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 $\overline{k}$ ?


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.