Let R be a commutative ring with identity.

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

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

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

However I feel that I cannot guarantee that $\displaystyle a_n$ 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 $\displaystyle b_i$ and that it has degree m) where p*q = 1.

So taking the product of the 2 polynomials:

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

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.

$\displaystyle a_nb_{n-1}+a_{n-1}b_n=0$

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

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

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