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Suppose I have a Cartesian square of morphisms of algebraic varieties over a field $K$ (apologies for the grotty diagram): $$\begin{array}{ccc} A & \to^\alpha & B \\ \downarrow^\beta & & \downarrow^\gamma \\ C & \to^\delta & D \end{array} $$ so $A$ is the fibre product of $B$ and $C$ over $D$. Suppose also that all four of $A,B,C,D$ are proper curves, and all the morphisms are finite and flat.

Is it then true that the two maps $K(C)^\times \to K(B)^\times$ given by $\alpha_* \beta^*$ and $\gamma^* \delta_*$ coincide?

(I am sure this must be standard, but I'm not a geometer and I don't really know where to look in the literature for this sort of thing.)

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Does $K(X)$ denote the rational function field of $X$? If so, what is the pushforward? I feel like this might be a special case of Theorem 6.2 in Fulton's "Intersection Theory", but I can't see how. – Mark Grant Jun 9 '12 at 13:38
Yes, $K(X)$ is just the rational function field of $X$ (or rather, rational function ring, since $A$ may not be irreducible even if $B,C,D$ are.) So $K(A)$ is a finite $K(B)$-algebra, via $\alpha^*$, and hence there is a norm map $K(A)^\times \to K(B)^\times$ which sends $x$ to the determinant of multiplication by $x$ regarded as a $K(B)$-linear map from $K(A)$ to itself. – David Loeffler Jun 9 '12 at 13:49

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