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I'm trying to prove that the function $\phi$ given $\phi(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ and $\phi(q)=q$ for all $q\in \mathbb{Q}$ is an automorphism of the extension field $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})}))/\mathbb{Q}$ but I can't prove the surjectivity. More precisely, I cannot find an element $x$ in the extension field such that $\phi(x)=\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ I try this: Assuming $\phi$ is a homomorphism (which I already proved) Let $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ and $\beta=\sqrt{(2-\sqrt{2})(3+\sqrt{3})})$ then $\alpha=\frac{\alpha \beta}{\beta}=\frac{\sqrt{2}(3+\sqrt{3})}{\phi(\alpha)}$ and i would like that ${\sqrt{2}(3+\sqrt{3})}=\phi(y)$ some $y\in\mathbb{Q}(\alpha)$ but I get circular reasoning ...

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eraldcoil
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1 Answers1

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Mainlines for an answer

Denote

$$\begin{cases} \pi &= \sqrt{2+ \sqrt 2}\\ \rho &= \sqrt{3+ \sqrt 3} \end{cases}$$

$\pi, \rho$ are respectively roots of the rational polynomials $$\begin{cases} p(x) &= x^4 - 4 x^2 +2\\ r(x) &= x^4 - 9 x^2 + 3 \end{cases}$$

$p,r$ are irreducible according to Eisenstein's criterion. The set of roots

  • of $p$ is $\{ \pi_1=\sqrt{2+ \sqrt 2}, \pi_2=\sqrt{2- \sqrt 2}, \pi_3=-\sqrt{2+ \sqrt 2},\pi_4=-\sqrt{2-\sqrt 2}\}$
  • the one of $r$ is $\{ \rho_1=\sqrt{3+ \sqrt 3}, \rho_2=\sqrt{3- \sqrt 3}, \rho_3=-\sqrt{3+ \sqrt 3},\rho_4=-\sqrt{3-\sqrt 3}\}$.

$\alpha, \beta$ belong to the finite field extension $\mathbb Q(\pi, \rho)/\mathbb Q$ and the field homomorphism $\phi$ satisfies $\phi(\alpha) = \beta$.

You suppose that $\beta$ belongs to $\mathbb Q(\alpha) / \mathbb Q $ which remains to be proved.

To do so, notice that $\alpha \notin \mathbb Q(\pi, \sqrt 3)$. If that would be the case, $\rho$ would also belong to $\mathbb Q(\pi, \sqrt 3)$. Which in turn implies the contradiction that $\{1, \sqrt 2, \sqrt 3, \sqrt 6\}$ would be linearly dependent over $\mathbb Q$. Similarly $\beta$ doesn't belong to $\mathbb Q(\pi, \sqrt 3)$.

Therefore $\mathbb Q(\alpha) = \mathbb Q(\beta) = \mathbb Q(\pi, \rho)$ and if $\phi(\alpha) = \beta$, $\phi$ can be extended into an homeomorphism of $\mathbb Q(\alpha)$ which is also an automorphism.

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  • May be I need more coffee but you lost me at a couple of points. Why does $\rho\in\Bbb{Q}(\pi,\sqrt3)$ imply that ${1,\sqrt2,\sqrt3,\sqrt6}$ is linearly dependent over the rationals? Why does $\alpha,\beta\notin\Bbb{Q}(\pi,\sqrt3)$ imply that $\Bbb{Q}(\alpha)=\Bbb{Q}(\beta)$? – Jyrki Lahtonen Nov 28 '20 at 06:14
  • What is the field homomorphism $\phi$ you refer to at the end of the first paragraph? How did you deduce it exists? – Jyrki Lahtonen Nov 28 '20 at 06:18
  • @JyrkiLahtonen As $\pi=\pi_1, \pi_2$ are both roots of the polynomial $p$ with rational coefficients, it exists a field homomorphism $\phi_1$ of $\mathbb Q(\pi)$ such that $\phi_1(\pi_1) = \pi_2$. You can extend $\phi_1$ into a field homomorphism of $\mathbb Q(\pi, \rho)$ such that $\phi(\rho)=\rho$ and therefore $\phi(\alpha)=\beta$. – mathcounterexamples.net Nov 28 '20 at 15:05
  • @JyrkiLahtonen And $\phi$ is the field homomorphism mentioned in the OP question. – mathcounterexamples.net Nov 28 '20 at 15:24
  • That only gives a homomorphism $\phi_1$ from $\Bbb{Q}(\pi_1)$ to $\Bbb{Q}(\pi_2)$. I do know that $\Bbb{Q}(\pi_1)$ is Galois over $\Bbb{Q}$, or equivalently $\pi_2\in\Bbb{Q}(\pi_1)$, so the two fields are equal. It just feels more than a bit incomplete as a solution to skip such details even though it is easy to see. If your goal was not to give a full solution, I apologize for reading it as such. – Jyrki Lahtonen Nov 28 '20 at 16:11
  • @JyrkiLahtonen I agree that some intermediate steps maybe missing... i'll add a comment! – mathcounterexamples.net Nov 28 '20 at 16:33