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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 a prime - see attached.

I follow the example down to the following statement:

" ... ... ... so the splitting field is precisely "

BUT ... then D&F write:

This field contains the cyclotomic field of roots of unity and is generated over it by , hence is an extension of at most p. It follows that the degree of this extension over is .

*** 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 over is .

I also find it hard to follow the statement:

" ... ... ... Since both and are subfields, the degree of the extension over 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

... ... "

*** Can someone please try to make the above clearer - why exactly is the degree of the extension over 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 where m and n are relatively prime: (n, m) = 1.

Then

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 is the set of all -th roots of unity. Additionally, . So, the -th roots of unity are solutions to . Obviously, 1 is a solution. But the other solutions can not be real so long as . We know that the multiplicative group of 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 (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, . The second factor is irreducible over the rationals, and it is also irreducible over . I hope this helps a little. If I have more time, I will try to write up a bit more explanation.

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 is isomorphic to not . So, if you look at the 8th roots of unity, they are generated by . That to the 4th power is -1, which is rational, even though there are 8 roots of unity.

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 . It's roots are where . And,

(Polynomial 1)

(Polynomial 2)

Notice how the numerators in the exponents of Polynomial 1 are precisely the elements of the multiplicative group of : 1,3. The numerators in the exponents of Polynomial 2 are precisely the elements of the multiplicative group of : 1,3,5,7.

In general, is factored linearly by terms of the form where is in the multiplicative group of and divides . This corresponds to a minimal polynomial whose terms are for all where . 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 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, , which counts the number of positive integers less than that are relatively prime to . So, for any prime .