Geometry

assets/tex/qingliu/Chapters/3/2/10.tex

@@ -1,4 +1,5 @@
 \mtexe{3.2.10}
 \begin{proof}
-
+	Recall that we previously showed (exercise 2.3.20) that if $A$ is a ring, $G$ is a finite group of automorphisms of $A$, $A^G$ is the invariant subring, and $p : \Spec A \to \Spec A^G$ the morphism induced by $A^G \hookrightarrow A$, then $p(x_1) = p(x_2)$ if and only if there is a $\sigma \in G$ such that $\sigma(x_1) = x_2$.
+	In our case, $G$ is the Galois group of $K/k$ and $A = L \otimes_k K$; in order to show that $G$ acts transitively on $\Spec A$, it thus suffices to show that $p(x_1) = p(x_2)$ for any $x_1,x_2 \in A$. So, we should compute $A^G = (L \otimes_k K)^G$. Suppose that $\sum_i \ell_i \otimes r_i \in A^G$. 
 \end{proof}