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Let $A$ be a local noetherian ring, $B$ and $C$ are finitely generated $A$-algebras and $M$ is a finitely generated $B$-module. Is the natural morphism $\mathrm{Hom}_B(M,B) \otimes_A C \to \mathrm{Hom}_B(M,B \otimes_A C)$ injective? If necessary, assume that $B$ is flat over $A$.

EDIT: Please note that the homomorphisms are as $B$-modules instead of $A$-modules as mentioned in this question.

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  • Please use the search function - this question and related questions have appeared a couple of times. A counterexample is (usually) $M = B^{\oplus \mathbb{N}}$. – Martin Brandenburg Feb 03 '16 at 15:52
  • @MartinBrandenburg Sorry, I forgot to mention, I assume $M$ is a finitely generated $B$-module. – user45397 Feb 03 '16 at 16:09
  • The more relevant difference between your question and the linked question is actually that you are assuming that $B$, rather than $C$, is flat. – Eric Wofsey Feb 04 '16 at 10:16
  • As the above comment seems to suggest, once you replace N by an A-algebra B, in the linked question, the natural homomorphism becomes one of B-modules. – user26857 Feb 04 '16 at 15:47
  • There is no duplicate now, so one can skip that part. – user26857 Feb 04 '16 at 16:57

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Here's a counterexample. Let $k$ be a field and take $A=B=k[x]/(x^2)$ and $M=C=k$ (considered as an $A$-algebra via $x\mapsto 0$). Then $\mathrm{Hom}_B(M,B)\cong k$ (spanned by $1\mapsto x$), so $\mathrm{Hom}_B(M,B) \otimes_A C\cong k$ as well. But $1\mapsto x$ maps to $0$ in $\mathrm{Hom}_B(M,B \otimes_A C)$, since $x=0$ in $B\otimes_A C=C$.

On the other hand, if you assume $C$ is flat, then the map always is injective (in fact, it is an isomorphism). Indeed, you can note that there is a natural isomorphism $\operatorname{Hom}_B(M,B)\otimes_A C\cong \operatorname{Hom}_B(M,B)\otimes_B (B\otimes_A C)$ that is compatible with the canonical maps to $\mathrm{Hom}_B(M,B \otimes_A C)\cong \mathrm{Hom}_B(M,B\otimes_B (B \otimes_A C))$, so you are reduced to this question over the ring $B$ with $N=B$ and $P=B\otimes_A C$ (which is flat over $B$ since $C$ is flat over $A$).

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