# Complete Intersection of Hypersurfaces are Fano varieties

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.

• Just note that the restriction of an ample divisor to any subvariety is also ample. Commented Apr 28, 2023 at 9:32
• @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.
– user1055643
Commented Apr 28, 2023 at 12:56
• 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. Commented Apr 28, 2023 at 13:40
• @Sasha Right. I should have noticed it. Thank you for the help. You can make it an answer and I will accept it.
– user1055643
Commented Apr 28, 2023 at 13:52

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 $$n$$.