From: Robin Chapman Subject: Re: units of the ring A[X] Date: Thu, 27 May 1999 13:41:45 GMT Newsgroups: sci.math In article <01bea81f$57e79580$37b939c2@buromath.ups-albi.fr>, "pascal ORTIZ" wrote: > Let A be a commutative ring with 1. > The units (ie invertible elements) of the ring A[X] are of type > a_0+a_1X+...+a_nX^n > where a_0 is a unit in A and the a_k's (k>0) are nilpotent elements of A. > This can be proved quickly by reducing the problem to the obvious case A > where > A is an integral domain and by use of the following well known result : > if a is in P for all prime ideal P of A then a is nilpotent. > I'm loocking for an elementary proof (avoiding the above result). The difficult bit is to show that for a unit f(x) = a_0 + a_1 x + ... + a_n x^n then a_1, ..., a_n are nilpotent. Obviously a_0 is a unit of A. As the sum of a nilpotent and a unit is a unit then all we need is to show that a_n is nilpoent. Let the inverse of f(x) be g(x) = 1 + b_1 x + ... + b_m x^m. Then a_n b_m = 0, a_n b_{m-1} + a_{n-1} b_m = 0 so muliplying by a_n gives a_n^2 b_{m-1} = 0. Keeping going eventually gives a_n^{m+1}b_0 = 0 and as b_0 is a unit, then a_n is nilpotent. Robin Chapman + "They did not have proper School of Mathematical Sciences - palms at home in Exeter." University of Exeter, EX4 4QE, UK + rjc@maths.exeter.ac.uk - Peter Carey, http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda --== Sent via Deja.com http://www.deja.com/ ==-- ---Share what you know. Learn what you don't.--- ============================================================================== From: "A. Caranti" Subject: Re: units of the ring A[X] Date: Thu, 27 May 1999 16:14:32 +0200 Newsgroups: sci.math pascal ORTIZ wrote: > Let A be a commutative ring with 1. > The units (ie invertible elements) of the ring A[X] are of type > a_0+a_1X+...+a_nX^n > where a_0 is a unit in A and the a_k's (k>0) are nilpotent elements of A. > This can be proved quickly [...] > I'm loocking for an elementary proof (avoiding the above result). The following (sketch of a) proof is hopefully correct, and possibly elementary. First, it is immediate that for a_0 + a_1 X + ... to be invertible, a_0 must be a unit. So you are looking at elements of the form 1 - b X, where b is another polynomial. If all coefficients of b are nilpotent, then so is b, and 1 - b X is clearly invertibile by the usual formula 1 = (1 - b X) * (1 + b X + (b X)^2 + ...) So now you want to go the other way: you have 1 - b X invertible, so there is 1 + c X such that 1 = (1 - b X) * (1 + c X) = 1 + (c - b) X - b c X^2. You get c = b, and thus b^2 = 0. So you are left with proving that if a polynomial b = b_0 + b_1 X + ..., satisfies b^2 = 0, then all of its coefficients are nilpotent. It is immediate that b_0^2 = 0. Therefore (b - b_0)^3 = 0. It follows that the polynomial b_1 + b_2 X + ... is nilpotent. Now take this as a model to set up an induction (b nilpotent implies b_0 nilpotent and b_1 + b_2 X + ... nilpotent) and you are done. Andreas