I'm kind of stuck trying to prove this particular exercise of Galois Theory, it came in my last exam and I don't see any way to reach the answer. It would be nice if you can give me some hint or a sketch of proof:

Let $L/k$ be a Galois Extention with $[L:k]=4 $. Show that if $\mathrm{Gal}(L/k) \cong \mathbb{Z}_2\times \mathbb{Z}_2$ then $L=k(\sqrt{a},\sqrt{b})$ for some $a,b \in k$

  • $\begingroup$ Use the correspondence, looking at subgroups order $2$, and at how a degree two extensions can be generated. $\endgroup$ – Arturo Magidin Jun 9 at 22:36
  • $\begingroup$ Can you give me more information, please? $\endgroup$ – Jorge Arturo Quiroz Jun 9 at 22:51
  • $\begingroup$ The result is not true if k is a finite field. $\endgroup$ – Dionel Jaime Jun 9 at 23:07
  • $\begingroup$ I think that as long as the characteristic is not 2, the result holds. $\endgroup$ – Alexandros Jun 9 at 23:16
  • $\begingroup$ EDIT: I think this is meant to be Dummit & Foote 14.2 15b $\endgroup$ – Theo C. Jun 9 at 23:18

$[L:K]=4$, and since we are dealing with a Galois extension, this means that the galois group is either $C_4$ or $C_2 \times C_2$. In this case you assume that it's $C_2 \times C_2$. By the main theorem of Galois theory, there's an (order reversing) one to one correspondence between then subgroups of the galois group and the intermediate fields.

As long as the characteristic of $k$ is not 2, any quadratic extension can be obtained by adjoining a square root, so say $k(\sqrt a)$ and $k(\sqrt b)$ are two of the intermediate fields.

Since $G=C_2 \times C_2$, then say $C_2 \times C_2= \langle \sigma, \tau, \sigma^2=\tau^2=id, \sigma\tau=\tau\sigma \rangle =\{id,\sigma,\tau,\tau\sigma\}$, where $\sigma(\sqrt a)=-\sqrt a$ and $\tau(\sqrt b)=-\sqrt b.$

Now, observe that all elements of $G$ send $\sqrt a+\sqrt b$ to something else, so $\sqrt a+\sqrt b$ is only fixed by the identity, so $L=k(\sqrt a + \sqrt b)$, and since $[L:k]=4$, and $k[\sqrt a + \sqrt b]$ is a subfield of $k[\sqrt a, \sqrt b]$, it follows that $L=k[\sqrt a,\sqrt b]$

  • 2
    $\begingroup$ I don’t think you need to do that much work; once you have the intermediate fields $k(\sqrt{a})$ and $k(\sqrt{b})$, observe that being distinct, $k(\sqrt{a},\sqrt{b})$ must be a nontrivial extension of $k(\sqrt{a})$, which is then of degree $2$, and so must equal $L$. $\endgroup$ – Arturo Magidin Jun 10 at 0:04

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