Finished 3.2.13

assets/tex/qingliu/Chapters/3/2/13.tex

@@ -10,5 +10,9 @@
 
 	Second, suppose that $X$ is not geometrically irreducible, so that $X_{\bar{k}}$ is not irreducible. If $X$ itself is not irreducible, we are done, so assume WLOG that $X$ is irreducible. Let $W_1,W_2$ be two distinct irreducible components of $X_{\bar{k}}$, and endow them with the induced reduced subscheme structure. By Lemma 3.2.6, there exist finite subextensions $K_1,K_2$ of $\bar{k}/k$ and reduced closed subschemes $Z_1 \subseteq X_{K_1}$ and $Z_2 \subseteq X_{K_2}$ such that $W_1 = (Z_1)_{\bar{k}}$ and $W_2 = (Z_2)_{\bar{k}}$. As noted in the proof of that lemma, we can replace $K_1,K_2$ with any finite extensions contained in $\bar{k}$, and so choosing, for example, the compositum allows us to assume $K_1 = K_2 = K$. But now again using that $\Spec\bar{k} \to \Spec K$ is surjective, we get that $W_i \to Z_i$ is surjective, and so each $Z_i$ is irreducible. Furthermore, I claim that each $Z_i$ is actually an irreducible component of $X_K$, which would complete the proof of this converse, since they are distinct (having different base changes in $X_{\bar{k}}$). Indeed, by similar reasoning to previous problems in this section, the generic points of $X_{\bar{k}},X_K$ all lie in the generic fiber. But the generic points of $W_1,W_2$ are generic points of $X_{\bar{k}}$, so they map to the unique generic point in $X$, and they also map to the generic points of $Z_1,Z_2$. So, the generic points of $Z_1,Z_2$ lie in the generic fiber and so must be generic points of $X_K$, i.e. they are full irreducible components.
 
-	Finally, suppose that $X$ is not geometrically connected, so that $X_{\bar{k}}$ is not connected. Then we can partition $X_{\bar{k}}$ into disjoint closed subsets $W_1,W_2$, and find some finite extension $K/k$ and closed subsets $Z_1,Z_2 \subseteq X_K$ with $(Z_i)_{\bar{k}} = W_i$ when each set is endowed with the appropriate reduced subscheme structure. Now I claim that $Z_1,Z_2$ partition $X_K$, which would finish the proof. The fact that $Z_1 \cup Z_2 = X_K$ is clear since the projection $X_{\bar{k}} \to X_K$ is surjective. 
+	Finally, suppose that $X$ is not geometrically connected, so that $X_{\bar{k}}$ is not connected. Then $\scO_{X_{\bar{k}}}(X_{\bar{k}})$ has a nontrivial idempotent $f$. But since $k \to \bar{k}$ is flat and $X$ is a Noetherian $k$-scheme, proposition 3.1.24 tells us that
+	\[ \scO_{X_{\bar{k}}}(X_{\bar{k}}) \cong \scO_X(X) \otimes_k \bar{k} \]
+	Under this isomorphism, we may write $f = \sum_{i=1}^n f_i \otimes b_i$ for some $f_1,\ldots,f_n \in \scO_X(X)$ and $b_1,\ldots,b_n \in \bar{k}$. As in the reduced case, we can then consider $K = k[b_1,\ldots,b_n]$ and conclude that
+	\[ f \in \scO_X(X) \otimes K \cong \scO_{X_K}(X_K) \]
+	so that $X_K$ is also disconnected.
 \end{proof}