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

Question

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.

Answer 1

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)

Answer 2

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.

Answer 3

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.

Answer 4

Let us introduce $k=\mathbf Q(\sqrt 2)$, $a=3+\sqrt 2$, $b=3-\sqrt 2$. By Kummer theory (or easy direct computation in the present case), $k(\sqrt a)=k(\sqrt b)$ if and only if $a$ and $b$ belong to the same class of $k^* /(k^*)^2$, if and only if $ab \in (k^*)^2$, if and only if $\sqrt 7 \in \mathbf Q(\sqrt 2)$ : impossible.

NB : the starting point is the same as @Lubin, but it seems to me that this question rather belongs to field theory, not to algebraic number theory.

Answer 5

This is essentially based on this nice answer and I use the following lemma proved there:

Lemma: If $F$ is a subfield of $\mathbb {R} $ and there is a real number $c\notin F$ and a positive integer $n$ such that $c^n\in F$ then $\text{tr} _F^{F(c)} (c) =0$.

Let $$a=\sqrt{3+\sqrt{2}},b=\sqrt{3-\sqrt{2}},F=\mathbb {Q} (\sqrt {2}),K=F(a)=\mathbb{Q} (a) ,L=F(a,b)$$ then $a, b$ are real numbers and $F, K, L$ are subfields of $\mathbb {R} $.

Further we have $a^2\in F, b^2\in F$ and both $a, b$ are of degree $2$ over $F$ (this follows from the fact that minimal polynomial for both $a, b$ over $\mathbb {Q} $ is of degree $4$).

Thus $$\text{tr}_F^{F(a)} (a) =0,\text{tr}_F^{F(b)}(b)=0$$ and since $L$ is a finite extension of both $F(a), F(b) $ we have $$\text{tr} _F^L(a) =\text{tr} _F^L(b) =0$$ Let us assume on the contrary that $b\in K=F(a) $ so that we have $$b=p+qa\tag{1}$$ for some $p, q\in F$. Viewing this equation as a relation between members of $L$ and applying trace $\text{tr} _F^L$ on it we see that $$0=np+q\cdot 0$$ where $n=[L:F] $. Thus we get $p=0$ and $ab=qa^2\in F$ or $$\sqrt{7}=t+u\sqrt{2}\tag{2}$$ for some rational $t, u$. Again applying trace $\text{tr} _{\mathbb{Q}} ^L$ on above equation we get $$0=2nt+u\cdot 0$$ ie $t=0$ and then $\sqrt{7/2}=u$ is rational which is absurd.

Therefore equation $(1)$ can't hold and $b\notin K= F(a) =\mathbb {Q} (a) $.