Consider extension $[\mathbb{Q}(\alpha\ ):\mathbb{Q}]$ where $\alpha\ $ is zero of $P(x) = x^4 + 9x^{2} + 15 $.

Question

Consider extension $[\mathbb{Q}(\alpha):\mathbb{Q}]$ where $\alpha$ is zero of $p(x) = x^4 + 9x^{2} + 15 $. Find $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2 + 3)]$.

My attempt: By Eisenstein's Criterion $p$ is irreducible over $\mathbb{Z}$ and by Gauss's lemma it is also over $\mathbb{Q}$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 4$. Now as $[\mathbb{Q}(\alpha):\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2 + 3)] [\mathbb{Q}(\alpha^2 + 3):\mathbb{Q}]$ we have $$ 4 = [\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2 + 3)] [\mathbb{Q}(\alpha^2 + 3):\mathbb{Q}] \, . $$ There are three possibilities: $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2 + 3)] = 1,2$ or $4$.

As $\alpha$ is zero of $p(x) = x^4 + 9x^{2} + 15$, I can write $p(x) = (x - \alpha)(x + \alpha)(x^2 + 9 + \alpha^2)$ on $\mathbb{Q}(\alpha)$. I got stuck here.

Any ideas would be appreciated.

Answer 1

You pretty much got an answer :)! However, I think it would be a little bit easier to calculate the degree extension of $\mathbb{Q}(\alpha^2+3)$ over $\mathbb{Q}$. We can proceed as following:

First of all note that $\alpha^2+3$ is not a rational number (why?), so $[\mathbb{Q}(\alpha^2+3):\mathbb{Q}]$ can't be 1. It remains for us find out what is the minimal polynomial of $\alpha^2+3$ over $\mathbb{Q}$. To find such polynomial, the natural thing to do is to set $$x=\alpha^2+3 \space$$ Or equivalently, $x-3=\alpha^2$ (*). Squaring both side of this equation, taking advantage of the fact that $\alpha$ is a zero of $x^4+9x^2+15$ to manipulate algebraically into $x^2+3x-3=0$. So consider $m(x)=x^2+3x-3$.

Now can you check that $m$ is indeed the minimal polynomial of $\alpha^2+3$ over $\mathbb{Q}$? This is a routine exercise. Once you have that, think of $\mathbb{Q}(\alpha^2+3)$ as $\mathbb{Q}[x]/(m(x))$. So it must be the case that $[\mathbb{Q}(\alpha^2+3):\mathbb{Q}]=2$

Answer 2

Using Eisenstein criterion to $x^4+9x^2+15=0$ with $p=3$, we have that the minimal polynomial is irreducible over $\mathbb{Q}$. So $\alpha\notin\mathbb{Q}(\alpha^2+3)$, for otherwise the minimal polynomial would be of degree 2. Note that the minimal polynomial of $\alpha$ over the field $\mathbb{Q}(\alpha^2+3)$ is $x^2-\alpha^2$. So $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2+3)]=2$.

Answer 3

9($\alpha^2$ + 3)= 12 -$\alpha^4$ imply $\alpha^4$ + 9a - 12 = 0 and this polynomial is irreducible (Eisenstein) hence the asked degree is 4.

Remark: Maybe I'm wrong and my polynomial is reducible (because of $\alpha^2$ + 3)....

Answer 4

To elaborate on my comment: Since $\alpha$ is a root of $x^4+9x^2 +15$, then $\alpha^2$ is a root of $x^2 + 9x + 15$. Then $$ \alpha^2 = \frac{-9 \pm \sqrt{21}}{2} $$ by the quadratic formula. Let $\beta = \alpha^2 + 3$. Then \begin{align*} \beta - 3 = \alpha^2 = \frac{-9 \pm \sqrt{21}}{2} &\implies 2\beta + 3 = \pm \sqrt{21} \implies 21 = (2\beta+3)^2 = 4\beta^2 + 12\beta +9\\ &\implies 0 = 4\beta^2 + 12\beta - 12 = 4(\beta^2 + 3\beta - 3) \, . \end{align*} Then $\beta$ satisfies $x^2 + 3x - 3$, which is irreducible by Eisenstein. Thus $[\mathbb{Q}(\beta) : \mathbb{Q}] = [\mathbb{Q}(\alpha^2 + 3) : \mathbb{Q}] = 2$.