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

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.

• Did the professor say for sure there was an example for $12?$ Maybe he wanted you to prove $12$ was not possible? Commented Mar 20, 2023 at 17:10
• @ThomasAndrews I remember he said that example could be derived with Galois Theory. Commented Mar 20, 2023 at 17:15
• You certainly don't require Galois theory, but finding examples is trickier. Commented Mar 20, 2023 at 19:20

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.

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 $$3.$$

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.

• Thank you very much, I appreciate such a complete answer! I was working through everything and it works out! Commented Mar 21, 2023 at 21:34

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.