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Suppose $P$ is a nontrivial prime ideal in an arbitrary ring. Can $P^2 = P?$ I suppose this might be true for some ring but I cannot think of an example...

Context: In a Dedekind domain, $P^2 \neq P$ as otherwise this would violate unique prime factorization of ideals. This is why I was wondering if equality could actually hold for some ring.

Watson
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green frog
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    Consider the valuation ring of $\mathbb{C}_p$, the completion of algebraic closure of $p$-adic number. This is in the same spirit as the answer below. – pisco Nov 01 '18 at 09:37
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    Related : https://math.stackexchange.com/questions/348789/can-the-square-of-a-proper-ideal-be-equal-to-the-ideal, https://math.stackexchange.com/questions/2820499/intersection-of-powers-of-prime-ideals-in-the-algebraic-integers – Watson Nov 01 '18 at 09:47
  • Typically, any prime in the ring of algebraic integers works (see second link above). – Watson Nov 01 '18 at 10:14
  • fyi: finitely generated idempotent ideals of commutative rings are principal. – Bill Dubuque Nov 01 '18 at 14:25

1 Answers1

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Here's a commutative ring with a non-zero maximal ideal $M$ with $M^2=M$. Let $k$ be a field and $R$ consist of "polynomials with rational exponents" over $k$, that is each element of $R$ is a formal $k$-linear combination of $x^r$ where $r\in\Bbb Q$ and $r\ge0$. The map $R\to k$ taking $a_0+\sum_{r>0}a_r x^r$ to $a_0$ is a surjective homomorphism with kernel $M$ generated by the $x^r$ for $r>0$. Each of these is a square of an element of $M$, so $M^2=M$.

Of course, this $R$ is non-Noetherian.

Angina Seng
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