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According to this question (or rather answer), a monomial ideal of a polynomial ring is primary iff every variable appearing in it has some power in $I$.

So for example the ideals $(y),(x^3,y^4)$ of $k[x,y]$ are ($P_i$-)primary for some prime ideals $P_i$.

That is, $Ass(k[x,y]/(y))=\{P_1\}$ and $Ass(k[x,y]/(x^3,y^4))=\{P_2\}$.

I looks clear for me that $(y)=Ann_{k[x,y]/(y)}(1)$. So $P_1=(y)$.

But how to find $P_2$? Is it $(x,y)$? If so, how to find an element of the quotient whose annihilator equals $(x,y)$? Is there a general procedure?

user557
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$P_2$ must be the radical of $(x^3, y^4)$, so $P_2 = (x,y)$. The annihilator of $x^2 y^3$ in the quotient $k[x,y]/(x^3,y^4)$ is $(x,y)$. You should be able to generalize this.

Ted
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  • Probably I don't know enough theory; is it always the case that an ideal $I$ from a primary decomposition is $rad(I)$-primary? – user557 Apr 02 '19 at 03:04
  • Yes, if $I$ is a primary ideal, then it is always $rad(I)$-primary. – Ted Apr 05 '19 at 15:39