Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $I$ be a proper ideal in $k[x_1,....,x_n]$, where $k$ is an algebraically closed field. Show that $\sqrt{I}= \cap M$, where $M$ runs through all maximal ideals containing $I$.

I am confused by what we can say is in the intersection of all maximal ideals.

For the proof, since k is algebraically closed, I think I would want to use Nullstellensatz, but once I have that $\sqrt{I}=I(V(I))$, I am not sure what to do next.

share|cite|improve this question
up vote 7 down vote accepted

The inclusion $\displaystyle \sqrt{I} \subseteq \bigcap_{m \supseteq I \text{ maximal}} m$ always holds. Suppose $\displaystyle f \in \bigcap_{m \supseteq I \text{ maximal}} m$. Since $k$ is algebraically closed, the maximal ideals containing $I$ are precisely the ideals of the form $(x_1 - a_1, \ldots, x_n - a_n)$, where $(a_1, \ldots, a_n)$ are the points of $V(I) \subseteq k^n$. This says that $f$ vanishes at every point of $V(I)$, so $f \in I(V(I)) = \sqrt{I}$, by the Nullstellensatz.

This in fact still holds even if $k$ is not algebraically closed, and says that $k[x_1, \ldots, x_n]$ (for any field $k$) is a Jacobson ring.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.