$p(x)$ is a unit in $R[x]$ where $R$ is a general unity ring if and only if it's constant term is a unit and all other coefficients are nilpotent

Question

I'm having a real hard time finding a the inverse for an arbitrary unit if this were the case.... Somehow I need to take advantage of getting every coefficient a high enough exponent to annihilate itself whilst still hitting the constant term with only it's inverse in $R$ where R is a general unitary ring... anyone got any tips?

Answer 1

The hard way round is to prove that if $f(X)=a_0+a_1X+\cdots+a_mX^m$ is invertible in $R[X]$ then $a_0$ is a unit in $R$ (that's easy) and each $a_i$ for $i\ge1$ is nilpotent (that isn't).

There are honest proofs of this implication, relying on manipulating polynomials, but here's a slick trick. Let $P$ be a prime ideal of $R$. Then $f(X)$ is invertible as a polynomial in $(R/P)[X]$. But $R/P$ is an integral domain, so that $f$ reduces to a constant in $(R/P)[X]$. Thus $a_1,\ldots,a_m\in P$. But then $a_1,\ldots,a_m$ lies in the intersection of the prime ideals of $R$, but that's the nilradical of $R$.

Answer 2

Here are the honest polynomial manipulations. Assume $p(x)$ is a unit, and write $p(x)=u+xr(x)$, where $u\in R$ is the constant term. Let $$q_n(x)=u^{-1}\sum_{k=0}^{n}(-u^{-1}xr(x))^k.$$ Using a telescoping series, you can show $$ p(x)q_n(x)=(1+u^{-1}xr(x))\sum_{k=0}^n (-u^{-1}xr(x))^k=1-(-uxr(x))^{n+1} $$ Note that the first $n+1$ coefficients of $p(x)q_n(x)$ are equal to the first $n+1$ coefficients of the polynomial $1$. Letting $q(x)$ be the inverse of $p(x)$, and letting $[x^i]f(x)$ be the coefficient of $x^i$ in $f(x)$, this means $$ [x^i]p(x)q_n(x)=[x^i]p(x)q(x)\qquad \text{for all }0\le i \le n $$ I claim that this implies $$ [x^i]q_n(x)=[x^i]q(x)\qquad\qquad\quad \text{for all }0\le i \le n $$ The proof is an easy exercise by induction. In particular, letting $n$ be the degree of $q(x)$, we must have that $q_m(x)=q(x)$ for all $m\ge n$. Since $q_{n+1}=q_n$, we must have $(-uxr(x))^{n+1}=0$, which implies $$ r(x)^{n+1}=0, $$ which quickly implies that all coefficients of $r(x)$ are nilpotent (consider the leading coefficient first).