# 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 &isin; A: f &isin; U(A) &lArr;&rArr; a0 &isin; U(R) and ai is nilpotent for i &ge; 1.
1
2012-10-31T08:48:58-0400
The &ldquo;if&rdquo; part is clear since, in anycommutative ring, the sum of a unit and a nilpotent element is always a unit.
For the &ldquo;only if&rdquo; 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 &ldquo;bare-hands&rdquo; approach, accessible to every beginning student in algebra.
Assume that f &isin; U(A). It is easy to see that a0&isin; U(R). We are done if we can show that an is nilpotent in casen &ge; 1 (for then f &minus; anxn is also a unit, andwe can repeat the argument). Let I = {b &isin; R : antb = 0 forsome t &ge; 1}.
Say fg = 1, where g = b0+ &middot; &middot; &middot; + bmxm &isin; A. We claim that bi &isin; I for every i. If this is the case, there exists t&ge; 1 such that antbi = 0 for all i,and so ant = f ant g =0, as desired. The claim is certainly true if i = m. Inductively,if bi+1, . . . , bm &isin; I, then, comparing the coefficients of xn+iin the equation fg =1, we have anbi + an&minus;1bi+1+ an&minus;2bi+2 + &middot; &middot; &middot;= 0. For s sufficiently large, we then have as+1nbi = &minus;an&minus;1asnbi+1&minus;&middot; &middot; &middot; = 0, so bi &isin; I as claimed.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!