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

In summary, the conversation discusses an example from Dummit and Foote's Section 13.4 on Splitting Fields and Algebraic Closures, specifically the example of the splitting field of x^p - 2 with p as a prime. The conversation addresses the statement made by D&F that the splitting field is precisely Q (sqrt[p]{2}, zeta_p) and explains why the degree of the extension over Q is at most p(p-1) using D&F's Corollary 22. There is also a discussion on why the degree of the extension is divisible by p and p-1, and why this is important. The conversation also mentions that the argument in the example depends on p being prime.
  • #1
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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.

PeterNote

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
 
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  • #2
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-abstract-algebra-14/automorphisms-splitting-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 $\Bbb Q$).
 
  • #3
Deveno said:
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-abstract-algebra-14/automorphisms-splitting-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
 
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  • #4
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: $\Bbb Q(-1) = \Bbb Q$.
 
  • #5
Deveno said:
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
 

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

What is a splitting field?

A splitting field is a field extension that contains all the roots of a given polynomial.

What is the significance of a splitting field?

A splitting field allows us to factor a polynomial into linear factors, making it easier to solve equations and understand the behavior of the polynomial.

How do you find the splitting field of a polynomial?

To find the splitting field of a polynomial, we first need to determine its irreducible factors. Then, we construct a field extension by adjoining the roots of those irreducible factors to the base field.

What is the importance of Example on page 541 in Dummit and Foote?

This example illustrates the concept of splitting fields for a polynomial of the form x^p - 2, where p is a prime number. It shows how to find the splitting field of this polynomial and how to use it to factor the polynomial into linear factors.

Why is p chosen to be a prime number in this example?

Choosing p as a prime number simplifies the algebraic structure of the polynomial and its splitting field, making it easier to understand and work with. Additionally, many important polynomials in mathematics have prime numbers as their coefficients or exponents.

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