Let $X$ be a noetherian integral separated scheme which is regular in codimension one. According to Hartshorne's Algebraic Geometry, Prop. II.6.6, The divisor class group $Cl(X)$ is isomorphic to $Cl(X \times \mathbb{A}^1)$. Is the statement true when $\mathbb{A}^1$ is replaced with $\mathbb{A}^n$?
Is the divisor class group $Cl(X \times \mathbb{A}^n)$ isomorphic to $Cl(X)$
algebraic-geometry
Related Solutions
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.
Let $X$ be any scheme, then I claim $\dim O_{x,X} = \operatorname{codim} \{x\}^-$ for any $x\in X$.
First step is to reduce to the case where $X$ is an affine scheme. Consider an affine open $\operatorname{Spec} A$ containing $x$, for each irreducible closed set $K$ of $X$ containing $x$, we obtain an irreducible closed set of $\operatorname{Spec} A$ containing $x$ by $K\mapsto K\cap \operatorname{Spec} A$. For each irreducible closed set of $\operatorname{Spec} A$ containing $x$, we obtain an irreducible closed set of $X$ containing $x$ by $C\mapsto C^-$, with closure taken inside $X$. We show this establishes a bijection. Clearly $C^-\cap \operatorname{Spec} A=C$, we have one sided inverse. For the other side, we need to show that $(K\cap \operatorname{Spec} A)^- = K$. Observe that $K - \operatorname{Spec} A$ and $(K\cap \operatorname{Spec} A)^-$ are two closed sets of $X$ whose union is $K$, and $(K\cap \operatorname{Spec} A)^-$ contains $x$ which is non-empty, we are thus done by irreducibility of $K$. (In fact this bijection works with same proof for any open set of $X$ containing $x$, not just $\operatorname{Spec} A$.)
Now let $P\in \operatorname{Spec} A$, we have $\dim O_{P} = \dim A_P = \operatorname{codim} P$. For each irreducible closed subset $K$ containing $P$, $K$ has a unique generic point $Q$, whence $K = \{Q\}^- = V(Q)$. Since $P\in V(Q)$ we have $P\supset Q$. Therefore an ascending chain of irreducible closed subset containing $P$ corresponds to a descending chain of prime ideals (the generic points of the irreducible closed subsets) from $P$. We thus conclude that $\dim O_P = \operatorname{codim} \{P\}^-$.
This solves question $1$ since any regular local ring of dimension $1$ is a DVR. For question two, we again reduce to an affine cover $\operatorname{Spec} A$ of $\eta$. Let $\xi$ be the generic point of $X$, we have $\xi \in \operatorname{Spec} A$ corresponds to the $0$ ideal since $A$ is a domain. Clearly the quotient ring of $A_\eta$ is $A_0 = K(O_{\xi,X}) = K(X)$.
Best Answer
Sure because
$Cl(X\times \mathbb{A}^n)=Cl((X\times \mathbb{A}^{n-1})\times \mathbb{A})\cong$
$\cong Cl(X\times \mathbb{A}^{n-1})\cong \dots \cong Cl(X)$