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Prove that $f = a_nx^n + \cdots + a_1x+a_0$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$ and $a_1, ..., a_n$ are nilpotent.

My argument is as follows.

Suppose that $f$ is a unit in $R[x]$ and consider the evaluation map $\varphi : R[x] \to R$ defined by $x \mapsto 1$. We therefore have that $\varphi(f) = (a_n + \cdots + a_1) + a_0$. We note that Question 1 gives us that the sum of a nilpotent element and a unit is a unit. We therefore conclude that since $\varphi$ must map a unit in $R[x]$ to a unit in $R$, $(a_n + \cdots + a_1) + a_0$ is a unit in $R$. Furthermore, with the some of nilpotent elements being nilpotent, this yields that $a_1, ..., a_n$ are nilpotent and $a_0$ is a unit in $R$.

My argument lacks rigour in showing that $(a_n + \cdots a_1)$ must be nilpotent and $a_0$ must be the unit. I simply state this, which I'm not satisfied with. Could I take another evaluation map, this time evaluating at $x=0$?

Furthermore, the other direction seems unclear given that I cannot consider something as simple as an evaluation map.

Also note that this question is not a duplicate. The induction argument that is used almost ubiquitously as a solution is not intuitive and therefore I want to prove it through this approach of evaluation maps.

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    (You need to specify the relation between $f$ and the $a_i$ -- which might be obvious, but still is part of what you're telling us. Some people use $a_i x^{n-i}$ and some use $a_ix^i$. If one already knows the solution to this exercise, then they will know which one it is, but still again please write it down for clarity.) – Pedro Jul 31 '16 at 03:45
  • The sentence "Furthermore, with the *sum of the nilpotent elements being nilpotent, this yields $a_{1}, \ldots, a_{n}$ are nilpotent and $a_{0}$ is a unit in $R$" does not follow from the fact you cite above. If you have some elements $r_{1}, \ldots, r_{n}$ of a ring $R$ whose sum is a unit, it is not true that $r_{1} + \cdots r_{n-1}$ must be nilpotent and $r_{n}$ a unit, no matter how you label things. Take $-1, 2 \in \mathbb{Z}$ ($-2$ is not nilpotent), for instance, or $[2], [3] \in \mathbb{Z}/6\mathbb{Z}$ (neither is a unit or nilpotent). – Alex Wertheim Aug 01 '16 at 00:12

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