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.

  • $\begingroup$ Indeed, you'd better assume integral. Otherwise consider $X = A^1 \coprod pt$ mapping to $A^1$ degree $1$, and $S$ a point mapping to the image of $pt$. Now I just need a finite morphism of integral schemes that isn't flat... $\endgroup$ Oct 9 '11 at 11:21
  • 1
    $\begingroup$ How do you map $X$ to $\mathbf{A}^1$ with degree 1? Won't one point in $\mathbf{A}^1$ have two points lying over it? For a finite morphism of integral schemes which isn't flat consider the normalization of k[x,y]/(xy), where k is a field. $\endgroup$
    – Taicho
    Oct 9 '11 at 11:37
  • 2
    $\begingroup$ How do you define the degree if you assume neither integral schemes nor $f$ flat? $\endgroup$
    – user2035
    Oct 9 '11 at 12:31
  • $\begingroup$ i don't know...I define the degree to be the degree of the finite field extension $K(Y) \subset K(X)$ if $X$ and $Y$ are integral. I see why one requires the flatness condition now. $\endgroup$
    – Taicho
    Oct 9 '11 at 12:45

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 $f_i=[B/{\mathfrak P}_i: A/ \mathfrak p]$): $$n=\sum e_if_i$$

  • $\begingroup$ The flatness condition is fulfilled in my situation so I'm happy. Thnx alot! $\endgroup$
    – Taicho
    Oct 9 '11 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.