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Is the following proof correct for showing that $\sqrt{3}\notin \mathbb{Q}(\sqrt[12]{2})$?

If $\xi$ is a primitive twelveth root of $1$, there are 12 morphisms $\mathbb{Q}(\sqrt[12]{2})\to {\mathbb{\bar Q}}$, given by $\sqrt[12]{2} \mapsto \xi^j\sqrt[12]{2}$ for $j=0,...,11$. If $\sqrt3 \in \mathbb Q (\sqrt[12]2)$, then we could write $$\sqrt{3} = \sum_{k=0}^{11}a_k\sqrt[12]{2^k}.$$ for some rational $a_k$. Now, applying each of the morphisms, we obtain $$\delta_j\sqrt{3} = \sum_{k=0}^{11}a_k\xi^{jk}\sqrt[12]{2^k}, \ j=0,...,11,$$ where each of the deltas is $\pm 1$. Adding for all $j,$ we get $$\sum_{j=0}^{11}\delta_j\sqrt{3} = 12a_0,$$ and since $\sqrt 3$ is irrational, $LHS = a_0 = 0$. If instead we add only for even $j,$ then we obtain $$\sum_{\text{even }j}\delta_j\sqrt{3} = 6a_6\sqrt{2};$$ but $\sqrt 3 \notin \mathbb Q (\sqrt 2)$, so again $LHS = a_6 = 0$. Finally, adding for $j$ a multiple of 4, $$\sum_{j=0,4,8}\delta_j\sqrt{3} = 3a_3\sqrt[4]{2} + 3a_9\sqrt[4]8.$$ Squaring both sides, $$3\bigg(\sum_{j=0,4,8}\delta_j\bigg)^2 = 9a_3^2 \sqrt 2 + 18a_9^2 \sqrt 2 + 36a_3a_9.$$ Irrationality of $\sqrt 2$ implies $a_3 = a_9 = 0$, but then $$\sum_{j=0,4,8}\delta_j = \pm 1 \pm 1 \pm 1 = 0,$$ which is impossible. Thus $\sqrt3 \notin \mathbb Q(\sqrt[12]2)$.

I'm a bit suspicious because the only values of $a_k$ used are those with $k$ a multiple of $3$.

Keplerto
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    What's the question? The header refers to $\sqrt[n] {2}$ but the text seems to set $n=12$. – lulu Jul 16 '22 at 00:00
  • Sorry, I only proved it for $n=12$, I've changed the title. – Keplerto Jul 16 '22 at 00:08
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    I'd look at ramification. Does $3$ ever ramify in $\mathbb Q(\sqrt[n] 2)$. – lulu Jul 16 '22 at 00:10
  • Thanks for the idea, though I'm not familiar with ramifications. – Keplerto Jul 16 '22 at 00:25
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    See https://math.stackexchange.com/q/3990185/72031 – Paramanand Singh Jul 16 '22 at 01:09
  • This is not a subject I am deeply familair with, so maybe I am just missing the needed background, but how do "we obtain $\delta_j\sqrt{3} = \sum_{k=0}^{11}a_k\xi^{jk}\sqrt[12]{2^k}, \ j=0,...,11$" from applying each of the morphisms? – Paul Sinclair Jul 17 '22 at 03:51
  • If $\sigma$ is the morphism sending $\sqrt[12] 2$ to $\xi^j\sqrt[12] 2$, then $\sqrt 3$ must go to a root of $x^2-3$ ($\sigma(\sqrt 3) = \pm \sqrt 3$), while $\sigma(\sqrt[12]{2^k})=\sigma(\sqrt[12] 2)^k = (\xi^j\sqrt[12] 2)^k=\xi^{jk}\sqrt[12]{2^k}$, and then using linearity. Do you see anything wrong there? – Keplerto Jul 17 '22 at 13:24
  • Thank you *paramanand-singh* for the reference, that's a nice proof of the general case. Now I'm wondering how the condition of primality of p and q could be relaxed for that argument to work, would it be enough that no power of $\alpha$ is a rational multiple of $\beta$, for example $\sqrt[3] 6$ and $\sqrt [3] {12}$? (In the notation of that answer) – Keplerto Jul 17 '22 at 15:45
  • Assuming of course that $\mathbb Q(\alpha)$ and $\mathbb Q(\beta)$ have degree m and n respectively – Keplerto Jul 17 '22 at 15:52
  • You can also use the following theorem: If $c\in\mathbb {R} \setminus \mathbb {Q} $ and $n$ is a positive integer such that $c^n\in\mathbb {Q} $ then the trace of $c$ as a member of $\mathbb {Q} (c) $ is $0$. Now consider $a=\sqrt{3},b=2^{1/12}$ and apply trace on your first equation (of the question) treating each side as member of $\mathbb {Q} (a, b) $. Repeated application of this idea will give each $a_i=0$. – Paramanand Singh Jul 18 '22 at 16:55
  • You may see the approach in last comment used in a more general setting at https://math.stackexchange.com/a/4304716/72031 – Paramanand Singh Jul 18 '22 at 17:01
  • I see now that for $\alpha=\sqrt[4]{12},\ \beta=\sqrt[4]{3}$, my conjecture fails since $[\mathbb Q(\alpha):\mathbb Q]=[\mathbb Q(\beta) : \mathbb Q] = 4$ but $[\mathbb Q(\alpha,\beta):\mathbb Q]=8<16$. It looks to me like this question you linked there is the best possible generalization. As a final remark, I think what I did in my proof can be expressed as taking traces first over $\mathbb Q$, then over $\mathbb Q(\sqrt 2)$, then over $\mathbb Q(\sqrt[4] 2)$. – Keplerto Jul 18 '22 at 19:10
  • Just need to use trace map of $\mathbb {Q} (\sqrt{3},2^{1/12})$ over $\mathbb {Q} $. In the proof you will need to make use of the fact that $2^{i/12}\sqrt{3}$ is irrational for each value of $i=0,1,2,\dots,11$. – Paramanand Singh Jul 19 '22 at 02:20
  • The proof by Paramanand Singh is simpler and more general; I just wanted to add that it was enough to show $\sqrt 3 \notin \mathbb Q(\sqrt [4] 2)$ and then use the degree theorem. Using the method of proof in the answer, however, this simplification is irrelevant – Keplerto Jul 21 '22 at 09:22

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We use the following

Lemma: Let $F\subseteq \mathbb {R} $ be a field and let $a$ be a real number and $m$ a positive integer such that $a\notin F, a^m\in F$. Then the trace $\operatorname {tr} _F^{F(a)} (a) =0$.

Let $a=\sqrt{3},b=\sqrt[12]{2},F=\mathbb {Q} $. Also let degree of extension $F(a, b) $ over $F$ be $n$.

Let us assume that $a\in F(b) $ so that there is a relation of the form $$a=a_0+a_1b+a_2b^2+\dots+a_{11}b^{11}\tag{1}$$ where $a_i\in F$. Note that each of $a, b^i$ with $i=1,2,\dots,11$ is such that it does not lie in $F$ but some power of it lies in $F$ and thus we can apply the lemma to get $$0=na_0+a_1\cdot 0+\dots+a_{11}\cdot 0$$ ie $a_0=0$.

Next we rewrite the equation $(1)$ as $$\frac{a} {b} =a_1+a_2b+\dots+a_{11}b^{10}$$ Applying trace map (and noting that $a/b\notin F$) we again get $0=na_1$ ie $a_1=0$. Continuing in this fashion we get all the $a_i=0$ which is absurd. It follows that $a\notin F(b) $.

Paramanand Singh
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