# Is there a unique way to make $\operatorname{Spec} R$ into an affine scheme?

Let $$R$$ be a commutative (noetherian) ring. Consider the set $$X=\operatorname{Spec}(R)$$ with Zariski topology. Let $$\mathcal O_X$$ be a sheaf of commutative rings on $$X$$ such that $$(X,\mathcal O_X)$$ is an affine scheme (notice that I'm not requiring $$\mathcal O_X$$ to be the standard structure sheaf yet).

My question is : Is $$\mathcal O_X$$ indeed the standard structure sheaf on $$\operatorname{Spec}(R)$$ ?

• So the question is whether two affine schemes are isomorphic if the underlying topological spaces are homeomorphic? Consider two nonisomorphic fields.
– JWL
Jul 6 '19 at 15:02
• What if $R$ is a field and $\mathcal O_X$ gives some other field? Jul 6 '19 at 15:03
• A next question would be if $Spec( \Bbb{Z}[x])$ is the only $\Bbb{Z}$-scheme whose underlying topological space is homeomorphic to the one of $Spec(\Bbb{Z}[x])$ ? Jul 6 '19 at 15:10
• @TorstenSchoeneberg: will they be isomorphic as schemes then ?
– uno
Jul 6 '19 at 15:18
• Who "they"? To explicate: $R=K$ (field) hence $X$ is a point, take some other field $L$ with structure sheaf $(Y, \mathcal{O}_Y)$, identify the points $X \simeq Y$, so take $(X, \mathcal{O}_Y)$ which is isomorphic to the affine scheme $(Y, \mathcal{O}_Y)$. Jul 6 '19 at 15:27

Let me just post JWL/Torstens answer, such that the question is answered and is not shown open anymore. All credit is due to them. They were suggesting to consider two non-isomorphic fields $$K$$ and $$L$$ and the affine schemes $$X = (\text{Spec}(K),\tilde{K})$$ and $$Y = (\text{Spec}(K),\tilde{L})$$. Then we obviously only have one map $$X \rightarrow Y$$ between the topological spaces, namely the identity sending the generic point $$\eta$$ to itself. On the level of sheaves we will never be able to find an isomorphism though, as we would need an isomorphism $$\varphi \colon \Gamma(\text{Spec}(K),\tilde{L}) = L \rightarrow K = \Gamma(\text{Spec}(K),\tilde{K}) = \Gamma(\text{id}^{-1}(\text{Spec}(K)),\tilde{K}).$$