Is the extension $ \mathbb{Q}(\sqrt{2}, \sqrt{3}) $ simple?

Question

Intuitively, it is not.

Can anyone give a proof?

Answer 1

It actually is, and the same can be said of any finite extension of $\mathbb{Q}$ per the primitive element theorem, which is indeed unintuitive and surprising. In other words, there does exist a $\beta \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$ such that $\mathbb{Q}(\beta) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. To find such a $\beta$, we can take advantage of the following lemma:

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Lemma: If $K/F$ is a Galois extension, then the minimal polynomial for any $a \in K$ has as its roots the elements in the orbit of $a$ under the action of $\text{Gal}(K/F)$. That is, if $S = \{ \phi(a) \ | \ \phi \in \text{Gal}(K/F) \}$, then $\displaystyle \min_a(x) = \prod_{u_k \in S} (x-u_k)$.

Proof: This is a consequence of the fact that a polynomial is irreducible $\iff$ its Galois group acts transitively on its roots. For a proof of this fact, see Theorem 2.9(b) here. Notice that $\displaystyle \min_a(x)$ will be an irreducible polynomial (by definition), and its Galois group will be a subgroup of $\text{Gal}(K/F)$. This latter fact is because, if $L \subseteq K$ is the splitting field of $\displaystyle \min_a(x)$, every $F$-automorphism of $L$ extends to an $F$-automorphism of $K$. $\qquad \blacksquare$

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For a candidate $\beta$ that we choose, we already have the inclusion $\mathbb{Q}(\beta) \subset \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Therefore, we just need to guarantee that $\mathbb{Q}(\beta)$ is a degree-$4$ extension of the rationals, or equivalently, that the minimal polynomial of $\beta$ is of degree $4$ -- this will imply that the two fields are equal. To do this, first determine the Galois group of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. Then, taking advantage of the above lemma, use some creativity to find an element that yields four distinct Galois conjugates when acted upon by this group.