Lemma 33.8.2. Let $X$ be a scheme over the field $k$. Let $k'/k$ be a field extension. Then $X$ is geometrically irreducible over $k$ if and only if $X_{k'}$ is geometrically irreducible over $k'$.

Proof. If $X$ is geometrically irreducible over $k$, then it is clear that $X_{k'}$ is geometrically irreducible over $k'$. For the converse, note that for any field extension $k''/k$ there exists a common field extension $k'''/k'$ and $k'''/k''$. As the morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of a surjective morphism between spectra of fields) we see that the irreducibility of $X_{k'''}$ implies the irreducibility of $X_{k''}$. Thus if $X_{k'}$ is geometrically irreducible over $k'$ then $X$ is geometrically irreducible over $k$. $\square$

There are also:

• 2 comment(s) on Section 33.8: Geometrically irreducible schemes

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).