# Picard group of a smooth projective curve

I have two (related) questions regarding the Picard group:

1) Are there examples of smooth projective curves with large Picard groups (say $Pic(X)\simeq\mathbb{Z}^n)$ for any $n$)?

2) In general, can we say what the Picard group is for a smooth projective curve of genus $g$? I was trying to compute it with the exponential sequence, but cannot quite make it work since I don't know the map $H^1(X,\mathcal{O}^*)\to H^2(X,\mathbb{Z})$.

• The Picard group of a smooth projective curve of genus $g \ge 1$ is $\mathbb{Z}$ times its Jacobian (en.wikipedia.org/wiki/Jacobian_variety), which is in particular uncountable. – Qiaochu Yuan Jun 5 '14 at 19:17
• What field are you working over? – Bruno Joyal Jun 5 '14 at 19:18
• The exponential sequence runs $1 \to H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_X^{\times}) \to H^2(X, \mathbb{Z}) \to 1$. The last map to $H^2$ is the degree, or equivalently the first Chern class, so its kernel is the degree-$0$ line bundles and the rest of the exact sequence tells you that this can be identified with $H^1(X, \mathcal{O}_X) / H^1(X, \mathbb{Z})$. This is the Jacobian, and in particular is a complex torus $\mathbb{C}^g / \Gamma$ of complex dimension $g$, once we use Serre duality to identify $H^1(X, \mathcal{O}_X)$ with $H^0(X, \Omega^1_X)$. – Qiaochu Yuan Jun 5 '14 at 19:26
• @Qiaochu I think you are missing a dual in your statement of Serre duality – Bruno Joyal Jun 5 '14 at 19:33
• Oops. Yes, that should be $H^0(X, \Omega_1^X)^{\ast}$. Thanks. – Qiaochu Yuan Jun 5 '14 at 19:35

Over an algebraically closed field, see Qiaochu's remark. Over a number field, it is always a finitely generated abelian group. If $g=1$ and the base field is a given number field, then it is believed there exist elliptic curves of arbitrarily large rank, but this is not known (I believe the largest rank known over $\mathbf Q$ is 28, due to Elkies).

• Minor nitpick: Certainly, for all $r$, there is a number field $K$ and an elliptic curve $E$ over $K$ of rank $r$. Of course, I know you are referring to the "belief" that there are elliptic curves over $\mathbb Q$ with arbitrarily large rank (although I have the feeling that not everyone seems to agree with this "belief", right?). – Ariyan Javanpeykar Jun 7 '14 at 12:27
• Dear @Ari: Of course, you are right. By "over a number field", I meant "over a fixed number field". Regarding your remark: by varying the field, it is obvious that one can obtain curves of arbitrarily large rank, but it doesn't seem obvious to me that one can obtain curves of arbitrary rank, on the nose. It is probably true, however. Cheers , – Bruno Joyal Jun 8 '14 at 0:02