How to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ [duplicate]

Is there a simple way to show that $\sqrt{3 - \sqrt{2}} \notin \mathbb{Q}(\sqrt{3 + \sqrt{2}})$ (assuming this is true), using Galois theory? I have tried to show this by computational means but the calculations are horrible.

Let $\,K\,$ denote $\,\mathbb Q(\sqrt{3+\sqrt2})\,$ for convenience, then observe that

$$x=\sqrt{3+\sqrt2}$$

$$\Rightarrow\quad x^2-3=\sqrt2\qquad\ \ \ $$

$$\Rightarrow\quad x^4-6x^2+7=0\quad$$

$$\Rightarrow\quad[\,K:\mathbb Q\,]=4\qquad\ \ \ $$

First, $\,(\sqrt{3+\sqrt2})^2-3=\sqrt2\in K$

Now if $\,\sqrt{3-\sqrt2}\in K,$ then $\,\sqrt{3+\sqrt2}\cdot\sqrt{3-\sqrt2}=\sqrt7\in K$

Thus $\,K=\mathbb Q(\sqrt2,\sqrt7)$,$\,$ and this means that

$$\qquad\qquad\exists\ a,b,c,d\in\mathbb Q\ :\ a+b\sqrt2+c\sqrt7+d\sqrt{14}=\sqrt{3+\sqrt2}\qquad(\#)$$

We must have $\,c=d=0\quad(\%)$

Hence $\,\sqrt{3+\sqrt2}\in\mathbb Q(\sqrt2)\,$, which is impossible because $$[\mathbb Q(\sqrt2):\mathbb Q]=2<\mathbb [\mathbb Q(\sqrt{3+\sqrt2}):\mathbb Q]$$

As a result, $\,\sqrt{3-\sqrt2}\notin K$

$\left.\right.$

$\textbf{Proof}$ of $(\%):$

Square both sides of $(\#)$, we get

$$(a^2+2b^2+7c^2+14d^2-3)+2(ab+7cd-1)\sqrt2+2(ac+2bd)\sqrt7+2(ad+bc)\sqrt{14}=0$$

Now we must have

\begin{align*} W&=a^2+2b^2+7c^2+14d^2-3=0\\ X&=ab+7cd-1=0\\ Y&=ac+2bd=0\\ Z&=ad+bc=0 \end{align*}

First, if $\,c=0,\,d\neq0$, we have $\,a=b=0\,$ by $Y$ and $Z$. Then $X$ will become $\,-1=0$, which is obviously a contradiction.

Next, if $\,c\neq0,\,d=0$, the same argument as above will lead to a contradiction.

Hence suppose $\,c,d\neq0$, then we have the following two cases:

(1) $b\neq0:$

$$c=-\frac{ad}b\ \ (\text{by }Z)\ \ \Rightarrow\ \ \frac db(a^2-2b^2)=0\ \ (\text{by }Y)\ \ \Rightarrow\ \ a=\pm\sqrt2b\notin\mathbb Q\ \ (\text{contradiction})$$

(2) $b=0:$

$$a=0\ \ (\text{by }Z)\ \ \Rightarrow\ \ c=\frac1{7d}\ \ (\text{by }X)\ \ \Rightarrow\ \ 98d^4-21d^2=-1\ \ (\text{by }W)\ \ \Rightarrow\ \ d\notin\mathbb Q\ \ (\text{contradiction})$$

As a result, we must have $\,c=d=0$

$\left.\right.$

(This answer has been improved by the suggestions from Kenny Wong)

Let $\alpha = \sqrt{3-\sqrt 2}$ and $\beta = \sqrt{3+\sqrt 2}$.

Your problem boils down to showing that the splitting field $L$ of $X^4 -6X^2 + 7$ over $\mathbb Q$ is an extension of degree strictly greater than $4$.

Suppose for contradiction that the degree of the extension is exactly $4$. Then there are two possibilities for the Galois group: either it is the cyclic group or it is the Klein group.

But $\alpha^2 = 3 - \sqrt{2}$ and $\alpha \beta = \sqrt 7$, so $\mathbb Q(\sqrt 2)$ and $\mathbb Q (\sqrt 7)$ are two distinct subfields of $L$. This rules out the cyclic group, which only has one subgroup of order two.

Since the quartic is irreducible, the Galois group must act transitively on the roots. Therefore, the only way the Galois group could be the Klein group is if the automorphisms act on the four roots of the quartic by the permutations $\{e, (12)(34), (13)(24), (14)(23) \}$. But if this is so, then $(\alpha + \beta)^2$ is fixed by all four automorphisms, hence it must be in $\mathbb Q$ by the Galois correspondence. Since $(\alpha + \beta)^2 = 6 + 2 \sqrt {7} \notin \mathbb Q$, we have a contradiction.

Here’s another argument, depending on the arithmetic of the field $k=\Bbb Q(\sqrt2\,)$ and Kummer theory.

The question asks whether $k\bigl((3+\sqrt2\,)^{1/2}\bigr)$ and $k\bigl((3-\sqrt2\,)^{1/2}\bigr)$ are the same quadratic extension of $k$. This is why not:

The ring of integers of $k$ is $R=\Bbb Z[\sqrt2\,]$, and as is well known (or easy to show), $R$ is a unique factorization domain. Since $2$ is a square modulo $7$, the prime $7$ splits in $R$, and indeed $7=(3-\sqrt2\,)(3+\sqrt2\,)$, product of two inequivalent primes $\pi_1$ and $\pi_2$ of $R$. (By “inequivalent” I mean that their quotient is not a unit.) Unique factorization says also that $\pi_1/\pi_2$ is not a square in $k$, and therefore (here comes Kummer) adjoining their square roots gives two different fields.