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I have a question about a remark from Liu's "Algebraic Geometry" at page 90:

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Our Lemma 2.6 states following:

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My QUESTION is how the author concretely deduces remark 3.1.12?

My ideas: Wlog $X_K =X$ (so $k=K$). Assume that $X_{\overline{k}}$ is reduced (or connected or irreducible) and $X$ not. The statement says that then there ixist an intermediate field extension $k=K \subset K'$ and a unique reduced (resp connected or irreducible) component $Z \subset X_{K'}$ with $Z_{\overline{k}}=X_{\overline{k}}$. Why is this a contradiction to assumption that $X_{\overline{k}}$ is reduced (or connected or irreducible)?

user267839
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    We can check reducesness affine locally, and then for a $k$-algebra $A$ we have $A_K \subset A_{\overline{k}}$ for all algebraic $K$, so certainly if the latter has no nilpotents then neither does $A_K$. For all the other statements you can use the fact that $\overline{k}/k$ is algebraic and so by the Lemma all irreducible (resp. connected) components are defined over a finite field extension $K/k$. (Strictly speaking we are using the fact that $X$ is quasi-compact so that there aren't infinitely many connected components defined over increasingly bigger extensions $K/k$). – Pol van Hoften Jul 11 '19 at 10:50
  • @user45878: Regarding the case that we assume that $X_{\overline{k}}$ is irreducible (resp. connected): Yes, you have quoted the statement of the Lemma but I don't understand how this lemma already imply that $X$ is irreducible (resp. connected). – user267839 Jul 11 '19 at 19:49
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    Suppose that $X_K$ is reducible, for some algebraic $K/k$. Then it follows that $X_K=X_1 \cup X_2$ with $X_1, X_2$ nonempty distinct closed subschemes of $X_K$. Therefore it follows that $X_{\overline{k}}= X_{1, \overline{k}} \cup X_{2, \overline{k}}$ and so $X_{\overline{k}}$ is also reducible. The argument for `connected' is similar. – Pol van Hoften Jul 12 '19 at 16:13
  • hmmm what me irritates is that for this argument it seems that you not use 2.6,right. Or do I oversee something and the Lemma is implicitely involved here? – user267839 Jul 13 '19 at 10:40

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