MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've learned when you have a integral smooth scheme line bundles are the same as Cartier divisors are the same as Weil divisors. My question is to what extent does this continue to hold (if at all) when you are talking about ind objects. Since things become more subtle in positive characteristic lets say we are working over an algebraically closed field of characteristic 0.


I think the situation can be quite different because smoothness, or algebraic smoothness is more complicated. In the non ind scheme case you can define smoothness of a point $p \in X$ as $S^q(m/m^2) \to m^q/m^{q+1}$ being an isomorphism for all $q\ge 0$, where $m$ is the maximal ideal of the local ring at $p$. In the ind case you have varieties $(X_n)_{n \ge 0}$ and $p \in X_n$ for all sufficiently large $n$ so you have $S_n^q(m_n/m_n^2) \to m^q_n/m^{q+1}_n$ for all sufficiently large $n$. Algebraic smoothness at $p$ means that the map

$\varprojlim S_n^q(m_n/m_n^2) \to \varprojlim m^q_n/m^{q+1}_n$

is an isomorphism for all $q\ge 0$


It is true that if $X$ is the union of $X_n$ and each $X_n$ is smooth then $X$ is algebraically smooth. However algebraic smoothness is not inductive in the sense that there are examples of $p \in X$ being algebraically smooth even when $p \in X_n$ is a singular point for all $X_n$ containing $p$. However line bundles are inductive in the sense that a line bundles on $X$ determines line bundles on each $X_n$ compatible under pull back and vice versa.

This suggests to me that you could have a collection of compatible Weil divisors in each $X_n$ which are not the zero section of any section of any line bundle. But I don't have an example.

share|cite|improve this question
up vote 5 down vote accepted

I don't know the general story, but I think the correspondence between Weil divisors and line bundles breaks down already for $\mathbf{A}^\infty$, which is pretty much the smoothest ind-variety out there. Using the standard ind-structure on $\mathbf{A}^\infty$, the example I have in mind is the union of all hyperplanes $x_n-x_1=0$. This is a codimension one closed ind-subscheme, but it's not the zero locus of any polynomial in $\mathcal{O}_{\mathbf{A}^\infty}$.

share|cite|improve this answer
Variant: fix a sequence $(a_n)_{n\in\mathbb{N}}$ in $k^\mathbb{N}$; for each $n$, put $X_n=\mathbb{A}^1=\mathrm{Spec}\,k[x]$ (with $\mathrm{Id}$ as transition map), and let $D_n$ be the Cartier divisor on $X_n$ defined by $\prod_{0\leq j\leq n} (x-a_n)$. – Laurent Moret-Bailly Sep 20 '11 at 7:37
That's a nice example of compatible inclusions that don't define a divisor on $X$ (I think) -- the union $D$ of all the $D_n$'s is not even a closed subset of $X$, since $D \cap X_n$ is not closed. – Dave Anderson Sep 20 '11 at 17:15
@Dave: right, but the same holds for your example: both your "ind-subscheme" (whatever this means) and mine are Zariski-dense. In fact, my example can be obtained from yours by intersecting with a suitable line. – Laurent Moret-Bailly Sep 21 '11 at 8:16
@Laurent, I may be making a silly mistake, but I think the intersection of my divisor with $X_n$ is the hyperplane arrangement with $x_i=x_1$ for $i=1,...,n$, plus $x_1=0$. (I'm using the standard ind-structure, where $X_n$ has $x_i=0$ for all $i>n$.) I agree that both are Zariski-dense, but I'm using the ind-topology on $\mathbf{A}^\infty$. (A subset is closed iff its intersection with each $X_n$ is.) – Dave Anderson Sep 21 '11 at 14:34
OK, then what I don't understand is what we should mean by "ind-subscheme" or "ind-divisor". – Laurent Moret-Bailly Sep 21 '11 at 15:43

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.