17
$\begingroup$

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

$\endgroup$
2
  • 4
    $\begingroup$ The wikipedia page mentioned in the question contains a counterexample. $\endgroup$ – lhf Dec 22 '11 at 15:27
  • $\begingroup$ statement holds for maximal ideal not general prime ideal. $\endgroup$ – love_sodam Feb 26 at 10:58
18
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $(x^3,x^2y)<F[x,y]$, 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.

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.