Let $E = \mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$. Find $\gamma \in E$ such that $E = \mathbb{Q}(\gamma)$

Question

As given in the title. Let $E = \mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7})$. Find $\gamma \in E$ such that $E = \mathbb{Q}(\gamma)$.

I am not really sure where to start. Help would be appreciated.

Answer 1

You're right to fear that things could get messy when squaring $\sqrt 2 + \sqrt 5 + \sqrt 7$, and all the more so when trying to generalize to arbitrary sums of square roots of primes. A less computational solution consists in using a bit of Kummer theory to deal only with the multiplicative structure, but since I don't know of your background, let me try to start from scratch (without being too messy). Preliminary observation : any quadratic extension of a field $F$ of characteristic $\neq 2$ is of the form $F(\sqrt a)$, with $a \in F^*$, $\notin {F^*}^2$, but actually $F(\sqrt a)$ depends only on the class $\bar a \in F^*/{F^*}^2$. It will be convenient in the sequel to view the group $F^*/{F^*}^2$ as a vector space over the finite field $\mathbf F_2$.

We want to prove by induction the property ($P_n$) : For $n \ge 2$, let $p_1, ..., p_n$ be $n$ pairwise distinct primes. Then the following assertions are valid : $(i_n)$ The extension $K_n:=\mathbf Q (\sqrt p_1,...,\sqrt p_n)$ is normal, with Galois group $\cong (\mathbf Z/2)^n$ ; $(ii_n)$ The subspace $V_n$ of $\mathbf Q^*/{\mathbf Q^*}^2$ generated by $\bar p_1,..., \bar p_n$ has $\mathbf F_2$-dimension $n$; $(iii_n)$ $K_n=\mathbf Q(\sqrt p_1+...+\sqrt p_n)$. Suppose that ($P_n$) holds and let us show that ($P_{n+1}$) holds:

$(i_{n+1})$ It suffices to show that $\sqrt p_{n+1} \notin K_n$. But the $2^n-1$ quadratic subextensions of $K_n$ are all of the form $\mathbf Q(\sqrt q)$, where $q$ is a product $p_{i_1}...p_{i_k}$, and $\mathbf Q(\sqrt q)=\mathbf Q(\sqrt p_{n+1})$ iff $q^{-1}p_{n+1} \in {\mathbf Q^*}^2$, but this is impossible since the primes $p_1,..., p_{n+1}$ are pairwise distinct

$(ii_{n+1})$ See the end of the proof of $(i_{n+1})$

$(iii_{n+1})$ By $(i_{n+1})$, the extension $K_{n+1}=K_n (\sqrt p_{n+1})$ is quadratic over $K_n$, and the element $(\sqrt p_1+...+\sqrt p_n)+\sqrt p_{n+1}$ of $K_{n+1}$ does not belong to $K_n$

It remains to prove ($P_2$), but in fact we have only to repeat the arguments already used in the recurrence.

NB: The 3 assertions in ($P_n$) are actually equivalent. The equivalence between $(i_n)$ and $(ii_n)$ is a straightforward consequence of Kummer theory. Logically speaking, we could have dispensed with the assertion $(ii_n)$, but it has the merit to show that only the multiplicative structure comes into play.