondence between prime ideals of \(A/\mathfrak{a}\) and prime ideals of \(A\) containing \(\mathfrak{a}\).

Proof

By the third result of [Algebraic Structures] §Quotient Rings and Ring Isomorphisms, ⁋Theorem 3, for \(\mathfrak{a}\subseteq \mathfrak{p}\subseteq A\),

\[A/\mathfrak{p}\cong \frac{A/\mathfrak{a}}{\mathfrak{p}/\mathfrak{a}}\]

holds, and then one can use the equivalent condition of [Algebraic Structures] §Field of Fractions, ⁋Proposition 9.

Hilbert Basis Theorem

So far, we have examined the basic properties of Noetherian modules. Since the rings we will deal with are mostly polynomial rings or their quotients, it is important to know whether the Noetherian property is preserved under the operation of taking polynomial rings. This is a classical result proved by Hilbert.

Theorem 12 (Hilbert basis theorem) For a Noetherian ring \(A\), the polynomial ring \(A[\x]\) is also Noetherian.

Proof

By Theorem 3, it suffices to show that in \(A[\x]\), any ideal \(I\) is finitely generated. For each integer \(n\geq 0\), let the set of leading coefficients of polynomials in \(I\) of degree \(n\), together with \(0\), be \(\mathfrak{a}_n\).

First, we check that \(\mathfrak{a}_n\) is an ideal of \(A\). If two degree \(n\) polynomials \(f,g\in I\) have leading coefficients \(a,b\) respectively, then \(f+g\in I\) has degree \(n\) with leading coefficient \(a+b\), or if \(a+b=0\), its degree is less than \(n\); in either case, \(a+b\in \mathfrak{a}_n\). Also, for any \(\lambda\in A\), the leading coefficient of \(\lambda f\in I\) is \(\lambda a\) or \(0\), so \(\lambda a\in \mathfrak{a}_n\). On the other hand, if \(f\in I\) has degree \(n\), then \(\x f\in I\) has degree \(n+1\) and the same leading coefficient, so \(\mathfrak{a}_n\subseteq \mathfrak{a}_{n+1}\) holds.

The ascending chain \(\mathfrak{a}_0\subseteq \mathfrak{a}_1\subseteq \cdots\) obtained in this way, since \(A\) is Noetherian, stabilizes at some \(N\), so for all \(n\geq N\) we have \(\mathfrak{a}_n=\mathfrak{a}_N\). Also, for each \(n\leq N\), the ideal \(\mathfrak{a}_n\) is also finitely generated; let its generators be \(a_{n,1},\ldots,a_{n,k_n}\), and for each \(a_{n,j}\), choose one degree \(n\) polynomial \(f_{n,j}\in I\) having it as leading coefficient.

Now we show that the finite set \(\{f_{n,j}\mid 0\leq n\leq N,\ 1\leq j\leq k_n\}\) generates \(I\), by induction on the degree of elements of \(I\). For a polynomial in \(I\), say \(f\), let its degree be \(d\) and its leading coefficient be \(c\). If \(d\leq N\), then \(c\in \mathfrak{a}_d\), so for some \(\lambda_j\in A\) we have \(c=\sum_j \lambda_j a_{d,j}\), and

\[f-\sum_j \lambda_j f_{d,j}\in I\]

has degree less than \(d\). If \(d>N\), then \(c\in \mathfrak{a}_d=\mathfrak{a}_N\), so \(c=\sum_j \lambda_j a_{N,j}\), and

\[f-\sum_j \lambda_j \x^{d-N} f_{N,j}\in I\]

also has degree less than \(d\). In both cases, the subtracted polynomial is a linear combination of the \(\{f_{n,j}\}\) over \(A[\x]\), so by the induction hypothesis the difference above is generated by the \(\{f_{n,j}\}\), and hence \(f\) itself is as well. (The case \(f=0\) is trivial.) Therefore \(I\) is finitely generated.

In particular, by applying this repeatedly, we see that the polynomial ring \(A[\x_1,\ldots,\x_n]=A[\x_1,\ldots,\x_{n-1}][\x_n]\) is also Noetherian, and from this we obtain the following.

Corollary 13 Over a Noetherian ring \(A\), a finitely generated \(A\)-algebra \(B\) is Noetherian.

Proof

Since \(B\) is a finitely generated \(A\)-algebra, for a suitable \(n\) there exists a surjective ring homomorphism \(A[\x_1,\ldots,\x_n]\rightarrow B\). That is, for some ideal \(I\), we have \(B\cong A[\x_1,\ldots,\x_n]/I\). Applying Theorem 12 (Hilbert basis theorem) repeatedly, we know that \(A[\x_1,\ldots,\x_n]\) is Noetherian.

Now it suffices to show that a quotient of a Noetherian ring is Noetherian. By [Algebraic Structures] §Quotient Rings and Ring Isomorphisms, ⁋Theorem 3, the ideals of \(B\) are in an inclusion-preserving one-to-one correspondence with the ideals containing \(I\) in \(A[\x_1,\ldots,\x_n]\). Thus, any ascending chain of ideals of \(B\) gives an ascending chain of ideals of \(A[\x_1,\ldots,\x_n]\), and since the latter stabilizes by the Noetherianness of \(A[\x_1,\ldots,\x_n]\), the former also stabilizes. Regarding the ring \(B\) as a \(B\)-module over itself, its submodules are precisely the ideals of \(B\), so by Theorem 3, \(B\) is Noetherian.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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