Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$

Solution 1:

In general, if $p_1,\dotsc,p_n$ is any list of integers, then the polynomial $$f=\prod_{e_1,\dotsc,e_n \in \{\pm 1\}} (x+e_1 \sqrt{p_1}+\dotsc+e_n \sqrt{p_n})$$ has coefficients in $\mathbb{Z}$, which follows by induction from the observation

$p \in \mathbb{Z},\,g \in \mathbb{Z}[x] \Rightarrow g(\sqrt{p}) \cdot g(-\sqrt{p}) \in \mathbb{Z}[x]$.

You can prove this using the automorphism group of $\mathbb{Z}[\sqrt{p}]$. But you can also just verify it via some direct calculation, which has the advantage that you really see why the $\sqrt{p}$-terms vanish and that you may compute $f$ faster.

Clearly, $f$ is monic, has degree $2^n$, and has $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ as a root. So this is quite elementary and for any fixed $n$, you can also compute $f$. What is not so easy to prove is that if $p_1,\dotsc,p_n$ are square-free and pairwise coprime integers, then $f$ is irreducible. Equivalently, the degree of $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ equals $2^n$. You can find the proof here ("square-root-extension.pdf").

For the numbers $p_1,p_2,p_3 = 2,3,5$, we get $f = x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$.

Solution 2:

The 'easy' way to get at the minimal polynomial in this case is to take the product of all the terms $x\pm\sqrt2\pm\sqrt3\pm\sqrt5$; this is, in essence, because the Galois group over $\mathbb{Q}$ in this case is just $(\mathbb{Z}/2\mathbb{Z})^3$, with each of the three copies of $\mathbb{Z}/2\mathbb{Z}$ corresponding to a sign change on one of the three square root terms. The simple approach (take the product of $(x-\alpha)$ over all of the possible $\alpha$ obtained by applying the automorphisms in the Galois group to $\sqrt2+\sqrt3+\sqrt5$) works here because $\sqrt2$, $\sqrt3$ and $\sqrt5$ are 'mutually irreducible' over $\mathbb{Q}$ (in the sense that $\sqrt2\not\in\mathbb{Q}(\sqrt{3}, \sqrt{5})$, etc.) , so we can compose extensions in a clean fashion.

Solution 3:

I started out by observing that $2+3 = 5$; I thought that might make it easier. Then

$$ x-\sqrt{5} = \sqrt{2}+\sqrt{3} $$ $$ x^2-2\sqrt{5}x+5 = 2\sqrt{6}+5 $$ $$ x^2-2\sqrt{5}x = 2\sqrt{6} $$ $$ x^4-4\sqrt{5}x^3+20x^2 = 24 $$ $$ x^4+20x^2-24 = 4\sqrt{5}x^3 $$ $$ x^8+40x^6+352x^4-960x^2+576 = 80x^6 $$ $$ x^8-40x^6+352x^4-960x^2+576 = 0 $$

(Or, of course, you can do it more systematically via the Galois theory!)