Subtle aspect of closed subscheme

Question

Let me define a closed subscheme of a scheme $X$ as:

An equivalence class of data in the form

$$(Z,Y,i,i^\sharp)$$

where $Z$ is a closed subset of $X$, $(i,i^\sharp): Y \to X$ is a morphism of schemes with $i$ giving a homeomorphism onto $Z$, and $i^\sharp : \mathcal{O}_X \rightarrow i_* \mathcal{O}_Y$ is surjective.

We say $(Z_1,Y_1,i_1,i_1^\sharp)$ and $(Z_2,Y_2,i_2,i_2^\sharp)$ are equivalent if there is an isomorphism of schemes $\rho$ from $Y_1$ to $Y_2$ such that $(i_1,i_1^\sharp)=\rho \circ (i_2,i_2^\sharp)$. Clearly, these two data are equivalent only if $Z_1$ and $Z_2$ are the same subset of $X$.

Can we have another closed subscheme of $X$ on the same exact closed subset $Z$, say $(Z,Y,j,j^\sharp)$, that is not equivalent to $(Z,Y,i,i^\sharp)$, with the same source scheme?

If yes, then is it possible to have an example where $j=i$ actually?

Thanks.

If $j=i$ then all data are the same, hence the two subschemes must be the same? — Servaes, Jun 05 '14 at 13:13

Martin Brandenburg · Accepted Answer · 2014-06-05T13:49:44.450

3

You seem to ask if there are two closed subschemes $Y$, $Y'$ of a scheme $X$ which are not isomorphic over $X$, but $Y \cong Y'$ as abstract schemes and $|Y|=|Y'|$ as subsets of $|X|$.

If $X=\mathrm{Spec}(A)$ is affine, this means that we have to find two ideals $I,J \subseteq A$ such that $I \neq J$, but $A/I \cong A/J$ as rings and $\sqrt{I} = \sqrt{J}$.

It suffices to find some ideal $I$ with $A/I \cong A/I^2$ and $I \neq I^2$.

Take $A=k[X_1,X_2,\dotsc]/(X_1^2,X_2^2,\dotsc)$ and $I=(X_1)$. Then $I^2 = 0 \neq I$, and $A/I = k[X_2,\dotsc]/(X_2^2,\dotsc) \cong A$.

I wonder if there is any example for $A$ noetherian.

edited Jun 05 '14 at 13:49

answered Jun 05 '14 at 13:17

Martin Brandenburg

163,620

Are the corresponding schemes attached to the ideal sheaves here, isomorphic as schemes? – user51197 Jun 05 '14 at 13:28
No. That is what Martin means by "different". – Zhen Lin Jun 05 '14 at 13:29
I want an example where the source schemes of the closed immersions are isomorphic. – user51197 Jun 05 '14 at 13:31
I've edited my answer. – Martin Brandenburg Jun 05 '14 at 13:34
I want to use the same common closed subset Z of X. Same meaning equal as subsets. – user51197 Jun 05 '14 at 13:35
Thank you. I would be interested in a Noetherian ring example (or any Noetherian scheme example). – user51197 Jun 05 '14 at 14:19
If yes, then is it possible to have an example where $j=i$ actually? Martin, your example answers my second question in the affirmative, since A has only one prime ideal. Thanks again.

– user51197 Jun 05 '14 at 15:27

Subtle aspect of closed subscheme

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