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?
Best Answer
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 $\epsilon \in \mathcal N(X)$.