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Let $X$ be an algebraic variety over a field $k$ (i.e. a scheme of finite type over $\text{Spec}(k)$.

According to Remark 3.2.11 of Qing Liu's book Algebraic geometry and arithmetic curves, we have the following property:

If $X_{\overline k}$ is connected (resp. reduced, irreducible), then $X_K$ is connected (resp. reduced, irreducible) for all algebraic extensions $K/k$.

following from Lemma 3.2.6:

Let $K/k$ be an algebraic extension. For any reduced closed subvariety $W\subset X_K$, there exists a finite subextension $K/K'/k$ and a unique (for fixed $K'$) reduced closed subvariety $Z$ of $X_{K'}$ such that $W=Z_K$.

I have some trouble to interpret this lemma and see how it implies the remark.

For example, suppose $X_K$ is not connected: $X_K=W_1\cup W_2$ for $W_1,W_2\subset X_K$ closed nonempty subvarieties. How would one get a covering of $X_{\overline{k}}$ from that ? From base-changing $X_K,W_1,W_2$ to $\overline{k}$ ?

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I also don't see how the Lemma applies here, but your approach is fine concerning connectedness & irreducibility: The base change $\pi: X_{\overline{k}}\to X_K$ is surjective, and given a surjective morphism of topological spaces with connected resp. irreducible domain, the codomain is also connected resp. irreducible.

For the reducedness, note that the structure maps ${\mathscr O}_{X_{K}}(U)\to {\mathscr O}_{X_{\overline{k}}}(\pi^{-1}(U))$ are injective (locally, they are the embeddings $A\otimes_k K\hookrightarrow A\otimes_k \overline{k}$), so ${\mathscr O}_{X_{K}}(U)$ is reduced if ${\mathscr O}_{X_{\overline{k}}}(\pi^{-1}(U))$ is.

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  • $\begingroup$ Excellent answer, dear Hanno, including the bit about Lemma 3.2.6 not being needed. $\endgroup$ – Georges Elencwajg Feb 27 '15 at 9:00

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