Ideal in polynomial ring which contains no non-zero prime ideal

Let $$J$$ be a non-zero ideal in $$\mathbb C[X,Y]$$ such that $$J$$ contains no non-zero prime ideal. Then is it true that $$J$$ has height $$1$$ ?

Possible approach: Since $$\mathrm{ht}(J^n)=\mathrm{ht}(J)$$ for every $$n>1$$ so $$\mathrm{ht}(J)=1$$ iff $$\mathrm{ht}(J^n)=1$$ iff $$J^n$$ is contained in a proper principal ideal ... don't know where to go from here.

For motivation see my comments to this question When the element-wise product of two ideals produces an ideal.

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Feb 13 at 14:47

I'd like to you acknowledge some help from Jenna Tarasova.

I'm going to prove the contrapositive statement:

(1) If $$J$$ has height 2, then it contains an irreducible polynomial $$f$$.

To start with, I'm going to reduce to the case of monomial ideals. Specifically, I'm going to show that (1) is implied by:

(2) If $$J$$ is a height 2 monomial ideal and $$w\in \Bbb N^2$$ is generic, then $$J$$ contains an irreducible polynomial $$f$$ which is also $$w$$-homogeneous.

Recall that $$w$$-homogeneous (of degree $$d$$) means that $$f(t^{w_1}X, t^{w_2}Y) = t^d f(X,Y)$$ for all $$t\neq0$$; equivalently, each monomial in $$f$$ has the same $$w$$-degree, where the $$w$$-degree of $$X^iY^j$$ is defined to be $$w_1i + w_2j$$.

Proof that (2) implies (1). If $$g$$ is a nonzero polynomial, let $$\operatorname{in}_w(g)$$ denote its initial form with respect to the weight $$w$$. By definition, this means that if $$g=\sum_{i,j} a_{ij} X^iY^j$$ has $$w$$-order $$d$$ (i.e. $$d = \max\{w_1i+w_2j : a_{ij}\neq 0\}$$), then $$\operatorname{in}_w(g) = \sum_{w_1i+w_2j=d} a_{ij} X^iY^j.$$ Define $$\operatorname{in}_w(J) = (\operatorname{in}_w(g) : g\in J)$$.

Basic Groebner basis theory implies that $$\operatorname{in}_w(J)$$ is a monomial ideal. Now use the following two facts (I encourage you to prove the second, also the first if you know Groebner basis things): (a) If $$g\in \operatorname{in}_w(J)$$ is $$w$$-homogeneous, then there exists an $$f\in J$$ with $$\operatorname{in}_w(f) = g$$; and (b) if $$\operatorname{in}_w(f)$$ is irreducible, then so is $$f$$. $$\quad\Box$$

Now that I've reduced us to (2), I'm going to reduce things even further to the following statement:

(3) Let $$n$$ be a positive integer, and let $$w\in \Bbb N^2$$ be generic. Then the ideal $$(X^n, Y^n)$$ contains an irreducible polynomial $$f$$ which is also $$w$$-homogeneous.

Proof that (3) implies (2). It suffices to show that a height $$2$$ monomial ideal $$J$$ contains some $$(X^n, Y^n)$$. Choose a (monomial) generating set $$m_1,\ldots,m_s$$ of $$J$$. If every $$m_i$$ is divisible by $$X$$, then $$J$$ is contained in $$(X)$$ and therefore has height at most $$1$$, a contradiction. Therefore, at least one of $$m_1,\ldots,m_s$$ is not divisible by $$X$$. Similarly, at least one of them is not divisible by $$Y$$. In other words, since the $$m_i$$'s are all nonconstant monomials, $$J$$ contains $$X^a$$ and $$Y^b$$ for some positive integers $$a,b$$. Choosing $$n\geq \max\{a,b\}$$, we get that $$J\ni X^n,Y^n$$, as claimed. $$\quad \Box$$

Finally, let's prove (3).

Proof of (3). Consider the $$w$$-homogeneous polynomial $$f = X^{w_2} + Y^{w_1}$$. By genericity, we may assume that $$w_1,w_2\geq n$$, so that $$f\in J$$. Also by genericity, we may assume that $$w_1,w_2$$ are distinct primes. (Suppose not. Then there exists a nonzero polynomial $$g(x,y)\in \Bbb C[x,y]$$ such that for all primes $$p\neq q$$, $$g(p,q)=0$$. Then every prime is a root of $$(y-p)g(p,y)$$, contradicting the fact that there are infinitely many primes.)

So I don't have to keep writing the subscripts, let's set $$p=w_1$$ and $$q=w_2$$, so that $$f=X^q + Y^p$$. This polynomial is irreducible: Thinking of $$f$$ as having coefficients in $$\Bbb C(Y)$$, let $$t$$ be a root of $$f$$ in some algebraic closure $$K$$ of $$\Bbb C(Y)$$. Then the roots of $$f$$ in $$K$$ are $$t, \zeta t, \ldots, \zeta^{p-1} t$$, where $$\zeta$$ is a primitive $$p$$th root of unity. But $$\zeta\in \Bbb C\subseteq \Bbb C(Y)$$, so for each $$k=0,\ldots,p-1$$, the map $$\alpha \mapsto \zeta^k \alpha$$ is an automorphism of the field extension $$\Bbb C(Y)(t)/\Bbb C(Y)$$ taking $$t$$ to $$\zeta^k\alpha$$. Thus, as $$t$$ is not itself in $$\Bbb C(Y)$$, we get that $$f$$ is irreducible (Dummit and Foote, Prop 14.2). $$\quad \Box$$

Remark

The same proof, with minor changes, works for the following generalization to any field and any number of variables:

Theorem. Let $$k[X_1,\ldots,X_n]$$ be a polynomial ring over a field $$k$$. If $$J$$ is a nonzero ideal in $$k[X_1,\ldots,X_n]$$ which contains no non-zero prime ideal, then $$\operatorname{ht}J=1$$.

• I will take time to carefully read the proof : some basic questions: what do you mean by generic element of $\mathbb N^2$ here ? That would clarify some things for me as in the usual sense, no homogeneous polynomial in two variable , of degree $>1$ is irreducible ... also, $f_1(X_1)+f_2(X_2)$ is always irreducible in $\mathbb C[X_1,X_2]$ as long as $f_1,f_2$ has co-prime degree .... I mentioned this in my comments to the question , so that takes care of your claim $(3)$ I guess after your reduction ? – user521337 Feb 11 at 22:15
• @user521337 The use of “generic” here means that there is a nonempty Zariski open subset $U$ of $\Bbb C^2$ such that the property holds for all $w\in U\cap \Bbb N^2$. (In fact, in this case, $U$ may be chosen to be the complement of a finite union of hyperplanes) – Avi Steiner Feb 11 at 22:31
• @user521337 also, I actually didn’t see your newer comments until just now. What you’re doing is indeed similar to what I’ve done here. – Avi Steiner Feb 11 at 22:34
• @user26857 You can still post your solution – Avi Steiner Feb 12 at 20:19
• Please give more details or a reference for the claim that the initial form ideal is monomial. – user26857 Feb 13 at 6:03