find out what $[\mathbb{Q}(a^2 + 2)/ \mathbb{Q} ] $ when $a$ satisfies $\alpha^6 - 3 \alpha^3 - 6 = 0$

Question

find out what $[\mathbb{Q}(a^2 + 2)/ \mathbb{Q} ] $ when $a$ satisfies $\alpha^6 - 3 \alpha^3 - 6 = 0$ using the tower law.

I have figured out that $[\mathbb{Q}(a)/ \mathbb{Q} ] = 6$, but I am unsure how to use the tower law. Is it true that $\mathbb{Q}(a^2+2) \subset \mathbb{Q}(a)$? if so why?

edit: using the towerlaw I can get that $[\mathbb{Q}(a)/ \mathbb{Q}] = 6 = [\mathbb{Q}(a)/ \mathbb{Q(a^2)}] [\mathbb{Q}(a^2) / \mathbb{Q}]$ Now I can argue that $[\mathbb{Q}(a^2) / \mathbb{Q}]$ is either $3$ or $6$, but I am unsure how to conclude if it's one of them.

Answer 1

We know that $$\mathbb Q(a) \supset \mathbb Q(a^2+2)\supset \mathbb Q$$ and hence, by the tower law, $$[\mathbb Q(a):\mathbb Q] = [\mathbb Q(a):\mathbb Q(a^2+2)][\mathbb Q(a^2+2):\mathbb Q]$$

or $$[\mathbb Q(a^2+2):\mathbb Q] = \frac{[\mathbb Q(a):\mathbb Q] }{[\mathbb Q(a):\mathbb Q(a^2+2)]}.$$

You've correctly calculated that $[\mathbb Q(a):\mathbb Q] = 6$. To work out $[\mathbb Q(a^2+2):\mathbb Q]$, you now need to calculate $[\mathbb Q(a):\mathbb Q(a^2+2)]$.

Can you find a polynomial in $\mathbb Q(a^2+2)$ which has $a$ as a root? You can then use this to finish up.

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A polynomial in $\mathbb Q(a^2+2)$ is an object $$f(X) = a_0 + a_1 X + a_2 X^2 + \cdots + a_nX^n$$where each $a_0,a_1,\ldots, a_n \in \mathbb Q(a^2+2)$. For example, we could have $$f(X) = 1 + \frac12 X + \frac9{12}(a^2+2)^3X^3.$$

How can we find a polynomial in $\mathbb Q(a^2+2)$ which has $a$ as a root? Let's try to mimic what we would to do find a polynomial over $\mathbb Q$ - say we want to find a polynomial in $\mathbb Q$ which has $\sqrt 3 + 1$ as a root. Set $x = \sqrt 3 + 1$. Then $$(x-1)^2 = 3$$ and since $3\in \mathbb Q$, we know that the polynomial $$f(X) = (X-1)^2 - 3 = X^2 - 2X - 2$$ will work.

We can do exactly the same in our case. Set $x = a$. Then $$x^2 + 2 = a^2 + 2$$ and $a^2 + 2 \in \mathbb Q(a^2+2)$.

So the polynomial $$f(X) = X^2 + 2 - (a^2 + 2) = X^2 -a^2$$will work, and you can show it is irreducible, as it has no roots in $\mathbb Q(a^2 + 2)$.

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CORRECTION: as pointed out by mercio, the polynomial constructed above is actually reducible, since $$a = \frac{1}{18}(a^{10}-15a^4)\in\mathbb Q(a^2 +2)$$

Answer 2

Note that $\Bbb Q(a^2+2) = \Bbb Q(a^2)$.

As you have already noticed, you either have $\Bbb Q(a) = \Bbb Q(a^2)$ and then $[\Bbb Q(a^2) : \Bbb Q] = 6$, or $[\Bbb Q(a) : \Bbb Q(a^2)] = 2$ and $[\Bbb Q(a^2) : \Bbb Q] = 3$.

So you might just want to look at the family $\{1,a^2,a^4,a^6\}$ and see if it's linearly dependant over $\Bbb Q$.

We know that a basis of $\Bbb Q(a)$ over $\Bbb Q$ is $\{1,a,a^2,a^3,a^4,a^5\}$, and in this basis our family is $\{1,a^2,a^4,3a^3+6\}$.

It is pretty clear then that those $4$ elements are independant over $\Bbb Q$ (because $\{1,a^2,a^3,a^4\}$ are), so there is no polynomial of degree $\le 3$ with rational coefficients that annihilates $a^2$.

Thus $[\Bbb Q(a^2) : \Bbb Q] = 6$ and $\Bbb Q(a) =\Bbb Q(a^2)$. And indeed, $a = \frac 1 {18} (a^{10} - 15 a^4) \in \Bbb Q(a^2)$