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Postmodern Algebra

Sam Macdonald

Section 18.2 Injective Modules in Noetherian Rings

So far, our observations on injectives are very general. For noetherian rings, there is a beautiful structure theory of injective modules due to Bass. To prepare for the first step, we note:

Proof.

Let \(\left\{E_{i}\right\}_{i \in I}\) be a family of injective modules. For an ideal \(\mathfrak{a}\) and any \(i, \operatorname{Hom}_{R}\left(R, E_{i}\right) \rightarrow\) \(\operatorname{Hom}_{R}\left(\mathfrak{a}, E_{i}\right)\) is surjective. Applying the previous lemma and the hypothesis that \(\mathfrak{a}\) is finitely generated, the composition
\begin{equation*} \operatorname{Hom}_{R}\left(R, \bigoplus_{i \in I} E_{i}\right) \cong \bigoplus_{i \in I} \operatorname{Hom}_{R}\left(R, E_{i}\right) \rightarrow \bigoplus_{i \in I} \operatorname{Hom}_{R}\left(\mathfrak{a}, E_{i}\right) \cong \operatorname{Hom}_{R}\left(\mathfrak{a}, \bigoplus_{i \in I} E_{i}\right) \end{equation*}
is surjective. By Baer’s criterion, \(\bigoplus_{i \in I} E_{i}\) is injective.

Associated Primes.

Evidently, the injective hulls of prime cyclic modules, \(E_{R}(R / \mathfrak{p})\text{,}\) are injective modules. They will be the building blocks in the structure theory. Both to that end and for our purposes later, we will want to understand their structure. We collect some of their key properties here.
Before we do this, we recall some facts about modules that are not necessarily finitely generated.
Recall that \(\mathfrak{p}\) is an associated prime of a module \(M\) if there is an embedding \(R / \mathfrak{p} \hookrightarrow M\text{;}\) we write \(\operatorname{Ass}(M)\) for the set of associated primes of \(M\text{.}\) For a general module,
  • \(\operatorname{Ass}(M)\) need not be finite, and
  • \(\bigcap_{\mathfrak{p} \in \operatorname{Ass}(M)} \mathfrak{p}\) does not necessarily agree with \(\sqrt{\operatorname{ann}_{R}(M)}\text{.}\)
However, \(M \subseteq N\) implies \(\operatorname{Ass}(M) \subseteq \operatorname{Ass}(N)\text{,}\) and if \(R\) is noetherian,
  • if \(M \neq 0\text{,}\) then \(\operatorname{Ass}(M) \neq \varnothing\text{,}\) and
  • \(\operatorname{Supp}(M)=\{\mathfrak{q} \in \operatorname{Spec}(R) \mid \mathfrak{q} \supseteq \mathfrak{p}\) for some \(\mathfrak{p} \in \operatorname{Ass}(M)\}\text{.}\)
Proof. For the first, take some nonzero \(m \in M ; \varnothing \neq \operatorname{Ass}(R m) \subseteq \operatorname{Ass}(M)\text{.}\) For the second, we note that \((R / \mathfrak{p})_{\mathfrak{q}} \neq 0\) if \(\mathfrak{q} \subseteq \mathfrak{p}\text{,}\) which yields " \(\supseteq\text{,}\)" and if \(x / 1 \neq 0\) in \(M_{\mathfrak{q}}\text{,}\) then \(\mathfrak{q}\) must contain an associated prime of \(R x\text{,}\) which is an associated prime of \(M\text{.}\)

\(I\)-torsion.

A module \(M\) is \(I\)-torsion if for every \(m \in M, \exists N\) such that \(I^{N} m=0\text{.}\) This does not necessarily imply that there is some \(N\) such that \(I^{N} M=0\text{.}\) If \(R\) is noetherian and \(I=\left(f_{1}, \ldots, f_{t}\right)\) then the following are equivalent:
  • \(M\) is \(I\)-torsion;
  • \(\forall m \in M\text{,}\) there is an \(N\) such that \(f_{i}^{N} m=0\) for all \(i\text{;}\)
  • \(\forall \mathfrak{p} \in \operatorname{Ass}(M), \mathfrak{p} \supseteq I\text{.}\)
Proof. For the equivalence of the first two, we note that \(I^{N t} \subseteq\left(f_{1}^{N}, \ldots, f_{t}^{N}\right) \subseteq I^{N}\text{,}\) so a large power of \(I\) kills an element iff large powers of each \(f_{i}\) do. This uses finite generation in a crucial way. The first implies the third since \(I\) contains a nonzerodivisor on the domain \(R / \mathfrak{q}\text{,}\) and any power of this element does not kill the generator of \(R / \mathfrak{q}\text{.}\) The third implies the first, since, for each \(x \in M\text{,}\) the associated primes of the cyclic module \(R x\) all are contained in \(I\text{,}\) so \(\sqrt{\operatorname{ann}(x)}=\bigcap_{\mathfrak{p} \in \operatorname{Ass}(R x)} \mathfrak{p} \subseteq\) \(I\text{.}\)

Proof.

We only need to see that an associated prime of \(N\) is also associated to \(M\text{.}\) If \(x\) generates a cyclic submodule of \(N\) isomorphic to \(R / \mathfrak{p}\text{,}\) then any nonzero multiple of \(x\) does as well; since \(x\) has a nonzero multiple in \(M\text{,}\) we are done.

Proof.

Proof. 4. This is immediate from the previous lemma and the discussion above it.
  1. By Example ??, \(R / \mathfrak{p} \subseteq \kappa\) is an essential extension, so \(E\) must contain a copy of \(\kappa\text{;}\) we identify \(\kappa\) with its isomorphic copy in \(E: R / \mathfrak{p} \subseteq \kappa \subseteq E\) are all essential. Since multiplication by \(x\) is one-to-one on \(\kappa\text{,}\) it must be one-to-one on its essential extension \(E\text{.}\) Then, the submodule \(x E \cong E\) is injective, so a direct summand of \(E\text{.}\) But, since \(\kappa \subseteq x E \subseteq E\) is essential, \(E\) cannot decompose as a nontrivial direct sum, so \(x E=E\text{.}\)
  2. By part (1), \(\operatorname{ann}_{E}(\mathfrak{p})=\operatorname{ann}_{E}\left(\mathfrak{p} R_{\mathfrak{p}}\right)\) is a \(\kappa\)-vector space, and contains the copy of \(\kappa\) we found earlier, so \(\kappa \subseteq \operatorname{ann}_{E}(\mathfrak{p})\) splits. Since the extensions \(\kappa \subseteq \operatorname{ann}_{E}(\mathfrak{p}) \subseteq E\) are essential, \(\operatorname{ann}_{E}(\mathfrak{p})\) cannot decompose as a nontrivial direct sum of \(\kappa\)-vector spaces.
  3. Since \(\kappa \subseteq E\) is essential as \(R\)-modules, it is clearly essential as \(R_{\mathfrak{p}}\)-modules. (Any nonzero element has a nonzero \(R\)-multiple, hence a nonzero \(R_{\mathfrak{p}}\)-multiple in the submodule.) To see that \(E\) is a maximal essential extension of \(\kappa\) as an \(R_{\mathfrak{p}}\)-module, let \(E \subseteq M\) be an essential extension of \(R_{\mathfrak{p}}\)-modules. For any nonzero \(m \in M\text{,}\) there is some nonzero multiple \((r / s) m\) in \(E\text{,}\) with \(s \notin \mathfrak{p}\text{.}\) Then \(r m\) is a nonzero multiple of \(m\) in \(E\text{,}\) so \(E \subseteq M\) is an essential extension as \(R\)-modules, so \(M=E\text{.}\)
  4. We recall that \(\operatorname{Hom}_{R_{\mathfrak{p}}}\left(R_{\mathfrak{p}} / \mathfrak{p} R_{\mathfrak{p}}, M\right)\) can be identified with the submodule of \(M\) consisting of elements killed by \(\mathfrak{p} R_{\mathfrak{p}}\text{.}\)
Then, the first isomorphism follows from part (2) applied in \(R_{\mathfrak{p}}\text{.}\)
Set \(E^{\prime}=E_{R}(R / \mathfrak{q})\text{.}\) By (4), \(E_{\mathfrak{p}}^{\prime} \neq 0\) only if \(\mathfrak{q} \subseteq \mathfrak{p}\text{.}\) By \((1), E_{\mathfrak{p}}^{\prime}=E_{R_{\mathfrak{p}}}\left(R_{\mathfrak{p}} / \mathfrak{q} R_{\mathfrak{p}}\right)\text{,}\) which by (4) has \(\mathfrak{q} R_{\mathfrak{p}}\) as its only associated prime; in particular, no element is killed by \(\mathfrak{p} R_{\mathfrak{p}}\text{,}\) so the hom is zero.
We can now prove the structure theorem for injective modules.

Proof.

By Zorn’s lemma, there is a submodule \(E^{\prime}\) (not a priori nonzero) of \(E\) that is
  • isomorphic to a direct sum of injective hulls of prime cyclic modules: \(E^{\prime} \cong \oplus_{\mathfrak{p}} E_{R}(R / \mathfrak{p})^{\eta_{\mathfrak{p}}}\text{,}\)
  • and no other submodule of \(E\) contains \(E^{\prime}\) and admits a decomposition of this form.
By Proposition ??, \(E^{\prime}\) is injective, so we can write \(E=E^{\prime} \oplus C\) for some \(C\text{.}\) If \(E^{\prime \prime} \neq 0\text{,}\) take \(\mathfrak{p} \in \operatorname{Ass}\left(E^{\prime \prime}\right)\text{.}\) The inclusion map \(R / \mathfrak{p} \rightarrow E^{\prime \prime}\) extends along the inclusion \(R / \mathfrak{p} \subseteq E_{R}(R / \mathfrak{p})\) to a map \(E_{R}(R / \mathfrak{p})\) since \(E^{\prime \prime}\) is injective; since the inclusion is essential, \(E_{R}(R / \mathfrak{p})\) embeds into \(E^{\prime \prime}\text{.}\) Since \(E_{R}(R / \mathfrak{p})\) is injective, it splits from \(E^{\prime \prime}\text{,}\) and we contradict the maximality of \(E^{\prime}\text{.}\) Thus, a decomposition as promised exists. Given such a decomposition, the formulas for \(\mu(\mathfrak{p})\) follow from part (5) of the previous theorem.
Thus, injective modules are uniquely determined by a multiplicity for each prime. This structural data, for the modules appearing in a minimal injective resolution of a module, are called Bass numbers.

Definition 18.19.

(Bass numbers). Let \(R\) be a noetherian ring, and \(M\) an \(R\)-module. Given a minimal injective resolution
\begin{equation*} 0 \rightarrow(M \rightarrow) E^{0} \rightarrow E^{1} \rightarrow E^{2} \rightarrow \cdots \end{equation*}
and decompositions
\begin{equation*} E^{i} \cong \bigoplus_{\mathfrak{p} \in \operatorname{Spec}(R)} E_{R}(R / \mathfrak{p})^{\mu(i, \mathfrak{p})} \end{equation*}
we call \(\mu(i, \mathfrak{p})\) the \(\mathrm{i}\)-th Bass number of \(M\) with respect to \(\mathfrak{p}\text{.}\)

Proof.

Given a minimal injective resolution
\begin{equation*} 0 \rightarrow(M \rightarrow) E^{0} \rightarrow E^{1} \rightarrow E^{2} \rightarrow \cdots \end{equation*}
we leave it as an exercise to check that
\begin{equation*} 0 \rightarrow\left(M_{\mathfrak{p}} \rightarrow\right) E_{\mathfrak{p}}^{0} \rightarrow E_{\mathfrak{p}}^{1} \rightarrow E_{\mathfrak{p}}^{2} \rightarrow \cdots \end{equation*}
is a minimal injective resolution of \(M_{\mathfrak{p}}\) as an \(R_{\mathfrak{p}}\)-module, and that the number of copies of \(E_{R}(R / \mathfrak{p})\) in \(E^{i}\) is the same as the number of copies of \(E_{R}(R / \mathfrak{p})\) in \(E_{\mathfrak{p}}^{i}\text{.}\) We compute the Ext modules in the statement from the latter injective resolution:
\begin{equation*} \operatorname{Ext}_{R_{\mathfrak{p}}}^{i}(\kappa(\mathfrak{p}), M)=H^{i}\left(0 \rightarrow \operatorname{Hom}_{R_{\mathfrak{p}}}\left(\kappa(\mathfrak{p}), E_{\mathfrak{p}}^{0}\right) \rightarrow \operatorname{Hom}_{R_{\mathfrak{p}}}\left(\kappa(\mathfrak{p}), E_{\mathfrak{p}}^{1}\right) \rightarrow \operatorname{Hom}_{R_{\mathfrak{p}}}\left(\kappa(\mathfrak{p}), E_{\mathfrak{p}}^{2}\right) \rightarrow \cdots\right) \end{equation*}
Claim. The differentials in the complex above are zero.
We need to show that for any \(R_{\mathfrak{p}}\)-linear \(\alpha: \kappa(\mathfrak{p}) \rightarrow E_{\mathfrak{p}}^{i}\text{,}\) the composition \(\delta^{i} \circ \alpha: \kappa(\mathfrak{p}) \rightarrow E_{\mathfrak{p}}^{i+1}\) is the zero map. Let \(x \in \kappa(\mathfrak{p})\text{,}\) and suppose that \(\alpha(x) \neq 0\) in \(E_{\mathfrak{p}}^{i}\text{.}\) The minimality of the injective resolution implies that \(\alpha(x)\) has a nonzero multiple in the image of \(\delta^{i-1}\text{.}\) Since \(\kappa\) is a field, this implies that \(\alpha(x)\) is in the image of \(\delta^{i-1}\text{,}\) and hence in the kernel of \(\delta^{i}\text{,}\) establishing the claim.
The proof of the theorem is now immediate, since the homology of the complex in the i-th spot is \(\operatorname{Hom}_{R_{\mathfrak{p}}}\left(\kappa(\mathfrak{p}), E_{\mathfrak{p}}^{i}\right)\text{,}\) which, by the previous theorem, is a \(\kappa(\mathfrak{p})\)-vector space of rank \(\mu(i, \mathfrak{p})\text{.}\)

Proof.

We can compute \(\mu(i, \mathfrak{p})\) by taking a minimal free resolution of \(\kappa(\mathfrak{p})\text{,}\) applying \(\operatorname{Hom}_{R_{\mathfrak{p}}}\left(-, M_{\mathfrak{p}}\right)\text{,}\) and computing homology. The complex we obtain consists of finitely generated modules, so its homology modules are finitely generated modules.

Example 18.22.

Example 2.30. Let’s compute the Bass numbers of \(\mathbb{Z}\) as a \(\mathbb{Z}\)-module. We claim that
\begin{equation*} \mathbb{Q} / \mathbb{Z} \cong \underset{p \text { prime }}{\bigoplus} \mathbb{Z}[1 / p] / \mathbb{Z} \cong \bigoplus_{\mathfrak{p} \in \operatorname{Spec}(\mathbb{Z}) \backslash\{(0)\}} E_{\mathbb{Z}}(\mathbb{Z} / p \mathbb{Z}) \end{equation*}
To see the first isomorphism, consider an element \([m / n] \in \mathbb{Q} / \mathbb{Z}\) with \(m / n\) in lowest terms. We want to see that \([m / n]\) has a unique expression as a sum of elements of \(\mathbb{Z}[1 / p] / \mathbb{Z}\text{.}\) We can discard any terms with denominators \(q\) not dividing \(n\text{,}\) since \((1 / q \cdot \mathbb{Z}) \cap(1 / n \cdot \mathbb{Z})=\mathbb{Z}\text{.}\) Write \(n=p_{1}^{e_{1}} \cdots p_{n}^{e_{n}}\text{.}\) Then, by the Chinese remainder theorem
\begin{equation*} \begin{aligned} & \frac{m}{n} \equiv \sum_{i} \frac{a_{i}}{p_{i}^{e_{i}}} \\ & m \equiv \sum_{i} a_{i} p_{1}^{e_{1}} \cdots \widehat{p_{i}^{e_{i}}} \cdots p_{n}^{e_{n}} \\ & (\bmod 1) \\ & (\bmod n) \\ & \left\{\begin{array}{c} m \equiv a_{1} \widehat{p_{1}^{e_{1}}} \cdots p_{i}^{e_{i}} \cdots p_{n}^{e_{n}} \quad\left(\bmod p_{1}^{e_{1}}\right) \\ \vdots \quad \vdots \quad \vdots \\ m \equiv a_{n} p_{1}^{e_{1}} \cdots p_{i}^{e_{i}} \cdots \widehat{p_{n}^{e_{n}}} \quad\left(\bmod p_{n}^{e_{n}}\right) \end{array}\right. \end{aligned} \end{equation*}
In each of these congruence equations, the product of prime powers on the RHS is a unit, so we can solve uniquely for each \(a_{i} \bmod p_{i}^{e_{i}}\text{;}\) that is, each \(\left[a_{i} / p_{i}^{e_{i}}\right] \in \mathbb{Z}[1 / p] / \mathbb{Z}\) is uniquely determined. This establishes the first isomorphism.
For the second, it suffices to note that \(\mathbb{Z}[1 / p] / \mathbb{Z}\) is divisible (check it!), that it is \(p\)-torsion (clear), and that the annihilator of \(p\) in this module is isomorphic to \(\mathbb{Z} / p \mathbb{Z}\text{.}\) From these observations, we see that the module is injective by Proposition ??, and that it is an essential extension of \(\mathbb{Z} / p \mathbb{Z}\) by Example ??.
We conclude that the Bass numbers of \(\mathbb{Z}\) are
\begin{equation*} \mu(0, \mathfrak{p})=\left\{\begin{array}{ll} 1 & \text { if } \mathfrak{p}=(0) \\ 0 & \text { if } \mathfrak{p}=(p) \neq(0) \end{array} \quad \text { and } \mu(1, \mathfrak{p})= \begin{cases}0 & \text { if } \mathfrak{p}=(0) \\ 1 & \text { if } \mathfrak{p}=(p) \neq(0)\end{cases}\right. \end{equation*}
Each Bass number here is either 0 or 1 . We caution that the "total first Bass number," the sum of the first Bass numbers, is infinite.
We note also that this 116 esque game of writing a fraction as a sum of fractions with pure-power denominators is very closely related to the calculation of local cohomology.
We now know that the indecomposable injectives of a noetherian ring are all of the form \(E_{R}(R / \mathfrak{p})\text{,}\) that every injective is (in a unique way) a direct sum of these, and finitely generated modules all have injective resolutions in which each injective has a finite number of copies of \(E_{R}(R / \mathfrak{p})\) in its direct sum decomposition. We also know many of the key structural properties of \(E_{R}(R / \mathfrak{p})\text{,}\) as collected in Theorem ??.
We will want to have a better understanding of \(E_{R}(k)\) for the residue field of a local ring in order to the develop the duality theory hinted at on the first day of class. We note that this encompasses any indecomposable injective \(E_{S}(S / \mathfrak{p})\) : by part (3) of Theorem ??, \(E_{S}(S / \mathfrak{p})=E_{R}(k)\) for \(R=S_{\mathfrak{p}}\) and \(k\) the residue field of \(R\text{.}\) We will pursue this with a little bit of redundancy. First, we will discuss a model for \(E_{R}(k)\) when \(R\) has a coefficient field (e.g., when \(R\) is complete) and the Matlis duality functor in that setting. Then, we will develop the duality theory in full generality.
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