1. ## Polynomial Rings

Let R be a commutative ring with identity.

Let $p(x)=a_nx^n+ a_{n-1}x^{n-1} . . . +a_2x^2+a_1x^1+a_0$ be a polynomial in R[x].

p(x) is a unit if and only if $a_0$ unit and $a_i for i>1$ are nilpotent.

So I can easily prove the first term, $a_0$ is a unit.

However I feel that I cannot guarantee that $a_n$ is nilpotent.

Since p is a unit, then there exists a q in R[x] (let's say the coefficients of q are of the form $b_i$ and that it has degree m) where p*q = 1.

So taking the product of the 2 polynomials:

$a_nb_mx^{n+m}+ . . . . . +a_0b_0=1$

and it must equal zero because a_0 is a unit, which implies that the last term = 1, and the rest equals 0.

So I can show that a_n is a zero divisor but I don't feel I can guarantee that it's nilpotent.

The hint says to look at the x^(n-1) term and multiply by a_n.

$a_nb_{n-1}+a_{n-1}b_n=0$
$a_n*a_nb_{n-1}+a_n*a_{n-1}b_n=0$

then the term on the right dissapears because a_nb_m =0.

$a_n^2b_{n-1}=0$

2. Originally Posted by seld
Let R be a commutative ring with identity.

Let $p(x)=a_nx^n+ a_{n-1}x^{n-1} . . . +a_2x^2+a_1x^1+a_0$ be a polynomial in R[x].

p(x) is a unit if and only if $a_0$ unit and $a_i for i>1$ are nilpotent.
suppose first that $a_n, \cdots , a_1$ are nilpotent and $a_0$ is a unit. then $p(x) - a_0$ is nilpotent and thus $p(x)=p(x) - a_0 + a_0$ is a unit becasue, in any commutative ring, nilpotent + unit is

a unit. for the converse, we'll use induction on $n,$ the degree of $p(x)$: it's clear if n = 0. so suppose the claim is true for any polynomial which is a unit and has degree < n.

let $p(x) = \sum_{j=0}^n a_jx^j, \ n \geq 1,$ be a unit. so there exists $q(x)=\sum_{j=0}^m b_jx^j$ such that $p(x)q(x)=1.$ if $m=0,$ then $a_n = \cdots = a_1=0$ and $a_0$ is a unit. so we may assume

that m > 0. clearly $b_0$ must be a unit and we also have:

$a_nb_m = 0, \ a_nb_{m-1}+ a_{n-1}b_m = 0, \ \cdots , a_nb_0 +a_{n-1}b_1 + \cdots = 0.$ write these relations in terms of matrices: $AX=0,$ where

$A=\begin{pmatrix}a_n & 0 & 0 & . & . & . & 0 \\ a_{n-1} & a_n & 0 & . & . & . & 0 \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ * & * & * & . & . & . & a_n \end{pmatrix}, \ \ X=\begin{pmatrix}b_m \\ b_{m-1} \\ . \\ . \\ . \\ b_0 \end{pmatrix},$ where $*$ means that we don't care what those entries are. the important point is that $A$ is a (lower) triangular matrix with $a_n$

on its main diagonal. now multiply $AX=0,$ from the left, by the adjoint matrix of $A,$ to get $a_n^m X =(\det A)X = \text{adj}(A)A X = 0.$ thus $a_n^mb_0=0,$ and hence $a_n^m = 0$ because $b_0$

is a unit. so $a_n$ and therefore $-a_nx^n$ is nilpotent. hence $p_1(x)=p(x) -a_nx^n$ is a unit, because, as i already mentioned, nilpotent + unit is a unit. now since $\deg p_1(x) < n,$ we can

apply the induction hypothesis to finish the proof.