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Let $R$ be a UFD (so $R[x]$ is as well).

Given $f, g \in R[x]$ coprime with $\deg(f), \deg(g) \geq 1$ and $a \in (f,g) \cap R$, can one always find $u,v \in R[x]$ with $\deg(u) < \deg(g)$ and $\deg(v) < \deg(f)$ such that $a = uf + vg$?

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    It is certainly not true if $f,g \in R$.2011-08-23
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    Also, $a$ has degree $0$, so all higher degree terms in $uf+vg$ would have to cancel...which is weird.2011-08-23
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    @Plop: True. Fixed.2011-08-24

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HINT $\: $ If $\rm\ a = u\:f + v\:g\:,\ f,g\:$ monic, the Division Algorithm $\rm\:\Rightarrow\ u = b\:g + \bar u,\ v = c\:f + \bar v\ $ with $\rm\:deg\ \bar u < deg\ g\ $ and $\rm\: deg\ \bar v < deg\ f\:.$

Google reduced resultant for literature on this topic.

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    Thanks. This does the trick.2011-08-24
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    Of course, if $f$ and $g$ are not monic, it's a different story.2011-08-24
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Let $R$ be the integers, and try $f(x)=2x+5$, $g(x)=2x+1$, $a=1$.

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    Is $a \in (f,g)$?2011-08-24
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    Yes.${}{}{}{}{}$2011-08-24