Let $a, b$ be rational numbers that are not perfect squares. Consider the set $S = \{\sqrt a + \sqrt b, \sqrt a - \sqrt b, - \sqrt a + \sqrt b, -\sqrt a - \sqrt b\}$.
If $p$ is a polynomial with rational coefficients, is it correct that if any element of $S$ is a root of $p$, then every element of $S$ is?
That is implied by the excellent post defining conjugates, and I want to confirm my understanding.
This is equivalent to show that $^4−2(+)^2+(−)^2 $ is irreducible in $\mathbb{Q}[X]$ You know that any non trivial divisor is of the form $(X -s_1) \dots (X-s_k)$ for $0<k<4$, $s_i$ all different and in $S$.
Let us first consider the case $k=1$, you obviously don't have a polynomial in $\mathbb{Q}[X]$.
The case $k=3$ follows from $k=1$.
For $k = 2$ you have various possibilities: $$(X-(\sqrt{a} + \sqrt{b})) (X-(\sqrt{a} - \sqrt{b})) = X^2 - 2 \sqrt{a} X + a-b$$ $$(X-(\sqrt{a} + \sqrt{b})) (X-(-\sqrt{a} + \sqrt{b})) = X^2 - 2 \sqrt{b} X + b-a$$ $$(X-(\sqrt{a} + \sqrt{b})) (X-(-\sqrt{a} - \sqrt{b})) = X^2 - (\sqrt{a } + \sqrt{b})^2$$
The other cases are deducible from these three. For the first two, it does not work, for the last it works iff $\sqrt{a} \sqrt{b} \notin \mathbb{Q}$ which proves our point.
In this case you have for $\mathbb{L} = \mathbb{Q}[X]/(^4−2(+)^2+(−)^2 ) $ : $$Gal(\mathbb{L}/\mathbb{Q}) \simeq \mathfrak{S} (S) \simeq \mathfrak{S}_4$$