0

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.

eulersnumber
  • 159
  • 7
  • 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

1 Answers1

1

Hint: If $fg=1$, then $\deg(f)+\deg(g)=0$. What does this imply about $\deg(f)$ and $\deg(g)$ ?

lhf
  • 208,399
  • 15
  • 224
  • 525
  • 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
  • 1
    @gaufler, also $f=g=4x+1$ in $R=\mathbb Z/8\mathbb Z$ – lhf Jul 23 '21 at 18:28