Base Change Morphism Etale

algebraic-geometry

I have a question about a step used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 116):

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The statement is that if $X$ is a geometrically integral affine curve over field $k$ and $L \vert k$ is a finite separable field extension then the base change morphism $X \times_k Spec(L)= X_L \to X$ is finite and etale over $X$.

Finiteness is ok since it is stable under basechange (right ?)

But what about etaleness: Let $P \in X$ (we interpret here $P$ as point and the corresponding prime ideal) then we have to verify that the $\kappa(P)$-algebra $\mathcal{O}(X_L)/P \mathcal{O}(X_L) = \kappa(P) \otimes_k L$ is a product of finite separable field extensions of $\kappa(P)$ as ring. Compare with definition in https://en.wikipedia.org/wiki/%C3%89tale_algebra

I don't see why this holds. Indeed we know that since $L$ is a finite dimensional $k$-vector space therefore $\kappa(P) \otimes_k L$ is a fd $\kappa(P)$-vspace so a "product" in the category of vector spaces.

But we want that $\kappa(P) \otimes_k L$ is a finite product of seprable extensions in the category of rings.

Does anybody see an appropriate argument?

Best Answer

To answer the actual question posed, note that since $L$ is finite and separable, by the primitive element theorem, $L=k[x]/(f)$ for some $f\in k[x]$ (irreducible, separable). Then $$\kappa(P)\otimes_k L=\kappa(P)\otimes_k k[x]/(f) = \kappa(P)[x]/(f),$$ and since $f$ was separable over $k$, it remains separable over $\kappa(P)$, factoring as $f=g_1\cdots g_r$ for distinct, separable, irreducible polynomials $g_i\in\kappa(P)[x]$. So by the chinese remainder theorem, $$\kappa(P)[x]/(f)\simeq \prod_{i=1}^r \kappa(P)[x]/(g_i),$$ and the fields $\kappa(P)[x]/(g_i)$ are finite, separable extensions of $\kappa(P)$.