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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?

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    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.2018-01-24

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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 $R$.

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    What do you mean by the other way around? There are certainly irreducible elements in $R[x]$ which are not constant.2012-05-14