MHB Splitting Fields - Beachy and Blair - Example 6.4.2

[Math Amateur]

I am reading Section 6.4: Splitting Fields in Beachy and Blair: Abstract Algebra.

I am currently studying Example 6.4.2 on page 290 which concerns the splitting field of $$ x^3 - 2 \text{ over } \mathbb{Q} $$ .

In Example 6.4.2, B&B show that the splitting field of $$ x^3 - 2 \text{ over } \mathbb{Q} $$ is $$ \mathbb{Q} ( \sqrt[3]{2} , \omega ) $$ where $$ \omega $$ is any complex number such that $$ \omega^3 = 1 $$.

Then we read the following:

"We claim that the degree of the splitting field over $$ \mathbb{Q}$$ is 6. Since $$ x^3 - 2 $$ is irreducible over $$ \mathbb{Q} $$, we have $$ [ \mathbb{Q} ( \sqrt[3]{2} \ : \ \mathbb{Q} ] = 3 $$. The polynomial $$ x^2 + x + 1 $$ is irreducible over $$ \mathbb{Q} $$ and stays irreducible over $$ \mathbb{Q} ( \sqrt[3]{2} ) $$ since it has no root in that field. The degree of $$ x^2 + x + 1 $$ is not a divisor of $$ [ \mathbb{Q} ( \sqrt[3]{2} \ : \ \mathbb{Q} ] $$. Since $$ \omega $$ is a root of $$ x^2 + x + 1 $$, this implies that $$ [ \mathbb{Q} ( \sqrt[3]{2} , \omega ) \ : \ \mathbb{Q} ( \sqrt[3]{2} ) ] = 2$$."

My question is as follows:

Why do B&B make the point that the degree of $$ x^2 + x + 1 $$ is not a divisor of $$ [ \mathbb{Q} ( \sqrt[3]{2}) \ : \ \mathbb{Q} ] $$?

Why is this important to check before you apply the result that $$ [ \mathbb{Q} ( \sqrt[3]{2}, \omega) \ : \ \mathbb{Q} ] = [ \mathbb{Q} ( \sqrt[3]{2}, \omega) \ : \ \mathbb{Q} ( \sqrt[3]{2}) ] [ \mathbb{Q} ( \sqrt[3]{2}) \ : \ \mathbb{Q} ] $$?

I would appreciate some help regarding this issue.

Peter

 

[Deveno]

Well, let's look at a slightly different polynomial:

$x^8 - 2$.

It is clear that this polynomial, too, is irreducible over $\Bbb Q$, and we have:

$[\Bbb Q(\sqrt[8]{2}):\Bbb Q] = 8$.

Next, let's look at a primitive 8-th root of 1, $\zeta$. What is the minimal polynomial of such a complex number?

Well, if $m(x)$ is that minimal polynomial, we have $m(x)|(x^8 - 1)$.

Now $x^8 - 1 = (x^4 + 1)(x^2 + 1)(x +1)(x - 1)$ and none of the roots of the last three factors have multiplicative order 8 in the complex numbers. Since $x^4 + 1$ is indeed irreducible over the rationals, this must be $m(x)$.

However, in THIS case, we see that its degree (4) DOES divide $[\Bbb Q(\sqrt[8]{2}):\Bbb Q]$. And I claim that:

$[\Bbb Q(\sqrt[8]{2},\zeta):\Bbb Q] = 16$ (and NOT 32).

One way to see this is to show that: $\Bbb Q(\sqrt[8]{2},\zeta) = \Bbb Q(\sqrt[8]{2},i)$ and the latter extension has degree 16 over the rationals (since $i \not\in \Bbb Q(\sqrt[8]{2}) \subseteq \Bbb R$), and $i$ satisfies a polynomial of degree 2 in $\Bbb Q(\sqrt[8]{2})[x]$). This boils down to showing:

$i \in \Bbb Q(\sqrt[8]{2},\zeta)$ and $\zeta \in \Bbb Q(\sqrt[8]{2},i)$.

Note that $\sqrt{2} = (\sqrt[8]{2})^4 \in \Bbb Q(\sqrt[8]{2})$ ,and that:

$z = \dfrac{\sqrt{2}}{2} + i\dfrac{\sqrt{2}}{2}$ is a root of $x^4 + 1$, which shows $\zeta \in\Bbb Q(\sqrt[8]{2},i)$ because we have:

$\zeta = z,z^3,z^5$ or $z^7$ (this is because all of the 8-th roots of unity form a finite cyclic subgroup of $\Bbb C-\{0\}$, and these are the only powers of a generator that can also be generators).

On the other hand, we have $[(\zeta)^2]^4 - 1 = 0$, and since $\zeta^4 \neq 1$ (since $\zeta$ is primitive), it follows that $\zeta^2$ is a root of $x^2 + 1$, that is:

$i = \pm \zeta^2$ so that $i \in \Bbb Q(\sqrt[8]{2},\zeta)$.

******************************

In the example in B&B, it is "obvious" that $\omega \not\in \Bbb Q(\sqrt{2})$, and so it must lie in some proper extension, and the smallest degree such an extension can have is 2, and since $\omega$ satisfies a rational quadratic, the largest degree such an extension can have is 2. The caveat is probably put there, to keep you from leaping to the conclusion that if $\alpha$ is an $n$-th primitive root of unity, with minimal polynomial of degree $k$, that the splitting field of:

$x^n - d$ (for $d$ such that this is irreducible over the rationals) that is: $\Bbb Q(\sqrt[n]{d},\alpha)$

has degree $kn$ over $\Bbb Q$, which as we have seen, is not always true.

 

[Math Amateur]

Well, let's look at a slightly different polynomial:

$x^8 - 2$.

It is clear that this polynomial, too, is irreducible over $\Bbb Q$, and we have:

$[\Bbb Q(\sqrt[8]{2}):\Bbb Q] = 8$.

Next, let's look at a primitive 8-th root of 1, $\zeta$. What is the minimal polynomial of such a complex number?

Well, if $m(x)$ is that minimal polynomial, we have $m(x)|(x^8 - 1)$.

Now $x^8 - 1 = (x^4 + 1)(x^2 + 1)(x +1)(x - 1)$ and none of the roots of the last three factors have multiplicative order 8 in the complex numbers. Since $x^4 + 1$ is indeed irreducible over the rationals, this must be $m(x)$.

However, in THIS case, we see that its degree (4) DOES divide $[\Bbb Q(\sqrt[8]{2}):\Bbb Q]$. And I claim that:

$[\Bbb Q(\sqrt[8]{2},\zeta):\Bbb Q] = 16$ (and NOT 32).

One way to see this is to show that: $\Bbb Q(\sqrt[8]{2},\zeta) = \Bbb Q(\sqrt[8]{2},i)$ and the latter extension has degree 16 over the rationals (since $i \not\in \Bbb Q(\sqrt[8]{2}) \subseteq \Bbb R$), and $i$ satisfies a polynomial of degree 2 in $\Bbb Q(\sqrt[8]{2})[x]$). This boils down to showing:

$i \in \Bbb Q(\sqrt[8]{2},\zeta)$ and $\zeta \in \Bbb Q(\sqrt[8]{2},i)$.

Note that $\sqrt{2} = (\sqrt[8]{2})^4 \in \Bbb Q(\sqrt[8]{2})$ ,and that:

$z = \dfrac{\sqrt{2}}{2} + i\dfrac{\sqrt{2}}{2}$ is a root of $x^4 + 1$, which shows $\zeta \in\Bbb Q(\sqrt[8]{2},i)$ because we have:

$\zeta = z,z^3,z^5$ or $z^7$ (this is because all of the 8-th roots of unity form a finite cyclic subgroup of $\Bbb C-\{0\}$, and these are the only powers of a generator that can also be generators).

On the other hand, we have $[(\zeta)^2]^4 - 1 = 0$, and since $\zeta^4 \neq 1$ (since $\zeta$ is primitive), it follows that $\zeta^2$ is a root of $x^2 + 1$, that is:

$i = \pm \zeta^2$ so that $i \in \Bbb Q(\sqrt[8]{2},\zeta)$.

******************************

In the example in B&B, it is "obvious" that $\omega \not\in \Bbb Q(\sqrt{2})$, and so it must lie in some proper extension, and the smallest degree such an extension can have is 2, and since $\omega$ satisfies a rational quadratic, the largest degree such an extension can have is 2. The caveat is probably put there, to keep you from leaping to the conclusion that if $\alpha$ is an $n$-th primitive root of unity, with minimal polynomial of degree $k$, that the splitting field of:

$x^n - d$ (for $d$ such that this is irreducible over the rationals) that is: $\Bbb Q(\sqrt[n]{d},\alpha)$

has degree $kn$ over $\Bbb Q$, which as we have seen, is not always true.

Thanks Deveno ... Most helpful, especially since we have an example to illustrate the points you make ...

Still reflecting and thinking over elements of your post ...

Thanks again,

Peter

 

[Math Amateur]

Well, let's look at a slightly different polynomial:$x^8 - 2$.It is clear that this polynomial, too, is irreducible over $\Bbb Q$, and we have:$[\Bbb Q(\sqrt[8]{2}):\Bbb Q] = 8$.Next, let's look at a primitive 8-th root of 1, $\zeta$. What is the minimal polynomial of such a complex number?Well, if $m(x)$ is that minimal polynomial, we have $m(x)|(x^8 - 1)$.Now $x^8 - 1 = (x^4 + 1)(x^2 + 1)(x +1)(x - 1)$ and none of the roots of the last three factors have multiplicative order 8 in the complex numbers. Since $x^4 + 1$ is indeed irreducible over the rationals, this must be $m(x)$.However, in THIS case, we see that its degree (4) DOES divide $[\Bbb Q(\sqrt[8]{2}):\Bbb Q]$. And I claim that:$[\Bbb Q(\sqrt[8]{2},\zeta):\Bbb Q] = 16$ (and NOT 32).One way to see this is to show that: $\Bbb Q(\sqrt[8]{2},\zeta) = \Bbb Q(\sqrt[8]{2},i)$ and the latter extension has degree 16 over the rationals (since $i \not\in \Bbb Q(\sqrt[8]{2}) \subseteq \Bbb R$), and $i$ satisfies a polynomial of degree 2 in $\Bbb Q(\sqrt[8]{2})[x]$). This boils down to showing:$i \in \Bbb Q(\sqrt[8]{2},\zeta)$ and $\zeta \in \Bbb Q(\sqrt[8]{2},i)$.Note that $\sqrt{2} = (\sqrt[8]{2})^4 \in \Bbb Q(\sqrt[8]{2})$ ,and that:$z = \dfrac{\sqrt{2}}{2} + i\dfrac{\sqrt{2}}{2}$ is a root of $x^4 + 1$, which shows $\zeta \in\Bbb Q(\sqrt[8]{2},i)$ because we have:$\zeta = z,z^3,z^5$ or $z^7$ (this is because all of the 8-th roots of unity form a finite cyclic subgroup of $\Bbb C-\{0\}$, and these are the only powers of a generator that can also be generators).On the other hand, we have $[(\zeta)^2]^4 - 1 = 0$, and since $\zeta^4 \neq 1$ (since $\zeta$ is primitive), it follows that $\zeta^2$ is a root of $x^2 + 1$, that is:$i = \pm \zeta^2$ so that $i \in \Bbb Q(\sqrt[8]{2},\zeta)$.******************************In the example in B&B, it is "obvious" that $\omega \not\in \Bbb Q(\sqrt{2})$, and so it must lie in some proper extension, and the smallest degree such an extension can have is 2, and since $\omega$ satisfies a rational quadratic, the largest degree such an extension can have is 2. The caveat is probably put there, to keep you from leaping to the conclusion that if $\alpha$ is an $n$-th primitive root of unity, with minimal polynomial of degree $k$, that the splitting field of:$x^n - d$ (for $d$ such that this is irreducible over the rationals) that is: $\Bbb Q(\sqrt[n]{d},\alpha)$has degree $kn$ over $\Bbb Q$, which as we have seen, is not always true.

Thanks again for your help Deveno ...

Just a clarification regarding your post ...

You write:"Now $x^8 - 1 = (x^4 + 1)(x^2 + 1)(x +1)(x - 1)$ and none of the roots of the last three factors have multiplicative order 8 in the complex numbers. Since $x^4 + 1$ is indeed irreducible over the rationals, this must be $m(x)$."

I am wondering about your statement that $x^4 + 1$ is $m(x)$?

What is the logic for this statement? Further, does the statement that "none of the roots of the last three factors have multiplicative order 8 in the complex numbers" come into the analysis?

Just to explain my unease … B&B define minimal polynomial in terms of a particular algebraic element as follows:

Definition: Let F be an extension field of K and let u be an algebraic element of F. The monic polynomial p(x) of minimal degree in K[x] such that p(u) = 0 is called the minimal polynomial of u over K. The degree of the minimal polynomial over K is called the degree of u over K.

So is there a minimal polynomial for $$x^8 - 1$$ - or only a minimal polynomial for a particular root of $$x^8 - 1$$?

Hope you can help.

Just a note to say that, in general, arguments in various texts about minimal polynomials often seem plausible to me but do not seem to measure up to a proof of minimality - for example that $$x^4 + 1$$ is the minimal polynomial of $$sqrt [4]{1}$$ seems entirely plausible but how does one prove this statement - how do we KNOW that there is not a polynomial with lower degree with $$sort [4]{1}$$ as a root? Peter

 

[Deveno]

Well arguments about minimal polynomials tend to go like this:

Suppose that $K$ is an extension of a field $F$, and that $\alpha \in K$. and that the minimal polynomial of $\alpha$ is $m(x) \in F[x]$. If for $p(x) \in F[x]$, we have $p(\alpha) = 0$, then $m(x)|p(x)$.

To see this, write:

$p(x) = q(x)m(x) + r(x)$, where $\text{deg}(r) < \text{deg}(m)$, or $r(x) = 0$.

Then

$0 = p(\alpha) = q(\alpha)m(\alpha) + r(\alpha) = q(\alpha)0 + r(\alpha) = r(\alpha)$.

Since by the minimal degree of $m$, we cannot have $\text{deg}(r) < \text{deg}(m)$, it must be $r(x) = 0$.

Now since for a primitive 8-th root of unity $\zeta$ we have by definition: $\zeta^8 = 1$, the minimal polynomial IS a factor of $x^8 - 1$. We can discard the rational roots -1, and 1, since they are 8-th roots of unity, but not PRIMITIVE roots.

We can also discard the 4-th roots of one (which include the two we discard earlier). In fact, we have this complete factorization:

$x^8 - 1 = (x^4 + 1)(x^2 + 1)(x + 1)(x - 1)$

$= (x - \zeta)(x - \zeta^3)(z - \zeta^5)(x - \zeta^7)(x + i)(x - i)(x + 1)(x - 1)$

(these are all powers of $\zeta$, the ones labeled differently are the powers $k$ for which $\text{gcd}(k,8) > 1$).

The group of 8-th roots of unity breaks down as follows:

$1$, the identity (order 1)
$-1$ the primitive square root of unity (of order 2)
$i,-i$ the primitive 4-th roots of unity (of order 4)
$\zeta,\zeta^3,\zeta^5,\zeta^7$ (primitive 8-th roots of unity, two square roots of $i$, two square roots of $-i$).

Note how these correspond to the factorization, roots of equal order are roots of the same irreducible factor.

I hope you can see that a primitive 8-th root of unity to the fourth power must be a primitive square root of 1:

$\zeta^8 = 1$ and $\zeta^k \neq 1$, for $k = 1,2,3,4,5,6,7$ means:

$1 = (\zeta^4)^2$ but $\zeta^4 \neq 1$, thus $\zeta^4 = -1$, so $\zeta^4 + 1 = 0$, that is:

$\zeta$ is a root of $x^4 + 1$, and this polynomial is irreducible over $\Bbb Q$, so IS the minimal polynomial of $\zeta$ (it's clearly monic).

Now it may be clear that $x^4 + 1$ has no ROOT in $\Bbb Q$ (for all rational numbers $q \neq 0$, we have:

$q^4 = (q^2)^2 > 0$ so $q^4 + 1 > 1 + 1 = 2 > 0$ and thus cannot be 0), but is it possible that:

$x^4 + 1 = (x + ax + b)(x + cx + d)$, for rational numbers $a,b,c,d$?

This would mean:

$a + c = 0$
$ac + b + d = 0$
$ad + bc = 0$
$bd = 1$

So we have $c = -a$ and $d = \dfrac{1}{b}$, and thus:

$-a^2 + b + \dfrac{1}{b} = 0$
$\dfrac{a}{b} - ab = 0$.

The second equation is equivalent to:

$a(1 - b^2) = 0$, so either $a = 0$, or $b^2 = 1$ (or both). Let's look at $a = 0$ first.

$(x^2 + b)\left(x^2 + \dfrac{1}{b}\right) = x^4 + \left(b + \dfrac{1}{b}\right)x + 1$, which means:

$b + \dfrac{1}{b} = 0 \implies b^2 = -1$ which has no (rational) solution.

If $b^2 = 1$, then $b = \pm 1$. Suppose $b = 1$. Then:

$x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1) = x^4 + (2 - a^2)x^2 + 1$, so:

$a^2 = 2$, which has no rational solution (note it DOES factor over $\Bbb Q(\sqrt{2})$, the "base field" matters).

A similar result holds if we take $b = -1$.

Hopefully, this convinces you that $x^4 + 1$ IS irreducible over $\Bbb Q$, and thus has no "smaller factor" $f(x)$ for which $f(\zeta) = 0$.************************

I note in passing that it is an (important) theorem that the $n$-th cyclotomic polynomial $\Phi_n(x)$ is in fact, irreducible, for every natural number $n$. This is the special case $n = 8$ (which is fairly easy, because 8 is a small number, and highly divisible).

 

[Math Amateur]

Well arguments about minimal polynomials tend to go like this:

Suppose that $K$ is an extension of a field $F$, and that $\alpha \in K$. and that the minimal polynomial of $\alpha$ is $m(x) \in F[x]$. If for $p(x) \in F[x]$, we have $p(\alpha) = 0$, then $m(x)|p(x)$.

To see this, write:

$p(x) = q(x)m(x) + r(x)$, where $\text{deg}(r) < \text{deg}(m)$, or $r(x) = 0$.

Then

$0 = p(\alpha) = q(\alpha)m(\alpha) + r(\alpha) = q(\alpha)0 + r(\alpha) = r(\alpha)$.

Since by the minimal degree of $m$, we cannot have $\text{deg}(r) < \text{deg}(m)$, it must be $r(x) = 0$.

Now since for a primitive 8-th root of unity $\zeta$ we have by definition: $\zeta^8 = 1$, the minimal polynomial IS a factor of $x^8 - 1$. We can discard the rational roots -1, and 1, since they are 8-th roots of unity, but not PRIMITIVE roots.

We can also discard the 4-th roots of one (which include the two we discard earlier). In fact, we have this complete factorization:

$x^8 - 1 = (x^4 + 1)(x^2 + 1)(x + 1)(x - 1)$

$= (x - \zeta)(x - \zeta^3)(z - \zeta^5)(x - \zeta^7)(x + i)(x - i)(x + 1)(x - 1)$

(these are all powers of $\zeta$, the ones labeled differently are the powers $k$ for which $\text{gcd}(k,8) > 1$).

The group of 8-th roots of unity breaks down as follows:

$1$, the identity (order 1)
$-1$ the primitive square root of unity (of order 2)
$i,-i$ the primitive 4-th roots of unity (of order 4)
$\zeta,\zeta^3,\zeta^5,\zeta^7$ (primitive 8-th roots of unity, two square roots of $i$, two square roots of $-i$).

Note how these correspond to the factorization, roots of equal order are roots of the same irreducible factor.

I hope you can see that a primitive 8-th root of unity to the fourth power must be a primitive square root of 1:

$\zeta^8 = 1$ and $\zeta^k \neq 1$, for $k = 1,2,3,4,5,6,7$ means:

$1 = (\zeta^4)^2$ but $\zeta^4 \neq 1$, thus $\zeta^4 = -1$, so $\zeta^4 + 1 = 0$, that is:

$\zeta$ is a root of $x^4 + 1$, and this polynomial is irreducible over $\Bbb Q$, so IS the minimal polynomial of $\zeta$ (it's clearly monic).

Now it may be clear that $x^4 + 1$ has no ROOT in $\Bbb Q$ (for all rational numbers $q \neq 0$, we have:

$q^4 = (q^2)^2 > 0$ so $q^4 + 1 > 1 + 1 = 2 > 0$ and thus cannot be 0), but is it possible that:

$x^4 + 1 = (x + ax + b)(x + cx + d)$, for rational numbers $a,b,c,d$?

This would mean:

$a + c = 0$
$ac + b + d = 0$
$ad + bc = 0$
$bd = 1$

So we have $c = -a$ and $d = \dfrac{1}{b}$, and thus:

$-a^2 + b + \dfrac{1}{b} = 0$
$\dfrac{a}{b} - ab = 0$.

The second equation is equivalent to:

$a(1 - b^2) = 0$, so either $a = 0$, or $b^2 = 1$ (or both). Let's look at $a = 0$ first.

$(x^2 + b)\left(x^2 + \dfrac{1}{b}\right) = x^4 + \left(b + \dfrac{1}{b}\right)x + 1$, which means:

$b + \dfrac{1}{b} = 0 \implies b^2 = -1$ which has no (rational) solution.

If $b^2 = 1$, then $b = \pm 1$. Suppose $b = 1$. Then:

$x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1) = x^4 + (2 - a^2)x^2 + 1$, so:

$a^2 = 2$, which has no rational solution (note it DOES factor over $\Bbb Q(\sqrt{2})$, the "base field" matters).

A similar result holds if we take $b = -1$.

Hopefully, this convinces you that $x^4 + 1$ IS irreducible over $\Bbb Q$, and thus has no "smaller factor" $f(x)$ for which $f(\zeta) = 0$.************************

I note in passing that it is an (important) theorem that the $n$-th cyclotomic polynomial $\Phi_n(x)$ is in fact, irreducible, for every natural number $n$. This is the special case $n = 8$ (which is fairly easy, because 8 is a small number, and highly divisible).

Hi Deveno … I really appreciate the help ...

Just working through this carefully now ...

Thanks so much ...

Peter