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I want to find the primary decomposition of $(x^2, xy^2)$ as an ideal of $k[x,y,z]$ where $k$ is some field. My guess is $(x^2, xy^2) = (x) \cap (x^2, y^2)$ however I am not 100% certain if $(x^2, y^2)$ is a primary ideal.

My approach to see this was to use the fact that $I$ is primary iff all zero divisors of $R/I$ are nilpotent. I argued that if we have some $p(x,y,z), q(x,y,z)$ whose images in $k[x,y,z]/(x^2,y^2)$ are zero divisors, then $p$ and $q$ can't have any $z$ terms (their product must be in $(x^2,y^2)$) or any constant terms either. Then any zero divisor is of the form $ax+by + cxy$ which is nilpotent since raising to a high enough power and expanding via the binomial theorem, every term will have at least $x^2$ or $y^2$ in it. Something about this seems too strong to me. This reasoning would prove that in $k[x,y]$ any ideal that contains $x^k$ and $y^l$ for some $k$ and $l$ is a primary ideal. Is that true?

Craig
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    Btw, why the primary decomposition? Can you find only one? (I suggest you to try to find at least another one.) –  Jul 05 '13 at 14:42
  • @YACP I struggled to find that one in fact. Is there some way to see these things that I haven't learned? – Craig Jul 05 '13 at 14:47
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    There are some tricks to find primary decompositions of monomial ideals, but in this case you can modify easily the above decomposition to find another one. –  Jul 05 '13 at 14:49
  • So what's your $R$? $k[x,y]$? – Bach Feb 29 '20 at 21:13

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The radical of ideal $(x^2,y^2)$ is $(x,y)$ which is maximal and this is enough to conclude that $(x^2,y^2)$ is primary. (This shows that any proper ideal containing $x^k$ and $y^l$ is primary, since its radical must contain the radical of $(x^k,y^l)$ which is also $(x,y)$.)