minimal polynomial given an algebraic number

Question

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 $g(\alpha)\neq 0$?

Answer 1

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 $4$.

Answer 2

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 $~N=R~T$.