Couple of questions about Picard group of $\mathbb{C}^*$

Question

I'd like to compute the Picard group of $\mathbb{P}^n\times \mathbb{C}^*$.

So using toric geometry I've easily found $\text{Cl}(\mathbb{P}^n\times \mathbb{C}^*)\simeq \text{Cl}(\mathbb{P}^n)\oplus\text{Cl}(\mathbb{C}^*)\simeq \mathbb{Z}$, but I'm a bit stuck for the Picard group for some reasons:

I don't know if in general $\text{Pic}(X\times Y)\simeq \text{Pic}(X)\oplus\text{Pic}(Y)$ (but I suspect is not, since otherwise I would have found these identity somewhere, but I really have no clue how to find a counterexample);
I'm not sure if $\mathbb{C}^*$ is smooth (otherwise I would conclude $\text{Pic}(\mathbb{C}^*)=0$, since this is a toric variety and thus $X$ smooth $\iff$ $\text{Cl}(X)=\text{Pic}(X)$). I strongly suspect it is smooth since its fan is given by $\{\{0\},e_1,e_2\}$ (I know the notation is not quite correct, I'm confusing a ray with its minimal generator), and since every cone of this fan can be extended to a $\mathbb{Z}$-basis, $\mathbb{C}^*$ is smooth.

So I'd like to see (using any technique you want, you don't have to use a toric argument) if $\mathbb{C}^*$ is indeed smooth and if you can help me with my starting problem (I thought it was not convenient to split the question in two sub-posts). Thanks in advance.

Yes. $\mathbf{C}^*$ is indeed smooth. It sits as an open subset of $\mathbf{A}^1_\mathbf{C}$. More precisely it is $\textrm{Spec}(\mathbf{C}[x,x^{-1}]).$ Now for your other case it is always true that if you have a smooth variety (integral in particular) then $Cl(X\times \mathbf{P}^n)\cong \mathbf{Z}\times Cl(X).$ The question you ask about the general Picard groups has an answer here https://math.stackexchange.com/questions/493005/picard-group-of-product-of-spaces — shubhankar, Apr 23 '20 at 16:50
Thank you both for your interesting arguments! thanks also for spotting the typo, you obviously right and I'll edit my question. Anyway unfortunately I still have no answer about what is $\text{Pic}(\mathbb{P}^1\times\mathbb{{C}^*$, so if some of you would spend some minutes writing a proper answer I'll surely approve it. Thanks again and enjoy your day! — , Apr 23 '20 at 17:26

score 3 · Accepted Answer · answered Apr 23 '20 at 17:52

If $X$ is a smooth projective variety with $H^1(X,\mathcal{O}_X)=0$, then for any variety $S$, $\operatorname{Pic}(X\times S)=\operatorname{Pic} X\times\operatorname{Pic} S$ (You can find a proof for example in Mumford's Abelian Varieties). In your case, $H^1(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})=0$.

score 0 · Answer 2 · answered Apr 28 '20 at 02:45

I would just like to add to Mohan's answer, and give a non-cohomological criteria in the OP's special case of taking the fibre product with Projective space over an algebraically closed field.

Note that for a smooth variety, the class group and the Picard group are canonically identified. Now the product of two smooth varieties remains smooth, since smoothness is stable under base change and composition.

Moreover if one takes fibre products of a smooth variety $X$ with $\mathbf{P}^n_k,$ then note that $$\mathbf{P}^n_k\times_k X\cong \mathbf{P}^n_{\mathbf{Z}}\times_{\mathbf{Z}} k \times_k X\cong \mathbf{P}^n_{\mathbf{Z}}\times_{\mathbf{Z}}X.$$

In this case the class group is easy to calculate and is just $$\textrm{Cl}(X\times\mathbf{P}^n_{\mathbf{Z}})\cong \textrm{Cl}(X)\times \mathbf{Z},$$ see, for example II Ex.6.1 in Hartshorne.

Then the same equality holds for the Picard group.

Thanks for the different but interesting approach, I'll definitely take a look at Hartshorne's exercise! — , Apr 28 '20 at 19:13

Couple of questions about Picard group of $\mathbb{C}^*$

2 Answers2