Remark 10.16.8. A fundamental commutative diagram associated to a ring map $\varphi : R \to S$, a prime $\mathfrak q \subset S$ and the corresponding prime $\mathfrak p = \varphi ^{-1}(\mathfrak q)$ of $R$ is the following

$\xymatrix{ \kappa (\mathfrak q) = S_{\mathfrak q}/{\mathfrak q}S_{\mathfrak q} & S_{\mathfrak q} \ar[l] & S \ar[r] \ar[l] & S/\mathfrak q \ar[r] & \kappa (\mathfrak q) \\ \kappa (\mathfrak p) \otimes _ R S = S_{\mathfrak p}/{\mathfrak p}S_{\mathfrak p} \ar[u] & S_{\mathfrak p} \ar[u] \ar[l] & S \ar[u] \ar[r] \ar[l] & S/\mathfrak pS \ar[u] \ar[r] & (R \setminus \mathfrak p)^{-1}S/\mathfrak pS \ar[u] \\ \kappa (\mathfrak p) = R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p} \ar[u] & R_{\mathfrak p} \ar[u] \ar[l] & R \ar[u] \ar[r] \ar[l] & R/\mathfrak p \ar[u] \ar[r] & \kappa (\mathfrak p) \ar[u] }$

In this diagram the arrows in the outer left and outer right columns are identical. The horizontal maps induce on the associated spectra always a homeomorphism onto the image. The lower two rows of the diagram make sense without assuming $\mathfrak q$ exists. The lower squares induce fibre squares of topological spaces. This diagram shows that $\mathfrak p$ is in the image of the map on Spec if and only if $S \otimes _ R \kappa (\mathfrak p)$ is not the zero ring.

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