minimal polynomial

[logistic_guy]

here is the question

Determine the minimal polynomial over $\displaystyle \mathbb{Q}$ for the element $\displaystyle \sqrt{2} + \sqrt{5}$.


my attemb
i can write the element in $\displaystyle 4$ different ways
$\displaystyle \sqrt{2} + \sqrt{5}$
$\displaystyle \sqrt{2} - \sqrt{5}$
$\displaystyle -\sqrt{2} + \sqrt{5}$
$\displaystyle -\sqrt{2} - \sqrt{5}$
so i think this mean the minimal polynomial will be of degree $\displaystyle \leq 4$
this are the factors
$\displaystyle (x - \sqrt{2} + \sqrt{5})(x - \sqrt{2} - \sqrt{5})(x - -\sqrt{2} + \sqrt{5})(x - -\sqrt{2} - \sqrt{5})$
how can i tell if this is the minimal polynomial without simplifying?

 

[fresh_42]

Please do not write $x -- c\,.$ This is $x+c \,.$

You want to find a polynomial $p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ with minimal $n$ such that $p\left(\sqrt{2}+\sqrt{5}\right)=0$ or algebraically written $\mathbb{Q}\left[\sqrt{2}+\sqrt{5}\right]=\mathbb{Q}[x]/\bigl\langle p(x) \bigr\rangle .$

The field $\mathbb{Q}\left[\sqrt{2}+\sqrt{5}\right]$ has to contain $\xi=\sqrt{2}+\sqrt{5}$ and therewith all its powers. Let's see how far we get:
$$\begin{array}{lll} \left(\sqrt{2}+\sqrt{5}\right)^1&=\sqrt{2}+\sqrt{5}\\[6pt] \left(\sqrt{2}+\sqrt{5}\right)^2&=7+2\sqrt{10}\\[6pt] \left(\sqrt{2}+\sqrt{5}\right)^3&=17\sqrt{2}+11\sqrt{5}\\[6pt] \left(\sqrt{2}+\sqrt{5}\right)^4&=89+28\sqrt{10}\\[6pt] \end{array}$$The elements $\left\{\xi^0\, , \,\xi^1\, , \,\xi^2\, , \,\xi^3\right\}$ are $\mathbb{Q}$-linearly independent (prove it!) but $\xi^4-14\xi^2+9\xi^0=0 .$ Hence $\dim_\mathbb{Q}\mathbb{Q}[\xi]=n=4.$

However, you need to show that $q(x)=\left(x-\sqrt{2}-\sqrt{5}\right)\left(x-\sqrt{2}+\sqrt{5}\right)\left(x+\sqrt{2}-\sqrt{5}\right)\left(x+\sqrt{2}+\sqrt{5}\right)\in \mathbb{Q}[x]$ to conclude $p(x)=q(x)$ (how?) so that you need to calculate the polynomial, i.e. multiply its factors.

 

[logistic_guy]

thank

i'll go directly to the something i understand

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

$\displaystyle =$

$\displaystyle (x^2 - \sqrt{2}x + \sqrt{5}x - \sqrt{2}x + 2 - \sqrt{10} - \sqrt{5}x + \sqrt{10} - 5)(x^2 + \sqrt{2}x + \sqrt{5}x + \sqrt{2}x + 2 + \sqrt{10} - \sqrt{5}x - \sqrt{10} - 5)$

$\displaystyle =$

$\displaystyle (x^2 - 2\sqrt{2}x - 3)(x^2 + 2\sqrt{2}x - 3)$

$\displaystyle =$

$\displaystyle x^4 + 2\sqrt{2}x^3 - 3x^2 - 2\sqrt{2}x^3 - 8x^2 + 6\sqrt{2}x - 3x^2 - 6\sqrt{2}x + 9$

$\displaystyle =$

$\displaystyle x^4 - 14x^2 + 9$

if this is irreducible then it's the minimal polynomial

what happen if it's reducible?

 

[fresh_42]

If it's irreducible, then you are done. No factor, no intermediate field in $\mathbb{Q}\subseteq \mathbb{Q}\left(\sqrt{2}+\sqrt{5}\right).$ Why is it irreducible?

 

[logistic_guy]

Why is it irreducible?

look in this factor
$\displaystyle (x^2 - 2\sqrt{2}x - 3)(x^2 + 2\sqrt{2}x - 3)$
i think it's irreducible because some of the coeeficients is irrational