The Stacks project

Lemma 33.8.10. Let $K/k$ be an extension of fields. Let $X$ be a scheme over $k$. For every irreducible component $\overline{T} \subset X_ K$ the image of $\overline{T}$ in $X$ is an irreducible component in $X$. This defines a canonical map

\[ \text{IrredComp}(X_ K) \longrightarrow \text{IrredComp}(X) \]

which is surjective.

Proof. Consider the diagram

\[ \xymatrix{ X_ K \ar[d] & X_{\overline{K}} \ar[d] \ar[l] \\ X & X_{\overline{k}} \ar[l] } \]

where $\overline{K}$ is the separable algebraic closure of $K$, and where $\overline{k}$ is the separable algebraic closure of $k$. By Lemma 33.8.7 the morphism $X_{\overline{K}} \to X_{\overline{k}}$ induces a bijection between irreducible components. Hence it suffices to show the lemma for the morphisms $X_{\overline{k}} \to X$ and $X_{\overline{K}} \to X_ K$. In other words we may assume that $K = \overline{k}$.

The morphism $p : X_{\overline{k}} \to X$ is integral, flat and surjective. Flatness implies that generalizations lift along $p$, see Morphisms, Lemma 29.25.9. Hence generic points of irreducible components of $X_{\overline{k}}$ map to generic points of irreducible components of $X$. Integrality implies that $p$ is universally closed, see Morphisms, Lemma 29.44.7. Hence we conclude that the image $p(\overline{T})$ of an irreducible component is a closed irreducible subset which contains a generic point of an irreducible component of $X$, hence $p(\overline{T})$ is an irreducible component of $X$. This proves the first assertion. If $T \subset X$ is an irreducible component, then $p^{-1}(T) =T_ K$ is a nonempty union of irreducible components, see Lemma 33.8.9. Each of these necessarily maps onto $T$ by the first part. Hence the map is surjective. $\square$

Comments (2)

Comment #7105 by WhatJiaranEatsTonight on

I think here should be the closure of the image of is an irreducible component?

Comment #7268 by on

If you trace back the references all the way to the proof of Lemma 5.8.15 you see that the bijection between the irreducible components of and is by taking images and similarly for and . Then in the second paragraph, the images of irreducible components are closed. So closure isn't necessary.

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  • 2 comment(s) on Section 33.8: Geometrically irreducible schemes

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