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I am doing exercise 2.2.3(b) in Hartshorne which asks to show that for any scheme $(X,\mathcal{O}_X)$, we have that $(X,({\mathcal{O}_X})_{red}^+)$ is a reduced scheme. By $({\mathcal{O}_X})_{red}$ I mean the the presheaf $U \mapsto \mathcal{O}_X(U)/\operatorname{nilrad}\mathcal{O}_X(U)$ and the superscript + denotes sheafification. Now to do this it is enough to show for any open affine cover $U_i$ of $X$ for which $(U_i,\mathcal{O}_X|_{U_i})$ is isomorphic to an affine scheme, that we also have $(U_i,({\mathcal{O}_X})_{red}^+|_{U_i})$ isomorphic to an affine scheme.

Now we have homeomorphisms $\varphi_i : \operatorname{Spec}(A_i) \to U_i$ and isomorphisms of sheaves $\varphi_i^\sharp : (\mathcal{O}_X|_{U_i}) \to (\varphi_i)_\ast \mathcal{O}_{\operatorname{Spec}(A_i)}$ for some rings $A_i$ since $X$ is a scheme. It follows for every $i$, $\varphi^\sharp_i$ descends to an isomorphism of presheaves $\psi_i : (\mathcal{O}_X)_{red}|_{U_i} \to (\varphi_i)_\ast {\mathcal{O}_{\operatorname{Spec}(A_i)}}_{red}.$ Now suppose I know that ${\mathcal{O}_{\operatorname{Spec}(A_i)}}_{red}$ is a sheaf on $A_i$. Then its pushforward under ${\varphi_i}_\ast$ is also a sheaf and so $(\mathcal{O}_X)_{red}|_{U_i} $ is a sheaf. Then because sheafification is local it follows $$(\mathcal{O}_X)_{red}|_{U_i} \cong (\mathcal{O}_X)_{red}|_{U_i}^+ \cong (\mathcal{O}_X)_{red}^+|_{U_i}$$ is a sheaf too.

My question is: I have shown that $(U_i,(\mathcal{O}_X)_{red}^+) \cong (\operatorname{Spec}(A_i),{\mathcal{O}_{\operatorname{Spec}(A_i)}}_{red})$. But I can't make sense of the right hand side: When I compute global sections I get $$\Gamma(\operatorname{Spec} A_i,{\mathcal{O}_{\operatorname{Spec}(A_i)}}_{red}) \cong A_i/\operatorname{nilrad} A_i$$ and not $A_i$, which is contrast to Proposition 2.2.2 (c) of Hartshorne. What am I misunderstanding about the proposition? It seems my result on global sections is correct because I want everything reduced. Also, is my approach above correct in general for such problems?

Edit: It seems the proposition does not apply because on the L.H.S. above I either have to change my sheaf or my ring. Does this mean that in order to complete the problem I need $$(U_i,(\mathcal{O}_X)_{red}^+) \cong (\operatorname{Spec}(A_i/\operatorname{nilrad}A_i),{\mathcal{O}_{\operatorname{Spec}(A_i)}}_{red})$$

and not as what I originally had above? But to me they are the same pair because the spectrum of a reduced ring is homeomorphic to the original ring. What am I misunderstanding in this subtle point here?

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  • $\begingroup$ I don't know much about the topic, so please disregard this if it's complete nonsense, but it doesn't look like what you have written quite matches up with Prop 2.2.2(c). Shouldn't the sheaf be $\mathcal{O}_{Spec(A_i)}$ in the LHS to apply the theorem? Or the ring be $A_i/{nil(A_i)}$? I would be perfectly content with the universe if what you had written was true! Also, don't you just have to find a cover by affine varities for the exercise, not a condition involving all covers of the original scheme? I would just try to show that something involving $A_i /nilrad\ A_i$ worked. $\endgroup$ – Zach L. Jul 11 '13 at 6:21
  • $\begingroup$ Dear @ZachL. you are correct, either the ring or the sheaf on the L.H.S. needs to be changed. $\endgroup$ – user38268 Jul 11 '13 at 16:06
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If $X=Spec(A)$, we have $X_ \text {red}=Spec(A_\text {red})$, where $A_{\text {red}}=A/\text{Nil(A)}$ so that:

$\Gamma(X,\mathcal O_X)=A$
but
$\Gamma(X_{\text {red}},\mathcal O_{X_ \text {red}})=A_{\text {red}}=A/\text{Nil(A)}$

No contradiction with Hartshorne's 2.2.2.(c)!

Edit: some details
Here are some statements which might help shed light on this subtle question.

a) Given a scheme $X$ we associate to it the quasi-coherent sheaf of ideal $\mathcal N\subset \mathcal O_X$ defined for an arbitrary open subset $U\subset X$ by $$\mathcal N(U)=\{f\in \mathcal O_X(U)\mid \forall x\in \mathcal O_{X,x },\; f_x \in \text {Nil}(\mathcal O_{X,x }) \}$$ b) The scheme $X_{\text {red}}$ has structure sheaf $\mathcal O_{X_{\text {red}}}=\mathcal O_X/\mathcal N$
c) For any affine subset $U=\text {Spec} (A)\subset X$, we have $ \text {Nil}(\Gamma(U,\mathcal O_X))=\mathcal N(U)=\text {Nil}(A)$
d) For any affine subset $U=\text {Spec} (A)\subset X$, we have $\mathcal O_{X_{\text {red}}}(U)=A_{\text {red}}=A/{\text {Nil}} (A)$
e) For a general open subset $U\subset X$ , we have $ \text {Nil}(\Gamma(U,\mathcal O_X))\subset \mathcal N(U)$ but the inclusion may be strict for non-affine $U$:

Let $X_m=\text {Spec}(\mathbb C[T]/T^m)=\text {Spec}(\mathbb C[\epsilon _m])$ and $X=\bigsqcup X_m$ (a non-affine scheme).
Then $\Gamma(X,\mathcal O_X)=\prod \Gamma(X_m,\mathcal O_{X_m})=\prod \mathbb C[\epsilon _m]$ and for $\epsilon=(\epsilon_1,\epsilon_2,\cdots)$ we have $\epsilon \notin \text {Nil}(\Gamma(X,\mathcal O_X))$ although $\epsilon \in \mathcal N(X)$.

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  • $\begingroup$ Thanks Georges for your answer. Is the approach I outlined above correct? Somehow I a confused because I think tthe conclusion $(U_i,(\mathcal{O}_X)_{red}^+) \cong (\operatorname{Spec}(A_i),{\mathcal{O}_{\operatorname{Spec}(A_i)}}_{red})$ is somehow wrong. Regards, $\endgroup$ – user38268 Jul 11 '13 at 7:02
  • $\begingroup$ Dear @Georges, I edited my question above. Could you help me with this new edit? Regards, $\endgroup$ – user38268 Jul 11 '13 at 16:25
  • $\begingroup$ Dear Benja, I think Hartshorne's presentation might not be optimal: thea association $U\mapsto \mathcal O(U)_{red}$ is already the sheaf $\mathcal O_{X_{red}}$ on the basis of $X$ consisting of affine subschemes $U$ of $X$, and there is no need for the sheafifying $ +$ construction. $\endgroup$ – Georges Elencwajg Jul 13 '13 at 14:02
  • $\begingroup$ Dear Benja, I have added a few points to my answer. $\endgroup$ – Georges Elencwajg Jul 13 '13 at 20:40
  • $\begingroup$ Georges, your method in the edit certainly seems like a potentially better way to approach the problem than the Hartshorne construction by sheafification. I was wondering though, why is it the case that $\mathcal{O}_{X_{red}}$ is the cokernel you give? Also, is there an elementary way to arrive at $d)$, or do we use e.g. knowledge about quasi-coherent $\mathcal{O}_X$ modules on affine schemes? I did the exercise rather explicitly and it seems a little long, I thought about cokernels but the "obvious" candidate $N(U)=\mathcal{O}_{X_{red}}$ needn't be a sheaf, so I gave up on the idea... $\endgroup$ – Tom Oldfield Feb 12 '15 at 23:10

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