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This question arises from following former thread of mine and refers to a remark from Liu's "Algebraic Geometry" at page 90:

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In remark 3.2.11 is stated if $X$ is an algebraic variety over field $k$ and it's base change $X_{\overline{k}}= X_{Spec(k)}Spec(\bar{k})$ wrt algebraic closure $k \subset \overline{k}$ verifies one of the properties reduced, connected, or irreducible, then Lemma 2.6 says that then $X_K$ verifies the same property for any algebraic extension $K/k$.

Here the lemma 2.6:

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My QUESTION is how the author concretely deduces remark 3.1.12 with help of 2.6?

I know that there are some ways to deduce the remark 3.2.11 without using explicitely Lemma 2.6 (see for example the comments of @user45878 in the linked thread).

But the main point of this question that makes me curious is concretely which role does lemma 2.6 in the proof? It seems quite redundant to me to consider a intermediate field $K'$ between algebraic extension $K/k$ having properties as stated.

Indeed:

Firstly, the "reduced" property is local so assume $X=Spec(A)$ and assume $A \otimes_k \bar{k}$ reduced. Assume, $K/k$ is an arbitrary algebraic extension of $k$. Then, since $A$ is a flat $k$-module we have the inclusion $A \otimes_k K \subset A \otimes_k \bar{k}$. So $A \otimes_k K$ reduced. So no usage of Lemma 2.6.

In case of irreducibility assume that $X_{\overline{k}}$ is irreducible but $X_K$ not. Wlog $X_K= X^1 \cup X^2$. Base change by $\bar{k}/K$ gives at least two irreducible components $X_{\bar{k}}= X^1 _{\bar{k}} \cup X^2 _{\bar{k}}$ (since $X^i _{\bar{k}}$ could after base change become also reducible). But nevertheless we obtain a contradiction to irreducibility of $X_{\overline{k}}$. For "connectness" the argument is similar.

And now the question agian, where do we here really need 2.6?

What do I oversee?

Or does anybody see what Liu had by usage of 2.6 in mind?

user267839
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  • I don't know what Liu has in mind. For any algebraic extension $k \subset K \subset \bar{k}$ we have that $X_{\bar{k}} \to X_{K}$ is a surjective morphism as it is the base change of a surjective morphism (so it remains surjective). Now the image of any connected/irreducible topological space under a continous map is connnected/irreducible. It doesn't getting any simpler than this. Now to prove that $X_{K}$ is reduced, your argument is exactly the same Liu uses to prove Lemma $2.6$. I would suggest not to get caught up in these sort of things. – Parthiv Basu Jul 26 '19 at 04:43
  • Also, why is $(X^1 \cup X^2){\bar{k}} =X{\bar{k}}^1 \cup X_{\bar{k}}^2$? This is an incorrect implication. – Parthiv Basu Jul 26 '19 at 05:17
  • Both $X^1$ and $X^2$ are only topological spaces. A priori they don't have any scheme structure on them (though it is possible to give any closed subset of a scheme a scheme structure). So I don't know what you mean by taking their base change when they are just topological spaces. – Parthiv Basu Jul 26 '19 at 05:32
  • @ParthivBasu: Yes, the base change argument is the correct one. Hmmm meanwhile I think that Lui refers by 2.6 only to the argumentation technique used in the proof of lemma 2.6. Otherwise I don't know why he really need the statment of 2.6 *as such* to settle $X_{\bar{k}} $ blabla implies $X_{K}$ blabla... – user267839 Jul 26 '19 at 11:40

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