# Proving a basic property of polynomial rings

I am learning ring theory in the Dummit & Foote's Abstract Algebra, and I am doing all the exercises to get as much experience as possible... but some of them just get me stuck for hours! Like this one :

Assume $R$ is a commutative ring with identity. Prove that $p(x)=a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in R[x]$ is a unit if and only if $a_0$ is a unit and $a_1, \dots, a_n$ are nilpotent in $R$.

Now I already have the $\Longleftarrow$ part, since I've proven in a previous exercise that the sum of nilpotent elements is nilpotent and the sum of a unit with a nilpotent element is a unit, hence $a_0 + (a_1 x + \dots + a_n x^n)$ has a sum of nilpotents between the parenthesis, and thus is written as the sum of a unit and a nilpotent element, thus is a unit. I can't deal with the converse though.

I've tried noticing that if $p(x)$ is a unit and $q(x)$ is its inverse then $p(x) q(x) = 1$ implies that $p(x)^m q(x)^m = 1$ (because $R$ is commutative), but that only gave me that if $p$ has degree $n$ and $q$ has degree $r$ with last coefficient $b_r$ then $a_n b_r$ is $0$, and got stuck there. I also tried to take a look at what $p(x) q(x)$ looks like : If $p(x) = \sum_{k=0}^n a_k x^k$ and $q(x) = \sum_{k=0}^n b_k x^k$ (just add zeros to get those two sums of same size), then $$a_0 b_0 =1 ,\quad a_1 b_0 + a_0 b_1 = 0, \quad a_2 b_0 + a_1 b_1 + a_0 b_2 = 0, \quad \cdots$$ and I thought I could get something out of those equations but all I have now is that $a_1$ is nilpotent if and only if $b_1$ is, which doesn't help me much.

Any hints? I don't need a full solution if there's a way to just point out a nice trick.

• possible duplicate of math.stackexchange.com/questions/82552/… – lhf Nov 20 '11 at 11:37
• I didn't find that duplicate myself at first, and even though the question was a duplicate (I can admit it is the same question!), the answer wasn't : Bill Dubuque's answer was totally what I was looking for and his answer was not a duplicate of the answers found at your link. So let's keep things that way. – Patrick Da Silva Nov 20 '11 at 11:46
• Sure, my point was simply that someone landing here should know about the other questions and the answers there. – lhf Nov 20 '11 at 15:06
• Oh, fine then!! – Patrick Da Silva Nov 20 '11 at 20:19

HINT $\$ If $\rm\:R\:$ is a domain then easily $\rm\:p(x)\:$ a unit $\rm\Rightarrow\ a_i = 0\:$ for $\rm\:i>0\:.\$ Now $\rm\ R\to R/P,\$ for $\rm\:P\:$ prime, reduces to the domain case, yielding that the $\rm\:a_i\:,\ i>0\:$ are in every prime ideal. But the intersection of all prime ideals is the nilradical, the set of all nilpotent elements.
Alternatively, more elementarily, successively examining the coefficients of $\rm\:f\:g\:$ one proves that $\rm\: a_n\:b_m = 0\ \Rightarrow\ a_n^2 b_{m-1} = 0\ \Rightarrow\ \ldots\: \Rightarrow\ a^{m+1}\:b_0 = 0\:.\:$ But $\rm\:b_0\:$ is a unit so $\ldots$
• I haven't read about domains/prime ideals yet, and they are not explained up to the point where the exercise is shown. So not only I got nothing of your hint but I'm not supposed to use it =(... I know that if $R$ would be an integral domain (is that the same thing?), the only units of $R[x]$ would be the units of $R$, because then $\deg(pq) = \deg p + \deg q$ hence they all must have degree $0$ for things to work. If you could explain your prime ideal reduction thing, or if you had another idea, I'd still be interested though. – Patrick Da Silva Nov 20 '11 at 8:33