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I'm trying to determine $U(\mathbb{R}[x])$, where $U(R)$ denotes the unit group of a ring $R$.

I think the answer is all non-zero constant polynomials, but I'm having trouble showing that these are the only units.

Bill Dubuque
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pmal
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1 Answers1

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If you look at the degree, if $PQ=1$ then $\deg(PQ)=\deg(P)+\deg(Q)=0$ (if $R$ is a domain) so $P$ and $Q$ are non-zero constant (because $\deg(P)=\deg(Q)=0$). And the constant has to be invertible.

And if $P(X)=a\in U(R)$, $P^{-1}(X)=a^{-1}$ is a multiplicative inverse.

So, if you are in a domain $U(R[X])=U(R)$.

user26857
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D.L.
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    Note that it's important for this argument that $\mathbf{R}$ is an integral domain; things are more complicated in the general case, because the product of the leading coefficients can be zero! In fact, there are rings over which there are nonconstant polynomials that are invertible, such as $(1-\epsilon x)(1 + \epsilon x + \epsilon^2 x)$ in any ring where $\epsilon^3 = 0$. –  Apr 02 '14 at 08:21
  • Yes, in this cas, we will have only $deg(P.Q)\leq deg(P)+deg(Q)$...

    But, I have red that we were talking about real, sorry...

    But if we have not a domain, it is very complicated, I think...

     – D.L. Apr 02 '14 at 08:27
  • Yes the coefficients are real numbers in my problem. – pmal Apr 03 '14 at 00:16