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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

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    What is your definition of "coprime" as applied to elements of a (general? polynomial?) ring?2012-02-10
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    @HenningMakholm: I mean they have no common factors.2012-02-10
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    Matt, are you looking at http://www.dpmms.cam.ac.uk/study/II/AlgebraicGeometry/2009-2010/ag2009_ex1.pdf by the way?2012-02-10
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    @hilbert: Yes, I am. I've done the second part of the question but my brain seems to have packed up and left.2012-02-10
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    It happens sometimes :-)2012-02-10
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    @Henning: all polynomial rings (over fields) are UFDs, so "coprime" can be taken to have the usual meaning. (Matt, for future reference, because of the existence of rings which are not UFDs, "coprime" sometimes means "generates the unit ideal" and the problem is trivial with this definition.)2012-02-10
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    @QiaochuYuan Thanks.2012-02-10
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    @QiaochuYuan What do you mean by "all" polynomial rings are UFDs? I know that if $R$ is a UFD then $R[x]$ is.2012-02-10
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    @Benjamin: I mean polynomial rings over fields. I'll be more precise.2012-02-10
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    @QiaochuYuan Thanks for clarifying.2012-02-10

1 Answers 1

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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?2012-02-10
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    @BenjaminLim $k(x)$ is a field. And for any field $F$, we have $F[x]$ is a euclidean domain.2012-02-10
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    @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 $k(x)[y]$?2012-02-10
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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 $k[x, y]$.2012-02-10
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    @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 $A = k[x]$.2012-02-10