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.

Any comments are welcome!


$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$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.

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  • $\begingroup$ 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 $D=\mathbb{Z}$". $\endgroup$ – user237522 Oct 31 '18 at 17:00
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    $\begingroup$ 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. $\endgroup$ – R. van Dobben de Bruyn Oct 31 '18 at 18:37
  • $\begingroup$ @R.vanDobbendeBruyn Fixed now. Thanks for pointing out this issue! $\endgroup$ – David E Speyer Oct 31 '18 at 21:25
  • $\begingroup$ @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? $\endgroup$ – user237522 Nov 1 '18 at 11:46
  • $\begingroup$ Refer and acknowledge. $\endgroup$ – David E Speyer Nov 1 '18 at 13:02

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