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The following is a statement from Eisenbud & Harris geometry of schemes

Let $(X, \mathcal{O})$ be any ringed space, and let $R = \mathcal{O}(X)$. For any $f \in R$ we can define a set $U_f \subset X$ as the set of points $x \in X$ such that $f$ maps to a unit of the stalk $\mathcal{O}_x$. If $(X, \mathcal{O})$ is an affine scheme we must have: $$\mathcal{O}(U_f) = R[f^{-1}].$$

I'm trying to do the following problem

Take $Z = \operatorname{Spec} \Bbb C[x]$, let $X$ be the result of identifying the two closed points $(x)$ and $(x − 1)$ of $|Z|$, and let $\varphi : Z \to X$ be the natural projection. Let $\mathcal{O}$ be $\varphi_∗\mathcal{O}_Z$, a sheaf of rings on $X$. Show that $(X, \mathcal{O})$ satisfies condition $(i)$ above for all elements $f \in \mathcal{O}(X) = \Bbb C[x]$, but does not satisfy condition $(ii)$. Note that there is no natural map $X \to |\operatorname{Spec} \Bbb C[x]|$.

For context the condition $(ii)$ is that the stalks $\mathcal{O}_x$ are local rings. So for the problem consider $f \in \mathcal{O}(X) = \mathbb{C}[x]$ and $U_f$. We now have that $$\begin{align*} \mathcal{O}(U_f) &= \varphi_*\mathcal{O}_Z(U_f) \\ &= \mathcal{O}_Z(\varphi^{-1}(U_f)).\end{align*}$$ And we would like to have $\mathcal{O}_Z(\varphi^{-1}(U_f)) = \mathcal{O}_Z(\varphi^{-1}(X))[f^{-1}].$ Does this hold just from $(Z, \mathcal{O}_Z)$ being an affine scheme or do we actually need to show the equality here?

To show that $(ii)$ does not hold we need to find a stalk $\mathcal{O}_x$ of $\mathcal{O}$ which has more than one maximal ideal. I guess that the sentence "Note that there is no natural map $X \to |\operatorname{Spec} \Bbb C[x]|$." is supposed to be a hint for this, but I'm not sure why this would be helpful?

Iman
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    Well, there is exactly one interesting point of $X$, so you should probably investigate the stalk at that point! – diracdeltafunk Apr 16 '23 at 19:15
  • I'm not entirely understanding $X$ being the space we get by identifying $(x)$ and $(x-1)$. Does $X$ have some quotient topology induced from $Z$'s Zariski topology? @diracdeltafunk – Iman Apr 16 '23 at 19:17
  • Yes, exactly! $X$ is the quotient of $\lvert Z \rvert$ by the subspace ${(x),(x-1)}$. – diracdeltafunk Apr 16 '23 at 19:18
  • So $X = \operatorname{Spec} \Bbb C[x]/ \sim$ where $\mathfrak{p} \sim \mathfrak{q}$ if $\mathfrak{p} = \mathfrak{q}$ or ${\mathfrak{p}, \mathfrak{q}} = {(x), (x-1) }$. The interesting point you mean is then which one? The two points identified? @diracdeltafunk – Iman Apr 16 '23 at 19:21
  • Okay apparrently this result is false due to https://math.stackexchange.com/questions/8331/exercise-from-eisenbud-harriss-the-geometry-of-schemes – Iman Apr 16 '23 at 19:29

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