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## #1 2009-01-03 06:55:27

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

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## #2 2009-01-03 08:16:17

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: Frobenius endomorphism

This theorem, which I have always heard as the "freshman calculus theorem", says that:

This property has important implications in Field theory and Galois theory.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #3 2009-01-03 08:27:06

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Frobenius endomorphism

Its a neat result.

By the way, the definition requires the characteristic of the ring

to be prime. What if the characteristic is a non-prime positive integer?
will still be a homomorphism, wont it?

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## #4 2009-01-03 09:49:18

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: Frobenius endomorphism

This theorem as I mentioned previously is mostly used in field and Galois theory (in my experience).  As such, we are always talking about integral domains and with any integral domain, the characteristic is always prime or 0 (i.e. infinite).

Characteristic also makes sense for rings with zero-divisors, so that's a good question to ask Jane.  The proof of the theorem involves use of the binomial theorem along with the fact that if d is not 0 or p, then p divides p choose d.  This simplifies the expression the binomial theorem provides considerably, leading directly to the conclusion.  This fact does not hold if p is not prime, which would lead one to believe that the theorem doesn't hold in cases where p is not prime.  The following counter-example justifies that thought:

In the integers modulo 4, (1+1)^4 = 2^4 = 16 = 0 (mod 4).  However, 1^4 + 1^4 = 1 + 1 = 2.

While this counter example shows that there are rings of non-prime characteristic where the theorem does not hold, is this true for all rings with non-prime characteristic?  This I'm not sure of, but I believe you can find a proof based off the fact that any such ring must have zero divisors.  In particular, if the characteristic of the ring is n = pq, then (p*1)(q*1) = (pq)*1 = n*1 = 0, and since p, q < n we have that p*1 and q*1 must both be nonzero.

Last edited by Ricky (2009-01-03 09:54:37)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #5 2009-01-03 09:52:46

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: Frobenius endomorphism

I see. Thanks for the explanation.

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## #6 2009-01-03 10:57:21

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: Frobenius endomorphism

I would like someone to check my work if possible, but here's what I have so far.

Let n = pq, p and q not necessarily primes.  The following are necessary conditions for the Frobenius map to be a homomorphism:

These are all found by expanding (1+1)^n, (p*1 +1)^n, and (q*1+1)^n by the binomial theorem.  What I find rather interesting is that (p*1 + q*1)^n = (p*1)^n + (q*1)^n, again found by expanding the binomial theorem.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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