Lemma 31.17.7. Let $\pi : X \to Y$ be a finite surjective morphism with $X$ and $Y$ integral and $Y$ normal. Then there exists a norm of degree $[R(X) : R(Y)]$ for $\pi$.

Proof. Let $\mathop{\mathrm{Spec}}(B) \subset Y$ be an affine open subset and let $\mathop{\mathrm{Spec}}(A) \subset X$ be its inverse image. Then $A$ and $B$ are domains. Let $K$ be the fraction field of $A$ and $L$ the fraction field of $B$. Picture:

$\xymatrix{ L \ar[r] & K \\ B \ar[u] \ar[r] & A \ar[u] }$

Since $K/L$ is a finite extension, there is a norm map $\text{Norm}_{K/L} : K^* \to L^*$ of degree $d = [K : L]$; this is given by mapping $f \in K$ to $\det _ L(f : K \to K)$ as in the proof of Lemma 31.17.6. Observe that the characteristic polynomial of $f : K \to K$ is a power of the minimal polynomial of $f$ over $L$; in particular $\text{Norm}_{K/L}(f)$ is a power of the constant coefficient of the minimal polynomial of $f$ over $L$. Hence by Algebra, Lemma 10.38.6 $\text{Norm}_{K/L}$ maps $A$ into $B$. This determines a compatible system of maps on sections over affines and hence a global norm map $\text{Norm}_\pi$ of degree $d$.

Property (1) is immediate from the construction. To see property (2) let $f \in A$ be contained in the prime ideal $\mathfrak p \subset A$. Let $f^ m + b_1 f^{m - 1} + \ldots + b_ m$ be the minimal polynomial of $f$ over $L$. By Algebra, Lemma 10.38.6 we have $b_ i \in B$. Hence $b_0 \in B \cap \mathfrak p$. Since $\text{Norm}_{K/L}(f) = b_0^{d/m}$ (see above) we conclude that the norm vanishes in the image point of $\mathfrak p$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).