# 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 '19 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 '19 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 '19 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 '19 at 22:34
• Please give more details or a reference for the claim that the initial form ideal is monomial. – user26857 Feb 13 '19 at 6:03
• @user26857 This follows from Theorem 2.7 of Mora and Rabiano’s article “The Gröbner Fan of an Ideal”, which states (after spending way too much time deciphering their notation) that for any given term order $\preceq$, the set of all $w\in \Bbb R^n$ for which $\operatorname{in}_\preceq(J)=\operatorname{in}_w(J)$ is an open cone of dimension $n$. I think Sturmfels’ book “Gröbner Based and Convex Polytopes” also discusses this, but I don’t have access to it at the moment. – Avi Steiner Feb 14 '19 at 17:39