Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to show that for smooth, nice variety $X$, $Pic(X \times A^{1})=Pic(X)$.

I only need to prove this in the generality where the Picard group and the divisor class group are the same.

The pullback homomorphism $p^{*}:Cl(X)\rightarrow Cl(X \times A^{1})$ of the projection is my best guess for an isomorphism.

Injectivity: Let $D$ be a divisor of $X$. $p^{*}(D)=(f) \in K(t)$. We need to show that $f$ is in $K$, but I am not sure how to do this.

Surjectivity: Here I have less of an idea. Is it enough to show that if $D_{1}$ and $D_{2}$ get sent to the same thing then there is an $f$ such that $D_{1}-D_{2}=div(f)$?

Disclaimer: I do not know about schemes.

share|improve this question
1  
This is Proposition II.6.6 in Hartshorne. I'm not sure if the proof can be written down without schemes. –  Martin Brandenburg Mar 26 '11 at 15:39
1  
What you've written under "surjectivity" seems to actually be "injectivity". –  Matt Mar 26 '11 at 20:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.