Consider a finite morphism $f:X\longrightarrow Y$ between two integral and Noetherian schemes. If $\operatorname {deg}(f)=[K(X):K(Y)]=n$, is it true that for every $y\in Y$ then $|f^{-1}(y)|\le n$? (with the notation $|\cdot|$ I mean the cardinality). I know that $|f^{-1}(y)|<\infty$ but what about its upper bound? If the statement is false as stated here, under which other hypothesis we have that $|f^{-1}(y)|\le n$?
Thanks in advance.
This is Asal Beag Dubh's answer.
Consider the finite ring map $k[x, y]/(y^2 - xy - x^3) \to k[t]$ with $x \mapsto t(t - 1)$ and $y \mapsto t^2(t - 1)$. Take Spec of this. Then $n = 1$ but the fibre over $(0, 0)$ has two points.
If $Y$ is normal, then the result does hold, but it isn't that easy to prove. One way to do it is to reduce to the case where the extension of function fields is Galois (say with group $G$; this reduction already takes a bit of work in case of inseparability) and then to show that the fibres of $X \to Y$ are acted on transitively by $G$ (in case $X$ is normal) as in one of the proofs of going down for finite over normal.