Lemma 33.8.9. Let $K/k$ be an extension of fields. Let $X$ be a scheme over $k$. For every irreducible component $T$ of $X$ the inverse image $T_ K \subset X_ K$ is a union of irreducible components of $X_ K$.
Proof. Let $T \subset X$ be an irreducible component of $X$. The morphism $T_ K \to T$ is flat, so generalizations lift along $T_ K \to T$. Hence every $\xi \in T_ K$ which is a generic point of an irreducible component of $T_ K$ maps to the generic point $\eta $ of $T$. If $\xi ' \leadsto \xi $ is a specialization in $X_ K$ then $\xi '$ maps to $\eta $ since there are no points specializing to $\eta $ in $X$. Hence $\xi ' \in T_ K$ and we conclude that $\xi = \xi '$. In other words $\xi $ is the generic point of an irreducible component of $X_ K$. This means that the irreducible components of $T_ K$ are all irreducible components of $X_ K$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: