Revision as of 15:58, 24 March 2009 by Achurley (Talk | contribs)


I was stuck on part a). Any ideas on how to start it? --Podarcze 13:27, 24 March 2009 (UTC)


What I did was use the Binomial Theorem. When you multiply out (x+y)^p, you notice the coefficients are all multiples of p. Since char(R)=p, then p times any element in R is 0. So, you end up with (x+y)^p = x^p+0+...+y^p=x^p+y^p.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood