# Is there an example of a power of prime ideal in a polynomial ring that is not primary?

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?

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.
• 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. – user25216 Oct 28 '19 at 15:39