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 $g(\alpha)\neq 0$?

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The degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is $4$. – André Nicolas Jul 25 '14 at 3:07
I'm not sure what you mean... – Lalaloopsy Jul 25 '14 at 3:10
I'm not sure about the proof of Nicolas's statement but I've factorized your polynomial via a guess: link. An intuitive way of thinking about it is that all corresponding roots with alternating signs must also be taken to make the radicals disappear. Much like complex conjugate roots of a real quadratic equation. – varun Jul 25 '14 at 3:12
A comment on another question has a proof that the degree of minimal polynomial is 4. – varun Jul 25 '14 at 3:24

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 $4$.

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I am currently taking a graduate number theory course. I've not done any abstract in a while, so please give me a bit. If I understand, $\mathbb{Q}$ is a field and we can extend the field if we consider numbers $a+b\sqrt{2}$, where $a,b$ are rational. The notation for this new extended field is $\mathbb{Q}(\sqrt{2})$. What do you mean that $\mathbb{Q}(\sqrt{2})$ has degree 2? Is that simply due to the fact that a minimal polynomial can not have degree less than two because of the extension with $\sqrt{2}$? – Lalaloopsy Jul 25 '14 at 3:31
Let $F$ be a field, and let $\beta$ be algebraic over $F$. Then the field $F(\beta)$ obtained by adjoining $\beta$ to $F$ can be viewed as a vector space over $F$. The dimension of that vector space is called the degree of $F(\beta)$ over $F$, and this is the degree of the minimal polynomial of $\beta$ over $F$. – André Nicolas Jul 25 '14 at 3:35
Yes, except that we have to either show that they are linearly independent over the rationals, or appeal to the general (and not difficult) theorem that the degree of $F(\alpha,\beta)$ over $F$ is the product of the degree of $F(\alpha)$ over $F$ and $F(\alpha,\beta)$ over $F(\alpha)$. Basically just linear algebra here, nothing hard. – André Nicolas Jul 25 '14 at 3:44
Yes, the degree is $8$. But to do that we need to show that we do get a genuine tower, that for example $\sqrt{5}$ is not in $\mathbb{Q}(\sqrt{2},\sqrt{3})$. – André Nicolas Jul 25 '14 at 3:46
Another way of showing that the degree $4$ thing you got is the minimal polynomial is to show it is irreducible over the rationals, or equivalently that it cannot be factored over the integers. For this it is more pleasant to work with the simpler polynomial in $y$. – André Nicolas Jul 25 '14 at 3:52

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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