My lecture notes state that an 'easy' result is

If $R$ is an integral domain then an irreducible element of $R$ remains irreducible in $R[x]$, and the units in $R$ and in $R[x]$ are the same.

I can't seem to get my head around why this is the case, and what a unit in $R[x]$ means intuitively because I don't see how the units can be the same if $R$ is the coefficients of the polynomials in $R[x]$. I.e. for an unit say $\alpha \in R$ then what is the 'corresponding' unit in $R[x]$? I is it $\alpha x$ or $\alpha x^2$... or am I getting the wrong end of the stick here?

  • 2
    $R$ embeds in $R[x]$ as the constant polynomials. A unit $\alpha$ in $R$ corresponds to the constant polynomial $\alpha$, which is a unit in $R[x]$. – Chris Eagle May 14 '12 at 11:07
  • [See here](http://math.stackexchange.com/a/19145/242) on units in general polynomial rings. – Bill Dubuque Apr 17 '14 at 00:13
  • 1
    If $f(x)$ is a unit and if $g(x)$ is its inverse, then $${\rm deg}(f(x)g(x)) = {\rm deg}(1) = 0.$$ This forces ${\rm deg}(f) ={\rm deg}(g) = 0.$ In other words a polynomial ring in one variable whose coefficients are drawn from an integral domain can have no units that are not constants. – student Jan 24 '18 at 01:33

1 Answers1


There is a natural way to inject $R$ into $R[x]$ by sending an element $a$ to the polynomial $a$ (so the coefficient of $x^i$ is 0 for all $i>0$. Now it should be easy to see that the image of a unit under this injection is again a unit. On the other hand, if some polynomial $p$ is a unit in $R[x]$ then there is another polynomial $q$ such that $pq = 1$, but if $p$ has degree $>0$ then so does $pq$ since this is over a domain. Thus, if $p$ is a unit then $p$ is actually constant, and clearly the only constants that are units are the ones that are already units in $R$.

Tobias Kildetoft
  • 20,063
  • 1
  • 59
  • 88
  • Ah thank you, well explained. Also, it's now clear to me that an irreducible element of $R$ remains irreducible in $R[x]$, but what about the other way round? –  May 14 '12 at 11:16
  • 4
    What do you mean by the other way around? There are certainly irreducible elements in $R[x]$ which are not constant. – Tobias Kildetoft May 14 '12 at 11:19