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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 [TEX] x^p - 2, p [/TEX] a prime - see attached.

I follow the example down to the following statement:

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

BUT ... then D&F write:

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


*** 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 [TEX] \mathbb{Q} ( \sqrt[p]{2}, \zeta_p )[/TEX] over [TEX] \mathbb{Q} [/TEX] is [TEX] \le p(p-1) [/TEX].

I also find it hard to follow the statement:

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

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

*** Can someone please try to make the above clearer - why exactly is the degree of the extension over [TEX] \mathbb{Q} [/TEX] 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 [TEX] [K_1 \ : \ F] = n, [ K_2 \ : \ F ] = m [/TEX] where m and n are relatively prime: (n, m) = 1.
Then [TEX] [K_1K_2 \ : \ F] = [K_1 \ : \ F] [ K_2 \ : \ F ] = nm [/TEX]

2. The above has also been posted on MHF
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
By the so-called "tower rule" if we have a tower of field extensions:

$F < E < K$, we know that:

$[K:F] = [K:E]\ast[E:F]$

whence it follows that both $[K:E]$ and $[E:F]$ divide $[K:F]$ (this is just a simple fact about divisibility of integers).

Note that this is true of ANY subfield $E$ of the field $K$ containing $F$.

As was noted in an earlier post:

$\Bbb Q(\sqrt[p]{2},\zeta_p) = \Bbb Q(\zeta_p)(\sqrt[p]{2})$.

But $\Bbb Q(\zeta_p)$ *IS*the cyclotomic field containing the $p\ p$-th roots of unity ($\zeta_p$ is a PRIMITIVE $p$-th root of unity, and all the other $p$-th roots of unity are powers of $\zeta_p$. To see this note that since $p$ is prime, and by definition of primitive $p$-th root of unity, $\zeta_p$ is a root of:

$x^p - 1 \in \Bbb Q[x]$.

By a corollary to Lagrange's theorem, we have that every non-identity element of $\langle \zeta_p \rangle$ is of order $p$ (since this is a cyclic group of prime order in $\Bbb C^{\ast}$ under complex multiplication), which gives $p-1$ roots, and clearly 1 is also a root, giving $p$ roots. Since a polynomial of degree $p$ over a field can have at MOST $p$ roots, this is indeed ALL of them and so $x^p - 1$ splits in $\Bbb Q(\zeta_p)$).

What is the degree of this extension $\Bbb Q(\zeta_p)$? It is $p-1$, since $\zeta_p$ satisfies the (irreducible) $p$-th cyclotomic polynomial:

$x^{p-1} + \cdots + x + 1 = \dfrac{x^p - 1}{x - 1} \in \Bbb Q[x]$

The irreducibility of cyclotomic polynomials has been discussed in this thread:

http://mathhelpboards.com/linear-ab...ng-field-m-th-cyclotomic-polynomial-6185.html

which you might find worthwhile to read.

Now $\sqrt[p]{2}$ satisfies the degree $p$ polynomial:

$x^p - 2 \in \Bbb Q(\zeta_p)[x]$, so its minimal polynomial must be of degree $\leq p$.

I have addressed part of the divisibility question in my opening remarks. Here is the rest of what you need to know:

Suppose $a|n$ and $b|n$ with $\text{gcd}(a,b) = 1$ for $a,b,n \in \Bbb Z^{+}$.

We then know that $n$ is a common multiple of $a$ and $b$, in particular:

$\text{lcm}(a,b)|n$. Since $\text{lcm}(a,b) = \dfrac{ab}{\text{gcd}(a,b)} = ab$, we have that $ab|n$.

Therefore, we have an integer $n$ which is at MOST $p(p-1)$ and is divisible by $p(p-1)$. The latter fact implies $n = k(p)(p-1)$, and the former fact then forces $k = 1$.

The primality of $p$ enters into it at the very start: with the field $\Bbb Q(\zeta_p)$. To see this consider what field is generated by a 4-th root of unity (one of these roots is often given a "special" name), and explain why:

$x^3 + x^2 + x + 1$ is not irreducible over $\Bbb Q$

(Hint: factor $x^4 - 1$ over $\Bbb Q$).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
By the so-called "tower rule" if we have a tower of field extensions:

$F < E < K$, we know that:

$[K:F] = [K:E]\ast[E:F]$

whence it follows that both $[K:E]$ and $[E:F]$ divide $[K:F]$ (this is just a simple fact about divisibility of integers).

Note that this is true of ANY subfield $E$ of the field $K$ containing $F$.

As was noted in an earlier post:

$\Bbb Q(\sqrt[p]{2},\zeta_p) = \Bbb Q(\zeta_p)(\sqrt[p]{2})$.

But $\Bbb Q(\zeta_p)$ *IS*the cyclotomic field containing the $p\ p$-th roots of unity ($\zeta_p$ is a PRIMITIVE $p$-th root of unity, and all the other $p$-th roots of unity are powers of $\zeta_p$. To see this note that since $p$ is prime, and by definition of primitive $p$-th root of unity, $\zeta_p$ is a root of:

$x^p - 1 \in \Bbb Q[x]$.

By a corollary to Lagrange's theorem, we have that every non-identity element of $\langle \zeta_p \rangle$ is of order $p$ (since this is a cyclic group of prime order in $\Bbb C^{\ast}$ under complex multiplication), which gives $p-1$ roots, and clearly 1 is also a root, giving $p$ roots. Since a polynomial of degree $p$ over a field can have at MOST $p$ roots, this is indeed ALL of them and so $x^p - 1$ splits in $\Bbb Q(\zeta_p)$).

What is the degree of this extension $\Bbb Q(\zeta_p)$? It is $p-1$, since $\zeta_p$ satisfies the (irreducible) $p$-th cyclotomic polynomial:

$x^{p-1} + \cdots + x + 1 = \dfrac{x^p - 1}{x - 1} \in \Bbb Q[x]$

The irreducibility of cyclotomic polynomials has been discussed in this thread:

http://mathhelpboards.com/linear-ab...ng-field-m-th-cyclotomic-polynomial-6185.html

which you might find worthwhile to read.

Now $\sqrt[p]{2}$ satisfies the degree $p$ polynomial:

$x^p - 2 \in \Bbb Q(\zeta_p)[x]$, so its minimal polynomial must be of degree $\leq p$.

I have addressed part of the divisibility question in my opening remarks. Here is the rest of what you need to know:

Suppose $a|n$ and $b|n$ with $\text{gcd}(a,b) = 1$ for $a,b,n \in \Bbb Z^{+}$.

We then know that $n$ is a common multiple of $a$ and $b$, in particular:

$\text{lcm}(a,b)|n$. Since $\text{lcm}(a,b) = \dfrac{ab}{\text{gcd}(a,b)} = ab$, we have that $ab|n$.

Therefore, we have an integer $n$ which is at MOST $p(p-1)$ and is divisible by $p(p-1)$. The latter fact implies $n = k(p)(p-1)$, and the former fact then forces $k = 1$.

The primality of $p$ enters into it at the very start: with the field $\Bbb Q(\zeta_p)$. To see this consider what field is generated by a 4-th root of unity (one of these roots is often given a "special" name), and explain why:

$x^3 + x^2 + x + 1$ is not irreducible over $\Bbb Q$

(Hint: factor $x^4 - 1$ over $\Bbb Q$).
Thanks Deveno. That has helped to clear up a number of issues for me.

However, I am still confused regarding the primality of p. I still cannot see why it is necessary. Let me explain my confusion on this matter.

You write that the prime nature of p has to do with the field \(\displaystyle \mathbb{Q} ({\zeta}_p) \) - that is, specifically, we need \(\displaystyle {\zeta}_p \) to be a primitive \(\displaystyle p^{th} \) root of unity. But surely this does not necessitate or entail p being prime???

Check D&F.

If we read D&F pages 539-540 (see attached we find the definition of a primitive \(\displaystyle n^{th} \) root of unity as follows: (see attachment - bottom of page 539)
----------------------------------------------------------------------------------
"Definition. A generator of the cyclic group of all \(\displaystyle n^{th} \) roots of unity is called a primitive \(\displaystyle n^{th} \) root of unity

Let \(\displaystyle {\zeta}_n \) denote a primitive \(\displaystyle n^{th} \) root of unity. The other primitive \(\displaystyle n^{th} \) roots of unity are then the elements
\(\displaystyle {{\zeta}_n}^a \) where \(\displaystyle 1 \le a \le n \) is an integer relatively prime to n, since there are other generators for a cyclic group of order n. In particular there are precisely \(\displaystyle \phi (n) \) primitive \(\displaystyle n^{th} \) roots of unity, where \(\displaystyle \phi (n) \) denotes the Euler \(\displaystyle \phi \) function."
-----------------------------------------------------------------------------------

Note that the requirement is that a is relatively prime to n - not that n is prime!

On the top of page 540 D&F continue as follows:

------------------------------------------------------------------------------------

"Over \(\displaystyle \mathbb{C} \) we can see all of this directly by setting

\(\displaystyle {\zeta}_n = e^{2 \pi i /n} \)

(the first \(\displaystyle n^{th} \) root of unity counterclockwise from 1). Then all the other roots of unity are powers of \(\displaystyle {\zeta}_n \):

\(\displaystyle e^{2 \pi k i /n} = {{\zeta}_n}^k \)

so that \(\displaystyle {\zeta}_n \) is one possible generator for the multiplicative group of \(\displaystyle n^{th} \) roots of unity. When we view the roots of unity in \(\displaystyle \mathbb{C} \) we shal usually use \(\displaystyle {\zeta}_n \) to denote this choice of a primitive \(\displaystyle n^{th} \) root of unity."

--------------------------------------------------------------------------------

As an example, for the case of the polynomial \(\displaystyle x^4 - 1 \) we can use

\(\displaystyle {{\zeta}_4}^1 = e^{2 \pi 1 i /4} = e^{ \pi i / 2 } = i \) and then this generates the 4 roots.

So I still do not see the reason for the primality of p - but I feel there must be something I am missing ???

Can you please clarify this issue for me.

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Yes, you are missing something...in general, the splitting field for:

$x^n - 1$ has degree $\phi(n)$ over $\Bbb Q$. It's not that we can't form this splitting field for non-prime $n$, it's just that its degree is typically much smaller than $n-1$ (the more composite $n$ is, the smaller $\phi(n)$ tends to be).

It is only when $n$ is prime that we can conclude $\phi(n) = n-1$. If you think about it, that's sort of the DEFINITION of a prime integer: every positive integer less than it, is co-prime to it (1 is a special case, and usually not counted as prime).

The whole of D&F's argument rests on proving the degree of a certain kind of extension, which they give an explicit formula for. For $n = 4$, you can see this formula fails, because:

$[\Bbb Q(i,\sqrt[4]{2}):\Bbb Q] = 8 \neq 12 = (n-1)n$, as we have seen in another thread.

The reason for this is 4 is not prime.

The motivation behind the proof is that it is a bit difficult to show DIRECTLY that:

$x^p - 2$ is irreducible over $\Bbb Q(\zeta_p)$: if $p$ is large there are a lot of possible factorizations to check. For example, for $p = 5$, we have to check:

1) that there are no roots in $\Bbb Q(\zeta_5)$ (this also eliminates the possibility of quartic factors)
2) that there are no quadratic (and hence cubic) factors.

Given that each element of $\Bbb Q(\zeta_5)$ is equivalent to solving for 5 "unknowns", this creates a system of 5 equations in 5 unknowns, which may in practice be quite difficult to solve.

When $p$ is prime we get ALL the primitive roots (any non-trivial (unity) one) by adjoining ANY root besides 1. This is not true for n = 4, for example: adjoining the root -1 does not enlarge our field: $\Bbb Q(-1) = \Bbb Q$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Yes, you are missing something...in general, the splitting field for:

$x^n - 1$ has degree $\phi(n)$ over $\Bbb Q$. It's not that we can't form this splitting field for non-prime $n$, it's just that its degree is typically much smaller than $n-1$ (the more composite $n$ is, the smaller $\phi(n)$ tends to be).

It is only when $n$ is prime that we can conclude $\phi(n) = n-1$. If you think about it, that's sort of the DEFINITION of a prime integer: every positive integer less than it, is co-prime to it (1 is a special case, and usually not counted as prime).

The whole of D&F's argument rests on proving the degree of a certain kind of extension, which they give an explicit formula for. For $n = 4$, you can see this formula fails, because:

$[\Bbb Q(i,\sqrt[4]{2}):\Bbb Q] = 8 \neq 12 = (n-1)n$, as we have seen in another thread.

The reason for this is 4 is not prime.

The motivation behind the proof is that it is a bit difficult to show DIRECTLY that:

$x^p - 2$ is irreducible over $\Bbb Q(\zeta_p)$: if $p$ is large there are a lot of possible factorizations to check. For example, for $p = 5$, we have to check:

1) that there are no roots in $\Bbb Q(\zeta_5)$ (this also eliminates the possibility of quartic factors)
2) that there are no quadratic (and hence cubic) factors.

Given that each element of $\Bbb Q(\zeta_5)$ is equivalent to solving for 5 "unknowns", this creates a system of 5 equations in 5 unknowns, which may in practice be quite difficult to solve.

When $p$ is prime we get ALL the primitive roots (any non-trivial (unity) one) by adjoining ANY root besides 1. This is not true for n = 4, for example: adjoining the root -1 does not enlarge our field: $\Bbb Q(-1) = \Bbb Q$.
Thanks Deveno, that really clarified some major issues I had!

Peter