Basic Notions

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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 have already seen after §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 examine a commutative ring $A$ and a module $M$ defined over it. Since any ideal $\mathfrak{a}$ of the ring $A$ can always be regarded as an $A$-module, in many cases we develop the theory of $A$-modules. 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, making such a distinction in notation causes more confusion rather than less, so in this category we do not make such a distinction.

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

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

On the other hand, for two ideals $\mathfrak{a},\mathfrak{b}$ of a ring $A$ 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, for two submodules $N_1,N_2$ of an $A$-module $M$ 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 $\ann(M)=(0:M)$.

On the other hand, in §Exact Sequences, ⁋Proposition 7 we examined two useful short exact sequences; it is worth adding the following short exact sequence to them:

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

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 the fact that

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

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 of $A/\mathfrak{a}$ generated by $a+\mathfrak{a}$.

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.

$M$ is noetherian.
Every submodule of $M$ is finitely generated.
Any 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 we can choose an arbitrary element $x_0\neq 0$ of $N$, and since $N$ is not finitely generated we have $N\neq \langle x_1\rangle$, so we can choose $x_2\in N\setminus \langle x_1\rangle$. Repeating this, we obtain an increasing sequence of submodules of $N$

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

which are also submodules of $M$, contradicting 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 $M'$ is finitely generated, so write $M'=\langle x_1,\ldots, x_n\rangle$. Now for each $i$, we can choose $k_i$ such that $x_i\in N_{k_i}$ holds, and the largest of these $k_i$ must then equal $M'$.

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

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

the collection of these must have a maximal element, so condition 1 holds.

Thus it is obvious that any submodule of a noetherian module is again 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 of the form $L/N$ for a suitable submodule $L$ of $M$, and now $L$ is finitely generated and the generators of $L$ generate $L/N$ under the canonical surjection, so this is obvious.

Proposition 5 For any $A$-module $M$ and any submodule $N$, $M$ being noetherian is equivalent to both $N$ and $M/N$ being noetherian.

Proof

One direction has already been proved. Thus it suffices to show that $M$ is noetherian assuming that $N$ and $M/N$ are noetherian. Fix an arbitrary submodule $L$ of $M$. Then the image $L/N$ of $L$ in $M/N$ is finitely generated, and $L\cap N$ is also a submodule of $N$ so it is finitely generated. Now let $x_1,\ldots, x_m\in L$ be elements whose images in $L/N$ generate $L/N$, and let $y_1,\ldots, y_n\in L\cap N$ generate $L\cap N$. Then for any $x\in L$ there exist $\alpha_i\in A$ such that

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

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 $M,N$, the direct sum $M\oplus N$ is a noetherian $A$-module.

Proof

Apply Proposition 5 to $M\oplus N$ and its submodule $M\oplus 0\cong M$.

The condition of a finitely generated $A$-module that we examined in the [Multilinear Algebra] category is that there exists an exact sequence

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

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

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

there may be infinitely many relations among the $x_i$. Since these relations are determined by the kernel of the surjection $A^{\oplus n} \rightarrow M$, we make the following definition.

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, then we obtain the 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 composing $A^{\oplus m} \rightarrow \ker \rightarrow A^{\oplus n}$, we obtain the 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.

$M$ is finitely generated.
$M$ is finitely presented.
$M$ is coherent.

Proof

The equivalence of conditions 1 and 2 has already been examined. 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 by applying Proposition 5 to the kernel of any given $A$-linear map $A^{\oplus n}\rightarrow M$, which is a submodule of $A^{\oplus n}$.

Prime Ideals

Finally we need the notion of prime ideal that we defined in §Field of Fractions, ⁋Proposition 8.

Definition 10 An ideal $\mathfrak{p}\subsetneq A$ of a ring $A$ 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 refine the fourth result of §Quotient Rings and Ring Homomorphisms, ⁋Theorem 3 to obtain the following.

Proposition 11 For any ideal $\mathfrak{a}$ of a ring $A$, 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 §Quotient Rings and Ring Homomorphisms, ⁋Theorem 3, for $\mathfrak{a}\subseteq \mathfrak{p}\subseteq A$ we have

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

and then using the equivalence condition of §Field of Fractions, ⁋Proposition 8 gives the result.

Hilbert Basis Theorem

So far we have examined the basic properties of noetherian modules. The rings we will deal with are mostly polynomial rings or quotients thereof, so it is important to know whether noetherianness is preserved under operations involving 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 any ideal $I$ of $A[\x]$ is finitely generated. For each integer $n\geq 0$, let $\mathfrak{a}_n$ be the set of leading coefficients of polynomials of degree $n$ belonging to $I$, together with $0$.

First we check that $\mathfrak{a}_n$ is an ideal of $A$. Let $a,b$ be the leading coefficients of two polynomials $f,g\in I$ of degree $n$. Then $f+g\in I$ has degree $n$ and leading coefficient $a+b$, or if $a+b=0$ then 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 stabilizes at some $N$ because $A$ is noetherian, so $\mathfrak{a}_n=\mathfrak{a}_N$ for all $n\geq N$. Also, for each $n\leq N$ the ideal $\mathfrak{a}_n$ is finitely generated; let its generators be $a_{n,1},\ldots,a_{n,k_n}$ and for each $a_{n,j}$ choose one polynomial $f_{n,j}\in I$ of degree $n$ having $a_{n,j}$ as its leading coefficient.

Now we show by induction on the degree of elements of $I$ that the finite set $\{f_{n,j}\mid 0\leq n\leq N,\ 1\leq j\leq k_n\}$ generates $I$. Let $f$ be a polynomial in $I$ of degree $d$ with leading coefficient $c$. If $d\leq N$, then $c\in \mathfrak{a}_d$ so there exist suitable $\lambda_j\in A$ with $c=\sum_j \lambda_j a_{d,j}$, 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 an $A[\x]$-linear combination of the $\{f_{n,j}\}$, so by the induction hypothesis the above difference is generated by the $\{f_{n,j}\}$, and hence $f$ itself is as well. (The case $f=0$ is obvious.) 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 A finitely generated $A$-algebra $B$ over a noetherian ring $A$ is noetherian.

Proof

Since $B$ is a finitely generated $A$-algebra, there exists a surjective ring homomorphism $A[\x_1,\ldots,\x_n]\rightarrow B$ for a suitable $n$. That is, $B\cong A[\x_1,\ldots,\x_n]/I$ for some ideal $I$. 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 §Quotient Rings and Ring Homomorphisms, ⁋Theorem 3, the ideals of $B$ are in one-to-one correspondence with the ideals of $A[\x_1,\ldots,\x_n]$ containing $I$, preserving inclusion. Thus any ascending chain of ideals of $B$ gives an ascending chain of ideals of $A[\x_1,\ldots,\x_n]$, and the latter stabilizes by the noetherianness of $A[\x_1,\ldots,\x_n]$, so the former also stabilizes. Regarding the ring $B$ as a $B$-module over itself, its submodules are exactly the ideals of $B$, so by Theorem 3 the ring $B$ is noetherian.

References

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

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Basic Definitions

Finiteness Condition

Prime Ideals

Hilbert Basis Theorem

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