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 [imath]x -- c\,. [/imath] This is [imath] x+c \,.[/imath]

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

The field [imath] \mathbb{Q}\left[\sqrt{2}+\sqrt{5}\right] [/imath] has to contain [imath] \xi=\sqrt{2}+\sqrt{5} [/imath] and therewith all its powers. Let's see how far we get:
[math]\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}[/math]The elements [imath] \left\{\xi^0\, , \,\xi^1\, , \,\xi^2\, , \,\xi^3\right\} [/imath] are [imath] \mathbb{Q} [/imath]-linearly independent (prove it!) but [imath] \xi^4-14\xi^2+9\xi^0=0 .[/imath] Hence [imath] \dim_\mathbb{Q}\mathbb{Q}[\xi]=n=4. [/imath]

However, you need to show that [imath] 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] [/imath] to conclude [imath] p(x)=q(x) [/imath] (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 [imath]\mathbb{Q}\subseteq \mathbb{Q}\left(\sqrt{2}+\sqrt{5}\right). [/imath] 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