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.

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 $\mathrm{Pic}(X \times \mathbb{A}^1) \not\cong \mathrm{Pic}(X)$.

share|improve this question

1 Answer 1

up vote 10 down vote accepted

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.

share|improve this answer
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

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.