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Is the Picard group of a scheme homotopy invariant in the sense that the projection $\pi : X \times \mathbb{A}^1 \to X$ induces an isomorphism $\mathrm{Pic}(X) \cong \mathrm{Pic}(X \times \mathbb{A}^1)$? Clearly it induces a split monomorphism, and it is an isomorphism iff the two sections $i_0,i_1 : X \to X \times \mathbb{A}^1$ induce the same homomorphism on Picard groups, i.e. $i_0^* \mathcal{L} \cong i_1^* \mathcal{L}$ for every line bundle $\mathcal{L}$ on $X \times \mathbb{A}^1$.

There are various special cases where this is true:

1) When $X$ is noetherian, separated, integral and locally factorial, it follows from Prop. 6.6 (homotopy invariance of the class group à la Weil) and Corollary 6.16 (isomorphism between the class group and the Picard group) in Hartshorne's book. I wonder if there is a more direct proof which doesn't take the detour with Weil divisors, but this is not my main question here.

2) It is also true when $X$ is affine and factorial (not necessarily noetherian), since one can check directly that the Picard group of a factorial domain vanishes. It then also follows when $X$ is covered by affine factorial schemes.

3) If $X$ is integral projective over an algebraically closed field with $H^1(X,\mathcal{O}_X)=0$, it is a special case of Ex. III.12.6 in Hartshorne's book. Perhaps someone can indicate a proof?

What about more general assumptions? What about integral schemes in general (then we can work with Cartier divisors)? Or does it even hold in general? Remark that I don't want to make any detour with class groups! If not, I would like to know specific examples for $X$ such that $\mathrm{Pic}(X \times \mathbb{A}^1) \not\cong \mathrm{Pic}(X)$.

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This is rather a question for mathoverflow. The canonical map $\pi^* : \mathrm{Pic}(X)\to\mathrm{Pic}(X\times\mathbb A^1)$ is an isomorphism when $X$ is normal, and there are counterexamples with $X$ local integral of dimension $1$ and of course non-normal (with no isomorphism between $\mathrm{Pic}(X)$ and $\mathrm{Pic}(X\times\mathbb A^1)$).

First suppose $X$ is normal. As $\pi^*$ is injective, it is enough to consider affine open subsets of $X$. Then descend to finitely generated $\mathbb Z$-algebras. Taking integral closure and because $\mathbb Z$ is excellent, we are reduced to the case $X$ noetherian and normal. Now apply EGA IV.21.4.11, page 360. Note that the same arguments apply for any non-empty open subset of $\mathbb A^1_\mathbb Z$, e.g., $\mathbb G_m$, instead of $\mathbb A^1$.

Now let us see a counterexample with $X$ non-normal. I will take the first integral non-normal example which comes to my mind: $X=\mathrm{Spec}(R)$ where $R$ is the local ring $$R=(k[u,v]/(u^2+v^3))_{(u,v)}$$ over a field $k$ of characteristic zero. Let $K=\mathrm{Frac}(R)$. As $X$ is local, $\mathrm{Pic}(X)$ is trivial. Denote by $Y:=X\times \mathbb A^1$. We want to show $\mathrm{Pic}(Y)$ is non-trivial.

Consider the polynomial $$f=1+vT^2\in R[T].$$ It is chosen in such a way that $f$ is not irreducible in $K[T]$ : $$f=(1+tT)(1-tT)=(v+uT)(1/v+(v/u)T),\quad t:=-u/v\in K$$ but is $f$ irreducible (I don't say prime) in $R[T]$ (I don't use explicitly this property, just to explaine where comes this $f$). Note that $Y$ is covered by the two affine open subsets $D(f)$ and the generic fiber $Y_K$. Let $L$ be the invertible sheaf on $Y$ given by $$L({D(f)})=(v+uT)R[T]_f, \quad L({Y_K})=K[T].$$ This is well defined because $(v+uT)$ is an invertible element of $O_Y(D(f)\cap Y_K)=K[T]_f$.

Admit for a moment that

$(R[T]_f)^{\star}=R^{\star}f^{\mathbb Z}$.

If $L$ is free, then there exist $\omega=af^{r}\in (R[T]_f)^\star$ and $\lambda\in (K[T])^\star=K^\star$ such that $(v+uT)\omega=\lambda$. This is impossible by comparing the degrees of both sides in $K[T]$. So $\mathrm{Pic}(Y)$ is non-trivial. The above fact on the units of $R[T]_f$ is (with the new $f$) easy to see. But I can post my solution if you want.

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Thanks a lot. I have some problems with the details. Why does $f$ factor that way? Perhaps there is a typo? And why is $Y_K$ open a priori? –  Martin Brandenburg Jul 3 '13 at 8:06
    
@MartinBrandenburg: for the factorization of $f$, use the relation $u^2+v^3=0$. $Y_K$ is open in $Y$ because the generic point of Spec($R$) is open. –  user18119 Jul 3 '13 at 10:59
    
Sorry I mixed up $u, v$ in some place. Now I consider a different form of $f$. –  user18119 Jul 3 '13 at 11:30
    
To prove the fact on the units of $R[T]_f$: if $P(T)\in R[T]$ is invertible in $R[T]_f$, we have $P(T)Q(T)=f^N$ for some $N\ge 0$ and $Q(T)\in R[T]$. This implies that $P(0)\in R^\star$ and $P(T)=a(1+tT)^{r_1}(1-tT)^{r_2}$ with $a=P(0)$ and $r_1,r_2 \ge 0$. We have $P(T)=f^r(1\pm tT)^s$. So $(1\pm tT)^s\in K[T]\cap R[T]_f=R[T]$ ($Y=D(f)\cup Y_K$). In characteristic $0$, this implies that $s=0$ and $P(T)\in R^\star f^{\mathbb Z}$. –  user18119 Jul 3 '13 at 11:47
    
Sorry, I still have some questions. 1) Why is the generic point open in $\mathrm{Spec}(R)$? This would mean that the intersection of all non-zero prime ideals of $R$ is non-zero. I cannot think of such an element. 2) Twice you mean $v+uT$ instead of $u+vT$, right? 3) Why is $Y$ the union of $D(f)$ and $Y_K=\mathrm{Spec}(K[T])$? This would mean $R[T]/(f)$ is an algebra over $K$. But I only see that $v,u$ are units. 4) In your last comment, why do we have $P(T)=f^r (1 \pm tT)^s$? –  Martin Brandenburg Jul 3 '13 at 12:12

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