Lemma 10.47.10. Let $K/k$ be a field extension. The following are equivalent

1. $K$ is geometrically irreducible over $k$, and

2. the induced extension $K(t)/k(t)$ of purely transcendental extensions is geometrically irreducible.

Proof. Assume (1). Denote $\Omega$ an algebraic closure of $k(t)$. By Definition 10.47.4 we find that the spectrum of

$K \otimes _ k \Omega = K \otimes _ k k(t) \otimes _{k(t)} \Omega$

is irreducible. Since $K(t)$ is a localization of $K \otimes _ k k(T)$ we conclude that the spectrum of $K(t) \otimes _{k(t)} \Omega$ is irreducible. Thus by Lemma 10.47.3 we find that $K(t)/k(t)$ is geometrically irreducible.

Assume (2). Let $k'/k$ be a field extension. We have to show that $K \otimes _ k k'$ has a unique minimal prime. We know that the spectrum of

$K(t) \otimes _{k(t)} k'(t)$

is irreducible, i.e., has a unique minimal prime. Since there is an injective map $K \otimes _ k k' \to K(t) \otimes _{k(t)} k'(t)$ (details omitted) we conclude by Lemmas 10.30.5 and 10.30.7. $\square$

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