Let $X,Y$ be integral $S$-schemes, and let $f:X\to Y$ be a proper morphism taking generic point to generic point $\xi_X\mapsto \xi_Y$. Consider the generic fibre:
$$
\require{AMScd}
\begin{CD}
X_\xi @>>> X\\
@V{f^\prime}VV @V{f}VV \\
\text{Spec}(k(\xi_Y)) @>>> Y
\end{CD}$$
Is $X_\xi$ an integral scheme, or are conditions needed? Being reduced can be checked on stalks, and since the stalks of $X_\xi$ can be identified with the stalks of points of $X$, which are reduced, it follows that $X_\xi$ is reduced.
If $[k(\xi_X):k(\xi_Y)]=n$ then since $R(X_\xi)=R(X)$ we should have that $X_\xi$ is a proper $k(\xi_Y)$-scheme of dimension $n$. If $n=1$ then I'm worried that $X_\xi$ is just the spectrum of a product of fields which are algebraic extensions of $k(\xi_Y)$, in which case it would be disconnected? So maybe $n\geq 2$ is needed, and I have to check that the point corresponding to $\xi_X\in X_\xi$ is the generic point of $X_\xi$?
Best Answer
You can check that the scheme $X_{\xi}$ is even locally integral (ie integral stalks), so you just need to check for global irreducibility. But this is easy as $X_{\xi}$ is clearly the closure (in itself) of the generic point of $X$.