Basis of the field $E$=$\mathbb{Q}(\sqrt{6}i-\sqrt{5})$.

Question

Let's take a field $E$=$\mathbb{Q}(\sqrt{6}i-\sqrt{5})$.

I wanted to find a basis of $E$ as a $\mathbb{Q}$-vectorspace. My first thought was to use the structure theorem for simply field extension. Therefore I needed to find an element $u\in E$ so that $u$ is algebraïc over $E$ and a $f$ so that $f(u)=0$. Then would $u,u^{2},...,u^{deg(f)-1}$ be a basis. I just don't know how I can do this correctly.

Second I wanted find a polynomial $f\in \mathbb{Q}[X]$ so that $E$ is the splitting field of $f$ over $\mathbb{Q}$. Because I'm stuck with my first thought I don't understand how I can find such an $f$.

EDIT:

FIRTS

• let's name $\alpha$=$\sqrt{6}i-\sqrt{5}$. We're looking for a polynomal $f\in \mathbb{Q}$ so that $f(\alpha)$=0.

$\alpha^{2}$=$-1-2i\sqrt{30}$

$\alpha^{3}$=$-i\sqrt{6}+\sqrt{5}+2\sqrt{+}\sqrt{30}+2i\sqrt{5}\sqrt{0}$

$\alpha^{4}$=$1+4i\sqrt{30}-120$=$-119+4i\sqrt{30}$

Then follows that $\alpha^{4}+2\alpha^{2}+121=0$

we name $f(x)$=$x^{4}+2x^{2}+121$

Because $f$ isn't $0$ and has no points $x$ so $f(x)=0$ , is $f$ irriducibel. This means that $f$ is the minimal polynomal of $\alpha$ over $E$. Now we use the theorem of structure for simply field extension and it follows that {$1,\alpha,\alpha^{2},\alpha^{3}$} is a basis.

SECOND

• I'm still stuck proving this

Answer 1

Hint

$$(\sqrt 6i-\sqrt 5)^2=-1-2\sqrt{30}i$$ and thus $$((\sqrt6i-\sqrt 5)^2+1)^2+120=0,$$ i.e. $$x^4+2x^2+121=0$$ is an annilhator polynomial. I let you conclude.

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An other way would be to show that $$\mathbb Q(\sqrt 6i,\sqrt 5)=\mathbb Q(\sqrt 6i-\sqrt 5),$$ and thus, a basis would be $$\{1,\sqrt 6i,\sqrt 5, i\sqrt{30}\}.$$

Answer 2

The minimal polynomial of $\sqrt 5$ over $\mathbb Q$ is $p(x)=x^2-5$. And the minimal polynomial of $\sqrt{6}i-\sqrt{5}$ over $\mathbb Q(\sqrt{5})$ is $q(x)= x^2+2\sqrt{5}x +11$.

You can then use telescopic basis theorem to get that a basis of $E$ over $\mathbb Q$ is

$$\{1, \sqrt{5}, \sqrt{6}i, \sqrt{30}i\}$$

Answer 3

For part two of your question. You already found the right polynomial, i.e. $f(x)=x^4+2x^2+121$. You know that $\sqrt{6}i-\sqrt{5}$ is a root of this polynomial. Since $\mathbb{Q}$ is a field and has characteristic 0, you know from Primitive element theorem that your Splitting field is $\mathbb{Q}(\sqrt{6}i-\sqrt{5})$.