# Is each power of a prime ideal a primary ideal?

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

• The wikipedia page mentioned in the question contains a counterexample. – lhf Dec 22 '11 at 15:27
• statement holds for maximal ideal not general prime ideal. – love_sodam Feb 26 at 10:58

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.
Alex Becker answered this question completely, but I would like to give another example, which shows the failure can be even more acute: there in fact exist principal prime ideals with powers that are not primary. Indeed, consider the ideal $$P=(\overline{x})$$ in the ring $$R:=F[x,y]\big/(x^2y)$$, where $$F$$ is your favorite field. $$P$$ is prime, since $$(x)$$ is a prime ideal of $$F[x,y]$$ containing $$(x^2y)$$. However, we claim the ideal $$I=P^3=(\overline{x^3})$$ is not primary. To see this, let $$a=\overline{x^2}$$ and $$b=\overline{y}$$. Then $$ab=\overline{0}$$ and thus lies in $$I$$. Note that $$a\notin I$$, since $$x^2$$ does not lie in the ideal $$(x^3,x^2y), which is the preimage of $$I$$ under the projection map $$F[x,y]\to R$$. (Every non-zero element of that ideal has degree at least $$3$$.) However, we also have $$b\notin\sqrt{I}=P$$, since $$y\notin (x)$$ in $$F[x,y]$$. Thus $$I$$ is a power of a principal prime ideal, but not primary, and so gives the desired counterexample. I find this behavior somewhat surprising (for instance, it cannot occur in integral domains), which I've why I've included this example.
Exercise: Show that, if $$R$$ is an integral domain, then every power of a principal prime ideal of $$R$$ is primary.
One may even find such an example in the polynomial ring $$k[x,y,z]$$: the ideal $$I=\langle x^3-yz, y^2-xz, z^2-x^2y\rangle$$ is prime but $$I^2$$ is not primary as explained in Northcott, Ideal Theory, Example 3, p. 29.