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Let $(A,\mathfrak{m})$ be an Artinian local ring. Is true that the only primary ideals of $A$ are the powers of $\mathfrak{m}$?

I'm not able to prove this or to disprove it by counterexample. Can you help me?

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Any proper ideal in a local Artinian ring is primary, this is because the radical of an ideal is the intersection of the prime ideals containing it, but in an Artinian local ring, there is only one prime ideal, so the radical of each ideal is maximal, hence the ideal is primary.
A different way to see this is to use another characterization of primary ideals: An ideal $I$ in a ring $A$ is primary iff every zero divisor of $A/I$ is nilpotent. Since quotients of local Artinian rings are again local Artinian, it suffices to show that local Artinian rings have the property that every zero divisor is nilpotent.
We can show even more: every non-unit is contained in the maximal ideal which is nilpotent because the ring is Artinian, so every element is either a unit or nilpotent.

So it suffices to give an example of a local Artinian ring with an ideal that is not the power of a maximal ideal. Consider $R=K[x,y]/(x^3,x^2y,y^2)$ for a field $K$, this is an Artinian local ring with the unique prime ideal $(x,y)$. We have $$(x,y)^2=(x^2,xy,y^2)=(x^2,xy)$$ $$(x,y)^3=(x^3,x^2y,xy^2,y^3)=0$$ Note that $(xy)$ is not equal to any of those powers of the maximal ideal.

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