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Let $X$ be a smooth complete intersection curve of genus $\geq 2$ sitting in $\mathbb P^n$. We know the canonical sheaf $\omega_X$ is $\mathcal O_X(\sum d_i - n -1)$ where the $d_i$ are the degree of the hypersurfaces cutting $X$. Is it true $N:=\sum d_i - n -1$ is equal to $2g-2$? How can we see this? [Edit: This is wrong! See the comments below].

I'd like to show $\omega_X$ is a very ample line bundle. $\omega_X$ determines a morphism into projective space by choosing a basis of global sections. Certainly all the monomials of degree $N := \sum d_i-n-1$ in $x_0, \dots, x_n$ span $\omega_X(X)$, but we might have to delete some to reach a basis. So it seems like this gives a "partial" $N$-uple embedding. How can we see this is an embedding?

Thank you!

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  • $\begingroup$ Are you trying to say that $\deg_X \mathscr{O}_X(1) = 1$? It seems like that can't be right. $\endgroup$
    – Hoot
    Commented Aug 2, 2016 at 0:42
  • $\begingroup$ @Hoot Are you referring to the degree of a line bundle? Does this mean the degree of a rational section on $\omega_X$? I don't think I am asserting this is $1$. $\endgroup$
    – hwong557
    Commented Aug 2, 2016 at 0:46
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    $\begingroup$ Degree of $\omega_X$ is $2g-2$. Degree of $\mathcal{O}_X(1)=\deg X$ which in your case is $\prod d_i$. So, $2g-2=N\prod d_i$, with $N$ as in the question. $\endgroup$
    – Mohan
    Commented Aug 2, 2016 at 0:49
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    $\begingroup$ @hwong557 Or degree of the line bundle, or degree of an associated divisor — it should all be the same. We always have $\deg \omega_X = 2g-2$ (as Mohan points out) and degree is a homomorphism, so if we are going to claim $\deg \mathscr{O}_X(N) = N$ then we'd better have $\deg \mathscr{O}_X(1) = 1$ and my hope that was that this would seem absurd. $\endgroup$
    – Hoot
    Commented Aug 2, 2016 at 1:45
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    $\begingroup$ This is just Bezout's theorem for complete intersections. If $X$ is a complete intersection of hypersurfaces of degree $d_i$ and $h$ is a hyperplane, then $h\cdot X=\prod d_i$. I would suggest that you write this out for $n=1,2$ and then it should become clear. $\endgroup$
    – Mohan
    Commented Aug 2, 2016 at 3:42

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$h^0(\mathcal O_X(N))=g \geq 2$. I claim $N>0$. For if $N \leq 0$ then $\mathcal O_X(N)$ has no global sections, and if $N = 0$, then $h^0(\mathcal O_X(0))=1$. (Alternately use Mohan's formula above to see $N>0$)

The line bundle $\mathcal O_X(N)$ corresponds to the morphism $$ X \overset{i}{\to} \mathbb P^n \overset{j}{\to} \mathbb P^M, $$ where $i$ is the embedding of $X$ into $\mathbb P^n$, and $j$ is the $N$-uple embedding, for $$ i^* j^* \mathcal O_{\mathbb P^M}(1) = i^* \mathcal O_{\mathbb P^n}(N) = \mathcal O_{X}(N). $$

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