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Let $R$ be a commutative ring with unity, and let $\mathfrak{p} \subset R$ be a prime ideal. If $ab \in \mathfrak{p}^2$, does one of the following hold?

  1. $a \in \mathfrak{p}^2$;
  2. $b \in \mathfrak{p}^2$;
  3. $a \in \mathfrak{p}$ and $b \in \mathfrak{p}$.

If this does not hold in generality, will it hold if $R$ is a domain? What if $R$ is Noetherian as well?

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1… – user26857 Feb 1 '13 at 21:01
@YACP I stared at the answer for that for far too long before recognizing it. – Alex Becker Feb 1 '13 at 21:07
up vote 3 down vote accepted

This is a classical example I have taken from Sharp's steps in commutative algebra. Let $k$ be a field and set $R=k[X,Y,Z]/(XY-Z^2)$ it can be verified with some computation that $XY-Z^2$ is irreducible so $R$ is an integral domain and its an image of a Noetherian ring, so also Noetherian. Let $x,y,z$ denote the images of $X,Y,Z$ in $R$. Then the ideal $\mathfrak p=(x,z)$ is prime because $R/\mathfrak p \cong K[y]$. I claim that $\mathfrak p^2$ is not primary in particular it does not satisfy any of your conditions. Then $xy=z^2 \in \mathfrak p^2$ but $y \notin \mathfrak p$ and $x,y \notin \mathfrak p^2$.

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2… – user26857 Feb 1 '13 at 20:59
It seems here that the problem is that powers of prime ideals are not always primary. This is true, however, if we specialize to maximal ideals, and I'm tempted to change the hypothesis of my question to add that $R$ have Krull dimension 1. This may be needlessly restrictive (or not restrictive enough). – A Walker Feb 1 '13 at 22:48

For an easier counter-example to 1) and 2):

Take $\mathfrak{m}^2=(x,y)^2=(x^2,xy,y^2) \subseteq k[x,y]$ and consider $xy \in \mathfrak{m}^2$.

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There are plenty of obvious counterexamples to 1) and 2), which together merely state that the square of a prime need not be a prime. For example, any prime ideal in the integers suffices. It's the third condition that makes things interesting. – Alex Becker Feb 1 '13 at 21:06

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