# Sufficient condition for counting degree n subfields

"Given that the automorphism group of $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$ is isomorphic to $\mathbb{Z}_2 \oplus\mathbb{Z}_2 \oplus\mathbb{Z}_2$, determine the number of subfields of $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$ that have degree 4 over $\mathbb{Q}$."

Looking at $\mathbb{Z}_2 \oplus\mathbb{Z}_2 \oplus\mathbb{Z}_2$, it is easy to see that every nontrivial element generates a subgroup of order 2, and thus having index 4. Considering that $\mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$ isn't the splitting field of a polynomial over $\mathbb{Q}$ we are not able to apply the Fundamental Theorem of Galois Theory here to obtain a one-to-one correspondence between subgroups of the Galois group and subfields of our extension field. However, is the existence of appropriate index subgroups in this case enough to give us these subfields of order 4?

Generally, when we're unable to use the FToGT, what basic facts do we still know to help us out?

Lastly, are these subfields always just the fixed fields corresponding to each subgroup of the Galois group?

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You have a misconception - it's the splitting field of $(x^2 - 2)(x^2 - 5)(x^2 - 7)$ and so is Galois. – user29743 Apr 28 '12 at 21:34
So, yes, you can just find the subfields by looking at all the subgroups of the Galois group. – user29743 Apr 28 '12 at 21:35

As was noted, your field actually is a Galois extension of $\mathbb{Q}$, as it is the splitting field of $(x^2-2)(x^2-5)(x^2-7)$.
An alternative way of verifying that it is Galois is to note that it has degree $8$ over $\mathbb{Q}$ (since $\sqrt{2}$, $\sqrt{5}$, and $\sqrt{7}$ are linearly independent), and so $|\mathrm{Aut}_{\mathbb{Q}}(K)| = [K:\mathbb{Q}]$, which in the finite case suffices to establish that $K$ is Galois over $\mathbb{Q}$.
So it is indeed enough to find the subgroups of index $4$ of the Automorphism group.
To answer your more general question: suppose $K/L$ is a separable extension, but not Galois. (In characteristic zero, separability always holds). Let $N$ be the Galois closure of $K$ over $L$ (in some fixed algebraic closure of $L$). Then by the Fundamental Theorem of Galois Theory, the intermediate fields of $K/L$ correspond to subgroups of $\mathrm{Gal}(N/L)$ that contain $K'=\mathrm{Aut}_L(K)$; you'd be looking for subgroups of $\mathrm{Gal}(N/L)$ that contain $\mathrm{Aut}_L(K)$ and have the appropriate index in $\mathrm{Gal}(N/L)$. So, yes, every intermediate field would correspond to a fixed field of an appropriate subgroup of the automorphism group of the Galois closure of $K$ over $L$.