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There are a few properties from Hartshorne IV on curves that I am trying to verify. Let $D$ be an effective divisor on a curve (integral scheme of dimension 1, proper over $k$, regular) $X$, $\dim |D| = 1$, and $f$ the induced morphism to $\mathbb{P}^1$.

  1. The degree of $D$ is the degree of $f$, where $\deg(f) = [K(X) : K(\mathbb{P}^1)]$.
  2. In lemma IV.4.2, Hartshorne seems to claim that all points in the support of $D$ are in the same fiber.
  3. Again in lemma IV.4.2, Hartshorne seems to claim that if $[K(X) : K(\mathbb{P}^1)]$ is Galois, then an automorphism of the Galois group permutes elements of the fiber.
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    $\begingroup$ By Hartshorne II.6.9 we have more generally $\deg f^∗E=\deg f\cdot \deg E$ for any $E$ on $P^1$. $\endgroup$
    – J.C. Ottem
    Commented Oct 10, 2011 at 14:40
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    $\begingroup$ Your first question is already answered. For the second, by definition, $f(P)=D'$ where $D'$ is the unique effective divisor linearly equivalent to $D$ with $D' \ge P$. So it is clear that $f(P)=D$ if $P$ is in the support of $D$. The third question is similarly straightforward. $\endgroup$
    – Felipe Voloch
    Commented Oct 10, 2011 at 17:54
  • $\begingroup$ Ah this makes so much more sense! I was not aware of the "definition" you mentioned; if it's in Hartshorne then I missed it. So for the third, the action of a Galois element fixes the elements of $\mathbb{P}^1$, which I'm identifying with effective divisors linearly equivalent to $D$. Since these divisors also define the fibers of my map, it must be that the fibers are permuted. $\endgroup$
    – rghthndsd
    Commented Oct 11, 2011 at 17:38

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