# Prove that $\sqrt{3\pm\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3\mp\sqrt{7}})$.

I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $$\sqrt{3+\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3-\sqrt{7}})$$ and $$\sqrt{3-\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3+\sqrt{7}})$$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem.

Any help is appreciated!

• $\sqrt{3+\sqrt{7}} = a+b\sqrt{3-\sqrt{7}} \implies 3+\sqrt{7} = a^2+2ab\sqrt{3-\sqrt{7}} + b(3-\sqrt{7})$ $\implies c+d\sqrt{7} = e\sqrt{3-\sqrt{7}} \implies c^2+2cd\sqrt{7}+7d^2 = e^2(3-\sqrt{7}) \implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction. – mathworker21 Jan 8 at 15:39
• @mathworker21 What is $a$ and $b$? $\{1,\sqrt{3-\sqrt{7}}\}$ is not a basis for $\mathbb{Q}(\sqrt{3-\sqrt{7}})$ over $\mathbb{Q}$ – mouthetics Jan 8 at 15:40
• @mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals) – mathworker21 Jan 8 at 15:41

Let $$x_{\pm}=\sqrt{3 \pm \sqrt{7}}$$.

It is easy to see that we have the following quadratic extensions:

$$\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}) \subset \mathbb{Q}(x_+)$$,

$$\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}) \subset \mathbb{Q}(x_-)$$.

Assume that $$x_+ \in K=\mathbb{Q}(x_-)$$. Then $$\sqrt{2} = x_+x_- \in K$$, thus $$L=\mathbb{Q}(\sqrt{2},\sqrt{7}) \subset K$$.

Since these fields have the same degree over $$\mathbb{Q}$$, $$K \subset L$$, ie $$x_+=a+b\sqrt{2}+c\sqrt{7}+d\sqrt{14}$$ for rationals $$a,b,c,d$$.

Taking squares, we get $$3+\sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)\sqrt{2} + (2da+2bc)\sqrt{14} + (2ca+4bd)\sqrt{7}$$.

Thus $$ab=-7cd$$, $$ad=-bc$$, $$2ca+4bd=1$$, $$a^2+2b^2+7c^2+14d^2=3$$.

Assume $$a=0$$: then $$d \neq 0$$ thus $$c=0$$ and $$bd=1/4$$, $$2b^2+14d^2=3$$. Usual quadratic theory yields then a contradiction.

Thus $$b=-7cd/a$$, and $$ad=7c^2d/a$$ thus $$a^2=7c^2$$ hence $$a=0$$. A contradiction, hence the result.

In general, in this kind of problem, it is better not to mix the two operations + and $$\times$$. Let me give an illustration here, using only $$\times$$. Introduce the quadratic field $$k=\mathbf Q(\sqrt 7)$$. Adopting the notation $$x_{\pm}=\sqrt {3 \pm \sqrt 7}$$ suggested by @Mindlack, let us write $$K_{\pm}=k(x_{\pm})$$. These are two extensions of $$k$$ of degree at most $$2$$ :

• if $$K_{+}$$ or $$K_{-} =k$$, i.e. $$(3 \pm\sqrt 7)\in {k^*}^2$$, norming down to $$\mathbf Q$$ shows that $$N(3\pm\sqrt 7)=2$$ is a square in $$\mathbf Q^*$$: impossible

• if both degrees are 2, $$K_{\pm}\subset K_{\mp}$$ iff $$K_{\pm}= K_{\mp}$$, iff $$k(x_{+})= k(x_{-})$$, iff $$2=(3+\sqrt 7)(3-\sqrt 7)\in {k^*}^2$$ (no specific calculation, this is rudimentary Kummer theory over $$k$$), iff $$\mathbf Q(\sqrt 2)=\mathbf Q(\sqrt 7)$$, iff $$2.7$$ is a square in $$\mathbf Q^*$$(again by Kummer): impossible because $$\mathbf Z$$ is a UFD ./.

HINT: Show that both have minimal polynomial $$f:=X^4-6X^2+2$$ over $$\Bbb{Q}$$, and hence that $$[\Bbb{Q}(\sqrt{3\pm\sqrt{7}}):\Bbb{Q}]=4$$, but that the splitting field of $$f$$ over $$\Bbb{Q}$$ has degree greater than $$4$$.

• My goal is to justify that the splitting field is $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking? – BBC3 Jan 8 at 16:27
• How to prove that the splitting field of $f$ over $\mathbb{Q}$ has degree greater than $4$? Any other method? – mouthetics Jan 8 at 16:28
• Do you mean $X^4 - 6X^2 + 2$? – Connor Harris Jan 8 at 17:08
• @ConnorHarris Indeed I do, edited. – Servaes Jan 8 at 22:46
• @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind. – Servaes Jan 8 at 22:48