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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.

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

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  • $\begingroup$ 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? $\endgroup$ – eatfood Oct 17 '19 at 12:20
  • $\begingroup$ @eatfood I have corrected it now. $\endgroup$ – Angina Seng Oct 17 '19 at 17:37
  • $\begingroup$ 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. $\endgroup$ – eatfood Oct 18 '19 at 6:13
  • $\begingroup$ Thank you very much, this is a very nice proof! $\endgroup$ – eatfood Oct 18 '19 at 6:14
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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.

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  • $\begingroup$ 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? $\endgroup$ – Arthur Oct 15 '19 at 7:58
  • $\begingroup$ Check Wikipedia, for example in the above case the extension is galois so the notions coincide $\endgroup$ – Teresa Lisbon Oct 15 '19 at 10:45
  • $\begingroup$ 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? $\endgroup$ – eatfood Oct 17 '19 at 12:25
  • $\begingroup$ 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 $\endgroup$ – Teresa Lisbon Oct 17 '19 at 13:14
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    $\begingroup$ Hi @астонвіллаолофмэллбэрг, thank you for the very neat alternative solution, however I accepted the other answer because it was more elementary. $\endgroup$ – eatfood Oct 18 '19 at 6:15
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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.

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