Lemma 33.8.6. Let X be an irreducible scheme over the field k. Let \xi \in X be its generic point. The following are equivalent
X is geometrically irreducible over k, and
\kappa (\xi ) is geometrically irreducible over k.
Lemma 33.8.6. Let X be an irreducible scheme over the field k. Let \xi \in X be its generic point. The following are equivalent
X is geometrically irreducible over k, and
\kappa (\xi ) is geometrically irreducible over k.
Proof. Assume (1). Recall that \mathcal{O}_{X, \xi } is the filtered colimit of \mathcal{O}_ X(U) where U runs over the nonempty open affine subschemes of X. Combining Lemma 33.8.5 and Algebra, Lemma 10.47.6 we see that \mathcal{O}_{X, \xi } is geometrically irreducible over k. Since \mathcal{O}_{X, \xi } \to \kappa (\xi ) is a surjection with locally nilpotent kernel (see Algebra, Lemma 10.25.1) it follows that \kappa (\xi ) is geometrically irreducible, see Algebra, Lemma 10.46.7.
Assume (2). We may assume that X is reduced. Let U \subset X be a nonempty affine open. Then U = \mathop{\mathrm{Spec}}(A) where A is a domain with fraction field \kappa (\xi ). Thus A is a k-subalgebra of a geometrically irreducible k-algebra. Hence by Algebra, Lemma 10.47.6 we see that A is geometrically irreducible over k. By Lemma 33.8.5 we conclude that X is geometrically irreducible over k. \square
Comments (2)
Comment #4583 by Wessel Bindt on
Comment #4763 by Johan on
There are also: