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 $k$. $\square$

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## Comments (2)

Comment #4583 by Wessel Bindt on

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