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I would like to ask about the relation $$\sqrt{I} \subseteq \bigcap_{\text{maximal } m \supseteq I} m$$

regarding ring $k[X_1,\dots,X_n]$, where $k$ is an arbitrary field and $I$ is a proper ideal in this polynomial ring. Based on the answer to this question I assume that should always hold. However I don't see why for an arbitrary maximal ideal $m \supseteq I$ one would obtain inclusion $\sqrt{I}\subseteq m.$

Thus, why $\sqrt{I}\subseteq m$ for arbitrary maximal ideal $m \supseteq I$?

Fallen Apart
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    It follows from the fact that $m$ is prime. – Mark Saving Dec 15 '21 at 17:05
  • ... and from the definition of the radical of an ideal. One property of the radical of an ideal is that it is the intersection of all prime ideals that contain it. The intersection of all maximal ideals that contain it is called the Jacobson radical. – hardmath Dec 15 '21 at 17:11
  • @MarkSaving Yeah, that will do. Thanks :D – Fallen Apart Dec 15 '21 at 17:22

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