The question may have been addressed in Math SE before; but my question is a little variation of it, coming because of a very common example for this question.

It is well known that powers of a prime ideal need not be primary and a standard example is to consider the quotient of $k[x,y,z]$ by the ideal $(xy-z^2)$, and in the quotient ring, take the prime ideal $(\bar{x},\bar{z})$; its square is not primary.

Now this example of prime ideal is in quotient of polynomial ring.

Q. Can we construct an example of a prime ideal within a polynomial ring, say $k[x,y]$ or $k[x,y,z]$, whose some power is not primary?


The following example is taken from Northcott, Ideal Theory.

Let $R=k[x,y,z]$ and $\mathfrak{p}=(f_1,f_2,f_3)$, where $f_1=y^2-xz$, $f_2=yz-x^3$, $f_3=z^2-x^2y$.
To see that $\mathfrak{p}$ is prime, consider $R/\mathfrak{p}$ as a $k[z]$-module. Since in this quotient we have the relations $y^2=xz, yz=x^3,z^2=x^2y$, a generating system for $R/\mathfrak{p}$ as a $k[z]$-module is given by $\{1,x,y,xy,x^2\}$. It follows that any $f \in k[x,y,z]$ can be written as $$f(x,y,z)=x^2A(z)+xyB(z)+xC(z)+yD(z)+E(z)+g(x,y,z),$$ where $A,B,C,D,E \in k[z]$ and $g(x,y,z) \in \mathfrak{p}$.
Now consider the ring homomorphism $\varphi:k[x,y,z] \to k[t], f(x,y,z) \mapsto f(t^3,t^4,t^5)$. We get that $\mathfrak{p} \subseteq \operatorname{ker}(\varphi)$. If $\varphi(f)=0$, then writing $f$ in the form as above, we get $$f(t^3,t^4,t^5)=t^6A(t^5)+t^7B(t^5)+t^3C(t^5)+t^4D(t^5)+E(t^5)=0.$$ This implies that $A=B=C=D=E=0$, because the degree $d$ of every term in $t^6A(t^5)$ satisfies $d \equiv 1 \pmod{5}$, terms in $t^7B(t^5)$ have degree $d$ with $d \equiv 2 \pmod{5}$, terms in $t^3C(t^5)$ have degree $d$ with $ d \equiv 3 \pmod{5}$ etc., so there can't be any cancellation. This shows that $f(x,y,z)=g(x,y,z) \in \mathfrak{p}$. So $\mathfrak{p}=\operatorname{ker}(\varphi)$ is a prime ideal.

Now, we show that $\mathfrak{p}^2$ is not primary. If $\mathfrak{p}^2$ was primary, it would be $\mathfrak{p}$-primary. $\mathfrak{p}^2$ contains $$f_2^2-f_1f_3=(yz-x^3)^2-(y^2-xz)(z^2-x^2y)=x (x^5 - 3 x^2 y z + x y^3 + z^3),$$ as $x$ is not in $\mathfrak{p}$, so if $\mathfrak{p}^2$ was $\mathfrak{p}$-primary it would contain $x^5-3x^2yz+xy^3+z^3$, but every polynomial in $\mathfrak{p}^2=(y^2-xz,yz-x^3,z^2-x^2y)^2$ contains no terms of degree less than $4$, so this is not the case.

  • 1
    $\begingroup$ Note that 3 is the smallest dimension for a polynomial ring over a field to have such an example, since in $k[x, y]$ every prime ideal is either maximal or principal. $\endgroup$ – user25216 Oct 28 '19 at 15:39

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.