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Math Help - Splitting Fields - Dummit and Foote - Example - page 541 - x^p - 2, p prime

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    Splitting Fields - Dummit and Foote - Example - page 541 - x^p - 2, p prime

    I am reading Dummit and Foote Section 13.4 Splitting Fields and Algebraic Closures

    In particular, I am trying to understand D&F's example on page 541 - namely "Splitting Field of  x^p - 2, p  a prime - see attached.

    I follow the example down to the following statement:

    " ... ... ... so the splitting field is precisely  \mathbb{Q} ( \sqrt[p]{2}, \zeta_p ) "

    BUT ... then D&F write:

    This field contains the cyclotomic field of  p^{th} roots of unity and is generated over it by  \sqrt[p]{2} , hence is an extension of at most p. It follows that the degree of this extension over  \mathbb{Q} is  \le p(p-1) .


    *** Can someone please explain the above statement and show formally and explicitly (presumably using D&F ch 13 Corollary 22 - see Note 1 below) why the degree of  \mathbb{Q} ( \sqrt[p]{2}, \zeta_p ) over  \mathbb{Q}  is  \le p(p-1) .

    I also find it hard to follow the statement:

    " ... ... ... Since both   \mathbb{Q} ( \sqrt[p]{2} )  and   \mathbb{Q} ( \zeta_p )  are subfields, the degree of the extension over  \mathbb{Q} is divisible by p and p - 1. Since both these numbers are relatively prime, it follows that the extension degree is divisible by p(p-1) so that we must have

     [\mathbb{Q} ( \sqrt[p]{2}, \zeta_p ) \ : \  \mathbb{Q}] = p(p - 1)    ... ... "

    *** Can someone please try to make the above clearer - why exactly is the degree of the extension over  \mathbb{Q} divisible by p and p - 1. What is the importance of "relatively prime" and why does equality hold in the statement regarding the degree of the extension?'

    *** Finally, we are told that p is a prime, but where does the argument in the example depend on p being prime. ["Relatively prime" is mentioned in the context of p and p-1 but they are consecutive integers and hence are coprime anyway]

    I would be grateful for some clarification of the above issues.

    Peter


    Note

    1. Corollary 22 (Dummit and Foote Section 13.2 Algebraic Extensions, page 529

    Suppose that  [K_1 \ : \ F] = n, [ K_2 \ : \ F ] = m where m and n are relatively prime: (n, m) = 1.
    Then  [K_1K_2 \ : \  F] = [K_1 \ : \ F] [ K_2 \ : \ F ]  = nm
    Last edited by Bernhard; October 3rd 2013 at 06:23 PM.
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    Re: Splitting Fields - Dummit and Foote - Example - page 541 - x^p - 2, p prime

    Do you know what they mean by a primitive root? That means the set \{\zeta_p^i \mid i \in \mathbb{N} \} is the set of all p-th roots of unity. Additionally, \zeta_p^p = 1. So, the p-th roots of unity are solutions to x^p-1. Obviously, 1 is a solution. But the other p-1 solutions can not be real so long as p>2. We know that the multiplicative group of \mathbb{F}_p is cyclic. This is why they use primes. If we used an arbitrary value, then it might not have any primitive roots of unity. For example, there are no primitive 8th-roots of unity. The 8th-roots of unity (as a group) are isomorphic to C_2\times C_4 (the product of a two-cycle and a four-cycle). Hence they cannot be generated by a single element. I don't have time to go into any more detail at the moment. I hope this helps clarify at least why they are using a prime number for the example. Also, x^p-1 = (x-1)(x^{p-1} + \ldots + 1). The second factor is irreducible over the rationals, and it is also irreducible over \mathbb{Q}(\sqrt[p]{2}). I hope this helps a little. If I have more time, I will try to write up a bit more explanation.
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    Re: Splitting Fields - Dummit and Foote - Example - page 541 - x^p - 2, p prime

    Ugh, I'm tired. I just read over my explanation, and that was terrible. The multiplicative group of \mathbb{F}_8 is isomorphic to C_2 \times C_2 not C_2 \times C_4. So, if you look at the 8th roots of unity, they are generated by e^{\tfrac{\pi i}{4}}. That to the 4th power is -1, which is rational, even though there are 8 roots of unity.
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    Re: Splitting Fields - Dummit and Foote - Example - page 541 - x^p - 2, p prime

    Ok, I have a little more time to try to explain what I was muddling through before.

    Let's look at x^8-2. It's roots are e^{2i\pi \tfrac{k}{n}}\sqrt[8]{2} where k=0,1,2,3,4,5,6,7. And,
    \begin{align*}x^8-1 & = (x^4-1)(x^4+1) \\ & = (x^2-1)(x^2+1)(x^4+1) \\ & = (x-1)(x+1)(x^2+1)(x^4+1)\end{align*}

    x^2+1 = (x - e^{i\pi\tfrac{1}{2}})(x - e^{i\pi\tfrac{3}{2}}) (Polynomial 1)
    x^4+1 = (x - e^{i\pi \tfrac{1}{4}})(x - e^{i\pi\tfrac{3}{4}})(x - e^{i\pi\tfrac{5}{4}})(x - e^{i\pi\tfrac{7}{4}}) (Polynomial 2)

    Notice how the numerators in the exponents of Polynomial 1 are precisely the elements of the multiplicative group of \mathbb{Z} / 4\mathbb{Z}: 1,3. The numerators in the exponents of Polynomial 2 are precisely the elements of the multiplicative group of \mathbb{Z} / 8\mathbb{Z}: 1,3,5,7.

    In general, x^n-1 is factored linearly by terms of the form \left( e^{2i\pi} \right)^{\tfrac{k}{d}} where k is in the multiplicative group of \mathbb{Z} / d\mathbb{Z} and d divides n. This corresponds to a minimal polynomial whose terms are (x-e^{2i \pi \tfrac{k}{n}}) for all 1\le k \le n where \mbox{gcd}(k,n)=1. This is called the cyclotomic polynomial corresponding to the n-th roots of unity. Since this polynomial is the minimal irreducible polynomial over the rationals with e^{2i\pi \tfrac{1}{n}} as a root, it is the order of the cyclotomic field of the n-th roots of unity over the rationals.

    This order is given by Euler's totient function, \varphi, which counts the number of positive integers less than n that are relatively prime to n. So, \varphi(p) = p-1 for any prime p.
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