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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.

Dubious
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    You say "I know that $|f^{-1}(y)|<\infty$", but that is false. Think about a birational map that contracts a curve: what is the degree of the field extension? Even if $f$ is finite the bound you want doesn't work: $f$ could be the normalisation of a nodal curve. –  Mar 20 '14 at 16:02
  • Finite morphism doesn't imply finite fibers? (finite morphisms are quasi-finite) – Dubious Mar 20 '14 at 16:16
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    But you did not say that your morphism is finite, just that it induces a finite extension of function fields. My point is exactly that this is weaker. (Search for "generically finite".) Also, the second example is finite. –  Mar 20 '14 at 16:20
  • yeah you're right I forgot to write that $f$ is finite. Is a typo. – Dubious Mar 20 '14 at 16:21
  • I've edited. Many thanks! – Dubious Mar 20 '14 at 16:22
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    On the other hand if $Y$ is normal, this should be true. See Shafarevich Volume 1, II.6.3 Theorem 3 for a proof in the case of varieties. –  Mar 20 '14 at 16:52
  • What is unsatisfactory about Asal Beag Dubh's comment? – Alex Youcis Mar 22 '14 at 21:50

1 Answers1

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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.

answer_bot
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    Thanks for writing this up, @answer_bot. (I am not sure how much sense it makes to thank a bot, but never mind.) –  Apr 02 '14 at 14:44