Lemma 10.35.9. Let $R \to S$ be a ring map. Let $\mathfrak m \subset R$ be a maximal ideal. Let $\mathfrak q \subset S$ be a prime ideal lying over $\mathfrak m$ such that $\kappa (\mathfrak m) \subset \kappa (\mathfrak q)$ is an algebraic field extension. Then $\mathfrak q$ is a maximal ideal of $S$.

Proof. Consider the diagram

$\xymatrix{ S \ar[r] & S/\mathfrak q \ar[r] & \kappa (\mathfrak q) \\ R \ar[r] \ar[u] & R/\mathfrak m \ar[u] }$

We see that $\kappa (\mathfrak m) \subset S/\mathfrak q \subset \kappa (\mathfrak q)$. Because the field extension $\kappa (\mathfrak m) \subset \kappa (\mathfrak q)$ is algebraic, any ring between $\kappa (\mathfrak m)$ and $\kappa (\mathfrak q)$ is a field (Fields, Lemma 9.8.10). Thus $S/\mathfrak q$ is a field, and a posteriori equal to $\kappa (\mathfrak q)$. $\square$

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