Galois group of an octic over $\mathbb{Q}(i)$

Question

Let $K=\mathbb{Q}(i)$ and $L=K(\beta)$ where $\beta$ is a root of $f(x)=27x^8+72x^4-16$. This problem comes from Silverman and Tate; $\beta$ is the $y$-coordinate of one of the 3-torsion points of $y^2=x^3+x$. Let \begin{equation*} \beta = \sqrt[4]{\frac{8\sqrt{3}-12}{9}}, \end{equation*} and if we set \begin{equation*} \alpha = \sqrt{\frac{2\sqrt{3}-3}{3}} = \frac{\beta^2\sqrt{3}}{2},\quad \alpha' = -\frac{i}{\alpha\sqrt{3}},\quad \beta' = \sqrt[4]{\frac{-8\sqrt{3}-12}{9}}, \end{equation*} then the roots of $f(x)$ are $\pm\beta, \pm i\beta, \pm\beta', \pm i\beta'$, and the 3-torsion points of $y^2=x^3+x$ are (together with the point at infinity) \begin{equation*} (\alpha, \pm\beta),\ (-\alpha, \pm i\beta),\ (\alpha', \pm \beta'),\ (-\alpha', \pm i\beta'). \end{equation*} I am trying to understand the Galois group of $L/K$ ($L$ is Galois over $K$). My first attempt was to show that the group is cyclic, with generator $\beta\mapsto \beta'$. However, $\beta\mapsto\beta'$ also maps $\alpha$ to $-\alpha'$, and $(-\alpha',\beta')$ is not a 3-torsion point of the curve. Thus this cannot be an automorphism of $L/K$, although I cannot see why it isn't. (Of note is that Sage seems to think that this is an automorphism of $L/K$.) The map $\beta\mapsto i\beta$ preserves the 3-torsion points, but has only order four.

So: 1) what is going on with $\beta\mapsto\beta'$? Is it an $L/K$-automorphism? If so, I must have done something wrong with my calculation of the image of $\alpha$ under that automorphism. And 2) if it is not, what is a second generator of the Galois group other than $\beta\mapsto i\beta$?

I'd appreciate not getting back a complete solution, as I'm really trying to understand this group in preparation for doing another exercise.

Answer 1

Assuming that $\beta$ is real and positive, and $\beta'$ has argument $\pi/4$, then I don't see the problem. If $\sigma$ is an automorphism with the property $\sigma(\beta)=\beta'$ then:

• $\sigma(\beta^4)=\beta'^4$ implying that $\sigma(\sqrt3)=-\sqrt3$, and hence
• $\sigma(\alpha)=-\sqrt3\beta'^2/2$, a complex number with argument $-\pi/2$, agreeing with the argument of $\alpha'$.

This suggests to me that $\sigma(\alpha)=\alpha'$, just as you would expect from the list of 3-torsion points. I think this settles your first question, and seems to make the second question moot, but the structure of the relevant Galois groups is not clear to me yet.