Question

Show that, if A = R[x] where R is a commutative ring, then for f =(sum: i)a_ix^i ∈ A: f ∈ U(A) ⇐⇒ a0 ∈ U(R) and ai is nilpotent for i ≥ 1.

Answer 1

The “if” part is clear since, in anycommutative ring, the sum of a unit and a nilpotent element is always a unit.
For the “only if” part, a rather slick proof using basic commutative algebra
(mainly the fact that the intersection of prime ideals in R is equal toNil(R)). For the sake of completeness, let us record below a proof thatuses a “bare-hands” approach, accessible to every beginning student in algebra.
Assume that f ∈ U(A). It is easy to see that a0∈ U(R). We are done if we can show that an is nilpotent in casen ≥ 1 (for then f − a_nxⁿ is also a unit, andwe can repeat the argument). Let I = {b ∈ R : a_n^tb = 0 forsome t ≥ 1}.
Say fg = 1, where g = b0+ · · · + b_mx^m ∈ A. We claim that bi ∈ I for every i. If this is the case, there exists t≥ 1 such that a_n^tb_i = 0 for all i,and so a_n^t = f a_n^t g =0, as desired. The claim is certainly true if i = m. Inductively,if b_i+1, . . . , b_m ∈ I, then, comparing the coefficients of xⁿ⁺ⁱin the equation fg =1, we have a_nb_i + a_n−1bⁱ⁺¹+ a_n−2bⁱ⁺² + · · ·= 0. For s sufficiently large, we then have a^s+1_nb_i = −a_n−1a^s_nb_i+1−· · · = 0, so b_i ∈ I as claimed.