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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 $uf + vg \in k[x]\backslash 0$?

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

Thanks

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What is your definition of "coprime" as applied to elements of a (general? polynomial?) ring? – Henning Makholm Feb 10 '12 at 1:15
@HenningMakholm: I mean they have no common factors. – Matt Feb 10 '12 at 1:17
Matt, are you looking at dpmms.cam.ac.uk/study/II/AlgebraicGeometry/2009-2010/… by the way? – hilbert Feb 10 '12 at 1:23
@hilbert: Yes, I am. I've done the second part of the question but my brain seems to have packed up and left. – Matt Feb 10 '12 at 1:26
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@Benjamin: I mean polynomial rings over fields. I'll be more precise. – Qiaochu Yuan Feb 10 '12 at 3:40
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1 Answer

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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 $uf+vg\in k[x]$.

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Side comment but how does one know that $k(x)[y]$ is an Euclidean domain? – BenjaLim Feb 10 '12 at 1:41
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@BenjaminLim $k(x)$ is a field. And for any field $F$, we have $F[x]$ is a euclidean domain. – Dimitri Surinx Feb 10 '12 at 1:55
@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 $k(x)[y]$? – BenjaLim Feb 10 '12 at 2:03
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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 2:11
@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 $A = k[x]$. – BenjaLim Feb 10 '12 at 2:20

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