# For a Noetherian scheme X, show that $X_{red}$ affine implies $X$ affine.

I have the following problem:

Let X be a Noetherian Scheme and suppose that $X_{red}$ is affine. Show that this implies that X is affine.

OK, so I know the "classical" proof of this using Serre's criterion for affineness and with cohomology. However, I encountered this in an early chapter of Görtz-Wedhorn's book where none of these concepts have thus far been defined. I have been trying to come up with an elementary proof but without much success. Clearly, we can assume that the ideal sheaf $\mathcal{N}$ satisfies $\mathcal{N}^2 = 0$.

I would be very grateful for help with this problem of any sort, ranging from hints to solutions.

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The tricky part is to prove that $\Gamma(X,O_X) \to \Gamma(X_{red},O_{X_{red}})$ is surjective i.e. $H^1(X,\mathcal{N}) = 0$ which I think can be done by hand. Once you have that you can deduce that $X \to Spec(\Gamma(X,O_X))$ is an homeomorphism (because it is so for $X_{red}$). Then check that $\Gamma(X,O_X)_{\mathfrak{p}} \to O_{X,\mathfrak{p}}$ is an isomorphism for each $\mathfrak{p}$. –  AFK Jun 23 '12 at 21:07
I posted a similar question in MathOverflow. –  Makoto Kato Apr 2 '14 at 0:16
@YBL Dear YBL, I wonder how you prove that the last homomorphism of local rings is an isomorphism. Regards, –  Makoto Kato Apr 5 '14 at 20:41

It suffices to prove the following proposition without using cohomology.

Proposition Let $X$ be a scheme. Suppose there exists a quasi coherent $\mathcal O_X$-ideal $\mathcal I$ such that $\mathcal I^2 = 0$ and $(X, \mathcal O_X/\mathcal I)$ is affine. Let $\mathcal F$ be a quasi-coherent $\mathcal O_X$-module. Then $\mathcal F$ is quasi-flasque.

Proof(without using cohomology): Consider the following exact sequence. $0 \rightarrow \mathcal I \mathcal F \rightarrow \mathcal F \rightarrow \mathcal F/\mathcal I\mathcal F \rightarrow 0$. Since $\mathcal I \mathcal F$ and $\mathcal F/\mathcal I\mathcal F$ are both quasi-coherent $\mathcal O_X/\mathcal I$-modules, they are quasi-flasque by my answer to this question. Hence $\mathcal F$ is quasi-flasque by jdc's answer to this question.

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