Is $\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}\left(\sqrt{2},i\sqrt[3]{2}\right)$ a Galois extension?

Question

Consider the field extension

$$\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}\left(\sqrt{2},i\sqrt[3]{2}\right)$$

Is it Galois?

I can't quite find the order of the automorphism group.

Answer 1

The comments that have been made seem to contain all that is needed to solve the problem: let me call $c$ the real cubic root of 2 and $K = Q(\sqrt{2})$, to ease notation. Since the element $i c$ is a root of $P(x)= x^6+4 = 0$, a polynomial with coefficients in $K$, the degree of the extension must divide $6$. On the other hand, as noted by Jyrki, $c$ is in the extension, so the degree must be a multiple of 3. If the degree were 3, then $K(c) = K(ic)$, because we have shown that the first is contained in the second and they have both the same degree. But this is a contradiction, because $K(c)$ is a real field, and $ic$ is not real. Thus the degree is $6$, and $P(x)$ is a minimal polynomial for $ic$ over $K$. From here we can conclude that the extension is NOT Galois, by showing that some root of this polynomial, or more generally some "algebraic combination of roots with coefficients in $K$" (i.e., some element in the splitting field of the polynomial) is not in $K(ic)$. If we call $z$ a primitive cubic root of unity, we see that $z \cdot ic$ is also a root of $P(x)$, and we divide by the root $ic$ to conclude that $z$ is in the splitting field. But since $z = (-1 + \sqrt{-3})/2$, we conclude that $\sqrt{-3}$ is also in the splitting field. Now it remains to check that $\sqrt{-3}$ is not in $K(ic)$ to conclude the proof that this is not a Galois extension of $K$. Maybe there is a shorter way to check this, but here is one way: suppose $\sqrt{-3}$ belongs to $K(ic)$. Since $c$ also belongs to $K(ic)$, comparing degrees we deduce $K(ic) = K(\sqrt{-3} , c)$. We can also assume that this is a Galois extension of $K$, since we are arguing by contradiction and this is what we want to conclude. We now use the fact, noted by Jyrki, that $i$ belongs to $K(ic)$. This contradicts the previous equalities of fields, because it gives too many quadratic fields inside the field $K(\sqrt{-3}, c)$. This is (by our assumption) a Galois extension of $K$, of degree 6 and given by a minimal polynomial with coefficients in $Q$, so it is Galois over $Q$ of degree $6$ or $12$ and because it is obtained as the compositum of $K$ and the splitting field of $P(x)$ as a polynomial over $Q$, its Galois group contains the quadratic fields $K$, $Q(\sqrt{-3})$, and also of course $Q(\sqrt{-6})$. Here we have used again that $\sqrt{-3}$ is in our field. In particular the degree over $Q$ must be $12$, it can not be $6$. Observe that there is no room for more quadratic fields (inside a Galois field of degree 12 containing already 3 quadratic fields) so $i$ is not in our field, and this gives the contradiction we wanted. Some help in giving a shorter proof that $\sqrt{-3}$ can not be contained in $K(ic)$? In any case, we conclude that the extension is not Galois.