How do I show that the units of $R[x] = $ the units of $R$ where $R$ is an integral domain? I understand that given $a,b\in R$, $a$ is a unit if $a\cdot b=1$. But I'm not really sure what this means as far as $R[x]$ is concerned. I'm pretty confused so if someone can talk about any of this in simple terms, that would be awesome.
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Also solved in a much more general question here http://math.stackexchange.com/q/19132/29335 . Found by searching "polynomial unit commutative". Try searching first next time, you might get answers faster. – rschwieb Mar 31 '17 at 12:30
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Hint: If $fg=1$, then $\deg(f)+\deg(g)=0$. What does this imply about $\deg(f)$ and $\deg(g)$ ?
lhf
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Why is it necessary for the ring to be an integral domain for this statement to hold true ? – gaufler Jul 23 '21 at 17:03

1@gaufler, $\deg(fg)=\deg(f)+\deg(g)$ need not hold in the presence of zero divisors. Consider $f=2x$ and $g=3x$ in $R=\mathbb Z/6\mathbb Z$. – lhf Jul 23 '21 at 18:21

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