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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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user138864
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  • Summed up with mathematical terminology, your question can be phrased as “if $k=\overline{k}$, prove that $k[x_1,\dots,x_n]$ is a Jacobson ring using the Nullstellensatz.” (Note that $k[x_1,\dots,x_n]$ is always a Jacobson ring regardless of whether $k$ is algebraically closed or not, by the first theorem here; but an easier proof is achieved assuming algebraic closedness and by means of the Nullstellensatz.) – Elías Guisado Villalgordo Mar 02 '23 at 16:55

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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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    $\def\frm{\mathfrak{m}}$The containment $\bigcap_{\frm\supset I}\frm\subset I(V(I))$ is always true regardless of whether $k$ is algebraically closed or not. So for an arbitrary field $k$, it holds $$ \sqrt{I}\subset\bigcap_{\frm\supset I}\frm\subset I(V(I)). $$ The conclusion follows then by assuming $k=\overline{k}$ and applying the Nullstellensatz. – Elías Guisado Villalgordo Mar 02 '23 at 17:02
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    $\def\frm{\mathfrak{m}}$Here is how one sees that $\bigcap_{\frm\supset I}\frm\subset I(V(I))$ for any field $k$: Pick $a\in V(I)$. Then $I\subset I(V(I))\subset I({a})$, but the last ideal equals $(x_1-a_1,\dots,x_n-a_n)$, which is maximal (prove this!). Thus $$ \bigcap_{\frm\supset I}\frm\subset\bigcap_{a\in V(I)}I({a})=I\left(\bigcup_{a\in V(I)}{a}\right)=I(V(I)). $$ – Elías Guisado Villalgordo Mar 02 '23 at 17:21