# Morphism of ringed spaces not induced by homomorphism of rings

I would like to see a bit more detailed explanation why the example in this thread is a example for a morphism of ringed spaces not induced by homomorphism of rings: Here we have $R = \mathbb{C}[T]_p$ as localisation at $p =(T)$. Let be $F = Frac(R) = \mathbb{C}(T)$. As topological spaces we choose $X,Y = Spec(R)= \{x_{(T)}, x_0\}$. Let $V = D(1), U = D(T)$. Obviosly $\Gamma(V, O_Y) =R, \Gamma(U, O_Y) =F$. Now there is a morphism of ringed spaces which should become a conterexample:

Because of inclusions $R \subset F \subset \mathbb{C}$ we define the ringhomomorphisms as folows: $\phi_V : \Gamma(V, O_Y) \to \Gamma(V, O_Y)$ for $\phi_V := R \subset F \subset \mathbb{C} \subset R$ and $\phi_U : \Gamma(U, O_Y) \to \Gamma(U, O_Y)$ for $\phi_U := F \subset \mathbb{C} \subset R\subset F$. That's clear that $R \xrightarrow{\phi_V} R \subset F = R \subset F \xrightarrow{\phi_U} F$, so with $f: X \xrightarrow{id} Y$ the pair $(f, \phi)$ would become a morphism between ringed spaces. That's clear to me that because of definition of $\phi_V$ as a composition through a field and the fact that $Spec(F) = {x_0}$ the induced morphism $Spec(\phi_V): Spec(R) \to Spec(R)$ maps $x_{(T)}, x_0 \to x_0$ but thats not clear why this construction provide a conterexample, so that it is not induced by a ringhomomorphism or aquivalent: there not exist a ringhomom $\psi: R \to R$ that for $U = D(s)$ the maps $\phi_U$ have the shape $R_s = \Gamma(U, O_Y) \xrightarrow{\phi_U} \Gamma(U, O_X), \frac{a}{s^n} \to \frac{\psi(a)}{\psi(s)^n}$.

I'm honestly getting a little lost in your argument. For one thing, at the beginning you include $\Bbb C(T)$ into $\Bbb C$, which is possible, but not natural and I don't see the benefit (nowhere do you need to perform such an unnatural operation to see the counterexample!). In fact, the map you're trying to write down is not the counterexample of the linked question at all: the counterexample map from the other question does not have the same domain and codomain, as you do here! So I'll try to explain the counterexample of the linked question in more detail, and while this won't address your argument explicitly, I hope it can clear up some of your confusion.

Let $(A,\mathfrak{m})$ be a DVR, so that $\operatorname{Spec} A = \{s,\eta\}$ ($s$ is the closed point corresponding to the ideal $\mathfrak{m}$, and $\eta$ is the generic point corresponding to $(0)$), and let $K = \operatorname{Frac}A$, so $\operatorname{Spec} K = \{\sigma\}$. As in the example in the other question, we define the map of ringed spaces $f : (\operatorname{Spec} K,\mathcal{O}_K)\to(\operatorname{Spec} A,\mathcal{O}_A)$ as follows: \begin{align*} f : \sigma &\mapsto s,\\ f^\sharp(\operatorname{Spec} A) : \mathcal{O}_A(\operatorname{Spec} A)\cong A&\to K\cong f_\ast\mathcal{O}_K(\operatorname{Spec} A)\quad\textrm{is the natural inclusion }\iota,\\ f^\sharp(\{\eta\}) : \mathcal{O}_A(\{\eta\})\cong K&\to K\cong f_\ast\mathcal{O}_K(\{\eta\})\quad\textrm{is the identity},\\ f^\sharp(\emptyset) : \mathcal{O}_A(\emptyset)\cong 0&\to 0\cong f_\ast\mathcal{O}_K(\emptyset). \end{align*} Because the restriction map on the structure sheaf of $\operatorname{Spec} A$ from $\operatorname{Spec} A$ to $\{\eta\}$ is the inclusion $\iota : A\to K$, we see that this is indeed a map of ringed spaces.

However, this is not a map of locally ringed spaces. Since the only open set containing $s$ is all of $\operatorname{Spec} A$, the stalk map $$f^\sharp_\sigma : \mathcal{O}_{A,f(\sigma)}\to\mathcal{O}_{K,\sigma}$$ is simply the map on global sections; i.e., $$f^\sharp_\sigma = f^\sharp(\operatorname{Spec} A) : A\to K.$$ This is a ring homomorphism, but the maximal ideal $\mathfrak{m}\subseteq A$ is not mapped into the maximal ideal $(0)\subseteq K$. Hence, the map of ringed spaces defined above cannot be induced by a ring morphism $\phi : A\to K$, because any ring homomorphism induces a morphism of locally ringed spaces $(\operatorname{Spec} K,\mathcal{O}_K)\to(\operatorname{Spec} A,\mathcal{O}_A)$.

Another way to argue that the map $f$ is not induced by a ring homomorphism $\phi : A\to K$ is to first show that if $f$ is induced by $\phi$, then the map of structure sheaves on global sections $f^\sharp(\operatorname{Spec} A) : \mathcal{O}_A(\operatorname{Spec} A)\to f_\ast\mathcal{O}_K(\operatorname{Spec} A)$ is in fact the map $\phi$. That is, the diagram $$\require{AMScd} \begin{CD} \mathcal{O}_A(\operatorname{Spec} A) @>{f^\sharp(\operatorname{Spec} A)}>> f_\ast\mathcal{O}_K(\operatorname{Spec} A);\\ @VVV @VVV \\ A @>{\phi}>> K; \end{CD}$$ whose vertical arrows are the natural isomorphisms commutes. However, the map on global sections $f^\sharp(\operatorname{Spec} A)$ for the map defined above is the inclusion, and the map on topological spaces induced by the inclusion sends $\sigma$ to $\eta$.