The Stacks project

Lemma 10.34.9. 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 $\kappa (\mathfrak q)$. $\square$

Comments (0)

There are also:

  • 5 comment(s) on Section 10.34: Jacobson rings

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00GA. Beware of the difference between the letter 'O' and the digit '0'.