# Units (invertibles) of polynomial rings over a field

If $R$ is a field, what are the units of $R[X]$?

My attempt: Let $f,g \in R[X]$ and $f(X)g(X)=1$. Then the only solution for the equation is both $f,g \in {R}$. So $U(R[X])=R$, exclude zero elements of $R$.

Is this correct ?

• The units of $R[X]$ certainly contain the units of $R$, yes? Can other elements work? Apr 8, 2013 at 3:16
• Sounds like a good start. Now equate coefficients. Apr 8, 2013 at 3:16
• Do you know that the deg$(f\cdot g)=$ deg$(f)+$ deg$(g)$? And if you want $f\cdot g =1$, compare degrees on both sides to see what you can conclude.. Apr 8, 2013 at 3:16
• Abstract duplicate of math.stackexchange.com/q/19132/264 Apr 8, 2013 at 3:17
• Duplicate of this question and others. Apr 8, 2013 at 3:17

## 1 Answer

If $f=a_0+a_1x+\cdots+a_mx^m$ has degree $m$, i.e. $a_m\ne 0$, and $fg=1$ for some $g=b_0+\cdots+b_n x^n$ (and $b_n\ne 0$), then observe that $$0=\deg(1)=\deg(fg)=\deg(a_0b_0+\cdots+a_mb_nx^{n+m})=m+n$$ as $a_mb_n\ne 0$. Hence $m+n=0$ and so $m=n=0$ as $m,n\ge 0$. Hence $f\in R^*$ and $R[x]^*=R^*$ (the star denotes the set of units).