# Is $\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$?

Is $$\sqrt{3-\sqrt{3}} \in L = \mathbb{Q}(\sqrt{3+\sqrt{3}})$$? I know that $$\frac{1}{\sqrt{3+\sqrt{3}}} = \frac{\sqrt{3-\sqrt{3}}}{\sqrt6}$$. So I just need to know whether $$\sqrt6 \in L$$. Since $$\sqrt 3 = (\sqrt{3+\sqrt{3}})^2 - 3$$, I only need to know if $$\sqrt 2 \in L$$. I would guess it is not, but how to show it? If I suppose that $$\sqrt 2 \in L$$ and aim for a contradiction:

It is clear that $$\mathbb{Q}(\sqrt 3) \subset L$$, and if $$\sqrt 2 \in L$$, then $$\mathbb Q(\sqrt 2) \subset L$$ as well. Then $$K = \mathbb Q (\sqrt 2, \sqrt 3) \subset L$$. Since both $$K$$ and $$L$$ are degree $$4$$ over $$\mathbb Q$$, this would imply they are equal. But then I'm back at square one.

Let $$K=\Bbb Q(\sqrt3)$$. Then for $$\alpha$$, $$\beta\in K$$, $$\sqrt\beta\in K(\sqrt\alpha)$$ if and only if $$\beta$$ or $$\alpha\beta$$ is a square in $$K$$.

In this example. $$\alpha=3+\sqrt3$$ and $$\beta=3-\sqrt3$$. Then $$\alpha$$ is not a square in $$K$$, since its norm to $$\Bbb Q$$ is $$6$$ which is not a square in $$\Bbb Q$$. Also $$\alpha\beta=6$$ and that is not a square in $$K$$: the rational numbers which are squares in $$K$$ are $$a^2$$ and $$3a^2$$ for $$a\in\Bbb Q$$.

In this case, $$\sqrt\beta\not\in\Bbb Q(\sqrt\alpha)$$.

• Hi @Lord Shark, sorry for the delay, its been a busy couple of days. I can't seem to prove the first line of your answer. I can understand the rest. Could you elaborate a bit more on that part? – eatfood Oct 17 '19 at 12:20
• @eatfood I have corrected it now. – Angina Seng Oct 17 '19 at 17:37
• Proof: $\sqrt \beta \in K(\sqrt \alpha) = \mathbb{Q} (\sqrt 3 \sqrt \alpha)$ if and only if $\sqrt\beta$ can be expressed as $\sqrt \beta = a + b\sqrt\alpha$ for some $a,b \in K$. Then the $(\impliedby )$ direction is quite clear. For the forward direction, if $b$ or $a$ is zero, then we have $\sqrt \beta = a$ or $\sqrt\beta = b\sqrt\alpha$, and from there it follows that either $\beta \in K$ or $\sqrt{\alpha\beta} \in K$. If $a, b$ both non-zero, then squaring gives $\beta = a^2 + b^2\alpha + 2ab\sqrt\alpha$, which gives $\sqrt\alpha \in K$, and so $\sqrt\beta \in K$ as well. – eatfood Oct 18 '19 at 6:13
• Thank you very much, this is a very nice proof! – eatfood Oct 18 '19 at 6:14

Use the tracial method, for example. The trace map on a number field $$Q(\alpha)$$ assigns to each $$\beta \in \mathbb Q(\alpha)$$ the quantity $$\sum_{\sigma} \sigma(\beta)$$ where $$\sigma(\beta)$$ are all the conjugates of $$\beta$$.

Suppose that $$\sqrt 2 \in L$$. Then $$\sqrt 3 \in L$$ implies that $$\mathbb Q(\sqrt{3+\sqrt 3}) = \mathbb Q(\sqrt 2 ,\sqrt 3)$$.

Write $$\sqrt{3 + \sqrt 3} = a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6$$. Taking the trace of both sides gives $$2a = 0$$(note that $$\sqrt{3+\sqrt 3}$$ has minimal polynomial $$x^4 - 6x^2+6$$ hence has trace zero) hence $$a = 0$$.

Next multiply by $$\sqrt 2$$ to get $$2b + c \sqrt 6 + 2d \sqrt 3 = \sqrt{6+2\sqrt 3}$$. Once again the trace of the RHS is seen to be zero(find the minimal polynomial, it is easy!), so $$b = 0$$ is obtained on taking trace.

We are left with $$\sqrt{3+\sqrt 3} = c \sqrt 3 + d \sqrt 6$$. Multiplying by $$\sqrt 3$$ gives $$\sqrt{9+3\sqrt 3} = 3c + 2d \sqrt 2$$. Again the trace of the LHS is zero(easy again!) so we get $$c = 0$$.

Finally we are left with $$\sqrt{3 + \sqrt{3}} = d \sqrt 6$$, here squaring gives $$\sqrt 3 = 6d^2 - 3$$, an obvious problem. Hence, we are done.

Note that $$\sqrt{3 - \sqrt 3}$$ is a conjugate of $$\sqrt{3+\sqrt 3}$$, therefore the above shows that this extension is not normal, hence not Galois.

• I thought the standard notion of the trace of $\beta$ was the trace of the linear map of the $\Bbb Q$-vector space $L$ given by multiplication by $\beta$. Do these two notions coincide? – Arthur Oct 15 '19 at 7:58
• Check Wikipedia, for example in the above case the extension is galois so the notions coincide – Teresa Lisbon Oct 15 '19 at 10:45
• This is a very cool method! It can also be used to show that $\mathbb{Q}[\sqrt p_1, \dots \sqrt p_n] = 2^n$ right? – eatfood Oct 17 '19 at 12:25
• Yes I would think so, by induction for example, but isn't there an easier method? There will be too many coefficients in that case – Teresa Lisbon Oct 17 '19 at 13:14
• Hi @астонвіллаолофмэллбэрг, thank you for the very neat alternative solution, however I accepted the other answer because it was more elementary. – eatfood Oct 18 '19 at 6:15

Let $$\alpha, \beta$$ be $$\sqrt{3+\sqrt{3}}$$ and $$\sqrt{3-\sqrt{3}}$$ respectively; they clearly have the same minimum polynomial on $$\mathbb{Q}$$ which is $$f(x) = x^4-6x^2+6$$, by direct calculation and using Eisenstein Criterion.

Now let $$E$$ be the splitting field of $$f$$ on $$\mathbb{Q}$$ and let $$G$$ be its Galois group. The order of $$G$$ may be $$4,8,12,24$$; I'll prove that it is exactly 8. In this way it can be shown that $$\beta$$ cannot be in $$\mathbb{Q}(\alpha)$$ because otherwise $$E = \mathbb{Q}(\alpha)$$ which has order 4 on $$\mathbb{Q}$$.

By Galois Correspondence Theorem there is at least one subgroup $$H$$ of $$G$$ with index 4 in it: that's the subgroup consisting of all $$\mathbb{Q}$$-automorphisms of E which fix the elements of $$\mathbb{Q}(\alpha)$$. Since the elements of H send roots of $$f$$ in roots of $$f$$, it's clear that the only possibility for $$H$$ is to be of the form $$H = \{id, \sigma\}$$ where $$\sigma$$ is the $$\mathbb{Q}$$-automorphism of E that sends $$\beta$$ to $$-\beta$$ and fixes $$\alpha$$ (and consequently fixes $$-\alpha$$).

Therefore $$G$$ contains a subgroup having 2 elements and index 4 in $$G$$; by Lagrange's Theorem follows that $$G$$ has order 8.