Does this Cartan-Eilenberg exercise contain a mistake?

Question

In Cartan and Eilenberg, Homological Algebra, Exercise 10, page 123, the last statement is the following: if $A\to B$ is a ring homomorphism such that $B$ is projective as an $A$-module, then for any $A$-module $M$ we have $$ \text{inj.dim}_B(B\otimes_AM)\leq \text{inj.dim}_A(M). $$ This immediately implies that if $B$ projective as an $A$-module and $M$ is an injective $A$-module, then $B\otimes_AM$ is an injective $B$-module.

However, in this stack exchange question, @EricWofsey gave a counter-example: let $A=k[x]$ the polynomial ring over a field $k$ and $B=l[x]$ where $l/k$ is a non-algebraic field extension. Since $A$ and $B$ are PID's, a module is injective if and only if that module is divisible. Let $M=k(x)$ then $M$ is an injective $A$-module. But $B\otimes_A M=l\otimes_k k(x)$ is the subring of $l(x)$ consisting of rational functions which can be written with a denominator in $k[x]$. In particular if $t\in l$ is not algebraic over $k$, then $\frac{1}{x-t}\notin l\otimes_k k(x)$ but $x-t\in l\otimes_k k(x)$. Therefore $l\otimes_k k(x)$ is not divisible hence not injective as an $l[x]$-module.

It seems that this gives a counter-example of Exercise 10. I'm not sure if that exercise is wrong or I misunderstand something.

Answer 1

The exercise indeed appears to have an error. Lest there be any doubt that my initial example was correct, here is a much simpler example. Let $A$ be a field, $M=A$, and $B$ be an $A$-algebra. Then $M$ is injective over $A$ and $B$ is projective over $A$ (indeed, any module over $A$ is both injective and projective). But $B\otimes_A M=B$ need not be injective over $B$. Indeed, most algebras over a field are not injective as modules over themselves.

It seems that the intended statement of the exercise had $\operatorname{Hom}_A(B,M)$ instead of $B\otimes_A M$ (in Cartan and Eilenberg's notation, these two objects differ only by changing a superscript to a subscript). Note that the functor $\operatorname{Hom}_A(B,-)$ does preserve injectives, since it is right adjoint to the forgetful functor from $A$-modules to $B$-modules. The exercise can then be proved by simply applying this functor to an injective resolution of $M$, which gives an injective resolution of $\operatorname{Hom}_A(B,M)$ (it is exact since $B$ is projective over $A$).