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Let $k$ be a field. Given $f, g \in k[x,y]$ coprime, why can we find $u,v \in k[x,y]$ such that $uf + vg \in k[x]\setminus\{0\}$?

I can do it for specific polynomials, but I'm struggling to structure a coherent proof. Any hints would be greatly appreciated!

Thanks

user26857
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Matt
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1 Answers1

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$k(x)[y]$ is a Euclidean domain, hence if $f,g$ are coprime in $k[x,y]$ they are coprime in $k(x)[y]$ and there are rational functions $U,V\in k(x)[y]$ such that $Uf+Vg=1$. Now multiply the denominators to get $uf+vg\in k[x]$.

Bonanza
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  • Side comment but how does one know that $k(x)[y]$ is an Euclidean domain? –  Feb 10 '12 at 01:41
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    @BenjaminLim $k(x)$ is a field. And for any field $F$, we have $F[x]$ is a euclidean domain. – sxd Feb 10 '12 at 01:55
  • @DimitriSurinx Ok sorry I missed that $k(x)$ is the fraction field of $k[x]$. However is hilbert claiming that the fraction field of $k[x,y]$ is $k(x)[y]$? –  Feb 10 '12 at 02:03
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    @Benjamin: no. hilbert is merely first working in $k(x)[y]$ and then observing that he can multiply by a common factor to move to $k[x, y]$. – Qiaochu Yuan Feb 10 '12 at 02:11
  • @QiaochuYuan Ok I get it: Write $k[x,y]$ as $k[x][y]$. Then we already know that if two polynomials are coprime in a $A[y]$, $A$ a PID then they are coprime in $\operatorname{Frac}(A)[y]$. The result follows by setting $A = k[x]$. –  Feb 10 '12 at 02:20