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 $\operatorname{Spec}(R)$ ?
Best Answer
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 $$\varphi \colon \Gamma(\text{Spec}(K),\tilde{L}) = L \rightarrow K = \Gamma(\text{Spec}(K),\tilde{K}) = \Gamma(\text{id}^{-1}(\text{Spec}(K)),\tilde{K}).$$
Very easy way to realize that situation more concretely if you want to is of course just by a cardinality argument if one takes a finite field and an infinite field for example.