<s>[INST] minimal polynomial given an algebraic number I am trying to find the minimal polynomial for the algebraic number $1+\sqrt{2}+\sqrt{3}$.  My original thought was just let $\alpha=1+\sqrt{2}+\sqrt{3}$.  The method I use though seems very complicated.  For example, 
$$\alpha^2=(1+\sqrt{2}+\sqrt{3})^2=1+\sqrt{2}+\sqrt{3}+\sqrt{2}+2+\sqrt{2}\sqrt{3}+\sqrt{3}+\sqrt{2}\sqrt{3}+3$$
$$\alpha^2=1+\sqrt{2}+\sqrt{3}+1+\sqrt{2}+\sqrt{3}+4+2\sqrt{2}\sqrt{3}$$
$$\alpha^2=4+2\alpha+2\sqrt{2}\sqrt{3}$$
If I substitute for $\sqrt{2}=\alpha-1-\sqrt{3}$ and $\sqrt{3}=\alpha-1-\sqrt{2}$, I'm still going to end up with $\sqrt{2}$ and $\sqrt{3}$.  So I figured I'd square again.  Moving the 4 over makes the calculation easier since binomials are easier than trinomials....
$$(\alpha^2-4)^2=(2\alpha+2\sqrt{2}\sqrt{3})^2$$
$$\alpha^4-8\alpha^2+16=4\alpha^2 +8\alpha\sqrt{2}\sqrt{3}+24$$
$$\alpha^4-12a^2-8=4\alpha(2\sqrt{2}\sqrt{3})$$
From the above calculation I see that $2\sqrt{2}\sqrt{3}=\alpha^2-2\alpha-4$
$$\alpha^4-12a^2-8=4\alpha(\alpha^2-2\alpha-4)$$
$$\alpha^4-4\alpha^3-4\alpha^2+16\alpha-8=0$$
But my question is, how do I KNOW this is the minimal polynomial?  Yes, it is true now, since I constructed it so, that if $f(x)=x^4-4x^3-4x^2+16x-8$, then $f(\alpha)=0$  Do I simply attempt to factor out an $(x-\alpha)$.  In that case, $f(x)=(x-\alpha)g(x)$.  Then I show that $g(\alpha)\neq 0$?
 [/INST] For $~P,Q,R,S,T\in\mathbb Q^\star,~$ irreducible, with $~R\neq T~$ both squarefree, we have

$$\begin{align}a~=~P~+~Q\sqrt R~+~S\sqrt T\quad&=>\quad(a-P)^2~=~\ldots~=~K+M\sqrt N~\not\in~\mathbb Q\\\\&=>\quad\Big[(a-P)^2-K\Big]^2=~M~^2\cdot N~\in~\mathbb Q\end{align}$$

where $~K=Q~^2R+S^2T,~$ $~M=2~Q~S,~$ and $~N=R~T$.
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