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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 $\operatorname{Spec}(R)$ ?

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    $\begingroup$ So the question is whether two affine schemes are isomorphic if the underlying topological spaces are homeomorphic? Consider two nonisomorphic fields. $\endgroup$
    – JWL
    Jul 6 '19 at 15:02
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    $\begingroup$ What if $R$ is a field and $\mathcal O_X$ gives some other field? $\endgroup$ Jul 6 '19 at 15:03
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    $\begingroup$ 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 $Spec(\Bbb{Z}[x])$ ? $\endgroup$
    – reuns
    Jul 6 '19 at 15:10
  • $\begingroup$ @TorstenSchoeneberg: will they be isomorphic as schemes then ? $\endgroup$
    – uno
    Jul 6 '19 at 15:18
  • $\begingroup$ 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 $(Y, \mathcal{O}_Y)$. $\endgroup$ Jul 6 '19 at 15:27
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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.

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