We are given that a polynomial f(x) has integer coefficients. The coefficient of x^4 being 1. One root of it is ($\sqrt{2}+\sqrt3$). How do we find the other roots?
I tried using long division, it was so long so i just did it until $x^2$, like I divided $x^4+ax^3+bx^2+cx+d$ by $(x-(\sqrt2+\sqrt3))$. The only good result that I got was that the coefficient of $x^3$ of the function obtained after dividing would be $1$.
To elaborate on the above answers, maybe it would be helpful to see how we come up with the polynomial without talking about automorphisms. Say we are trying to solve your problem for $\sqrt{a}+\sqrt{b}$ for $a,b$ integers. Set
$$x = \sqrt{a}+\sqrt{b}$$ and after squaring both sides we get: $$x^2 = a + 2\sqrt{ab} + b$$ Move the $a+b$ to the left and square the equation again:
$$x^4 + 2(a+b)x^2+ (a+b)^2 = 4ab$$
Therefore the polynomial $f(x) = x^4+2(a+b)x^2 + (a+b)^2 - 4ab $ will have the desired properties.
Setting $a=2$ and $b=3$ gives you the same result as the previous answers.