Definition 47.5.1. Let $R$ be a ring. A injection $M \to I$ of $R$-modules is said to be an injective hull if $I$ is a injective $R$-module and $M \to I$ is an essential injection.
47.5 Injective hulls
In this section we briefly discuss injective hulls.
Injective hulls always exist.
Lemma 47.5.2. Let $R$ be a ring. Any $R$-module has an injective hull.
Proof. Let $M$ be an $R$-module. By More on Algebra, Section 15.55 the category of $R$-modules has enough injectives. Choose an injection $M \to I$ with $I$ an injective $R$-module. Consider the set $\mathcal{S}$ of submodules $M \subset E \subset I$ such that $E$ is an essential extension of $M$. We order $\mathcal{S}$ by inclusion. If $\{ E_\alpha \} $ is a totally ordered subset of $\mathcal{S}$, then $\bigcup E_\alpha $ is an essential extension of $M$ too (Lemma 47.2.3). Thus we can apply Zorn's lemma and find a maximal element $E \in \mathcal{S}$. We claim $M \subset E$ is an injective hull, i.e., $E$ is an injective $R$-module. This follows from Lemma 47.3.5. $\square$
Lemma 47.5.3. Let $R$ be a ring. Let $M$, $N$ be $R$-modules and let $M \to E$ and $N \to E'$ be injective hulls. Then
for any $R$-module map $\varphi : M \to N$ there exists an $R$-module map $\psi : E \to E'$ such that
commutes,
if $\varphi $ is injective, then $\psi $ is injective,
if $\varphi $ is an essential injection, then $\psi $ is an isomorphism,
if $\varphi $ is an isomorphism, then $\psi $ is an isomorphism,
if $M \to I$ is an embedding of $M$ into an injective $R$-module, then there is an isomorphism $I \cong E \oplus I'$ compatible with the embeddings of $M$,
In particular, the injective hull $E$ of $M$ is unique up to isomorphism.
Proof. Part (1) follows from the fact that $E'$ is an injective $R$-module. Part (2) follows as $\mathop{\mathrm{Ker}}(\psi ) \cap M = 0$ and $E$ is an essential extension of $M$. Assume $\varphi $ is an essential injection. Then $E \cong \psi (E) \subset E'$ by (2) which implies $E' = \psi (E) \oplus E''$ because $E$ is injective. Since $E'$ is an essential extension of $M$ (Lemma 47.2.2) we get $E'' = 0$. Part (4) is a special case of (3). Assume $M \to I$ as in (5). Choose a map $\alpha : E \to I$ extending the map $M \to I$. Arguing as before we see that $\alpha $ is injective. Thus as before $\alpha (E)$ splits off from $I$. This proves (5). $\square$
Example 47.5.4. Let $R$ be a domain with fraction field $K$. Then $R \subset K$ is an injective hull of $R$. Namely, by Example 47.3.6 we see that $K$ is an injective $R$-module and by Lemma 47.2.4 we see that $R \subset K$ is an essential extension.
Definition 47.5.5. An object $X$ of an additive category is called indecomposable if it is nonzero and if $X = Y \oplus Z$, then either $Y = 0$ or $Z = 0$.
Lemma 47.5.6. Let $R$ be a ring. Let $E$ be an indecomposable injective $R$-module. Then
$E$ is the injective hull of any nonzero submodule of $E$,
the intersection of any two nonzero submodules of $E$ is nonzero,
$\text{End}_ R(E, E)$ is a noncommutative local ring with maximal ideal those $\varphi : E \to E$ whose kernel is nonzero, and
the set of zerodivisors on $E$ is a prime ideal $\mathfrak p$ of $R$ and $E$ is an injective $R_\mathfrak p$-module.
Proof. Part (1) follows from Lemma 47.5.3. Part (2) follows from part (1) and the definition of injective hulls.
Proof of (3). Set $A = \text{End}_ R(E, E)$ and $I = \{ \varphi \in A \mid \mathop{\mathrm{Ker}}(\varphi ) \not= 0\} $. The statement means that $I$ is a two sided ideal and that any $\varphi \in A$, $\varphi \not\in I$ is invertible. Suppose $\varphi $ and $\psi $ are not injective. Then $\mathop{\mathrm{Ker}}(\varphi ) \cap \mathop{\mathrm{Ker}}(\psi )$ is nonzero by (2). Hence $\varphi + \psi \in I$. It follows that $I$ is a two sided ideal. If $\varphi \in A$, $\varphi \not\in I$, then $E \cong \varphi (E) \subset E$ is an injective submodule, hence $E = \varphi (E)$ because $E$ is indecomposable.
Proof of (4). Consider the ring map $R \to A$ and let $\mathfrak p \subset R$ be the inverse image of the maximal ideal $I$. Then it is clear that $\mathfrak p$ is a prime ideal and that $R \to A$ extends to $R_\mathfrak p \to A$. Thus $E$ is an $R_\mathfrak p$-module. It follows from Lemma 47.3.3 that $E$ is injective as an $R_\mathfrak p$-module. $\square$
Lemma 47.5.7. Let $\mathfrak p \subset R$ be a prime of a ring $R$. Let $E$ be the injective hull of $R/\mathfrak p$. Then
$E$ is indecomposable,
$E$ is the injective hull of $\kappa (\mathfrak p)$,
$E$ is the injective hull of $\kappa (\mathfrak p)$ over the ring $R_\mathfrak p$.
Proof. By Lemma 47.2.4 the inclusion $R/\mathfrak p \subset \kappa (\mathfrak p)$ is an essential extension. Then Lemma 47.5.3 shows (2) holds. For $f \in R$, $f \not\in \mathfrak p$ the map $f : \kappa (\mathfrak p) \to \kappa (\mathfrak p)$ is an isomorphism hence the map $f : E \to E$ is an isomorphism, see Lemma 47.5.3. Thus $E$ is an $R_\mathfrak p$-module. It is injective as an $R_\mathfrak p$-module by Lemma 47.3.3. Finally, let $E' \subset E$ be a nonzero injective $R$-submodule. Then $J = (R/\mathfrak p) \cap E'$ is nonzero. After shrinking $E'$ we may assume that $E'$ is the injective hull of $J$ (see Lemma 47.5.3 for example). Observe that $R/\mathfrak p$ is an essential extension of $J$ for example by Lemma 47.2.4. Hence $E' \to E$ is an isomorphism by Lemma 47.5.3 part (3). Hence $E$ is indecomposable. $\square$
Lemma 47.5.8. Let $R$ be a Noetherian ring. Let $E$ be an indecomposable injective $R$-module. Then there exists a prime ideal $\mathfrak p$ of $R$ such that $E$ is the injective hull of $\kappa (\mathfrak p)$.
Proof. Let $\mathfrak p$ be the prime ideal found in Lemma 47.5.6. Say $\mathfrak p = (f_1, \ldots , f_ r)$. Pick a nonzero element $x \in \bigcap \mathop{\mathrm{Ker}}(f_ i : E \to E)$, see Lemma 47.5.6. Then $(R_\mathfrak p)x$ is a module isomorphic to $\kappa (\mathfrak p)$ inside $E$. We conclude by Lemma 47.5.6. $\square$
Proposition 47.5.9 (Structure of injective modules over Noetherian rings). Let $R$ be a Noetherian ring. Every injective module is a direct sum of indecomposable injective modules. Every indecomposable injective module is the injective hull of the residue field at a prime.
Proof. The second statement is Lemma 47.5.8. For the first statement, let $I$ be an injective $R$-module. We will use transfinite recursion to construct $I_\alpha \subset I$ for ordinals $\alpha $ which are direct sums of indecomposable injective $R$-modules $E_{\beta + 1}$ for $\beta < \alpha $. For $\alpha = 0$ we let $I_0 = 0$. Suppose given an ordinal $\alpha $ such that $I_\alpha $ has been constructed. Then $I_\alpha $ is an injective $R$-module by Lemma 47.3.7. Hence $I \cong I_\alpha \oplus I'$. If $I' = 0$ we are done. If not, then $I'$ has an associated prime by Algebra, Lemma 10.63.7. Thus $I'$ contains a copy of $R/\mathfrak p$ for some prime $\mathfrak p$. Hence $I'$ contains an indecomposable submodule $E$ by Lemmas 47.5.3 and 47.5.7. Set $I_{\alpha + 1} = I_\alpha \oplus E_\alpha $. If $\alpha $ is a limit ordinal and $I_\beta $ has been constructed for $\beta < \alpha $, then we set $I_\alpha = \bigcup _{\beta < \alpha } I_\beta $. Observe that $I_\alpha = \bigoplus _{\beta < \alpha } E_{\beta + 1}$. This concludes the proof. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (4)
Comment #1189 by Felipe Zaldivar on
Comment #1204 by Johan on
Comment #7554 by Tao on
Comment #7677 by Stacks Project on