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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.

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1 Answer 1

up vote 6 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.

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