Is the degree of a finite morphism stable by base change - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:29:56Z http://mathoverflow.net/feeds/question/77594 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77594/is-the-degree-of-a-finite-morphism-stable-by-base-change Is the degree of a finite morphism stable by base change Taicho 2011-10-09T11:03:48Z 2011-10-10T13:01:55Z <p>Let $f:X\longrightarrow Y$ be a finite morphism of schemes of degree $n$. Let $S\to Y$ be a morphism of schemes.</p> <p>Is the degree of the finite morphism $X\times_Y S \longrightarrow S$ equal to $n$?</p> <p>If not, what conditions should be put on $X$ and $Y$?</p> <p>If it helps, you can assume all the schemes to be integral.</p> http://mathoverflow.net/questions/77594/is-the-degree-of-a-finite-morphism-stable-by-base-change/77598#77598 Answer by Georges Elencwajg for Is the degree of a finite morphism stable by base change Georges Elencwajg 2011-10-09T13:02:29Z 2011-10-10T13:01:55Z <p>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 $$dim_{\kappa (y)} \mathcal O(X_y)= deg(f) \quad (\star)$$</p> <p>if and only if $f$ is flat (cf. Qing Liu's book, page176).<br> So non flat morphisms will give you counterexamples by taking for $S$ a point of $Y$.<br> 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 $k$.<br> More generally, normalizations of non-normal varieties are <em>never</em> flat and will yield any number of countereamples.</p> <p>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.</p> <p><strong>A well-known formula</strong> Here is an arithmetically flavoured illustration of the above.<br> Let <em>A</em> 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 $A$.<br> That ring $B$ is flat over $A$ (because for Dedekind rings flat=without torsion) and is a Dedekind domain, <em>finite over $A$</em> (Krull-Akizuki).<br> We can apply the considerations above above to the associated morphism $f:Spec(B)=X\to Y=Spec(A)$.<br> Take a nonzero prime $\mathfrak p =y \in Y$ and write ${\mathfrak p}B=\prod {\mathfrak P}_i^{e_i}$.<br> Since $X_y=Spec(B/{\mathfrak p}B)$, the formula $(\star )$ translates into the very classical formula of algebraic number theory (where $f_i=[B/{\mathfrak P}_i: A/ \mathfrak p]$): $$n=\sum e_if_i$$</p>