## 47.3 Injective modules

Some results about injective modules over rings.

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

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 10.14.3 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 10.107.14. $\square$

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 10.14.4. $\square$

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

1. $E$ is injective, and

2. 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 15.55).

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

$M' \xrightarrow {\psi |_{M'}} E + K \to (E + K)/K = E$

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 10.25.1). 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

$R \to E \to S^{-1}E \xrightarrow {f^{-1}} S^{-1}E$

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.

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

2. 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 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 $J$. Set $J' = \{ a \in R \mid ax' \in E \}$ so $J \subset J'$. Conversely, we have $a \in J'$ if and only if $ax' \in E$ if and only if $f^ m a x' = 0$ for some $m \geq 0$. But then $f^ m a x' = f^{m + n} a x$ implies $ax \in E$, i.e., $a \in J$. Hence $J = J'$. Thus $J = J' = \text{Ann}(x')$, 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

$(\ldots ((I[f_1^\infty ])[f_2^\infty ])\ldots )[f_ t^\infty ]$

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

$0 \to E[x] \to \mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) \to \mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) \to 0$

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. $\square$

Comment #3443 by Sebastian Bozlee on

There are two minor typos in Lemma 08XW: The definition of $J$ should read $J = \\{ a \in R \mid ax \in E \\}$. Later, $Rx'$ should not be equal to 0.

May I suggest the following rephrasing? 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 $J = \\{ a \in R \mid ax' \in E \\} = \mathrm{Ann}(x')$, so $Rx' \cap E = 0$. Hence $E'$ is not an essential extension of $E$, a contradiction.

I'm hoping to prevent the reader from misreading that $J$ is a maximal ideal of $R$ (I did), remind the reader what the $I[f^n]$ notation means, and clarify how the maximality of $J$ figures into the argument.

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