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I want to show that each power of a prime ideal is a primary ideal or I have to think about a counterexample?

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    $\begingroup$ The wikipedia page mentioned in the question contains a counterexample. $\endgroup$ – lhf Dec 22 '11 at 15:27
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Just to clear this from the list of unanswered questions, here's a fleshed-out version of the counterexample on Wikipedia:

Consider the ideal $P=(x,z)$ in $k[x,y,z]/(xy-z^2)$. I will denote equivalence of elements in this ring by $\equiv$ and equivalence in the ring $k[x,y,z]$ by $=$. $P$ is prime, since $$\frac{k[x,y,z]/(xy-z^2)}{(x,z)}\cong \frac{k[x,y,z]}{(x,z,xy-z^2)}\cong k[y]$$ is an integral domain. However, $P^2=(x^2,xz,z^2)$ contains $xy\equiv z^2$ but does not contain $x$, as if $$x=fx^2+gxz+hz^2+p(xy-z^2)$$ then setting $x=0$ shows that $h-p=qx$ so $$x=fx^2+gxz+pxy+qxz^2=(fx+gz+py+qxz)x\implies fx+gz+py+qxz=1$$ which can be see to be impossible by evaluating at $x=y=z=0$. It also does not contain $y^n$ for any $n$, as if $$y^n=fx^2+gxz+hz^2+p(xy-z^2)$$ then setting $x=z=0$ gives a contradiction. Thus $P^2$ is a power of a prime which is not primary.

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