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How can I do the problem 1.2.8 c in "Algebraic Geometry and Aritmetic Curves". Namely, let $A$ be a Noetherian ring, $M$ a finitely generated $A$-module, and $N$ an $A$-module. Let $B$ be a flat $A$-algebra. Let us consider the canonical homomorphism $$\rho:\operatorname{Hom}_A(M,N)\otimes_AB\to\operatorname{Hom}_B(M\otimes_AB,N\otimes_AB).$$ Let $0\to K\to L\to M\to 0$ be an exact sequence of $A$-modules. How to show that if $L$ is a free module of finite rank then $\rho$ is injective, and the injectivity of $K$ gives $\rho$ to be an isomorphism?

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  • $\begingroup$ You have to see what's going on when replace $M$ by $L$, respectively by $K$ (in the first case I suppose $\rho$ is an isomorphism), and then make a diagram starting from the exact sequence. $\endgroup$ – user26857 Jan 20 '15 at 16:47
  • $\begingroup$ In the problem it was said that $\rho$ is just homomorphism, not necessary isomorphism. It is an isomorphism if $M$ is free of finite rank by 1.2.8 a. But does the sequence $0\to K\to L\to M\to 0$ together with $L$ free of finite rank implies that $M$ is a free of finite rank? $\endgroup$ – Jaakko Seppälä Jan 20 '15 at 16:56
  • $\begingroup$ No. But if $K$ is an injective $A$-module, the exact sequence splits, so that $M$ is a direct factor of $L$, hence is a finitely generated projective $A$-module. $\endgroup$ – Bernard Jan 20 '15 at 18:45
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$\require{AMScd}$ $\DeclareMathOperator{\Hom}{Hom}$ Let me write $M'$ for $M \otimes_A B$ and $\rho_M\colon \Hom(M,N)' \to \operatorname{Hom}(M',N')$. The point is that there's a commutative diagram

\begin{CD} 0 @>>> {\Hom(M,N)'} @>>> {\Hom(L,N)'} @>>> {\Hom(K,N)'}\\ @. @V{\rho_M}VV @V{\rho_L}VV @V{\rho_K}VV\\ 0 @>>> {\Hom(M',N')} @>>> {\Hom(L',N')} @>>> {\Hom(K',N')}\\ \end{CD}

You should check the commutativity (just do one square). The rows are exact (why?) and $\rho_L$ is an isomorphism. From there I hope you can see the end. Make notice of where you use Noetherianity.

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