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, for any field extension $k''/k$ there exists a common field extension $k'''/k'$ and $k'''/k''$, see Fields, Lemma 9.6.9. 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 $k$. $\square$
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