The Stacks project

Lemma 37.35.1. Let $R = \mathop{\mathrm{colim}}\nolimits R_ i$ be colimit of a directed system of rings whose transition maps are faithfully flat. Then the number of minimal primes of $R$ taken as an element of $\{ 0, 1, 2, \ldots , \infty \} $ is the supremum of the numbers of minimal primes of the $R_ i$.

Proof. If $A \to B$ is a flat ring map, then $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ maps minimal primes to minimal primes by going down (Algebra, Lemma 10.39.19). If $A \to B$ is faithfully flat, then every minimal prime is the image of a minimal prime (by Algebra, Lemma 10.39.16 and 10.30.7). Hence the number of minimal primes of $R_ i$ is $\geq $ the number of minimal primes of $R_{i'}$ if $i \leq i'$. By Algebra, Lemma 10.39.20 each of the maps $R_ i \to R$ is faithfully flat and we also see that the number of minimal primes of $R$ is $\geq $ the number of minimal primes of $R_ i$. Finally, suppose that $\mathfrak q_1, \ldots , \mathfrak q_ n$ are pairwise distinct minimal primes of $R$. Then we can find an $i$ such that $R_ i \cap \mathfrak q_1, \ldots , R_ i \cap \mathfrak q_ n$ are pairwise distinct (as sets and hence as prime ideals). This implies the lemma. $\square$


Comments (0)


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Your email address will not be published.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 0CB3. Beware of the difference between the letter 'O' and the digit '0'.