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 $g(\alpha)\neq 0$?
 A: Consider the polynomial
$$(y-\sqrt{2}-\sqrt{3})(y-\sqrt{2}+\sqrt{3})(y+\sqrt{2}-\sqrt{3})(y+\sqrt{2}+\sqrt{3}).$$
The first two terms have product $y^2-2\sqrt{2}y-1$ and the next two have product $y^2+2\sqrt{2}y-1$. Multiply. We get $y^4-10y^2+1$.
We can now see that $\alpha$ is a zero of the polynomial $(x-1)^4-10(x-1)^2+1$. 
To show this is minimal, some algebra is useful. One can show that the field $\mathbb{Q}(\sqrt{2})$ has degree $2$ over the rationals, since $\sqrt{2}$ is irrational. And $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $2$ over $\mathbb{Q}(\sqrt{2})$, since $\sqrt{3}$ cannot be expressed as $s+t\sqrt{2}$ with $s$ and $t$ rational. Thus $\mathbb{Q}(\alpha)$ has degree $4$ over the rationals, so the minimal polynomial has degree $4$.
A: 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 $~N=R~T$.
A: If just want the know the answer, without knowing why it is the answer, you can use a verified implementation of algebraic numbers. For instance, if you load the Isabelle library Algebraic_Numbers/Algebraic_Numbers_Test.thy from the Archive of Formal proofs, you can just enter
value (code) "show (1 + sqrt 2 + sqrt 3)"

and will get your polynomial as output: the output will always deliver the minimal polymial. Moreover, if you want to test whether a given polynomial is minimal, i.e., irreducible, you can similarly invoke a verified polynomial factorization algorithm to test irreducibility.
value (code) "show_sf_factorization (factorize_int_poly [: -8, 16, -4, -4, 1 :])" 

As a side remark, at the time of writing, the implementation uses resultants and Berlekamp-Zassenhaus factorization to compute the minimal polynomial.
