# Is $\sqrt{7} \in \mathbb{Q}\left(\sqrt{3+\sqrt{2}}\right)\;$? [duplicate]

Let $u = \sqrt{3+\sqrt{2}}\;$. Is $\sqrt{7} \in \mathbb{Q}\left(u\right)$? Equivalently, is $\mathbb{Q}(u)$ a splitting field of $u$ over $\mathbb{Q}\,$?

The original question is whether or not $\mathbb{Q}\left(u\right)$ is a splitting field of $u$ over $\mathbb{Q}$. The minimal polynomial of $u$ is $x^4-6x^2+7$, and the four roots are given by the four values of $\pm\sqrt{3\pm\sqrt{2}}\;$. Then the hard part of showing that it's a splitting field comes down to showing $\sqrt{3-\sqrt{2}} \in \mathbb{Q}\left(u\right)$. We get that

$$\sqrt{3-\sqrt{2}} \;\;=\;\; \sqrt{3-\sqrt{2}} \left(\frac{u}{u}\right) \;\;=\;\; \frac{\sqrt{7}}{u} \;\;.$$

The inverse of $u$ is in there because it's a field, but I can't decide whether or not $\sqrt{7} \in \mathbb{Q}\left(u\right)\;$.

I will keep it as elementary as possible, only using the Galois correspondence, i.e. the fundamental theorem of galois theory.

Fix some notations: $u = \sqrt{3+\sqrt 2}, v=\sqrt{3-\sqrt 2}$, note that $uv = \sqrt 7$.

Note that the conjugates of $u$ are $u,-u,v,-v$, i.e. any field homomorphism will map $u$ to one of these 4 guys. The same holds for $v$.

Let us assume $\sqrt 7 \in \mathbb Q(u)$, i.e. $v \in \mathbb Q(u)$. We deduce $\mathbb Q(\sqrt 2, \sqrt 7) \subset \mathbb Q(u)$. Both extensions are of degree $4$ over $\mathbb Q$, hence equality holds.

Let $G$ be the galois group of $\mathbb Q(\sqrt 2, \sqrt 7)/\mathbb Q$. $G$ is well known and there is some $\sigma \in G$, which fixes $\sqrt 2$ and maps $\sqrt 7$ to $-\sqrt 7$. From here, I will derive a contradiction.

• Note that $u$ and $v$ are both primitive elements of $\mathbb Q(u)=\mathbb Q(\sqrt 2, \sqrt 7)$, hence they are not fixed by $\sigma \neq \operatorname{id}$. This leaves us with the possibilities $\sigma(u) \in \{-u,v,-v\}$ and $\sigma(v) \in \{-v,u,-u\}$

• Note that $u^2=3+\sqrt{2}$ and $v^2=3-\sqrt 2$ are both fixed by $\sigma$, since $\sigma$ fixes $\sqrt 2$. This implies that $\sigma(u)= \pm v$ is not possible, since we would obtain $\sigma(u^2)=v^2\neq u^2$. This yields $\sigma(u)=-u$ and similiarly we get $\sigma(v)=-v$. Thus we compute

$$-\sqrt 7=\sigma(\sqrt 7)=\sigma(uv)=\sigma(u)\sigma(v)=(-u)(-v)=uv=\sqrt 7,$$ clearly a contradiction!

Well, the minimal polynomial of $u=\sqrt{3+\sqrt{2}}$ is $X^4-6X^2+7$. According to SageMath, its splitting field is of degree $8$, with defining polynomial $x^8 - 40x^6 + 428x^4 - 784x^2 + 196$. Therefore $v=\sqrt{3-\sqrt{2}}\notin\mathbb{Q}(u)$, because the Galois conjugates of $u$ are $\pm u$ and $\pm v$ and $-u\in\mathbb{Q}(u)$. Consequently, $\sqrt{7}\notin\mathbb{Q}(u)$.

The code is (for the curious):

X = polygen(ZZ)
(X^4-6*X^2+7).splitting_field('u')

• I don't know whether proofs by SageMath's NT library count :)
– yo'
Jun 7, 2016 at 19:52
• We do not need SageMath to compute the splitting field, since there is a precise theorem how the splitting field and galois group of a bi-quadratic polynomial $X^4+aX^2+b$ looks like in terms of $a$ and $b$. But as i said, ths uses galois theory of course.
– MooS
Jun 7, 2016 at 19:54
• @MooS Yeah I know, but this is "somehow simpler" (it's a two-liner in Sage, I should actually show it for the curious :) )
– yo'
Jun 7, 2016 at 19:55