Let $X$ be an integral scheme of finite type over a field $k$. If $k'\supseteq k$ is a field extension, then $X' = X\otimes_k k'$ is not necessarily integral.
Why does each irreducible component of $X'$ have the same dimension of $X$?
$\begingroup$
$\endgroup$
5
asked Feb 15, 2018 at 14:59
-
6$\begingroup$ After a finite field extension, the irreducible components of the base change become geometrically irreducible, precisely, after base change to the algebraic closure $\kappa$ of $k$ in the function field $k(X)$. Thus, assume that $k'/k$ is finite. Then $X'\to X$ is finite and faithfully flat. So now you can apply the Going Up and Going Down Theorems (Cohen-Seidenberg). $\endgroup$– Jason StarrCommented Feb 15, 2018 at 15:06
-
$\begingroup$ I don't get your reduction step: why can I suppose that $k'/k$ is finite? By 'algebraic closure' you mean 'integral closure'? $\endgroup$– Davide Cesare VenianiCommented Feb 15, 2018 at 16:11
-
4$\begingroup$ For a field extension $k\to K$, e.g., for $K=k(X)$, the algebraic closure of $k$ in $K$ is the set of all elements of $K$ that satisfy a nontrivial polynomial equation over $k$. This is a subextension $\kappa$ of $K/k$, and it is finite over $k$ if $K/k$ is finitely generated (often proved via Noether normalization). Every irreducible component $Y$ of $X\times_{\text{Spec}\ k} \text{Spec}\ \kappa$ has pure dimension equal to $\text{dim}(X)$ (Cohen-Seidenberg). As $\kappa$ is algebraically closed in $\kappa(Y)$, $Y$ is a geometrically irreducible $\kappa$-scheme of dimension $\text{dim}(X)$. $\endgroup$– Jason StarrCommented Feb 15, 2018 at 16:33
-
2$\begingroup$ If you are willing to assume that $\kappa/k$ is Galois, then this is quite easy to visualise. The Galois group of this extension acts transitively on the irreducible components over the algebraic closure, from which it follows that they all have the same dimension. $\endgroup$– Daniel LoughranCommented Feb 15, 2018 at 17:45
-
$\begingroup$ The dimension of the local ring at a closed point of $X$ is constant (equal to the transcendence degree of $k(X)$ over $k$). This remains true after extension of the base field. $\endgroup$– anonCommented Feb 16, 2018 at 0:35
Add a comment
|