Lemma 10.47.10. Let K/k be a field extension. The following are equivalent
K is geometrically irreducible over k, and
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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