The Stacks project

47.5 Injective hulls

In this section we briefly discuss injective hulls.

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.

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

  1. for any $R$-module map $\varphi : M \to N$ there exists an $R$-module map $\psi : E \to E'$ such that

    \[ \xymatrix{ M \ar[r] \ar[d]_\varphi & E \ar[d]^\psi \\ N \ar[r] & E' } \]


  2. if $\varphi $ is injective, then $\psi $ is injective,

  3. if $\varphi $ is an essential injection, then $\psi $ is an isomorphism,

  4. if $\varphi $ is an isomorphism, then $\psi $ is an isomorphism,

  5. 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

  1. $E$ is the injective hull of any nonzero submodule of $E$,

  2. the intersection of any two nonzero submodules of $E$ is nonzero,

  3. $\text{End}_ R(E, E)$ is a noncommutative local ring with maximal ideal those $\varphi : E \to E$ whose kernel is nonzero, and

  4. 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

  1. $E$ is indecomposable,

  2. $E$ is the injective hull of $\kappa (\mathfrak p)$,

  3. $E$ 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. $\square$

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. $\square$

Comments (4)

Comment #1189 by Felipe Zaldivar on

For Proposition 43.5.9 two corrections: The name of the Proposition should be : Structure of injective modules over Noetherian rings" (the word "of" is missing). Also, in the proof, in the first line the phrase "For the second statement,..." should change to "For the first statement,..."

Comment #7554 by Tao on

A typo: in the proof of lemma Lemma 47.5.6 Line 3, there is no map in the definition of .

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