This question asks: If $f,g \in k[x,y]$ are two algebraically dependent polynomials over an arbitrary field $k$, is it true that there exists a polynomial $h \in k[x,y]$ such that $f,g \in k[h]$, namely, $f=u(h)$ and $g=v(h)$ for some $u(t),v(t) \in k[t]$; the answer is positive.

Is it possible to replace the field $k$ by an integral domain $D$? Namely: If $f,g \in D[x,y]$ are two algebraically dependent polynomials over an arbitrary integral domain $D$, is it true that there exists a polynomial $h \in D[x,y]$ such that $f,g \in D[h]$?

Denote the field of fractions of $D$ by $Q(D)$. It is clear that if $f,g \in D[x,y] \subset Q(D)[x,y]$ are two algebraically dependent polynomials over $D$, then from the above question there exists a polynomial $h \in Q(D)[x,y]$ such that $f,g \in Q(D)[h]$, namely, $f=u(h)$ and $g=v(h)$ for some $u(t),v(t) \in Q(D)[t]$.

I do not see why, for example, $D[x][y] \ni f=u(h)=u_mh^m+\cdots+u_1h+u_0$ should imply that $h \in D[x,y]$ and $u_j \in D$ (changing variables does not seem to help, namely if the leading term is $cy^l$, with $c \in Q(D)$).

Any comments are welcome.

  • $\begingroup$ I guess that there exists also a counterexample with $D=k[t^2,t^3]$, the simplest non-normal integral domain. It seems that $f=(tx+ty)^2=t^2(x+y)^2 \in D[x,y]$ and $g=(tx+ty)^3=t^3(x+y)^3 \in D[x,y]$ is a counterexample. $\endgroup$ – user237522 Nov 1 '18 at 0:37
  • $\begingroup$ Another example in $k[t^2,t^3][x,y]$: $f=t^2x+t^3y$, $g=t^3x+t^4y$. We have, $\operatorname{Jac}(f,g)=\operatorname{Jac}(t^2x+t^3y,t^3x+t^4y)=t^2t^4-t^3t^3=0$. Notice that $g=tf$, so there is no $h \in k[t^2,t^3][x,y]$ such that $f,g \in k[t^2,t^3][h]$. $\endgroup$ – user237522 Nov 1 '18 at 17:08

No. Choose a field $k$ and $D=k[u^2,u^3,v^2,v^3,uv]\subset k[u,v]$, so $D$ is a noetherian domain. In $D[x,y]$, choose $f=(ux+vy)^2$ and $g=(ux+vy)^3$; they are clearly algebraically dependent (but $ux+vy\notin D[x,y]$). Write $K=k(u,v)=\mathrm{Frac}(D)$.

Claim: there is no $P\in D[x,y]$ such that $f,g\in D[P]$.

By contradiction, let $P\in D[x,y]$ such that $f,g\in D[P]$. So $f=r_1(P)$, $g=r_2(P)$ with $r_1,r_2\in D[t]$. Since $f^3=g^2$ and $K[t]$ is a UFD, there exists $r\in K[t]$ such that $r_1=r^2$ and $r_2=r^3$. So $r(P)^2=f$. So $r(P)=\pm (ux+vy)$; up to change $(u,v)$ to $(-u,-v)$, let us suppose $r(P)=ux+vy$. This implies that $r\in K[t]$ and $P\in K[x,y]$ have degree 1. Write $r=at+b$ and $P=cx+dy+e$; then $r(P)=acx+ady+ae+b=ux+vy$. So $ac=u$, $ad=v$, $ae+b=0$.

So $c,d$ are nonzero; then $c/d=u/v$, hence $cv=du$. Since $c,d\in k[u,v]$, we can write $c=uq$ and $d=vq$ with $q\in k[u,v]$. We have $aq=1$. Since $r^2\in D[t]$, we have $a^2\in D\subset k[u,v]$. So $a^2$ is invertible in $k[u,v]$, and hence $a\in k^*$. So $u=a^{-1}c\in D$, a contradiction.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much! What if $D$ is an integral domain, for example $D=\mathbb{Z}$? $\endgroup$ – user237522 Dec 3 '17 at 13:58
  • $\begingroup$ @user237522 my $D$ is an integral domain... $\endgroup$ – YCor Dec 3 '17 at 14:03
  • $\begingroup$ Oh sorry I meant a UFD. $\endgroup$ – user237522 Dec 3 '17 at 14:04
  • $\begingroup$ Still my $D$ is indeed not a normal domain, which sounds like a natural requirement to look at. UFD is a stronger requirement, and PID even stronger. Non-normality sounds central in my argument... $\endgroup$ – YCor Dec 3 '17 at 14:05
  • $\begingroup$ I even do not mind to assume in my question that $D=\mathbb{Z}$, the simplest UFD I know; can you find a counterexample for $\mathbb{Z}$? $\endgroup$ – user237522 Dec 3 '17 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.