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The Jordan-Hölder Theorem
We now make the following definition.
Definition 1 An \(A\)-module \(M\) is called simple if \(M\neq 0\) and the only submodules of \(M\) are \(0\) and \(M\).
It is obvious that a simple module must be generated by a single element \(x\in M\), but the converse does not hold, as can be seen by considering \(\mathbb{Z}/6\mathbb{Z}\). On the other hand, if a simple module \(M\) is generated by \(x\), then the \(A\)-module homomorphism \(A \rightarrow M\) sending \(1\mapsto x\) yields an isomorphism
\[A/\ann(M)=A/\ann(x)\cong M.\]If \(\ann(M)\) were not a maximal ideal, then for a maximal ideal \(\mathfrak{m}\) of \(A\) containing \(\ann(M)\), the quotient \(\mathfrak{m}/\ann(M)\) would be a submodule of \(M\); thus it is also obvious that \(\ann(M)\) must be a maximal ideal of \(A\).
Definition 2 Fix an \(A\)-module \(M\). A decreasing sequence of submodules of \(M\)
\[M=M_0\supsetneq M_1\supsetneq \cdots\supsetneq M_n=0\]is called a chain of length \(n\). This chain is called a composition series if \(M_k/M_{k+1}\) is a simple module for every \(k\). The smallest length among such composition series is called the length of \(M\), denoted \(\length(M)\). If no composition series of \(M\) exists, we define \(\length(M)=\infty\).
Then the following theorem can be regarded as the module version of the Jordan-Hölder theorem.
Theorem 3 An \(A\)-module \(M\) has a finite composition series if and only if \(M\) is both artinian and noetherian. Assume that this condition is satisfied and that a composition series of length \(n\)
\[M=M_0\supsetneq M_1\supsetneq \cdots\supsetneq M_n=0\]is given. Then the following hold.
- Any chain of submodules of \(M\) of length at most \(n\) admits a refinement to a composition series.
- For the collection of maximal ideals for which there exists a \(k\) such that \(M_k/M_{k+1}\cong A/\mathfrak{m}\), there is an isomorphism \(M\cong\bigoplus_{\mathfrak{m}}M_\mathfrak{m}\).
- If \(\mathfrak{p}^k\) annihilates \(M\) for some \(k\), then \(M=M_\mathfrak{p}\).
Proof
First, suppose that \(M\) is both artinian and noetherian. From the condition that \(M\) is noetherian, we can find some maximal proper submodule \(M_1\) of \(M\). On the other hand, since it is obvious that a submodule of a noetherian module must be noetherian, we can repeat this process to find a maximal proper submodule \(M_{k+1}\) of \(M_k\). However, the chain defined in this way
\[M=M_0\supsetneq M_1\supsetneq \cdots\]has finite length by the artinian condition, and since \(M_{k+1}\) is a maximal proper submodule of \(M_k\), we see that this chain is a composition series.
The first result is proved in the same way as the Jordan-Hölder theorem, so we omit the proof. Once we accept this, given any chain we can refine it to a composition series, and therefore we can also prove the converse of the preceding equivalence.
We now prove the second result. From the finiteness of the given chain, we know that there are only finitely many maximal ideals satisfying the condition, and hence \(\bigoplus_\mathfrak{m} M_\mathfrak{m}\) can be viewed as \(\prod_\mathfrak{m} M_\mathfrak{m}\); the given map is then obtained by applying the universal property of the direct product to the maps \(M \rightarrow M_\mathfrak{m}\). To show that this map is an isomorphism, it suffices to look at the localization at a maximal ideal and apply §Properties of Localization, ⁋Proposition 4.
For this, we first note that if \(M\cong R/\mathfrak{m}\), then for any maximal ideal \(\mathfrak{m}'\),
\[M_{\mathfrak{m}'}=\begin{cases}M&\text{if $\mathfrak{m}=\mathfrak{m}'$,}\\0&\text{otherwise.}\end{cases}\]From this, if a composition series of \(M\)
\[M=M_0\supsetneq M_1\supsetneq \cdots\supsetneq M_n=0\]is given, then localizing at a maximal ideal \(\mathfrak{m}\) yields
\[M_\mathfrak{m}=(M_0)_\mathfrak{m}\supsetneq (M_1)_\mathfrak{m}\supsetneq \cdots\supsetneq (M_n)_\mathfrak{m}=0.\]Since the localization functor is exact (§Properties of Localization, ⁋Proposition 2) and by the computation just above, we have
\[(M_k)_\mathfrak{m}/(M_{k+1})_\mathfrak{m}\cong (M_k/M_{k+1})_\mathfrak{m}=\begin{cases}M_k/M_{k+1}&\text{if $M_k/M_{k+1}\cong A/\mathfrak{m}$,}\\0&\text{otherwise}\end{cases}\]From this we obtain the second result, and the third result can be proved similarly.
Properties of Artinian and Noetherian Rings
Theorem 4 The following are all equivalent for a ring \(A\).
- \(A\) is noetherian and every prime ideal is maximal.
- \(A\) has finite length as an \(A\)-module.
- \(A\) is artinian.
Proof
First, assume the first condition and prove the second. Suppose for contradiction that \(A\) satisfies the first condition but does not have finite length. Let \(\mathfrak{a}\) be a maximal element among the ideals of \(A\) such that \(A/\mathfrak{a}\) does not have finite length. Then if \(ab\in \mathfrak{a}\) and \(a\not\in \mathfrak{a}\), we have the short exact sequence
\[0\longrightarrow A/(\mathfrak{a}:a)\overset{a}{\longrightarrow}A/\mathfrak{a}\longrightarrow A/(\mathfrak{a}+(a))\longrightarrow 0\](§Basic Notions, §§Basic Definitions). Since \(\mathfrak{a}+(a)\) is an ideal strictly containing \(\mathfrak{a}\), by definition \(A/(\mathfrak{a}+(a))\) must have finite length. On the other hand, \(\mathfrak{a}\subseteq (\mathfrak{a}:(a))\) by definition, and if \(b\not\in \mathfrak{a}\) then \(\mathfrak{a}\subsetneq (\mathfrak{a}:(a))\), so again by the definition of \(\mathfrak{a}\), \(A/(\mathfrak{a}:a)\) must have finite lenngth. Hence, by combining these composition series we obtain a composition series for \(A/\mathfrak{a}\), which contradicts the assumption that \(A\) does not have finite length; therefore \(b\in \mathfrak{a}\). That is, \(\mathfrak{a}\) is a prime ideal. Now, by the assumption of condition 1, \(\mathfrak{a}\) is maximal, so \(A/\mathfrak{a}\) is a field, which again contradicts the assumption that \(A/\mathfrak{a}\) does not have finite length; hence we obtain the desired result.
Now if we assume the second condition, the third follows immediately from Theorem 3. Thus it suffices to assume the third condition and prove the first. To this end, consider the collection of ideals that are products of maximal ideals of \(A\). Since \(A\) is artinian, there exists a minimal ideal \(\mathfrak{a}\) in this collection. Then \(\mathfrak{a}=0\), and therefore the zero ideal can be written as a product of maximal ideals \(0=\mathfrak{m}_1\cdots\mathfrak{m}_k\).
To prove this claim, we first observe that the minimality of \(\mathfrak{a}\) means that \(\mathfrak{m}\mathfrak{a}=\mathfrak{a}\) for every maximal ideal \(\mathfrak{m}\); that is, \(\mathfrak{a}\subseteq \mathfrak{m}\). By the same logic, \(\mathfrak{a}^2\) is also a product of maximal ideals, so \(\mathfrak{a}^2=\mathfrak{a}\) also holds. Now assume for contradiction that \(\mathfrak{a}\neq 0\). Then there exists an ideal \(\mathfrak{b}\) such that \(\mathfrak{a}\mathfrak{b}\neq 0\), and again by the artinian assumption we can choose \(\mathfrak{b}\) to be minimal among ideals satisfying this property. Then
\[(\mathfrak{b}\mathfrak{a})\mathfrak{a}=\mathfrak{b}\mathfrak{a}^2=\mathfrak{b}\mathfrak{a}\neq 0,\]so by the minimality of \(\mathfrak{b}\) we must have \(\mathfrak{b}\mathfrak{a}=\mathfrak{b}\).
Now, by the definition of \(\mathfrak{b}\), there must exist some \(y\in \mathfrak{b}\) with \(y \mathfrak{a}\neq 0\), and by the minimality of \(\mathfrak{b}\) we must have \(\mathfrak{b}=(y)\). From the above equality \(\mathfrak{b}\mathfrak{a}=\mathfrak{b}\), there must exist some \(x\in \mathfrak{a}\) such that \(xy=y\). That is, \((1-x)y=0\). But since \(x\in \mathfrak{a}\) and \(\mathfrak{a}\) is contained in every maximal ideal, \(1-x\) is not contained in any maximal ideal. Hence \(1-x\) is a unit, which implies \(y=0\), contradicting the definition of \(\mathfrak{b}\). Therefore \(\mathfrak{a}=0\), and so the zero ideal can be written as a product of maximal ideals \(0=\mathfrak{m}_1\cdots\mathfrak{m}_k\).
Now for each \(l=1,\ldots, k-1\), viewing \(\mathfrak{m}_1\cdots\mathfrak{m}_l/\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\) as an \(A/\mathfrak{m}_{l+1}\)-vector space, its submodules are of the form \(\mathfrak{a}/\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\), that is, they can be regarded as ideals of \(A\) containing \(\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\); since \(A\) is artinian, this implies that \(\mathfrak{m}_1\cdots\mathfrak{m}_l/\mathfrak{m}_1\cdots\mathfrak{m}_{l+1}\) is finite-dimensional. Collecting all of these gives a composition series for \(A\), and therefore by Theorem 3, \(A\) is noetherian.
To show the remaining condition, take any prime ideal \(\mathfrak{p}\) of \(A\). From
\[\mathfrak{m}_1\cdots\mathfrak{m}_k=0\subseteq \mathfrak{p}\]we know that \(\mathfrak{m}_l\subseteq \mathfrak{p}\) for some \(l\), and hence \(\mathfrak{m}_l=\mathfrak{p}\). Thus the first condition holds.
From this one can show that any artinian ring is a finite product of local artinian rings. For noetherian rings, the following holds.
Theorem 5 For a noetherian ring \(A\), \(A\) is a finite product of domains if and only if \(A_\mathfrak{m}\) is a domain for every maximal ideal \(\mathfrak{m}\).
Proof
First, suppose that \(A\) is a finite product of domains \(A=\prod A_i\). Then any prime ideal of \(A\) cannot contain the unit \(e_i\) of \(A_i\), so it lies in the multiplicative subset, and since this element \(e_i\) annihilates \(A_j\) for \(i\neq j\), we have \(A=(A_i)_\mathfrak{p}\). (§Localization, ⁋Proposition 5)
Conversely, assume that the localization at every maximal ideal \(\mathfrak{m}\) of \(A\) is a domain, and let \(\{\mathfrak{q}_i\}\) be the minimal prime ideals of \(A\). Then by §Associated Primes, ⁋Theorem 7, which will be proved in the next post, \(\{\mathfrak{q}_i\}\) is a finite set. Accepting this, there is a natural map
\[A \rightarrow \prod_{i\in I} A/\mathfrak{q}_i\]to a finite direct product of domains. We now show that this map is an isomorphism. By §Properties of Localization, ⁋Proposition 4, it suffices to show that the localized map
\[A_\mathfrak{m} \rightarrow \left(\prod_{i\in I} A/\mathfrak{q}_i\right)_\mathfrak{m}\]is an isomorphism for every maximal ideal. On the other hand, by §Localization, ⁋Proposition 8, there is a one-to-one correspondence between the minimal prime ideals of \(A\) contained in \(\mathfrak{m}\) and the minimal prime ideals of \(A_\mathfrak{m}\). But by assumption \(A_\mathfrak{m}\) is a domain, so \(A_\mathfrak{m}\) has a unique minimal prime ideal \((0)\); hence exactly one of the \(\mathfrak{q}_i\)’s is contained in \(\mathfrak{m}\). From this,
\[\left(\prod_{i\in I} A/\mathfrak{q}_i\right)_\mathfrak{m}=(A/\mathfrak{q}_i)_\mathfrak{m}=A_\mathfrak{m}\]so the above map is an isomorphism.
Corollary 6 For a noetherian ring \(A\) and a finiely generated \(A\)-module \(M\), the following are all equivalent.
- \(M\) has finite length.
- Some product of maximal ideals \(\prod_{i=1}^n \mathfrak{m}_i\) annihilates \(M\).
- Every prime ideal containing \(\ann(M)\) is maximal.
- \(A/\ann(M)\) is artinian.
Proof
First, if we assume condition 1, then condition 2 follows immediately from the second and third results of Theorem 3. Now assume condition 2. Then for any prime ideal \(\mathfrak{p}\) containing \(\ann(M)\), we have \(\mathfrak{p}\supseteq\prod \mathfrak{m}_i\), so \(\mathfrak{p}\supseteq \mathfrak{m}_i\) for some \(i\), and hence \(\mathfrak{p}=\mathfrak{m}_i\). That condition 3 implies condition 4 follows from the equivalence of the first and third conditions in Theorem 4, and finally, if we assume condition 4, then \(A/\ann(M)\) has finite length as an \(A/\ann(M)\)-module by condition 2 of Theorem 4, and since \(M\) is a finitely generated \(A/\ann(M)\)-module, we obtain the desired result.
Then the following holds.
Corollary 7 Fix a noetherian ring \(A\), a finitely generated \(A\)-module \(M\), and a prime ideal \(\mathfrak{p}\) containing \(\ann(M)\). Then \(M_\mathfrak{p}\) has finite length as an \(A_\mathfrak{p}\)-module if and only if \(\mathfrak{p}\) is minimal among the primes containing \(\ann(M)\).
Proof
If \(\mathfrak{p}\) is a minimal prime ideal containing \(\ann(M)\), then the \(A_\mathfrak{p}\)-module \(M_\mathfrak{p}\) has finite length. Indeed, considering the localization \(A_\mathfrak{p}\), its prime ideals correspond by §Localization, ⁋Proposition 8 to the prime ideals of \(A\) contained in \(\mathfrak{p}\); then by the minimality of \(\mathfrak{p}\), the only prime ideal containing \(\ann(M)A_\mathfrak{p}\) is \(\mathfrak{p}A_\mathfrak{p}\), which is the (unique) maximal ideal of the local ring \(A_\mathfrak{p}\).
Conversely, suppose that \(M_\mathfrak{p}\) has finite length as an \(A_\mathfrak{p}\)-module. Then by Corollary 6, every prime ideal containing the annihilator \(\ann(M)A_\mathfrak{p}\) of \(M_\mathfrak{p}\) is maximal, and these again correspond via §Localization, ⁋Proposition 8 to the prime ideals of \(A\) containing \(\ann(M)\) and contained in \(\mathfrak{p}\); thus the above argument can be reversed.
In particular, for any ideal \(\mathfrak{a}\) of \(A\), considering the \(A\)-module \(A/\mathfrak{a}\) we have \(\ann(A/\mathfrak{a})=\mathfrak{a}\), and from this we obtain the following.
Corollary 8 For a noetherian ring \(A\), an arbitrary ideal \(\mathfrak{a}\), and a prime ideal \(\mathfrak{p}\) containing \(\mathfrak{a}\), the following are all equivalent.
- \(\mathfrak{p}\) is minimal among the prime ideals containing \(\mathfrak{a}\).
- \(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p}\) is artinian.
- In the localization \(A_\mathfrak{p}\), we have \((\mathfrak{p}A_\mathfrak{p})^n\subseteq \mathfrak{a}A_\mathfrak{p}\) for all sufficiently large \(n\).
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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