Is it correct that $f \in R[X]$ is a unit, iff $a_0$ is a unit.

Question

Let $R$ be ring and $R[X]$ its polynomial ring in one indeterminate $X$. Then $$ f \in R[X] \mbox{ is a unit } \Leftrightarrow a_0 \mbox{ is a unit } $$ for $f = a_0 + a_1 X + \ldots + a_n X^n$.

If $f \cdot g = 1$ with $g = b_0 + b_1 X + \ldots b_n X^n$, then $a_0 b_0 = 1$, hence $a_0$ is a unit. Conversely, suppose $a_0$ is a unit, and choose $b_0$ such that $a_0 b_0 = 1$ and define \begin{align*} b_1 & := b_0(-a_1 b_0) \\ b_2 & := b_0(-a_2 b_0 - a_1 b_1 ) \\ b_3 & := b_0(-a_3 b_0 - a_2 b_1 - a_1 b_2 ) \\ \vdots \\ b_n & := b_0(-a_n b_0 - a_{n-1} b_1 - \ldots - a_1 b_{n-1} ). \end{align*} Then every sum of the form $$ \sum_{i + j = k} a_i b_j $$ for $k > 0$ vanishes, hence the polynomial $g = b_0 + b_1 X + \ldots + b_n X^n$ is an inverse for $f$.

I am just asking if this is correct, or if I have overlooked something. I am asking because I nowhere find this characterisation stated for general rings (just find a variant for fields), and for rings I just find the result: $$ f \in R[X] \mbox{ is a unit } \Leftrightarrow a_0 \mbox{ is a unit and } a_1, \ldots, a_n \mbox{ are nilpotent } $$ which proofs could be found on this page, for example here, here or here. But nowhere do I find it that the condition of nilpotency could be dropped in the implication from right to left.

What happens in your argument if say $f(x)=1+x$ in $\mathbb{Q}[x]$? — carmichael561, Aug 18 '17 at 15:30
If $R$ has no zero divisors, then $\deg (f\cdot g) = \deg f + \deg g$, so only constants can be units. — Daniel Fischer, Aug 18 '17 at 15:30
What is the case that over a commutative ring $R$, $f=a_0+a_1x+\cdots\in R[[x]]$ is a unit in $R[[x]]$ iff $a_0$ is a unit in $R$, where $R[[x]]$ is the formal power series ring. — Angina Seng, Aug 18 '17 at 15:48
Thanks for your comments, I hope I have interpreted them correctly and compiled and answer. — StefanH, Aug 18 '17 at 15:57
Your result as stated is false. However, it's true in the ring $R[[x]]$. — Xam, Aug 18 '17 at 16:40

score 1 · Accepted Answer · answered Aug 18 '17 at 15:57

I just considered the coefficients in the result $f\cdot g$ that have a term of the form $a_0 b_j$ $j=1,\ldots,n$ in them, and those will vanish. But I totally forgot about the other coefficients like $a_1 b_n + \ldots + a_n b_1$ and so on up $i + j = n + n$ with $g = b_0 + b_1 X + \ldots b_n X$, $f = a_0 + a_1 X + \ldots a_n X$.

Thanks to @charmichael561 and @DanielFischer for their comments!

But if we contine this process indefinitely, we can build a series $g \in R[[x]]$ and for this $f\cdot g = 1$ if I interpret @LordSharkTheUnknown's comment correctly. But we can find no polynom.

Is it correct that $f \in R[X]$ is a unit, iff $a_0$ is a unit.

1 Answers1