How to find a polynomial of which this field is a splitting field?

Question

Consider the field $Q(\sqrt{2} + i \sqrt{5})$. I've proven that $\mathbb{Q}(\sqrt{2} + i \sqrt{5}) = \mathbb{Q}(\sqrt{2}, i \sqrt{5})$. Also, by the degree product rule for extension degrees, I have $$ [\mathbb{Q}(\sqrt{2}, i \sqrt{5}) : \mathbb{Q}] = 4. $$

I'm now being asked to give a polynomial $f \in \mathbb{Q}[X]$ such that $Q(\sqrt{2} + i \sqrt{5})$ is the splitting field of $f$ over $\mathbb{Q}$.

I don't know how to handle this problem. I tried setting $x = \sqrt{2} + i \sqrt{5}$ and squaring etc. but I cannot get a polynomial over $\mathbb{Q}$. The problem gives me a hint, saying I should look for a suitable field $E$ such that $$ \mathbb{Q} \subset E \subset Q(\sqrt{2} + i \sqrt{5})$$ but I'm not sure how this will help me. I know that $\mathbb{Q}(\sqrt{2}, i \sqrt{5}) = (\mathbb{Q}(\sqrt{2})(i\sqrt{5})$. So then I would maybe let $E = \mathbb{Q}(\sqrt{2})$. Then $x^2 + 5$ is the minimal polynomial of $i \sqrt{5}$ over $\mathbb{Q}(\sqrt{2})$. But how to find a polynomial from this over $\mathbb{Q}$?

Answer 1

The setting $x=\sqrt{2}+i\sqrt{5}$ and squaring, etc, approach works: $$ (x-i\sqrt{5})^2 = 2\\x^2-7=2i\sqrt{5}x \\ (x^2-7)^2 = -20x^2\\ x^4+6x^2+49=0$$ This gives the minimal polynomial for $\sqrt{2}+i\sqrt{5},$ whose roots are $\pm(\sqrt{2}\pm i\sqrt{5})$ and whose splitting field is $\mathbb{Q}(\sqrt{2},i\sqrt{5}) = \mathbb{Q}(\sqrt{2}+i\sqrt{5}).$

Answer 2

Let $\alpha = \sqrt 2 + \mathrm i \sqrt 5$ and consider successive powers:

\begin{eqnarray*} \alpha &=& 0+1\sqrt 2 + \mathrm i \sqrt 5 + 0\sqrt{10}\\ \\ \alpha^2 &=& -3+0\sqrt 2 + 0\sqrt 5 + \mathrm i \sqrt{10} \\ \\ \alpha^3 &=& 0-13\sqrt 2+\mathrm i \sqrt 5 + 0\sqrt{10}\\ \\ \alpha^4 &=& -31+0\sqrt 2 + 0\sqrt 5 -12\mathrm i \sqrt{10} \end{eqnarray*}

Putting this into a matrix equation gives $$\left[\begin{array}{c} \alpha \\ \alpha^2 \\ \alpha^3 \\ \alpha^4 \end{array}\right] = \left[\begin{array}{cccc} 0 & 1 & \mathrm i & 0 \\ -3 & 0 & 0 & 2\mathrm i \\ 0 & -13 & \mathrm i & 0 \\ -31 & 0 & 0 & -12\mathrm i \end{array}\right] \left[\begin{array}{c} 1 \\ \sqrt 2 \\ \sqrt 5 \\ \sqrt{10} \end{array}\right]$$ The four-by-four matrix has non-zero determinant, and hence: $$\frac{1}{98}\left[\begin{array}{cccc} 0 & -12 & 0 & -2 \\ 7 & 0 & -7 & 0 \\ -91\mathrm i & 0 & -7 \mathrm i & 0 \\ 0 & -31\mathrm i & 0 & 3\mathrm i \end{array}\right] \left[\begin{array}{c} \alpha \\ \alpha^2 \\ \alpha^3 \\ \alpha^4 \end{array}\right] = \left[\begin{array}{c} 1 \\ \sqrt 2 \\ \sqrt 5 \\ \sqrt{10} \end{array}\right]$$ Expanding the first row gives $-\frac{12}{98}\alpha^2-\frac{2}{98}\alpha^4=1$, i.e. $$2(\alpha^4 + 6\alpha^2 + 49) = 0$$

The polynomial in question is then $x^4+6x^2+49$.

Answer 3

Look at the minimal polynomial of the matrix for the multiplication by $\alpha=\sqrt 2 + i\sqrt 5$ map. This will give you the minimal polynomial for $\alpha$, in particular it will be one for the field. If you don't care about irreducibility, just use $(x^2-2)(x^2+5)$.

Answer 4

Regarding the approach you tried to take....

$\mathbb{Q}(\sqrt{2} + i \sqrt{5})$ is a four-dimensional $\mathbb{Q}$-vector space, spanned (for example) by the basis $\{ 1, \sqrt{2}, \sqrt{-5}, \sqrt{-10} \}$. If you set $x = \sqrt{2} + i \sqrt{5}$, then $\{ 1, x, x^2, x^3, x^4 \}$ must be a linearly dependent set — you can find a $\mathbb{Q}$-linear combination of them that sums to zero. And you can find it by doing ordinary linear algebra.

Sometimes you can do tricks; e.g. start with $ (x - \sqrt{2})^2 = -5 $, and then isolate $\sqrt{2}$ so you can square again.

You could, just immediately write down the minimal polynomial since you know all of its roots:

$$ (x - \sqrt{2} - i \sqrt{5})(x - \sqrt{2} + i \sqrt{5}) (x + \sqrt{2} - i \sqrt{5})(x + \sqrt{2} + i \sqrt{5}) $$

although it's a pain to expand it and compute the coefficients if you want to know them.