Find all the p-th roots in a field of characteristic p. Factorise $\displaystyle x^q-1$ over a field K of order q.

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- Feb 3rd 2009, 11:14 PMAmanda1990[SOLVED] Find all p-th roots in field of characteristic p
Find all the p-th roots in a field of characteristic p. Factorise $\displaystyle x^q-1$ over a field K of order q.

- Feb 4th 2009, 04:25 AMThePerfectHacker
- Feb 6th 2009, 02:23 AMAmanda1990
I'm afraid I don't really get why this hint works, although I speant a lot of thought on it. Here's what I'm thinking:

For the first part of the question:

I can prove that the size of the field must be p^n for some n. If n = 1, then we could use Fermat's little theorem and the fact that the field must be isomorphic to Zp to show that the only root is 1. But firstly this only deals with the case n=1 (we would need to show that the field is isomorphic to "something else" to get the general case to work, but I'm not sure if this is a good idea). Secondly I'd much rather not appeal to Fermat's little theorem anyway. I'm sure there must be a nicer way to do this relying just on standard field theory.

For the second part of the question, I've seen the Frobenius automorphism and it looks as though it may be applicable here, though I cannot see how.

Also, I was aware of the results given in the hint in the post above, but really cannot see how it helps (though it does seem to point in the direction of the Frobenius automorphism). - Feb 6th 2009, 08:20 AMThePerfectHacker
Suppose we are working in $\displaystyle \mathbb{F}_3$, and we want to compute $\displaystyle (a+b)^3$. If we use the binomial theorem we get $\displaystyle a^3 + 3\cdot a^2b+ 3\cdot ab^2 + b^3$. Just remember when we write $\displaystyle 3\cdot x$ we mean $\displaystyle x+x+x$. Now the characheristic of $\displaystyle \mathbb{F}_3$ is $\displaystyle 3$ therefore $\displaystyle 3\cdot x =0$ and so $\displaystyle (a+b)^3 = a^3+b^3$.

Now, if $\displaystyle p$ is a prime and $\displaystyle 0 < k <p$ then $\displaystyle p$ divides $\displaystyle {p\choose k}$.

Let $\displaystyle a,b\in \mathbb{F}_q$ where $\displaystyle q=p^m$ and so $\displaystyle \text{char}(\mathbb{F}_q)=p$.

This means $\displaystyle (a+b)^p = \sum_{k=0}^p {p\choose k}a^kb^{p-k}$.

But if $\displaystyle 0<k<p$ then all the terms become zero, so, $\displaystyle (a+b)^p = a^p+b^p$.

Thus, $\displaystyle (a+b)^{p^2} = \left( (a+b)^p \right)^p = (a^p+b^p)^p = a^{p^2}+b^{p^2}$.

And, $\displaystyle (a+b)^{p^3} = \left( (a+b)^{p^2} \right)^p = \left(a^{p^2}+b^{p^2} \right)^p = a^{p^3}+b^{p^3}$

In general, $\displaystyle (a+b)^{p^n} = a^{p^n}+b^{p^n}$.

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The mapping $\displaystyle \sigma: \mathbb{F}_q\to \mathbb{F}_q$ by $\displaystyle \sigma(x) = x^q$ is an (Frobenius) automorphism because $\displaystyle \sigma(ab)=\sigma(a)\sigma(b)$ and $\displaystyle \sigma(a+b)=\sigma(a)+\sigma(b)$ by above. It remains to show that $\displaystyle \sigma$ is a bijection. However, since $\displaystyle \mathbb{F}_q$ is finite it means all we have to show is that $\displaystyle \sigma$ is an injection by the pigeonhole principle. Thus, say $\displaystyle \sigma(x) = \sigma(y) \implies \sigma(x-y) = 0 \implies (x-y)^q =0 \implies x=y$. - Feb 9th 2009, 03:01 AMAmanda1990
OK...I think I'm getting somewhere now. We can use your hint to show that the only p-th root of 1 in a field of characteristic p is 1.

Using the same idea, it seems like we can show that the only root of $\displaystyle x^q-1$ in a field of order q is 1, since it is a standard result that $\displaystyle q=p^n$.

So it seems to me that $\displaystyle x^q-1$ factorises only as far as $\displaystyle (x-1)(x^ {q-1} +x^{q-2}...+1)$. Does this seem right??

EDIT: Actually I thought this looked a bit too easy to be true. I forgot that just because a polynomial doesn't have a root in a field K, it can still factorise as a product of two (or more) irreducibles. So I'm still quite stuck (and probably more confused than before...) - how do we factorise this?

EDIT 2: Sorry. I've just realised how silly I've been! Thanks for the help!