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Let $f: A \to B$ be a finite ring homomorphism and $N$ a $B$-module. $N$ can be considered as an $A$-module if we define $A \times N \to N$, $(a, n) \mapsto f(a)n$. Therefore we have a map $f_{!}: K(B) \to K(A)$.

Let $M$ be a $B$ module. $B$ can be considered as an $A$ module if we define $A \times B \to B$, $(a, b) \mapsto f(a)b$. Let $M_{B}=B\otimes_{A} M$. Then $M_{B}$ is a $B$-module. The action of $B$ on $M_{B}$ is given by $(b', b\otimes m) \mapsto (b'b)\otimes m$. Therefore we have a map $f^{!}: K_{1}(A) \to K_1(B)$, where $K_1(A)$ is the Grothendieck group obtained from the set of all isomorphism classes of finitely generated flat $A$-modules.

On Page 88 of the book Introduction to commutative algebra by Atiyah and Macdonald, Exercise 27(v), it is said that $f_{!}(f^{!}(x)y)=xf_{!}(y)$ for $x \in K_1(A), y \in K(B)$. How to prove this result? I think that $f_{!}(f^{!}(x)y)=f_{!}((B\otimes_{A} x)y)$. But why $f_{!}((B\otimes_{A} x)y) = xf_{!}(y)$? Thank you very much.

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LJR
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It's clearly enough to prove that $\left(B\otimes_A M\right) \otimes_B N \cong M \otimes_A N$ as $A$-modules for every $A$-module $M$ and every $B$-module $N$.

Well, notice that $N \cong B \otimes_B N$ as $A$-module. Thus, $M \otimes_A N \cong M \otimes_A \left(B \otimes_B N\right) \cong \left(M \otimes_A B\right) \otimes_B N \cong \left(B \otimes_A M\right) \otimes_B N$. If I am not doing something wrong, this should be enough; no naturality needs to be checked (although the isomorphisms I have sketched are natural), and no statements about preservation of exact sequences need to be proven (those were already incorporated in the definition of the $K_1\left(A\right)$-structure on $K\left(B\right)$).

However, I am a bit surprised about part ii) of the problem. How do we know that $0\cdot y = 0$ for every $y\in K\left(A\right)$ ? Does tensoring an exact sequence of flat $A$-modules with an arbitrary $A$-module always give an exact sequence?

darij grinberg
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    The answer to your last question is yes. In fact, if $0 \to M \to F \to N \to 0$ is exact with $F$ flat then $0 \to M \otimes M' \to F \otimes M' \to N \otimes M' \to 0$ is exact for all $M'$ iff $N$ is flat. See Lam's First Course, Theorem 24.23. The proof of the relevant direction is obtained by choosing a short exact sequence $0 \to K' \to F' \to M' \to 0$ with $F'$ free and performing a diagram chase in the $3 \times 3$-diagram obtained by tensoring all the terms of the two short exact sequences. – Martin Jun 14 '13 at 10:28
  • @darij, thank you very much. But why $(B⊗AM)⊗_BN≅M⊗_AN$ is the same as $f{!}((B⊗AM)N)=Mf{!}(N)$? – LJR Jun 14 '13 at 12:46
  • The multiplication in $K$ (or $K_1$) comes from the tensor product over the respective ground ring, and $f_!$ comes from restriction of scalars (which is suppressed in my notation). – darij grinberg Jun 14 '13 at 16:06
  • @darij, thank you very much. But we do not have tensor product in $Mf_{!}(N)$. I am confused. – LJR Jun 15 '13 at 02:32
  • @LJR: What do you mean by that? – darij grinberg Jun 15 '13 at 23:00