# $f,g \in \mathbb{Z}[x,y]$ satisfying: $\operatorname{Jac}(f,g)=0$ and $f,g \notin \mathbb{Z}[h]$ for every $h \in \mathbb{Z}[x,y]$?

Is it possible to find $$f,g \in \mathbb{Z}[x,y]$$ (with $$\deg(f),\deg(g) \geq 1$$) such that the following two conditions are satisfied:

(1) $$\operatorname{Jac}(f,g)=f_xg_y-f_yg_x = 0$$.

(2) There exist no $$h \in \mathbb{Z}[x,y]$$ such that $$f,g \in \mathbb{Z}[h]$$.

Please see the answer to this question, in which it is shown that, if in the above question we replace $$\mathbb{Z}$$ by some non-normal integral domain, then the answer is positive.

$$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$$No. As explained in this question, in $$\mathbb{Q}[x,y]$$, the condition $$\operatorname{Jac}(f,g)=0$$ implies that there exists an $$h \in \mathbb{Q}[x,y]$$ such that $$f$$ and $$g$$ are in $$\mathbb{Q}[h]$$. We now need some lemmas that are basically variants of Gauss's lemma, with multiplication replaced by composition.

Recall that a polynomial with coefficients in $$\ZZ$$ is called primitive if the set of its coefficients have $$GCD=1$$. I'll also define a polynomial to be very primitive if the GCD of the coefficients other than the constant term is $$1$$.

Lemma 1 Let $$a \in \ZZ[t]$$ be primitive and $$b \in \ZZ[x,y]$$ be very primitive. Then $$a \circ b$$ is primitive.

Proof: Suppose to the contrary that $$p$$ is a prime dividing every coefficient $$a \circ b$$. Let $$\bar{a}$$ and $$\bar{b}$$ denote the reductions modulo $$p$$, so these are polynomials in $$(\ZZ/p)[t]$$ and $$(\ZZ/p)[x,y]$$ respectively. The polynomial $$\bar{a}$$ is nonzero, $$\bar{b}$$ is not a constant, and $$\ZZ/p$$ is a field, so $$\bar{a} \circ \bar{b}$$ is nonzero. But the hypothesis is that $$a \circ b$$ is $$0$$ modulo $$p$$, and composition commutes with reduction modulo $$p$$. $$\square$$

Lemma 2: Let $$b \in \ZZ[x,y]$$ be very primitive, let $$c \in \QQ[t]$$ and suppose that $$c \circ b \in \ZZ[x,y]$$. Then $$c \in \ZZ[t]$$.

Proof: Write $$c(t) = \tfrac{p}{q} a(t)$$ with $$a$$ primitive and $$p$$ and $$q \in \ZZ$$ relatively prime. Then $$c(b(x,y,)) = \tfrac{p}{q} a(b(x,y))$$ and, by Lemma 1, $$a(b(x,y))$$ is primitive. So $$q$$ divides every coefficient of a primitive polynomial, and we deduce that $$q=1$$. So $$c(t) = p a(t) \in \ZZ[t]$$. $$\square$$

We now prove your result. Let $$Jac(f,g)=0$$. So there is $$h \in \QQ[x,y]$$ and $$a$$ and $$b \in \QQ[t]$$ such that $$f=a \circ h$$ and $$g = b \circ h$$. The case where $$h$$ is constant is clear, so we assume it is not.

Subtracting a constant from $$h$$, (and changing $$a$$ and $$b$$ appropriately) we may assume that the constant term of $$h$$ is $$0$$. Rescaling $$h$$ by an appropriate element of $$\QQ$$ (and rescaling the coefficients of $$a$$ and $$b$$ correspondingly), we may assume that $$h$$ is primitive. A primitive polynomial with constant term $$0$$ is very primitive. So Lemma 2 tells us that $$a$$ and $$b \in \ZZ[t]$$, and we are done. $$\square$$

This argument generalizes immediately to any UFD, and with a bit more work to any normal ring.

• Thank you very much! Seems exactly what I was looking for. Notice that in mathoverflow.net/questions/287610/… YCor mentions in one of his comments: "Non-normality sounds central in my argument... " and I mention: "I strongly suspect that there is no conterexample for $D=\mathbb{Z}$". – user237522 Oct 31 '18 at 17:00
• There is a small problem with lemma 1. Consider $b = 2x + 3$ and $a = t - 3$; then $a \circ b = 2x$ is not primitive. I think some version of it should work if you rule out constant coefficients, but you need to be a bit careful. – R. van Dobben de Bruyn Oct 31 '18 at 18:37
• @R.vanDobbendeBruyn Fixed now. Thanks for pointing out this issue! – David E Speyer Oct 31 '18 at 21:25
• @DavidESpeyer, Please, I would like to use your above answer in a paper of mine. Which you prefer: (1) Referring to your answer+acknowledge. (2) Co-authorship? – user237522 Nov 1 '18 at 11:46
• Refer and acknowledge. – David E Speyer Nov 1 '18 at 13:02