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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.

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  • $\begingroup$ possible duplicate of math.stackexchange.com/questions/82552/… $\endgroup$ – lhf Nov 20 '11 at 11:37
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    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Nov 20 '11 at 11:46
  • $\begingroup$ Sure, my point was simply that someone landing here should know about the other questions and the answers there. $\endgroup$ – lhf Nov 20 '11 at 15:06
  • $\begingroup$ Oh, fine then!! $\endgroup$ – Patrick Da Silva Nov 20 '11 at 20:19
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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.

See also my post here on reduction to domains by factoring out prime ideals.

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 $\ldots$

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  • $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Nov 20 '11 at 8:33
  • $\begingroup$ I took a look at your "my post here" but it seems a little too advanced for me to understand something... sorry =( $\endgroup$ – Patrick Da Silva Nov 20 '11 at 8:41
  • $\begingroup$ In ring theory jargon, "domain" is always short for "integral domain", usually with the rationale that there isn't any other use of the word "domain" that the prefix "integral" could possibly help disambiguate from. I suppose the people who like that rationale don't do much complex analysis (or category theory, or ...) $\endgroup$ – Henning Makholm Nov 20 '11 at 8:45
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    $\begingroup$ @Pat I'm surprised to learn that the exercise has been posed before any exposition of integral domains and prime ideals. The approach I hinted at is a very natural way to proceed - one which lends much more insight than the alternative ad-hoc approach needed without such structural foundations. I highly recommend that you revisit this answer after learning such. $\endgroup$ – Bill Dubuque Nov 20 '11 at 9:03
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    $\begingroup$ @Pat I've added a hint to my answer. $\endgroup$ – Bill Dubuque Nov 20 '11 at 9:35

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