Is there any example of a $m$-primary ideal $I$ in a noetherian local domain $(R, m)$ such that $I^2=mI\not=m^2 $?
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$\begingroup$ A related post on MathOverflow: Square of primary ideals. $\endgroup$– Martin SleziakApr 25, 2019 at 16:11
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Here's an example: Take $R = \mathbb{Q}[x,y,z,w]_{(x,y,z,w)}/(x^2-zw, z^2-yw, y^3-xw, w^3-xy^2z)$. One can check (e.g. in Macaulay2) that $R$ is a domain (with $\dim R = 1$). Let $m = (x,y,z,w)$ be the maximal ideal, and set $I = (x,y,z)$. Then the first $3$ relations in $R$ guarantee that $I^2 = mI$, while the last relation gives that $w^3 \in I^2$, so $m^3 \subseteq I$ and $I$ is $m$-primary, but in fact $w^2 \in m^2 \setminus I^2$.
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$\begingroup$ Very interesting example. One also sees $m^3 = m^2I$. How did you come up with the example? $\endgroup$– YoungsuJun 29, 2014 at 8:23
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$\begingroup$ @Youngsu: I just played around in Macaulay2 for a bit. It was natural to try an ideal $I$ which had one less generator than $m$, and then take a minimal prime over the ideal obtained by forcing products of that generator with the ones in $I$ to be in $I^2$. The key point is to ensure that $m^2 \ne mI$ after quotienting, which happened to work in this example because of the exponent $3$ $\endgroup$– zcnJun 29, 2014 at 8:30