## 22.17 Injective modules over algebras

In this section we discuss injective modules over algebras analogous to More on Algebra, Section 15.55. This section should probably be moved somewhere else.

Let $R$ be a ring and let $A$ be an $R$-algebra, see Section 22.2 for our conventions. For a right $A$-module $M$ we set

$M^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(M, \mathbf{Q}/\mathbf{Z})$

which we think of as a left $A$-module by the multiplication $(a f)(x) = f(xa)$. Namely, $((ab)f)(x) = f(xab) = (bf)(xa) = (a(bf))(x)$. Conversely, if $M$ is a left $A$-module, then $M^\vee$ is a right $A$-module. Since $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group (More on Algebra, Lemma 15.54.1), the functor $M \mapsto M^\vee$ is exact (More on Algebra, Lemma 15.55.6). Moreover, the evaluation map $M \to (M^\vee )^\vee$ is injective for all modules $M$ (More on Algebra, Lemma 15.55.7).

We claim that $A^\vee$ is an injective right $A$-module. Namely, given a right $A$-module $N$ we have

$\mathop{\mathrm{Hom}}\nolimits _ A(N, A^\vee ) = \mathop{\mathrm{Hom}}\nolimits _ A(N, \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(A, \mathbf{Q}/\mathbf{Z})) = N^\vee$

and we conclude because the functor $N \mapsto N^\vee$ is exact. The second equality holds because

$\mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N, \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(A, \mathbf{Q}/\mathbf{Z})) = \mathop{\mathrm{Hom}}\nolimits _\mathbf {Z}(N \otimes _\mathbf {Z} A, \mathbf{Q}/\mathbf{Z})$

by Algebra, Lemma 10.12.8. Inside this module $A$-linearity exactly picks out the bilinear maps $\varphi : N \times A \to \mathbf{Q}/\mathbf{Z}$ which have the same value on $x \otimes a$ and $xa \otimes 1$, i.e., come from elements of $N^\vee$.

Finally, for every right $A$-module $M$ we can choose a surjection $\bigoplus _{i \in I} A \to M^\vee$ to get an injection $M \to (M^\vee )^\vee \to \prod _{i \in I} A^\vee$.

We conclude

1. the category of $A$-modules has enough injectives,

2. $A^\vee$ is an injective $A$-module, and

3. every $A$-module injects into a product of copies of $A^\vee$.

Comment #499 by m.o. on

After Lemma 22.12.1 and before Lemma 22.12.2, there is a formula to show that if $M$ is a right differential graded $(A,\text{d})$-module then $M^{\vee}$ is a left differential graded $(A,\text{d})$-module. However since $x$ has degree $-n-m-1$ and not $-n-m$, two parts of the signs in the formla change, and I am wondering the conclusion might be that
$M^{\vee}$ is a left differential graded $(A,-\text{d})$-module. On the other hand, if $M$ is a left differential graded $(A,\text{d})$-module then $M^{\vee}$ is a right differential graded $(A,\text{d})$-module. The reason of this (strange looking ?) phenomenon seems to come from the fact that the differential $\text{d}$ always act from the left, but I am not an expert in this field at all, and I am not quite sure these things. Do I misunderstand or overlook some important part ?

Comment #502 by on

Thank you very much for this comment. I agree the sign is wrong and I agree with your explanation. I corrected this here and I also tried to explain a little bit what you said.

Note: I cannot currently update the web server because we are making a change in the way things work but in a week or so this change will percolate to the webinterface. Thanks again.

Comment #506 by m.o. on

Thank you for your quick reply. After I read your tex file, I noticed the point where I made a mistake: let $M$ be a right differential graded $A$-module, and let $a\in A^m$, $f\in (M^\vee)^n$, and $x\in M^{-n-m}$. Then the correct left action of $A$ on $M^\vee$ would be given by $(af)(x)=(-1)^{m^2}f(xa)$. The reason of this definition is as follows. Denote by $\text{Hom}^*(M, \mathbb{Q}/\mathbb{Z})$ the differential graded module with the usual differential as before (i.e, the one which does not depend on the (left or right) $A$-module structure). We are given a homomorphism $M\otimes_R A\to M$ of differential graded modules. Then we have $\text{Hom}^*(M,\mathbb{Q}/\mathbb{Z})\to \text{Hom}^*(M\otimes_R A,\mathbb{Q}/\mathbb{Z})$, and this corresponds to a homomorphism $\text{Hom}^*(M,\mathbb{Q}/\mathbb{Z})\otimes M\otimes_R A\to \mathbb{Q}/\mathbb{Z}$ of graded differential modules. Now we have the commutativity: $A\otimes_R \text{Hom}^*(M,\mathbb{Q}/\mathbb{Z})\otimes M\cong \text{Hom}^*(M,\mathbb{Q}/\mathbb{Z})\otimes M\otimes_R A$, which is the origin of the sign $(-1)^{m^2}$ if $a\in A^m$. Thus we have $A\otimes_R \text{Hom}^*(M,\mathbb{Q}/\mathbb{Z})\otimes M\to \mathbb{Q}/\mathbb{Z}$, and this corresponds to the desired map $A\otimes_R \text{Hom}^*(M,\mathbb{Q}/\mathbb{Z})\to \text{Hom}^*(M, \mathbb{Q}/\mathbb{Z})$ of differential graded modules. By the same way, the evaluation map $ev:M\to M^{\vee\vee}$ will be given by $ev(m)(f)=(-1)^{n^2}f(m)$ if $m\in M^n$ and $f\in (M^\vee)^n$. These sign changes agree with your suggestion that the definition of left action is wrong if $a\in M^m$ and $m$ is odd. Am I right ? (I am very sorry for the undefined control sequence.)

Comment #507 by on

Yes, that is exactly right and this is what I was thinking too. But I decided we should not want to change the $A$-action on $M^\vee$ in the graded case and we should not have different $A$-actions depending on whether $M$ is a graded module or a differential graded module, so I suggest we leave it like it is now (I mean in the commit I linked to above).

Thank you very much once again (and please realize that everybody is an expert on this stuff in the sense that once the material is written out clearly, then we can all decide whether some formula is right or wrong).

Comment #513 by m.o. on

Now I understand the problem well, and I also think that we should leave it like it is now. It's fun to read the Stacks project. Thank you.

Comment #4648 by on

The comments above will soon be obsolete as I am changing the sign rules in this chapter to be more uniform which causes the "problem" above to be fixed.

Comment #4794 by on

Let me just add that the discussion of signs is now in Section 22.18.

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