# Answer to Question #16870 in Algebra for sanches

Question #16870

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.

Expert's answer

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

Assume that

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*a*0*∈**U(**R*). We are done if we can show that*an is nilpotent in casen ≥*1 (for then*f − a*is also a unit, andwe can repeat the argument). Let_{n}x^{n}*I*=*{b**∈**R*:*a*= 0 forsome_{n}^{t}b*t ≥*1*}.*

Say*fg*= 1, where*g*=*b*0+*· · ·*+*b*_{m}x^{m}*∈**A*.*We claim that bi**∈**I for every i*. If this is the case, there exists*t≥*1 such that*a*= 0 for all_{n}^{t}b_{i}*i*,and so*a*=_{n}^{t }*f a*=0, as desired. The claim is certainly true if_{n}^{t}g*i*=*m*. Inductively,if*b*_{i}_{+1}*, . . . , b*_{m}*∈**I*, then, comparing the coefficients of*x*^{n}^{+i}in the equation*fg*=1, we have*a*+_{n}b_{i}*a*_{n−}_{1}*b*^{i}^{+1}+*a*_{n−}_{2}*b*^{i}^{+2}+*· · ·*= 0*.*For*s*sufficiently large, we then have*a*^{s}^{+1}*=*_{n}b_{i}*−a*_{n−}_{1}*a*^{s}_{n}b_{i}_{+1}*−· · ·*= 0, so*b*_{i}*∈**I*as claimed.Need a fast expert's response?

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