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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. | 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. | ||
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| + | Using that theorem and induction i should get part B but I am not sure if simply (x+y)^p =x^p+y^p=x^p^n+y^p^n is enough to prove this. ANy suggestions?--[[User:Lmiddlet|Lmiddlet]] 20:35, 25 March 2009 (UTC) | ||
Latest revision as of 16:35, 25 March 2009
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.
Using that theorem and induction i should get part B but I am not sure if simply (x+y)^p =x^p+y^p=x^p^n+y^p^n is enough to prove this. ANy suggestions?--Lmiddlet 20:35, 25 March 2009 (UTC)
