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In every post in this category, a ring means a commutative ring. Also, an arbitrary \(A\)-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 to think of an \(A\)-algebra in what follows 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 any ideal \(\mathfrak{a}\) of a ring \(A\) can always be regarded as an \(A\)-module, in many cases we will 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 also regard \(\mathfrak{a}\) as an \(A\)-module, this kind of notational distinction causes more confusion rather than less, so in this category we do not make such a distinction.
Definition 1 For an arbitrary \(A\)-module \(M\), the annihilator \(\ann(M)\) of \(M\) is defined by the formula
\[\ann(M)=\{a\in A\mid aM=0\}\]Next, for two ideals \(\mathfrak{a},\mathfrak{b}\) of a ring \(A\), the ideal quotient \((\mathfrak{a}:\mathfrak{b})\) is defined 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 \(M\),
\[(N_1:N_2)=\{a\in A\mid aN_2\subseteq N_1\}\]The ideal quotient \((\mathfrak{a}:\mathfrak{b})\) can be roughly thought of as \(\mathfrak{a}/\mathfrak{b}\), and for any \(A\)-module \(M\) we have \(\ann(M)=(0:M)\).
Meanwhile, we saw two useful short exact sequences in [Multilinear Algebra] §Exact Sequences, ⁋Proposition 7; 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 the fact that this is well defined is obvious from
\[y\in (\mathfrak{a}:(a))\iff ay\in \mathfrak{a}\]The second map \(A/\mathfrak{a} \rightarrow A/(\mathfrak{a}+(a))\) is defined by
\[x+\mathfrak{a}\mapsto x+(\mathfrak{a}+(a))\]and one can check that it is surjective and its kernel is exactly the submodule of \(A/\mathfrak{a}\) generated by \(a+\mathfrak{a}\).
Finiteness condition
In many cases we will assume some kind of finiteness. For example, in the posts of [Multilinear Algebra] we assumed that a given module was a finitely generated \(A\)-module and, by choosing a basis, reduced many computations to matrix computations. In a similar vein we define the notions of finiteness that we will use frequently.
Definition 2 An \(A\)-module \(M\) is said to satisfy the ascending chain condition if, whenever an increasing sequence of submodules of \(M\)
\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]is given, there exists a suitable \(k\) such that \(M_k=M_{k+1}=\cdots\). Similarly, \(M\) is said to satisfy the descending chain condition if, whenever a decreasing sequence of submodules of \(M\)
\[M_0\supseteq M_1\supseteq M_2\supseteq\cdots\]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 \(A\) is noetherian or artinian if it is noetherian or artinian as a module over itself.
Then the following holds.
Theorem 3 For an arbitrary \(A\)-module \(M\), the following are all equivalent.
- \(M\) is noetherian.
- Every submodule of \(M\) is finitely generated.
- Every collection of submodules of \(M\) 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 pick any 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 pick \(x_2\in N\setminus \langle x_1\rangle\). 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. Suppose an ascending chain of submodules of \(M\)
\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]is given and let \(M'=\bigcup M_k\). Then \(M'\) is finitely generated, so write \(M'=\langle x_1,\ldots, x_n\rangle\). For each \(i\), we can choose \(k_i\) such that \(x_i\in M_{k_i}\), and then for the largest of these \(k_i\) we have \(M_{k_i}=M'\).
Now we show that condition 1 and condition 3 are equivalent. First, if condition 1 is satisfied, then for any collection of submodules of \(M\) the hypothesis of [Set Theory] §Axiom of Choice, ⁋Theorem 4 is satisfied by ACC, so 3 is obvious. Conversely, if condition 3 is satisfied, then when an ascending chain of submodules of \(M\)
\[M_0\subseteq M_1\subseteq M_2\subseteq\cdots\]is given, this collection 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 canonical surjection maps generators of \(L\) to generators of \(L/N\), so this is obvious.
Proposition 5 For an arbitrary \(A\)-module \(M\) and any submodule \(N\), \(M\) is noetherian if and only if both \(N\) and \(M/N\) are noetherian.
Proof
One direction has already been proved. Thus it suffices to assume that \(N\) and \(M/N\) are noetherian and show that \(M\) is 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 \(N\), hence 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 writing this again using the generators of \(L\cap N\) gives the desired result.
Therefore 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 saw in the [Multilinear Algebra] category is the existence of an exact sequence
\[A^{\oplus n} \rightarrow M \rightarrow 0\]and then the images \(x_1,\ldots, x_n\) of a basis of \(A^{\oplus n}\) 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 among 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. Indeed, if \(M\) is a finitely generated \(A\)-module, 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 \(A^{\oplus m} \rightarrow \ker u \rightarrow A^{\oplus n}\), 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 an \(A\)-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
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 by applying Proposition 5 to the kernel, which is a submodule of \(A^{\oplus n}\), when an arbitrary \(A\)-linear map \(A^{\oplus n}\rightarrow M\) is given.
Prime Ideals
Finally, we need the notion of a prime ideal defined in [Algebraic Structures] §Field of Fractions, ⁋Proposition 8.
Definition 10 An ideal \(\mathfrak{p}\subsetneq A\) of a ring \(A\) is a prime ideal if, whenever \(ab\in \mathfrak{p}\), then necessarily \(a\in \mathfrak{p}\) or \(b\in \mathfrak{p}\).
Then we can refine the fourth result of [Algebraic Structures] §Quotient Rings, 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 [Algebraic Structures] §Quotient Rings, 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 afterwards we can use the equivalence condition of [Algebraic Structures] §Field of Fractions, ⁋Proposition 8.
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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