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I am currently studying Algebraic Geometry, and the wikipedia page of Fano Variety says the following "a smooth complete intersection of hypersurfaces in n-dimensional projective space is Fano if and only if the sum of their degrees is at most n".

I wonder how this statement can be proved. A fano variety is one which $-K_X$ is ample, where $X$ is a smooth projective variety and $K_X$ is the canonical divisor. I have already computed that $K_X=\mathfrak{O}_X(\sum d_i-n-1)$, where $d_i$ are the degree of the hypersurfaces. How should I use this to prove the statement regarding $X$ being Fano? Thanks in advance.

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    $\begingroup$ Just note that the restriction of an ample divisor to any subvariety is also ample. $\endgroup$
    – Sasha
    Commented Apr 28, 2023 at 9:32
  • $\begingroup$ @Sasha Thank you for the response. I am not so sure how this would help me. To clarify, I am most interested in proving X Fano implies the sum of degrees is at most n. $\endgroup$
    – user1055643
    Commented Apr 28, 2023 at 12:56
  • $\begingroup$ If the sum is $n + 1$ or $\ge n + 2$ your formula proves that the canonical class of $X$ is trivial or ample. But the negative of trivial or ample class cannot be ample. $\endgroup$
    – Sasha
    Commented Apr 28, 2023 at 13:40
  • $\begingroup$ @Sasha Right. I should have noticed it. Thank you for the help. You can make it an answer and I will accept it. $\endgroup$
    – user1055643
    Commented Apr 28, 2023 at 13:52

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I hope this answer could help you. By definition, a Fano variety is a variety such that its anticanonical divisor is ample. Now as you have computed the canonical divisor of a complete intersection $X$ on $\mathbb{P}^n$ is $K_X=\mathcal{O}_X(\sum d_i-n-1)$. Now we know that the ample divisor on projective spaces $\mathbb{P}^n$ are divisors of positive degree and by the commentary of Sasha $-K_X=\mathcal{O}_X(-\sum d_i+n+1)$ is ample if and only if $-\sum d_i+n+1\geq 0$. So, $X$ is Fano if and only if the sum of their degrees is at most $n$.

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