Let $f:X\longrightarrow Y$ be a finite morphism of schemes of degree $n$. Let $S\to Y$ be a morphism of schemes.


Is the degree of the finite morphism $X\times\_Y S \longrightarrow S$ equal to $n$?


If not, what conditions should be put on $X$ and $Y$?


If it helps, you can assume all the schemes to be integral.



Is the degree of a finite morphism stable by base change 
I shall assume that $X,Y$ are integral, locally noetherian schemes and that $f$ is dominant. Then the degree of $f$ is the degree of the corresponding extension of fields, namely
$$deg(f)=[Rat(X):Rat(Y)]$$. 
We have for the fibers $X\_y \; (y\in f(X))$ of $f$ the interesting result: 
$$dim\_{\kappa (y)} \mathcal O(X\_y)\geq deg(f)$$ 
with equality for all fibers 
$$ dim\_{\kappa (y)} \mathcal O(X\_y)= deg(f) \quad (\star)$$


if and only if $f$ is flat (cf. Qing Liu's book, page176).  

So non flat morphisms will give you counterexamples by taking for $S$ a point of $Y$.  

For an explicit counterexample, consider the case where $Y$ is a node, $X$ the affine line (both over a field $k$) and $f$ the normalization morphism. This is a finite morphism of degree one, but the fiber of the singular point has degree $2$ over $k$.  

More generally, normalizations of non-normal varieties are *never* flat and will yield any number of countereamples.


Also if $f$ is flat the criterion will tell you, since flatness is preserved under base-change, that the degree of $f$ will be preserved under some reasonable assumptions on the morphism $S\to Y$, the most obvious one being that $S$ should be locally noetherian and integral too.


**A well-known formula** Here is an arithmetically flavoured illustration of the above.  

 Let *A* be a Dedekind domain with fraction field $K$ and $L$ a separable field extension of $K$ of degree $[L:K]=n$. Let $B$ be the ring of elements in $L$ integral over $A$.  

That ring $B$ is flat over $A$ (because for Dedekind rings flat=without torsion) and is a Dedekind domain, *finite over $A$* (Krull-Akizuki).  

 We can apply the considerations above above to the associated morphism $f:Spec(B)=X\to Y=Spec(A)$.  

Take a nonzero prime $\mathfrak p =y \in Y $ and write ${\mathfrak p}B=\prod {\mathfrak P}\_i^{e\_i}$.  

Since $X\_y=Spec(B/{\mathfrak p}B) $, the formula $(\star )$ translates into the very classical formula of algebraic number theory (where $f\_i=[B/{\mathfrak P}\_i: A/ \mathfrak p]$): 
$$n=\sum e\_if\_i$$


