Skip to main content
6 events
when toggle format what by license comment
Feb 16, 2018 at 0:35 comment added anon 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.
Feb 15, 2018 at 17:45 comment added Daniel Loughran 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.
Feb 15, 2018 at 16:33 comment added Jason Starr 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 $\text{dim}(X)$.
Feb 15, 2018 at 16:11 comment added Davide Cesare Veniani I don't get your reduction step: why can I suppose that $k'/k$ is finite? By 'algebraic closure' you mean 'integral closure'?
Feb 15, 2018 at 15:06 comment added Jason Starr 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).
Feb 15, 2018 at 14:59 history asked Davide Cesare Veniani CC BY-SA 3.0