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Let $f:X\to Y$ be a finite dominant morphism of nonsingular varieties over some algebraically closed field. Let $P\in X$ be a point of codimension one, then $f(P)$ is also of codimension one. We can define the ramification index $e_P$ of $P$ in the same way one would do it if $X$ and $Y$ were curves, my main reference here is Hartshorne IV.2. My question is somewhat about the correct generalization of Proposition II.6.9 in Hartshorne.

From Liu's Book (Algebraic Geometry and Arithmetic Curves; in particular Thm. 7.2.18 and Exc. 7.2.3) I conclude that for any $Q\in Y$ of codimension one, we should obtain the equality

$\deg(f)=\displaystyle\sum_{P\in f^{-1}(Q)} e_P\cdot [k(P):k(Q)]$.

while

$\deg(f^\ast Q) = \displaystyle\sum_{P\in f^{-1}(Q)} e_P$

Now, I really want $\deg(f^\ast Q)$ to be constant for all $Q$ of codimension one - is this always the case? If not, can you name additional assumptions to make this work? Clearly, the condition $[k(P):k(Q)]=1$ for all $f(P)=Q$ comes to mind - in other words, the restriction of $f$ to any prime divisor of $X$ is an isomorphism. Does this hold under certain assumptions about $f$?

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  • $\begingroup$ Do you care primarily about curves, or do you need to say something about higher dimensional varieties? Hagen Knaf points this out as well. Also, I think the condition $[k(P) : k(Q)] = 1$ does NOT mean that the restriction of $f$ to any prime divisor of $X$ is an isomorphism. It means that the restriction of $f$ to any prime divisor induces a birational map to its image (of course, for curves, these are the same things). $\endgroup$
    – Karl Schwede
    Commented Apr 26, 2011 at 11:43
  • $\begingroup$ I am mostly interested in higher dimensional varieties. You're right about everything else, though. $\endgroup$
    – Jesko Hüttenhain
    Commented Apr 26, 2011 at 11:52
  • $\begingroup$ If $f$ is a Galois cover, then for all $P\in Y$ of codimension $1$ we have $r_P e_P f_P =\mathrm{deg}(f)$, where $r_P =|f^{-1}(P)$, $e_P$ is the ramification index of some (and thus all) point $Q\inf^{-1}(P)$ and $f_P =[k(Q):k(P)]$ for some (and thus all) $Q\inf^{-1}(P)$. Your requirement then is equivalent to $f_P$ as a function of $P$ is constant. If in addition $\mathrm{deg}(f)$ is a prime $p$, then either $f_P=p$ for all $P$ or $f_P=1$ for all $P$. Statements like this are treated in algebraic number theory, but I have'nt seen something like that for varieties. $\endgroup$
    – Hagen
    Commented Apr 27, 2011 at 10:29
  • $\begingroup$ That sounds interesting, what exactly is a Galois cover and why is the degree $[k(Q):k(P)]$ identical for all points in the fiber? $\endgroup$
    – Jesko Hüttenhain
    Commented Apr 28, 2011 at 7:52

1 Answer 1

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In the situation you consider the degree $\mathrm{deg}(f)$ is equal to the degree $[K(X):K(Y)]$ of the extension of the function fields of $X$ and $Y$. The precise requirements are that $X$ and $Y$ are normal varieties and that $f$ is finite. The base field is not required to be algebraically closed.

For varieties of dimension $\geq 2$ there is no degree function for Weil divisors comparable to the one in the case of curves. Your definition(?) of $\mathrm{deg}(f^\ast Q)$ looks at least strange to me. How would you define the degree of an arbitrary Weil (prime) divisor of $X$? There should then be no reference to $f$ ...

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  • $\begingroup$ I would have defined the degree of any Weil divisor to be the sum of its coefficients, but that is not really my point anyways. Mostly I would like $\sum_P e_P$ to be constant, and I would like to know when this is the case. $\endgroup$
    – Jesko Hüttenhain
    Commented Apr 26, 2011 at 11:45
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    $\begingroup$ Dear Jesko, You can't define the degree of a Weil divisor in this way (well you can, of course, but it is a meaningless quantity, not preserved under linear equivalence, for example). You have to choose a projective embedding (or more-or-less equivalently, an ample divisor) and define degree relative to this embedding. Regards, Matthew $\endgroup$
    – Emerton
    Commented Apr 26, 2011 at 13:03
  • $\begingroup$ Dear Matthew, I am terribly sorry about my poor choice of words. I completely understand that my terminology was flawed there, if not even wrong. My intuition comes from the one-dimensional case and I should probably have given it more thought - However, I would really just like to know under what conditions the quantity $\sum_{P\in f^{-1}(Q)} e_P$ is a constant for all $Q$ of codimension one. It does not matter to me whether it is the degree of anything, to be frank. $\endgroup$
    – Jesko Hüttenhain
    Commented Apr 26, 2011 at 14:01

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