Let $\alpha$, $\beta$ be roots of $X^3-2$ and $X^3-5$ respectively, find $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$

Question

Given $\alpha,\beta \in \mathbb{C}$. Suppose $\alpha$ is a root of $X^3-2$ and $\beta$ is a root of $X^3-5$. Find the degrees $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$.

I know that $3=[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]|[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$ and $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]<=9$

So $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = 3,6$ or $9$

But how to determine which one is the degrees?

Thanks a lot.

Answer 1

Hint:

$$\beta\notin\Bbb Q(\alpha)\implies [\Bbb Q(\alpha,\beta):\Bbb Q(\beta)]=3$$

Answer 2

You have that: $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]\leq 9$, as you clearly stated.

So $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=3$ or $9$ (it cannot be 6, as you are adjoining, $\beta$, so a cube to $\mathbb{Q}(\alpha).$

So the degree is 3 if $\beta$ is in $\mathbb{Q}(\alpha)$ (as you're adjoining something which already exists in your field, or it's 9 if $\beta$ is not in $\mathbb{Q}(\alpha)$, as the minimal polynomial of $\beta$ over $\mathbb{Q}(\alpha)$ is $X^3-5$, a cube.

If you want to prove it from first principles, I am afraid that this is a tedious computation.

So you start by supposing that $\beta \in \mathbb{Q}(\alpha)$, in which case you may write:

$\beta = a + b\alpha + c\alpha^2$, for $a,b,c \in \mathbb{Q}$,

Now you'd have to cube both sides, and since $\{1,\alpha, \alpha^2\}$ is a $\mathbb{Q}$-basis for $\mathbb{Q}(\alpha)$.

You can then "Equate coefficients$ to get:

$a^3 + 2b^3 + 4c^3 + 12abc = 5 $

$3a^2b + 6ac^2 + 6b^2c = 0$

$3a^2c + 3ab^2 + 6bc^2 = 0$

From second and third equations you get that $a(2c^3-b^3)=0$, since implies that $a=0$.

So second equation becomes $b^2c=0$, ie $b=0$ or $c=0$

$b=0$ gives $4c^3=5$, which is a contradiction.

$c=0$ gives $2b^3=5$, which is again a contradiction.

Answer 3

Here’s moderately advanced argument that shows that $\Bbb Q(\alpha)=K_1$ and $\Bbb Q(\beta)=K_2$ are different fields, which is equivalent to the statement that $[\Bbb Q(\alpha,\beta):\Bbb Q]=9$.

For each prime $p$, $\Bbb Q$ is a canonical subfield of $\Bbb Q_p$, the field of $p$-adic numbers. The important prime here is $p=11$, for over the prime field $\Bbb F_{11}$, the polynomial $X^3-5$ factors as $(X-3)(X^2+3X+9)$. Then Hensel’s Lemma in any form shows then that $\Bbb Q_{11}$ has a cube root of $5$, and thus $K_2$ can be embedded in $\Bbb Q_{11}$.

On the other hand, $X^3-2$ has no root in $\Bbb F_{11}$ (indeed, $2$ is a generator of the multiplicative group there, so can not be an $n$-th root for any $n$ prime to $10$). This says that $X^3-2$ is irreducible over $\Bbb F_{11}$, hence irreducible over $\Bbb Q_{11}$, and $K_1$ can not be embedded into that $p$-adic field.