Let $f:X\to Y$ be a finite, dominant morphism of projective varieties. I suspect that $\deg(f)\cdot\deg(Y)=\deg(X)$ always holds, where $\deg(f)=[K(X):K(Y)]$. If required, we may assume that $X$ and $Y$ are projective over an algebraically closed field (of characteristic zero even). My question is whether this is true and if so, I would love to have a reference.
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4$\begingroup$ What is the degree of an abstract variety? $\endgroup$– Qiaochu YuanCommented Apr 30, 2011 at 16:46
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$\begingroup$ I don't know, but these are projective varieties. $\endgroup$– Jesko HüttenhainCommented Apr 30, 2011 at 16:59
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$\begingroup$ I might still explain what I mean by $\deg(X)$: It is $\dim(X)!$ times the leading coefficient of the Hilbert Polynomial of its homogeneous coordinate ring. For hypersurfaces $X=Z(f)$ and $f$ an element of degree $d$, one can show that $d=\deg(X)$. This is about all the intuition I have about degrees of projective varieties, maybe someone else can shed some more light. $\endgroup$– Jesko HüttenhainCommented Apr 30, 2011 at 17:02
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4$\begingroup$ The homogeneous coordinate ring is not a property of a projective variety; it is a property of an embedding of a projective variety into projective space. In particular, the same variety can embed into projective space via embeddings of different degrees. $\endgroup$– Qiaochu YuanCommented Apr 30, 2011 at 17:16
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2$\begingroup$ A simple example of what can go wrong is to let $X = Y = \mathbb{P}^1$ but to embed the first copy as a conic in $\mathbb{P}^2$ and to embed the second copy via an isomorphism in $\mathbb{P}^1$. Then the "degree" of $X$ is $2$, the "degree" of $Y$ is $1$, and there is an isomorphism $f : X \to Y$ which therefore has degree $1$. $\endgroup$– Qiaochu YuanCommented Apr 30, 2011 at 17:37
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1 Answer
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This is not true in general.
In fact, let $H_X$ be a hyperplane section of $X$ and $H_Y$ be a hyperplane section of $Y$, with respect to the fixed embeddings $X \subset \mathbb{P}^N$ and $Y \subset \mathbb{P}^M$. Then
$\deg X= (H_X)^n, \quad \deg Y =(H_Y)^n$,
where $n= \dim X = \dim Y$.
Now requiring
$\deg X = (\deg f) \cdot \deg Y \quad (*)$
is equivalent to require $(H_X)^n=(\deg f) \cdot (H_Y)^n$. This is true for instance if the map $f \colon X \to Y$ is induced by a subsystem of the complete linear system $|H_X|$, otherwise it is false in general.
For instance, let $X \subset \mathbb{P}^5$ be the Veronese surface and $Y=\mathbb{P}^2$. Since $X$ and $Y$ are isomorphic, there is a map $f \colon X \to Y$ of degree $1$, but $\deg X=4$ and $\deg Y=1$.
Analogously, let us consider $X=\mathbb{P}^1 \times \mathbb{P}^1 \subset \mathbb{P}^3$. Then $X$ is a quadric and any $2$-dimensional, base-point free subsystem of the complete linear system $|H_X|$ induces a finite morphism $f \colon X \to \mathbb{P}^2$ of degree $2$, so in this case $(*)$ is satisfied. However, you can compose $f$ with any isomorphism $g \colon X' \to X$, where $X' \subset \mathbb{P}^N$ is a Segre embedding of $\mathbb{P}^1 \times \mathbb{P}^1$ with $N >3$, obtaining a map $f' \colon X' \to \mathbb{P}^2$ of degree $2$ which does not satisfy $(*)$, since $\deg X' >2$.
answered Apr 30, 2011 at 17:11
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$\begingroup$ Thanks very much, the last example made Qiaochu Yan's point perfectly clear. $\endgroup$ Commented May 1, 2011 at 13:44