This is the problem I am facing:

Compute the intermediate fields of the extension $K | \mathbb{Q}(i)$ where $K = \mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i)$ and find the intermediate fields $M$ such that $M | \mathbb{Q}(i)$ is a Galois extension.

What I have done:

I start by thinking about the grade of the extension $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(i)]$

$[\mathbb{Q}(\sqrt{3},i) \, : \, \mathbb{Q}(i)] = 2$ because the irreducible polinomial of $\sqrt{3}$ over $\mathbb{Q}(i)$ is $x^2 -3$ (since $\sqrt{3} \notin \mathbb{Q}(i)$

$[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(\sqrt{3},i)] = 3$ because the irreducible polinomial of $\sqrt[3]{2}$ over $\mathbb{Q}(\sqrt{3},i)$ is $x^3 -2$ (*)

Hence, $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(i)] =[\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, : \, \mathbb{Q}(\sqrt{3},i)] * [\mathbb{Q}(\sqrt{3},i) \, : \, \mathbb{Q}(i)] = 3 * 2 = 6$

Now, a $\mathbb{Q}(i) $-base for $\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i)$ is $\{ 1, \sqrt[3]{2} \, \xi , \sqrt[3]{2} \, \xi^2, \sqrt{3} ,\sqrt{3} \sqrt[3]{2} \, \xi , \sqrt{3} \sqrt[3]{2} \, \xi^2 \}$

The non-linear elements whose images fix every $\sigma \in \textrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i))$ are:

$$ \sqrt{3} \to \pm \sqrt{3}$$ $$ \sqrt[3]{2} \to \{\sqrt[3]{2}, \sqrt[3]{2} \, \xi, \sqrt[3]{2} \, \xi ^2 \}$$

The extension $\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i)$ is a Galois extension (it is normal(splitting field of $(x^3-2)(x^2-3)$, separable (char($\mathbb{Q}(i)$) = 0) and finite) so $| \textrm{Gal}(\mathbb{Q}(\sqrt[3]{2}, \sqrt{3},i) \, | \, \mathbb{Q}(i)) | = 6$ which means that all the options are valid.

I am stuck here, how can I determine the Galois group, and the normal subgroups (and then, how to extract the matching fields for those subgroups?)

(*) How to show justify this?

  • $\begingroup$ Also, $Gal(K/\mathbb{Q}[i]) \cong S_3$. You've listed your automorphisms, and none of them generate the Galois group, so it cannot be cyclic. Remember that the only groups of order $6$ are $\mathbb{Z}_6$ and $S_3$. $\endgroup$ – Kaj Hansen Jul 17 '14 at 21:27
  • 2
    $\begingroup$ $\mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6$ since $gcd(2, 3) = 1$. See here: proofwiki.org/wiki/Group_Direct_Product_of_Cyclic_Groups $\endgroup$ – Kaj Hansen Jul 17 '14 at 21:30

To address the (*):

To show that $f(x) = x^3 - 2$ is irreducible over $\mathbb{Q}(\sqrt{3}, i)$, it suffices to show that none of its roots are in this field (since $\deg(f) \leq 3$). Indeed, $\sqrt[3]{2}, \ \omega\sqrt[3]{2}$, and $\omega^2\sqrt[3]{2}$ are not elements of $\mathbb{Q}(\sqrt{3}, i)$. If they were, we would have a tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\sqrt{3}, i)$, which would imply $\Big[ \mathbb{Q}(\sqrt{3}, i):\mathbb{Q} \Big] = \Big[ \mathbb{Q}(\sqrt{3}, i): \mathbb{Q}( \sqrt[3]{2}) \Big] \cdot \Big[ \mathbb{Q}(\sqrt[3]{2}):\mathbb{Q} \Big]$. Contradiction! Look at the degrees of the extensions: $3$ does not divide $4$.

To address the crux of the question:

You're close! You've done a good bit of the work already.

Remember that the only groups of order $6$ are $\mathbb{Z}_6$ and $S_3$. However, you've found $6$ automorphisms and none of them generate the entire group. Therefore, it must be the case that $\operatorname{Gal}(K/\mathbb{Q}(i)) \cong S_3$.

Next, the fundamental theorem of Galois theory tells us that there is a one-to-one correspondence between subgroups of $\operatorname{Gal}(K/\mathbb{Q}(i))$ and the intermediate fields between $K$ and $\mathbb{Q}(i)$ that they fix. Let's list the subgroups of $S_3$. They are:

  • $1$ copy of the trivial subgroup
  • $3$ copies of $\mathbb{Z}_2$
  • $1$ copy of $\mathbb{Z}_3$
  • The whole group

You should check this yourself! Hint: Lagrange's theorem tells us that the proper subgroups are of prime order, and hence are cyclic. First, the trivial subgroup corresponds to the entire field $K$, and the whole group corresponds to the base field $\mathbb{Q}(i)$.

Next, let's call the two 'generator' automorphisms that you found $\phi$ and $\tau$ such that $\phi$ sends $\sqrt{3} \mapsto \sqrt{3}$ while fixing the other basis elements, and likewise $\tau$ sends $\sqrt[3]{2} \mapsto \sqrt[3]{2}\xi$.

You should check that the copies of $\mathbb{Z}_2$ are:

  • $\{\text{id}, \phi \}$
  • $\{\text{id}, \tau \phi \}$
  • $ \{\text{id}, \tau^2 \phi \} $

And the copy of $\mathbb{Z}_3$ is $\{\text{id}, \tau, \tau^2\}$.

Some proper intermediate fields between $K$ and $\mathbb{Q}(i)$ are $\mathbb{Q}(i, \sqrt[3]{2})$, $\mathbb{Q}(i, \sqrt{3})$, $\mathbb{Q}(i, \sqrt[3]{2}\xi)$, and $\mathbb{Q}(i, \sqrt[3]{2}\xi^2)$. How do we know this is a complete list? Keeping in mind that $\displaystyle \xi = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$, I'll leave it to you match the subgroups with the intermediary fields they fix.

Finally, to determine which intermediate fields are Galois over $\mathbb{Q}(i)$, simply apply the fact than an intermediate field $E$ is Galois if and only if its corresponding subgroup is a normal subgroup of $\operatorname{Gal}(K/\mathbb{Q(i)})$. What are the normal subgroups of $S_3$?


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