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*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*_{n}x^{n} is also a unit, andwe can repeat the argument). Let *I *= *{b **∈** R *: *a*_{n}^{t}b = 0 forsome *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*_{n}^{t}b_{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 *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.

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

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