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Is there an example of a scheme X whose reduction X_red is affine but X is not affine?

I got a question, which may be very easy, but I didn't figure out it. Let $X$ be a scheme such that $X_{\mathrm{red}}$ is an affine scheme. Could we conclude that $X$ itself must be affine also? This question came into my mind because I am thinking if the natural morpihsm $ i: X_{\mathrm{red}} \rightarrow X$ has such a property that for any open affine subset $U$ of $X_{\mathrm{red}}$, $i(U)$ is an open affine subset of $X$. Notice that $i$ is an homeomorphism, hence $i(U)$ is an open subset of $X$.

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    $\begingroup$ It's always true, see mathoverflow.net/questions/95/… $\endgroup$ – Philipp Hartwig Jul 14 '11 at 15:55
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    $\begingroup$ Dear unknown (google): Don't be afraid to put a complete-sentence question into the titles of your posts. MathOverflow allows for titles that can be up to 240 characters, which is longer than a tweet, and longer than this comment. $\endgroup$ – Theo Johnson-Freyd Jul 14 '11 at 18:25
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If $X$ is noetherian the answer is yes.

Indeed, in this case $X$ is affine if and only if $X_{\textrm{red}}$ is affine.

See [Hartshorne, Algebraic Geometry], Exercise 3.1 p. 216.

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Here is a direct argument for $X$ satisfies the following condition: () there is a covering of $X$ by finitely many open affines $(U_i)$ such that each intersection $U_i\cap U_j$ is quasi-compact. ()holds if the underlying space of $X$ is noetherian or $X$ is quasi-compact and quasi-separated.

Let $A:=\Gamma(X,O_X)$. Then we have a map of schemes $f:X\to {\rm Spec(A)}$. Now since $X_{red}$ is affine, we have a collection of elements $(f_i)\subseteq A$ such that each $X_{f_i}$ (the set of points of $X$ in which $f_i$ is invertible) is affine and $(f_i)$ generate the unit ideal.

The reason that I can choose such $f_i$ is that I can cover $X$ by open affines and I can cover these affines by the basis in $X_{red}$ (note that topologically $|X_{red}|=|X|$).

We have $X_{f_i}\to {\rm Spec(A_{f_i})}$ is an isomorphism for each $i$ because one can check easily under the assumption (*) that for any $a\in\Gamma(X,O_X)$ we have the canonical map $\Gamma(X,O_X)_a\to\Gamma(X_a,O_{X_a})$ is an isomorphism. Since $f^{-1}({\rm Spec(A_{f_i})})=X_{f_i}$ we know $f$ is an isomorphism.

[edit]As brunoh has pointed out in the comment, the above proof is not complete. I need to show $A_{red} \to \Gamma(X_{red},O_{X_{red}})$ is surjective. The following clever proof is taken from EGA 1, 5.19.1, I find it nice and want to share it with others.

Assumption: the ideal sheaf $\mathcal{I}$ of $X'\hookrightarrow X$ satisfies $\mathcal{I}^n=0$ for some $n>0$. $X'$ affine.

Factorize $\Gamma(X,O_X)\to \Gamma(X,O_X/\mathcal{I})$ as $\Gamma(X,O_X)=\Gamma(X,O_X/\mathcal{I}^n)\to \Gamma(X,O_X/\mathcal{I}^{n-1})\to\cdots\to \Gamma(X,O_X/\mathcal{I})$, we may assume $n=2$ (here we also need my above argument with $X'$ replace $X_{red}$). But then $\mathcal{I}=\mathcal{I}/\mathcal{I}^2$ is a quasi-coherent $O_{X'}$-module. Thus $H^1(X,\mathcal{I})=H^1(X',\mathcal{I})=0$. This shows the surjectivity.

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  • $\begingroup$ @Lei i am very sorry but I find your "proof" very confusing. For instance, you are right when you say that because X is qcqs at any $a$ the canonical map is an isomorphism. But it does not mean at all that your $f_i$ are scheme isomorphisms except if you already know that each $X_{f_i}$ is affine which is precisely what you are trying to show ! I do not think there is such an easy proof of this result : especially the hypothesis that X is noetherian (stronger than X qcqs) is crucial to show that the nilpotent radical ideal is indeed nilpotent, a key point in my opinion. $\endgroup$ – brunoh Jul 15 '11 at 9:34
  • $\begingroup$ Hi brunoh, thank you for your comment. I choose $X_{f_i}$ so that each of them is in an affine subscheme of $X$, this makes $X_{f_i}$ affine. This is possible because $|X_{red}|=|X|$ and $(X_{red})_{f_i}$ form a basis. $\endgroup$ – Lei Jul 15 '11 at 10:31
  • $\begingroup$ @Lei Hi Lei, forgive me again for being persistent and for bothering you ! Your answer to my comment is precisely the other point I found confusing ! If you want to cover $X$ with affine of the type $Xred_{f_i}$ you can do it but using $(f_i)$ belonging to $Ared$ where $Xred=Spec(Ared)$ but you do not know yet that $Ared=A/nil(A)$ which is the difficult part to prove in my opinion. Therefore you only get that your $Xred_{f_i}$ are indeed open affines of $X$ but not of the form $X_{f_I}$ ! $\endgroup$ – brunoh Jul 15 '11 at 10:44
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    $\begingroup$ You are completely right, I overlooked something which maybe crucial in the argument. I shoud show that $\Gamma(X,O_X)_{red}\to \Gamma(X_{red},O_{X_{red}})$ is, at least, surjective. I haven't thought it through. Thank you very much for pointing this out! $\endgroup$ – Lei Jul 15 '11 at 11:43
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    $\begingroup$ I totally agree with brunoh. This moring, I have thought this question. It looks like that we get an open affine covering $U_i$ of $X$, each $U_i$ is a principal open affine subset of an (not necessary principal ) open affine subset $V$ of $X$, which may not be of a form $X_f$ for some global section $f$ of $O_X$ $\endgroup$ – user565739 Jul 15 '11 at 11:52

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