Calculate $irr(a,\mathbb{Q})$ where $a=\sqrt{2}+\sqrt{3}\in\mathbb{R}$

Question

Consider the element $a=\sqrt{2}+\sqrt{3}\in\mathbb{R}$. Calculate $irr(a,\mathbb{Q})$.

What I did:
Calculate powers of $a$. $a^2=5+2\sqrt{6},a^3=11\sqrt{2}+9\sqrt{3},a^4=49+20\sqrt{6}$. I wish to find some relationships between powers of $a$, but I have no idea where to begin. I believe there should be a better way than just trial-and-error.

It is shown in the hint that we need to consider vectors $v_0=(1,0,0,0), v_1=(0,1,1,0), v_2=(5,0,0,2),v_3=(0,11,9,0), v_4=(49,0,0,20)$, but I have no clue how to use the given hints and how can we even use vectors in our problem.

Thanks in advance for the helps!

Edit:
I have made some attempt and I observed that $v_4=-v_0+10v_2$ and by following the pattern (a very naive way) to replace $v_4$ by $a^4$, $v_2$ by $a^2$ and $v_0$ by 1, I got $a^4-10a^2+1=0$.

Is this progress useful in solving the problem? Is the "pattern" just a coincidence or is there any explanation to it? How can then I proceed in finding $irr(a,\mathbb{Q})$?

Answer 1

Hint:

$$x=\sqrt2+\sqrt3\implies x^2=5+2\sqrt6\implies (x^2-5)^2=(2\sqrt6)^2\ldots$$

Answer 2

The point is that naively you expect to only get numbers involving rationals, rational multiples of $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{6}$ in your calculations. In otherwords the set of numbers you could encounter as a whole under these operations is a $4$-dimensional vector space over $\mathbb{Q}$.

Here I am bypassing the fact that you are working in a field but if you know this stuff then you are working in the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ which has degree $4$ over $\mathbb{Q}$.

Now what do you know about n-dimensional vector spaces over $\mathbb{Q}$? They are all isomorphic to $\mathbb{Q}^n$, once a basis has been chosen. So using the basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ we can label numbers of the form $\alpha + \beta\sqrt{2} + \gamma\sqrt{3} + \delta\sqrt{6}$ as the vector $(\alpha,\beta,\gamma,\delta)$.

Now our element $a = \sqrt{2} + \sqrt{3}$ and its powers upto the $4$th correspond EXACTLY with the vectors $v_0,v_1,v_2,v_3,v_4$. We are in a $4$-dim vector space so these must be linearly dependent (there are $5$ of them!). Finding a linear combination giving $(0,0,0,0)$ is equivalent to finding a polynomial which $a$ satisfies over $\mathbb{Q}$. Showing what you get is irreducible is equivalent to showing linear independence all of the v's of smaller index than the highest one you used.