# Trivial Picard group of projective variety

Assume X is an irreducible smooth projective variety. I am trying to prove Pic(X)=0 iff X is a point.

I am completely new to divisor, so even have no idea for such problem which may be trivial. So is this similar to $\mathbb P^n$? By the way, I am learning this stuff without using sheaves and schemes, so could someone help me without using sheaves? (I have found a similar question with an answer using sheaves, but unfortunately I am not able to understand it.)

Thanks!

• What definition of the Picard group are you using? You say you're trying to learn about this without using "sheaves and schemes," but the definition of the Picard group I'm aware of uses those concepts. Dec 14, 2015 at 23:57
• @Patrick Presumably this group is defined using Weil divisors and linear equivalence. In the case described, all the standard definitions coincide. Dec 15, 2015 at 0:03
• @Slade Yes, I see. In that case, could we say something like the following? if the space isn't a point, take an irreducible (smooth?) proper subvariety $V$. If this is linearly equivalent to zero, then there's a holomorphic function on $X$ vanishing at $V$ with no poles. But this can't be, because any non-vanishing holomorphic function is constant. Dec 15, 2015 at 0:04
• @Patrick Yes, this works; somehow I was trying to think of something more complicated. I would replace "holomorphic" with "regular" since the base field is not specified to be $\mathbb{C}$. Dec 15, 2015 at 0:06
• Oh, it is defined as the quotient of all locally principal divisors and the principal divisions. I hope this help.
– user198206
Dec 15, 2015 at 0:06

If the space isn't a point, fix an irreducible proper subvariety $V$. If this is linearly equivalent to zero, then there's a regular function on $X$ vanishing at $V$ with no poles. But this can't be, because any non-vanishing regular function is constant.
• @Learning Recall the definition of div(f). It is constructed by taking the zeros of f minus the poles. Here $V$ appears here with positive coefficient $+1$ and no terms with negative coefficients appear. So there can't be any poles. Dec 15, 2015 at 1:05