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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.

SRobertJames
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  • Is the polynomial $X^4 -2 (a+b) X^2 + (a-b)^2$? ( Just to be sure) – julio_es_sui_glace Jun 23 '23 at 13:00
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    You also need that $ab$ is not a perfect square, or equivalently that $\sqrt{a}$ is not a rational multiple of $\sqrt{b}$. For example $a = 18$, $b = 50$ makes $\sqrt{a} + \sqrt{b}$ satisfy $x^2 - 128 = 0$, which $\sqrt{a}-\sqrt{b}$ does not satisfy. – eyeballfrog Jun 23 '23 at 13:01
  • If you know the roots, show that for any $k < 4$, and $k$ different roots in $S$, $(X-s_1) \cdots (X-s_k)$ is not with rational coefficients you will easily find that the only condition is $(\sqrt{a}, \sqrt{b}$ free as elements of a $\mathbb{Q}$-vector space. – julio_es_sui_glace Jun 23 '23 at 13:07
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    Simpler counterexample: $\sqrt{2}+\sqrt{2}$ is a root of $x^2-8$ but $\sqrt{2}-\sqrt{2}$ isn't. – coiso Jun 23 '23 at 13:47

1 Answers1

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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 ) $ : $$Gal(\mathbb{L}/\mathbb{Q}) \simeq \mathfrak{S} (S) \simeq \mathfrak{S}_4$$

julio_es_sui_glace
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