I'm new to all this scheme stuff, so I'm trying to get a handle of them with what I think is supposed to be an easy example.
Let $K$ be a field, and let $R = K[x]_{(x)} = \{\frac{f(x)}{g(x)} | f(x), g(x) \in K[x], g(0) \neq 0\}.$ I know that Spec $R$ has two points, being the ideals $(0)$ and $(x).$ The topological space is also pretty small, containing the sets $\emptyset, \{(0)\},$ and $\{(0), (x)\} =$ Spec $R.$ However, I don't really understand the structure sheaf. I know that $\mathcal{O}_{\text{Spec } R}(\text{Spec } R) = R,$ but I don't know why. I also need help understanding why $\mathcal{O}_{\text{Spec } R}(\{(0)\}) = K(x).$ The definition I'm working with is to let $\mathcal{O}_{\text{Spec } R}(D(f))$ be the localization of the $R$ at the multiplicative set of all functions that do not vanish outside of $V(f).$ With that in mind, how can I calculate the structure sheaves here?
$D(x)$ is an open set containing the prime ideals that don't contain $x$. In this case, $D(x) = \{(0)\}$. Therefore, $O_{Spec(R)}(0) = O_{Spec(R)}(D(x)) = R_x = (K[x]_{(x)})_x$.
Can you show this is isomorphic to $K(x)$ now?