**Proof.**
Consider any prime $\mathfrak q' \subset S$ which corresponds to a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak p})$. This means that $\mathfrak p' = R \cap \mathfrak q'$ is contained in $\mathfrak p$. Here is a picture

\[ \xymatrix{ \mathfrak q' \ar@{-}[d] \ar@{-}[r] & ? \ar@{-}[r] \ar@{-}[d] & S \ar@{-}[d] \\ \mathfrak p' \ar@{-}[r] & \mathfrak p \ar@{-}[r] & R } \]

Assume (1) and (2)(a). By going up there exists a prime $\mathfrak q'' \subset S$ with $\mathfrak q' \subset \mathfrak q''$ and $\mathfrak q''$ lying over $\mathfrak p$. By the uniqueness of $\mathfrak q$ we conclude that $\mathfrak q'' = \mathfrak q$. In other words $\mathfrak q'$ defines a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak q})$.

Assume (1) and (2)(b). By going down there exists a prime $\mathfrak q'' \subset \mathfrak q$ lying over $\mathfrak p'$. By the uniqueness of primes lying over $\mathfrak p'$ we see that $\mathfrak q' = \mathfrak q''$. In other words $\mathfrak q'$ defines a point of $\mathop{\mathrm{Spec}}(S_{\mathfrak q})$.

In both cases we conclude that the map $\mathop{\mathrm{Spec}}(S_{\mathfrak q}) \to \mathop{\mathrm{Spec}}(S_{\mathfrak p})$ is bijective. Clearly this means all the elements of $S - \mathfrak q$ are all invertible in $S_{\mathfrak p}$, in other words $S_{\mathfrak p} = S_{\mathfrak q}$.
$\square$

## Comments (0)

There are also: