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In Eisenbud's Commutative Algebra, Chapter 3.9, it says,

"If $P$ is a maximal ideal of $R$ and $I$ is any proper ideal containing a power of $P$, then $I$ is $P$-primary: For in this case $P$ is the only prime containing the annihilator $I$ of $R/I$."

If the power $n$ is 1, it's trivial by definition, but for $n>1$ I don't see why this is true. First why $P^n\subseteq I$ implies $I\subseteq P$? And since $R$ is not necessarily local ring, why it's not possible that there is another maximal ideal $Q$ so that $P^n\subseteq I \subseteq Q$ for $n>1$?

Bamqf
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2 Answers2

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The answer to your first and second question are the same. $I$ must be contained in some maximal ideal $Q$, so $P^n\subseteq Q$. Suppose $x\in P\setminus Q$, and note that $x^n\in Q$. By primality of $Q$, we have either $x^{n-1}\in Q$ or $x\in Q$. Proceeding by induction shows that $x\in Q$, thus no such $x$ exists, so we must have $P=Q$.

Alex Becker
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To answer your first question, take radicals: $\sqrt{P^n} \subseteq \sqrt{I}$, but $\sqrt{P^n} = P$ is a maximal ideal. Therefore, $P \subseteq \sqrt{I}$, but again by the maximality of $P$, it must be the case that $P = \sqrt{I}$. So $I$ is $P$-primary.

To answer your second question, you should again think about taking radicals of the inclusion $P^n \subseteq I \subseteq Q$, and see what goes wrong if $P \neq Q$.

David Mehrle
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