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?