Find the possible values for $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ and provide an example of $(\alpha,\beta)$ for each possible value

Question

Let $\alpha, \beta \in \mathbb{C}$ s.t. $[\mathbb{Q}(\alpha):\mathbb{Q}] = [\mathbb{Q}(\beta):\mathbb{Q}] = 4$

Find the possible values for $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ and provide an example for each possible value.

We know that

$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] \leq [\mathbb{Q}(\alpha):\mathbb{Q}] [\mathbb{Q}(\beta):\mathbb{Q}] = 4 \cdot 4 = 16$

On the other hand,

$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}] = k \cdot 4 = 4k$

We got that $4|[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ and $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] \leq 16$, implying the possible values are in the set $\{4,8,12,16\}$

These are the examples I have

• $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 4$

Take $\alpha = \beta = \sqrt[4]{2}$

Here we have that $m_\alpha(\mathbb{Q}) = x^4 - 2$, then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$.

Since $\beta = \sqrt[4]{2} \in \mathbb{Q}(\sqrt[4]{2}) = \mathbb{Q}(\alpha)$, we have $m_\beta(\mathbb{Q}(\alpha)) = x - \sqrt[4]{2}$, then $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] = 1$

Therefore, $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] [\mathbb{Q}(\alpha):\mathbb{Q}] = 1 \cdot 4 = 4$

• $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 8$

Take $\alpha = \sqrt[4]{2}, \beta = \sqrt[4]{2} i$

Here we have that $m_\alpha(\mathbb{Q}) = x^4 - 2$, then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$.

Let $x=\sqrt[4]{2} i$. Squaring, $x^2 = - (\sqrt[4]{2})^2$, then $x^2 + (\sqrt[4]{2})^2 = 0$.

Since $(\sqrt[4]{2})^2 \in \mathbb{Q}(\sqrt[4]{2}) = \mathbb{Q}(\alpha)$, we have $m_\beta(\mathbb{Q}(\alpha)) = x^2 + (\sqrt[4]{2})^2$, then $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] = 2$.

Therefore, $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] [\mathbb{Q}(\alpha):\mathbb{Q}] = 2 \cdot 4 = 8$

• $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 16$

Take $\alpha = \sqrt[4]{2}, \beta = \sqrt[4]{3}$

Here we have that $m_\alpha(\mathbb{Q}) = x^4 - 2$, then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$.

Since $\beta = \sqrt[4]{3} \notin \mathbb{Q}(\sqrt[4]{2}) = \mathbb{Q}(\alpha)$, we have that $deg(m_\beta(\mathbb{Q}(\alpha))) > 1$.

Clearly $\beta = \sqrt[4]{3}$ solves $x^4 - 3 = 0$.

We now look for possible quadratic or cubic factors on $\mathbb{Q}(\sqrt[4]{2})[x]$.

$x^4 - 3 = (x^2 - \sqrt{3})(x^2 + \sqrt{3})$. But $\sqrt{3} \notin \mathbb{Q}(\sqrt[4]{2})$.

$x^4 - 3 = (x-\sqrt[4]{3})(x^3+\sqrt[4]{3}x^2+(\sqrt[4]{3})^2x+(\sqrt[4]{3})^3)$. But $\sqrt[4]{3} \notin \mathbb{Q}(\sqrt[4]{2})$.

This implies $m_\beta(\mathbb{Q}(\alpha)) = x^4 - 3$, then $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] = 4$.

Therefore, $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)] [\mathbb{Q}(\alpha):\mathbb{Q}] = 4 \cdot 4 = 16$

I got this problem a while ago in an Abstract Algebra II midterm, on which the professor later decided that the $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 12$ example wasn't required to earn full credit, given its extreme difficulty level.

I would like to know an example for $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 12$, for the sake of curiosity.

Answer 1

If $\left[\mathbb Q[\alpha,\beta]:\mathbb Q[\alpha]\right]=3,$ then $\beta$ must have a minimal polynomial some cubic monic polynomial $p(x)$ in $\mathbb Q[\alpha][x].$

But $x$ has a minimal polynomial of degree $4,$ $q(x)\in\mathbb Q[x]\subset\mathbb Q[\alpha][x].$

So $p(x)\mid q(x)$ in $\mathbb Q[\alpha][x].$ So $q(x)=p(x)m(x)$ for some monic polynomial in $\mathbb Q[\alpha][x],$ which means $q(x)$ must have exactly one root in $\mathbb Q[\alpha].$

So we require a polynomial of degree $4$ where each root is not in the field extension of the other roots, we can pick $\alpha$ one root, and $\beta$ another root.

So any polynomial of degree $4$ and splitting field of degree $24$ will work.

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Example without knowing splitting field result

For example, if $\alpha,\beta$ are two distinct roots of $x^4+x+1=0,$ let $s=\alpha+\beta, p=\alpha\beta.$

Then we can factor:

$$x^4+x+1 =(x^2-sx+p)(x^2+sx+p^{-1})$$ which tells us: $$s^2=p+\frac1p,\\s\left(p-\frac1p\right)=1,$$ and solving this for $w=p+\frac1p$ this gives a cubic equation:

$$w^3-4w-1=0$$

It is easy to see this has no rational root, and thus $x^3-4x-1$ is irreducible in $\mathbb Q[x].$

But $p\in\mathbb Q[\alpha,\beta],$ so $\mathbb Q[p+\frac1p]$ is a subfield of $\mathbb Q[\alpha,\beta]$ and is of degree $3$ over $\mathbb Q,$ and thus $\mathbb Q[\alpha,\beta]$ must be of degree a multiple of $3.$

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This will work for any $\alpha,\beta$ distinct roots of $x^4+ax^2+bx+c$ where $x^3-ax^2-4cx+(4ac-b^2)=0$ has no rational roots.

That $4ac-b^2$ looks awful fishy.

Answer 2

Take a polynomial like $f(x)=x^4+x+1$. It is known that the result of adjoining all four roots of $f$ to $\Bbb Q$ (the so-called splitting field of $f$) has order $24$ over $\Bbb Q$, as the Galois group of this extension is $S_4$. It is not difficult to use this fact to prove that adjoining two of those roots to $\Bbb Q$ yields an order $12$ extension, as each successive root adjoining must give a strictly smaller degree extension.