Suppose we have the extension field $\mathbb{Q}(\sqrt2, \sqrt[3]2)$.

For simplicity, let $\alpha = \sqrt2$ and $\beta = \sqrt[3]2$

First, I can show that $\mathbb{Q}(\alpha, \beta) = \mathbb{Q}(\alpha + \beta)$ by subset arguments, so to find an irreducible polynomial (which would be irreducible by the rational zeros theorem) I can do the following:

$x = \alpha + \beta$

$x - \alpha = \beta$

$(x - \alpha)^3 = 2$

$x^3 + 4x - 2\alpha(x^2 + 1) = 2$

$x^3 + 4x - 2 = 2\alpha(x^2 + 1)$


$f(x) = (x^3 + 4x -2)^2 - 8(x^2 + 1)^2$

Which is of degree $6$, and so my extension is a degree $6$ extension.

I can also verify this using the tower law for extensions, since:

$[\mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\alpha)] \cdot [\mathbb{Q}(\alpha) : \mathbb{Q}] = 3 \cdot 2 = 6$

So the Galois group for this extension will be isomorphic to a subgroup of $S_6$. That much is clear. Where I need help is in constructing the automorphisms. Besides the identity automorphism, I can have the following:

$\pi_1: \alpha \rightarrow -\alpha$

$\pi_2: \beta \rightarrow \beta^2$

$\pi_3: \alpha \rightarrow -\alpha, \beta \rightarrow \beta^2$

But if $\pi_2(\beta) = \beta^2$, then $\pi_2(\beta^2) = \beta^4 = 2\beta$ which seems weird. Something is wrong with what I am doing.

And what about something like this?

$\pi_4: \beta \rightarrow -\beta$

$\pi_5: \beta \rightarrow -\beta^2$

  • $\begingroup$ That is not a Galois extension. $\endgroup$ – Lord Shark the Unknown Jul 18 '17 at 11:28
  • 1
    $\begingroup$ Okay, I figured there was something very wrong. In order for it to be one, we'd need to also adjoin the 3rd root of unity, correct? $\endgroup$ – Yabbadule Jul 18 '17 at 14:37

Just keep in mind that an element of the Galois group over the rationals must take a root of a rational polynomial to another root. So how many choices do you get for $\sqrt{2}$ (which is a root of $x^{2} - 2$), and how many for $\sqrt[3]{2}$ (which is a root of $x^{3} - 2$)?


You may want to check what are the other roots of $x^{3} - 2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.