R: Integral Domain with Characteristic p>0

[kathrynag]

Show that if R is an integral domain with characteristic p>0, then for all a,b in R, we must have (a+b)^p=a^p+b^p. Show by induction that we must also have (a+b)^p^n=a^p^n+b^p^n for all positive integers n.


R is an integral domain, so ab=0 implies a=0 or b=0.
The smallest positive integer n such that n*1=0 is called the characteristic of R.

 

[snipez90]

You know you could actually demonstrate some thought about this problem as opposed to providing basic definitions that anyone who could conceivably solve this would already know.

Look up the binomial theorem and figure out why it holds in R.

 

[kathrynag]

You know you could actually demonstrate some thought about this problem as opposed to providing basic definitions that anyone who could conceivably solve this would already know.

Look up the binomial theorem and figure out why it holds in R.

 

[micromass]

And why does the binomial theorem hold in R?

 

[kathrynag]

Proof by induction.

Base Case (n = 1):
We show that (a + b)^p = a^p + b^p.

This follows from the Binomial Theorem.
(a + b)^p
= Σ(n=0 to p) C(p,n) a^(p-n) b^n
= a^p + [Σ(n=1 to p-1) C(p,n) a^(p-n) b^n] + b^p
= a^p + 0 + b^p, since p|C(p,n) for n = 1,2,...p-1.
= a^p + b^p.

Inductive step.
Assuming the at the claim is true for n = k:
(a + b)^(p^(k+1))
= (a + b)^(p^k * p)
= [(a + b)^(p^k)]^p
= (a^(p^k) + b^(p^k))^p, by inductive hypothesis
= (a^(p^k))^p + (b^(p^k))^p, by the base case
= a^(p^(k+1)) + b^(p^(k+1)), as required.