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It is well-know that flat base change in general does not preserve injective modules, i.e. let $M$ be an injective $R$-module and $R\to S$ be a flat morphism, then it is not necessary that $M\otimes_R S$ is an injective $S$-module.

Now let's consider a more special case:

Let $R$ be a commutative noetherian algebra over a base field $k$ and $l/k$ be a (non-finite) field extension. Let $M$ be an injective $R$-module. Then is it necessary that $M\otimes_kl$ is an injective $R\otimes_kl$-module?

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No. For instance, suppose $l$ is not algebraic over $k$ and let $R=k[x]$ and $M=k(x)$. Then $R\otimes_k l= l[x]$ and $M\otimes_k l$ 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 transcendental over $k$, then $\frac{1}{x-t}$ is not in $M\otimes_k l$. It follows that $M\otimes_k l$ is not an injective $l[x]$-module (since it is not divisible).

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  • $\begingroup$ If we assume $l/k$ is an algebraic extension, do we have a affirmative answer? $\endgroup$
    – Zhaoting Wei
    May 24, 2017 at 15:25
  • $\begingroup$ I don't know, but I would guess so. $\endgroup$
    – Eric Wofsey
    May 24, 2017 at 16:30

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