Let $f,g \in \mathbb{Q}[x,y]$ be coprime. I want to show that there are only finitely many common roots, i.e. only finitely many pairs $(a,b) \in \mathbb{C}^2$ with $f(a,b)=g(a,b)=0$. As a hint I should consider the gcd of $f,g$ over $\mathbb{Q}(x)[y]$.

If I could show that $gcd(f,g)=1$ considering $f, g$ in $\mathbb{Q}(x)[y]$ I can say that $res_y(f,g) \neq 0$. As $res_y(f,g) \in \mathbb{Q}(x)$ I have finitely many possibilities for the second component to get a root of $f$ and $g$. In the same way I can say that there are only finitely many possibilities for the first component by considering $\mathbb{Q}(y)[x]$ and the statement follows as all the common roots considered in $\mathbb{Q}[x,y]$ also have to be common roots in $\mathbb{Q}(x)[y]$.

My question is how I could show that $gcd(f,g)=1$ if I consider $f$ and $g$ in $\mathbb{Q}(x)[y]$? If I assume $gcd(f,g)=d \in \mathbb{Q}[x,y]$ (which I can as all non-zero polynomials $p \in \mathbb{Q}[x]$ are units in $\mathbb{Q}(x)[y]$), I get by definition $d \: | \: f, g$ in $\mathbb{Q}(x)[y]$. However I do not know how to get $d \: | \: f, g$ in $\mathbb{Q}[x,y]$.

  • $\begingroup$ $d|f$ in $\mathbb{Q}(x)[y]$ implies, $d|p(x)f$ in $\mathbb{Q}[x,y]$ for some $0\neq p(x)\in\mathbb{Q}[x]$. Can you finish the proof now? $\endgroup$ – Mohan Jan 29 '18 at 15:42
  • $\begingroup$ If we assume that $d(x,y)$ really depends on $y$, then $d(x,y) | f$, a contradiction (as the same holds for $g$ with the same $p(x)$ if we choose it right). So $d \in \mathbb{Q}[x]$ is a unit in $\mathbb{Q}(x)[y]$. However I used that d is a prime element in $\mathbb{Q}[x][y]$ which I am not sure about. $\endgroup$ – blablablup Jan 29 '18 at 15:56
  • $\begingroup$ If $d(x,y)$ did not depend on $y$, wouldn't it be a unit? $\endgroup$ – Mohan Jan 29 '18 at 16:55
  • $\begingroup$ Yes, but I don't really understand how this helps saying that $d(x,y) | p(x) f(x,y)$ implies $d(x,y) | f(x,y)$. Thanks for your help! $\endgroup$ – blablablup Jan 29 '18 at 17:17

Your Answer

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

Browse other questions tagged or ask your own question.