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Let x and y belong to a commutative ring R with char(R)- p-0- a- Show.docx

Let x and y belong to a commutative ring R with char(R)= p?0.
a. Show that {(x+y)^p}= (x^p)+(y^p).
b. Show that, for all positive integers n, (x+y)^(p^n)= (x^(p^n))+ (y^(p^n)).
c. Find elements x and y in a ring of characteristic 4 such that (x+y)^4? (x^4)+(y^4).
Solution
Observe that according to Euler\'s totient formula, let a be any integer coprime to p, then (a, p) = 1. Then a^(phi(p)) = 1 (mod p). Now since p is prime, all a in Z/PZ is coprime to p and phi(p) = (p-1). Then (x+y)^p = (x+y)(x+y)^(p-1) (mod p) = (x+y)(1) (mod p). Then we have (x+y)^p = (x+y) (mod p), hence (x+y)^p = x (mod p) + y (mod p). Multiply back through by (x+y)^(p-1) gives us (x+y)^p(x+y)^(p-1) = x^p + y^p (mod p) and since (x+y)^(p-1) = 1 (mod p) we have (x+y)^p = x^p + y^p (mod p) as we wished to prove.
.

Let x and y belong to a commutative ring R with char(R)= p?0.
a. Show that {(x+y)^p}= (x^p)+(y^p).
b. Show that, for all positive integers n, (x+y)^(p^n)= (x^(p^n))+ (y^(p^n)).
c. Find elements x and y in a ring of characteristic 4 such that (x+y)^4? (x^4)+(y^4).
Solution
Observe that according to Euler\'s totient formula, let a be any integer coprime to p, then (a, p) = 1. Then a^(phi(p)) = 1 (mod p). Now since p is prime, all a in Z/PZ is coprime to p and phi(p) = (p-1). Then (x+y)^p = (x+y)(x+y)^(p-1) (mod p) = (x+y)(1) (mod p). Then we have (x+y)^p = (x+y) (mod p), hence (x+y)^p = x (mod p) + y (mod p). Multiply back through by (x+y)^(p-1) gives us (x+y)^p(x+y)^(p-1) = x^p + y^p (mod p) and since (x+y)^(p-1) = 1 (mod p) we have (x+y)^p = x^p + y^p (mod p) as we wished to prove.
.

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