Lemma 47.3.1. Let $R$ be a ring. Any product of injective $R$-modules is injective.

## 47.3 Injective modules

Some results about injective modules over rings.

**Proof.**
Special case of Homology, Lemma 12.27.3.
$\square$

Lemma 47.3.2. Let $R \to S$ be a flat ring map. If $E$ is an injective $S$-module, then $E$ is injective as an $R$-module.

**Proof.**
This is true because $\mathop{\mathrm{Hom}}\nolimits _ R(M, E) = \mathop{\mathrm{Hom}}\nolimits _ S(M \otimes _ R S, E)$ by Algebra, Lemma 10.13.3 and the fact that tensoring with $S$ is exact.
$\square$

Lemma 47.3.3. Let $R \to S$ be an epimorphism of rings. Let $E$ be an $S$-module. If $E$ is injective as an $R$-module, then $E$ is an injective $S$-module.

**Proof.**
This is true because $\mathop{\mathrm{Hom}}\nolimits _ R(N, E) = \mathop{\mathrm{Hom}}\nolimits _ S(N, E)$ for any $S$-module $N$, see Algebra, Lemma 10.106.14.
$\square$

Lemma 47.3.4. Let $R \to S$ be a ring map. If $E$ is an injective $R$-module, then $\mathop{\mathrm{Hom}}\nolimits _ R(S, E)$ is an injective $S$-module.

**Proof.**
This is true because $\mathop{\mathrm{Hom}}\nolimits _ S(N, \mathop{\mathrm{Hom}}\nolimits _ R(S, E)) = \mathop{\mathrm{Hom}}\nolimits _ R(N, E)$ by Algebra, Lemma 10.13.4.
$\square$

Lemma 47.3.5. Let $R$ be a ring. Let $I$ be an injective $R$-module. Let $E \subset I$ be a submodule. The following are equivalent

$E$ is injective, and

for all $E \subset E' \subset I$ with $E \subset E'$ essential we have $E = E'$.

In particular, an $R$-module is injective if and only if every essential extension is trivial.

**Proof.**
The final assertion follows from the first and the fact that the category of $R$-modules has enough injectives (More on Algebra, Section 15.54).

Assume (1). Let $E \subset E' \subset I$ as in (2). Then the map $\text{id}_ E : E \to E$ can be extended to a map $\alpha : E' \to E$. The kernel of $\alpha $ has to be zero because it intersects $E$ trivially and $E'$ is an essential extension. Hence $E = E'$.

Assume (2). Let $M \subset N$ be $R$-modules and let $\varphi : M \to E$ be an $R$-module map. In order to prove (1) we have to show that $\varphi $ extends to a morphism $N \to E$. Consider the set $\mathcal{S}$ of pairs $(M', \varphi ')$ where $M \subset M' \subset N$ and $\varphi ' : M' \to E$ is an $R$-module map agreeing with $\varphi $ on $M$. We define an ordering on $\mathcal{S}$ by the rule $(M', \varphi ') \leq (M'', \varphi '')$ if and only if $M' \subset M''$ and $\varphi ''|_{M'} = \varphi '$. It is clear that we can take the maximum of a totally ordered subset of $\mathcal{S}$. Hence by Zorn's lemma we may assume $(M, \varphi )$ is a maximal element.

Choose an extension $\psi : N \to I$ of $\varphi $ composed with the inclusion $E \to I$. This is possible as $I$ is injective. If $\psi (N) \subset E$, then $\psi $ is the desired extension. If $\psi (N)$ is not contained in $E$, then by (2) the inclusion $E \subset E + \psi (N)$ is not essential. hence we can find a nonzero submodule $K \subset E + \psi (N)$ meeting $E$ in $0$. This means that $M' = \psi ^{-1}(E + K)$ strictly contains $M$. Thus we can extend $\varphi $ to $M'$ using

This contradicts the maximality of $(M, \varphi )$. $\square$

Example 47.3.6. Let $R$ be a reduced ring. Let $\mathfrak p \subset R$ be a minimal prime so that $K = R_\mathfrak p$ is a field (Algebra, Lemma 10.24.1). Then $K$ is an injective $R$-module. Namely, we have $\mathop{\mathrm{Hom}}\nolimits _ R(M, K) = \mathop{\mathrm{Hom}}\nolimits _ K(M_\mathfrak p, K)$ for any $R$-module $M$. Since localization is an exact functor and taking duals is an exact functor on $K$-vector spaces we conclude $\mathop{\mathrm{Hom}}\nolimits _ R(-, K)$ is an exact functor, i.e., $K$ is an injective $R$-module.

Lemma 47.3.7. Let $R$ be a Noetherian ring. A direct sum of injective modules is injective.

**Proof.**
Let $E_ i$ be a family of injective modules parametrized by a set $I$. Set $E = \bigcup E_ i$. To show that $E$ is injective we use Injectives, Lemma 19.2.6. Thus let $\varphi : I \to E$ be a module map from an ideal of $R$ into $E$. As $I$ is a finite $R$-module (because $R$ is Noetherian) we can find finitely many elements $i_1, \ldots , i_ r \in I$ such that $\varphi $ maps into $\bigcup _{j = 1, \ldots , r} E_{i_ j}$. Then we can extend $\varphi $ into $\bigcup _{j = 1, \ldots , r} E_{i_ j}$ using the injectivity of the modules $E_{i_ j}$.
$\square$

Lemma 47.3.8. Let $R$ be a Noetherian ring. Let $S \subset R$ be a multiplicative subset. If $E$ is an injective $R$-module, then $S^{-1}E$ is an injective $S^{-1}R$-module.

**Proof.**
Since $R \to S^{-1}R$ is an epimorphism of rings, it suffices to show that $S^{-1}E$ is injective as an $R$-module, see Lemma 47.3.3. To show this we use Injectives, Lemma 19.2.6. Thus let $I \subset R$ be an ideal and let $\varphi : I \to S^{-1} E$ be an $R$-module map. As $I$ is a finitely presented $R$-module (because $R$ is Noetherian) we can find an $f \in S$ and an $R$-module map $I \to E$ such that $f\varphi $ is the composition $I \to E \to S^{-1}E$ (Algebra, Lemma 10.10.2). Then we can extend $I \to E$ to a homomorphism $R \to E$. Then the composition

is the desired extension of $\varphi $ to $R$. $\square$

Lemma 47.3.9. Let $R$ be a Noetherian ring. Let $I$ be an injective $R$-module.

Let $f \in R$. Then $E = \bigcup I[f^ n] = I[f^\infty ]$ is an injective submodule of $I$.

Let $J \subset R$ be an ideal. Then the $J$-power torsion submodule $I[J^\infty ]$ is an injective submodule of $I$.

**Proof.**
We will use Lemma 47.3.5 to prove (1). Suppose that $E \subset E' \subset I$ and that $E'$ is an essential extension of $E$. We will show that $E' = E$. If not, then we can find $x \in E'$ and $x \not\in E$. Let $J = \{ a \in R \mid ax \in E\} $. Since $R$ is Noetherian, we may choose $x$ so that $J$ is maximal among ideals of this form. Again since $R$ is Noetherian, we may write $J = (g_1, \ldots , g_ t)$ for some $g_ i \in R$. By definition $E$ is the set of elements of $I$ annihilated by powers of $f$, so we may choose integers $n_ i$ so that $f^{n_ i}g_ ix = 0$. Set $n = \mathrm{max}\{ n_ i \} $. Then $x' = f^ n x$ is an element of $E'$ not in $E$ and is annihilated by $J$. Then by maximality of $J$ we have $J = \{ a \in R \mid ax' \in E \} = \text{Ann}(x')$, so $Rx' \cap E = 0$. Hence $E'$ is not an essential extension of $E$, a contradiction.

To prove (2) write $J = (f_1, \ldots , f_ t)$. Then $I[J^\infty ]$ is equal to

and the result follows from (1) and induction. $\square$

Lemma 47.3.10. Let $A$ be a Noetherian ring. Let $E$ be an injective $A$-module. Then $E \otimes _ A A[x]$ has injective-amplitude $[0, 1]$ as an object of $D(A[x])$. In particular, $E \otimes _ A A[x]$ has finite injective dimension as an $A[x]$-module.

**Proof.**
Let us write $E[x] = E \otimes _ A A[x]$. Consider the short exact sequence of $A[x]$-modules

where the first map sends $p \in E[x]$ to $f \mapsto fp$ and the second map sends $\varphi $ to $f \mapsto \varphi (xf) - x\varphi (f)$. The second map is surjective because $\mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) = \prod _{n \geq 0} E[x]$ as an abelian group and the map sends $(e_ n)$ to $(e_{n + 1} - xe_ n)$ which is surjective. As an $A$-module we have $E[x] \cong \bigoplus _{n \geq 0} E$ which is injective by Lemma 47.3.7. Hence the $A[x]$-module $\mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x])$ is injective by Lemma 47.3.4 and the proof is complete. $\square$

## 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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #3443 by Sebastian Bozlee on

Comment #3497 by Johan on