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

1. ## 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 $[K_1K_2 \ : \ F] = [K_1 \ : \ F] [ K_2 \ : \ F ] = nm$

2. ## 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.

3. ## 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.

4. ## 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 $p$.