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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.

StefanH
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1 Answers1

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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.

StefanH
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