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 $\displaystyle x^p - 2, p $ a prime - see attached.

I follow the example down to the following statement:

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

BUT ... then D&F write:

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

I also find it hard to follow the statement:

" ... ... ... Since both $\displaystyle \mathbb{Q} ( \sqrt[p]{2} ) $ and $\displaystyle \mathbb{Q} ( \zeta_p ) $ are subfields, the degree of the extension over $\displaystyle \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

$\displaystyle [\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 $\displaystyle \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 $\displaystyle [K_1 \ : \ F] = n, [ K_2 \ : \ F ] = m $ where m and n are relatively prime: (n, m) = 1.

Then $\displaystyle [K_1K_2 \ : \ F] = [K_1 \ : \ F] [ K_2 \ : \ F ] = nm $