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 Subject: "Units in Polynomial Rings" Previous Topic | Next Topic
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Michael Klipper guest
Aug-17-04, 08:06 PM (EST)

"Units in Polynomial Rings"

 Hello everyone. I've been rereading one of my algebra textbooks and got stuck on this problem. I am told to assume that R is a commutative ring with an identity (for this problem it is safe to assume 1 is not 0 in R). I want to prove that p(x) = a_n * x^n + a_{n-1} * x^{n-1} + ... + a_0 is a unit of R[x] if and only if a_0 is a unit in R and a_1, a_2, ... a_n are nilpotent in R.In case definitions are helpful: An element r of R is a unit if there exists a two-sided inverse s in R, i.e. r * s = s * r = 1. An element r of R is nilpotent if there exists a positive integer k for which r^k = 0.It's clear that each a_i for i >= 1 must be a zero divisor or 0, but I don't know how to show any of them are nilpotent. If I can show that one of them is nilpotent, then this gives me the way to form an inductive proof of what I want to show. I already have proven that the sum of a unit and a nilpotent element is a unit, and this theorem gives me one direction of the proof I seek.I was considering formal series at one point, since in the formal series ring R[[x]], a polynomial is a unit if and only if its constant term is a unit in R. I came up with a way to describe the coefficients of the inverse of a unit in R[[x]], but the best I can think of is a formula that shows the "i"th coefficient of the inverse in terms of partitions of i. I was hoping to use the fact that R[x] is just the subring of R[[x]] in which the sequence of coefficients has finite support, but this hasn't seemed to help much.Any help would be appreciated. I'm pretty sure I've gone further than I need to and that there is some trick under my nose.

alexb Charter Member
2330 posts
Aug-25-04, 04:57 PM (EST)    1. "RE: Units in Polynomial Rings"
In response to message #0

 >Hello everyone. I've been rereading one of my algebra >textbooks and got stuck on this problem. I am told to >assume that R is a commutative ring with an identity (for >this problem it is safe to assume 1 is not 0 in R). I want >to prove that p(x) = a_n * x^n + a_{n-1} * x^{n-1} + ... + >a_0 is a unit of R[x] if and only if a_0 is a unit in R >and a_1, a_2, ... a_n are nilpotent in R. >>It's clear that each a_i for i >= 1 must be a zero divisor >or 0, but I don't know how to show any of them are >nilpotent. To claim this, you have probably assumed that p(x) is a unit, which would imply existence of q(x) such thatp(x)·q(x) = 1Assuming q(x) = bncn + ...make an assumption about n, i.e., the degree of q. For example, for n = 1, you'll get three equations:a0·b0 = 1,a0·b1 + a1·b0 = 0, and a1·b1 = 0.The first is solved for b0, the second for b1. Substitute what you found into the third one. You'll geta12a0-2 = 0,which impliesa12 = 0

rik
Member since Aug-29-04
Aug-29-04, 12:50 PM (EST)     2. "RE: Units in Polynomial Rings"
In response to message #0

 alexb's solution is perfectly fine but here's another way of doing it more structurely:This requires the following fact: In a commutative ring R with identity, intersection of all prime ideals of R is equal to the set of all nilpotents of R, called the nilradical of R and is denoted Nil(R). Note that one implication is easy: That Nil(R) is contained in the intersection of all prime ideals of R. The other direction is harder and it requires an axiom of choice, or equivalently and more useful here Zorn's lemma. I don't know how much you know but if you don't know these, I encourage you to look them up. Here's a sketch of proof for the other direction: suppose x is in the intersection of all prime ideals and x is not a nilpotent. Then consider the set S = {x, x^2, x^3, ...} which is multiplicative and contains no 0. Let T be the set of all ideals of R which does not intersect S. Note that T is nonempty because {0} in particular is in T. Partial order T by the usual set inclusion and verify that for any chain in T, there is an upper bound, so hypotheses of Zorn's lemma is satisfied. So there is a maximal ideal M that does not intersect S. Now show that M is prime ideal (in fact every maximal ideal in a ring R such that R^2 = R (which is clearly true if R has unity) is a prime ideal). This you can do by assuming ab is in M then considering M + (a) and M + (b) and their product (In the case of non-commutative ring, you'd consider ideals AB < M and M + A, M + B). So M is a prime ideal which does not intersect S. This contradicts the assumption that x is in the intersection of all prime ideals.Now getting back to your problem. I only do one implication: that if p(x) is a unit in R then a_0 is a unit and a_1, a_2, ... are nilpotents, as this is the harder implication. Let P be any prime ideal. Then there is a canonical epimorphism (onto homomorphism) t: R -> R/P. It induces an epimorphism t_ : R -> R/P in an obvious way (just reduce the coefficients of polynomials in R modulo P). Note that R/P is an integral domain because P is prime ideal. As p(x) is a unit in R, t_(p(x))also a unit in R/P. Because R/P is an integral domain, it follows that t(a_1), t(a_2), ... t(a_n) are all zero elements in R/P, i.e. a_1, a_2, ..., a_n are in P. Since P was an arbitrary prime ideal, it follows that a_1, a_2, ..., a_n are in the intersection of all prime ideals. But we just proved that this intersection is Nil(R) ! So a_1, a_2, ..., a_n are nilpotents. Now p(x) = a_0 + a_1*x + ... + a_n*x^n. Then a_0 = p(x) - a_1*x - a_2*x - ... - a_n*x^n = unit - (nilpotent), which is a unit (verify!).-Henry Spencer guest
Jan-28-09, 08:07 AM (EST)

3. "RE: Units in Polynomial Rings"
In response to message #2

 So I follow the rest of your argument, but I'm having a hard time seeing why chains in T must be bounded. Is it because they are bounded by R itself? alexb Charter Member
2330 posts
Jan-28-09, 09:25 AM (EST)    4. "RE: Units in Polynomial Rings"
In response to message #3

 No, but because the union of the ideals in T is an ideal.

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