4
$\begingroup$

Let $(X_n)_{n \geq 0}$ be a family of schemes. Let $$X_0 \to X_1 \to X_2 \to \dotsc$$ be a sequence of closed immersions (which therefore gives rise to an ind-scheme). Under which (necessarly and/or sufficient) conditions is there a scheme $S$ and a family of closed immersions $X_n \to S$ such that

$\hspace{13em}\begin{array}{ccc} X_n\!\!\!\!\!\! && \longrightarrow && \!\!\!\!\!\!\!\!X_{n+1} \\ & \searrow && \swarrow &\\ &&S &&\end{array}$

commutes? This is not always satisfied (see the comments).

Improve this question
$\endgroup$
6
  • $\begingroup$ If all your schemes are affine then you have a bunch of ring surjections, and the projective limit surjects onto all these things, so you're Ok in this case -- let me just clarify though -- you're not asking for $S$ to be universal in any way? You just want one $S$ to exist? I guess closed immersions are affine so you could now just try and globalise that construction -- what goes wrong? Did you try this yourself? $\endgroup$
    – znt
    Nov 15, 2016 at 7:48
  • $\begingroup$ My instinct is that projective limits won't commute with localisation so there will be a glueing issue. $\endgroup$
    – znt
    Nov 15, 2016 at 7:50
  • 3
    $\begingroup$ Probably you can make a counterexample starting with Hironaka's example of a smooth $3$-dimensional algebraic space $S$ that is not a scheme (described in the appendices to Hartshorne's "Algebraic Geometry"). Begin with the exceptional surface $X_0$ with its coherent ideal sheaf $\mathcal{I}$ on $S$. For every $n\geq 0$, define $X_n\subset S$ to be the closed algebraic subspace defined by the ideal sheaf $\mathcal{I}^{n+1}$. $\endgroup$
    – Jason Starr
    Nov 15, 2016 at 12:10
  • 1
    $\begingroup$ If we do not bound the dimension of $X_n$ from above can't we easily construct examples without compatible closed immersions $X_n\to S$. For example, pick $X_n=P^{i_n}$ and closed immersions $\phi _n:X_n\to X_{n+1}$ such that $\phi _n^*\mathcal O _{X_{n+1}}(1)=\mathcal O _{X_n}(l_n)$ with $l_n>1$. Let $U$ be an affine open subset of $S$ and suppose that $U\cap X_1\ne \emptyset$. Then, if $U_n=X_n\cap U$ and $H_n=X_n\setminus U_n$ is an ample divisor of degree $d_n$. Then I think $d_n=l_nd_{n+1}$ which implies $l_1\cdots l_n$ divides $d_1$ for $n\gg 0$ impossible. $\endgroup$
    – Hacon
    Nov 15, 2016 at 21:17
  • 1
    $\begingroup$ So maybe it is reasonable to assume that the dimensions are bounded (and hence passing to a subsequence fixed). If each $X_n$ is a smooth projective variety, then (similarly to above) I think we get that $X_n\to X_{n+1}$ is birational (and then an isomorphism) for $n\gg 0$. So maybe most bad examples are non-reduced as in pointed out by JS above. $\endgroup$
    – Hacon
    Nov 15, 2016 at 21:22

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.