# $\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2})$

$$\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2})$$ where $$p_1,p_2,q_1,q_2$$ are all distinct primes

My original question is actually one more: $$\mathbb{Q}(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}) = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3})$$. But I think first proving the above would help the original problem. I'm pretty sure we use degree argument for this. Could you give any hints?

• Are you familiar with Galois theory? – paul blart math cop Nov 23 '20 at 7:46
• @paulblartmathcop Yes. but I don't know how to apply that theory here – love_sodam Nov 23 '20 at 10:03

## 1 Answer

EDIT: As pointed out in the comments, I swapped $$q_i$$ and $$p_i$$. After this edit, this change is at least consistent throughout.

First, let $$K = \mathbb Q(\sqrt[q_1]{p_1}, \dots, \sqrt[q_n]{p_n})$$, $$\alpha = \sum \sqrt[q_i]{p_i}$$. $$\mathbb Q$$ has characteristic 0, so this extension $$K/\mathbb Q$$ is separable. We can then consider a field extension $$L/K$$ called the Galois closure of $$K/\mathbb Q$$. We require that $$L/\mathbb Q$$ is the smallest Galois extension containing $$\mathbb Q$$. Essentially, we just adjoin all the conjugates of the $$\sqrt[q_i]{p_i}$$. We can describe it even more explicitly, as the conjugates of $$\sqrt[q_i]{p_i}$$ are of the form $$\zeta^k \sqrt[q_i]{p_i}$$ where $$\zeta$$ is a primitive $$q_i^{th}$$ root of unity. Then $$L = \mathbb Q(\sqrt[q_1]{p_1}, \dots, \sqrt[q_n]{p_n}, \zeta_1, \dots, \zeta_n)$$ where $$\zeta_i$$ is a primitive $$q_i^{th}$$ root of unity.

Anyways, the point of doing this is that our initial extension $$K/\mathbb Q$$ was not Galois (unless all $$q_i = 2$$), so we are trying to replace it with the Galois closure and work within that. The reason we want a Galois extension is the following:

Lemma. Let $$K/F$$ be a finite Galois extension and let $$\alpha \in K$$. Then $$K=F(\alpha)$$ iff $$\sigma(\alpha) \neq \alpha$$ for all $$\sigma \in G(K/F)$$ other than $$\sigma = id$$.

Proof: Let $$H = G(K/F(\alpha))$$. For $$\sigma \in G(K/F)$$, $$\sigma(\alpha)=\alpha$$ iff $$\sigma \in H$$. Then $$[K:F(\alpha)] = |H|$$ and $$F(\alpha) = K$$ iff $$|H| = 1$$.

Now, our extension $$K/\mathbb Q$$ is not Galois so we cannot immediately try to apply this result. However, it gives us the idea of applying every element of $$G(L/\mathbb Q)$$ to $$\alpha$$ and observing where it is fixed. To do this, we need to understand this Galois group $$G = G(L/\mathbb Q)$$ a bit better so let me first define $$K_i = \mathbb Q(\sqrt[q_i]{p_i})$$ and $$L_i = K_i(\zeta_i)$$ be the Galois closure. Then $$K = K_1 \cdots K_n$$ and $$L = L_1 \cdots L_n$$. We'll work first with these slices $$L_i/K_i/\mathbb Q$$.

Let's first consider $$G_i = G(L_i/\mathbb Q)$$. I won't show this, but $$G_i$$ is generated by the following two elements:

$$\sigma_i: \begin{cases} \sqrt[q_i]{p_i} \mapsto \zeta_i \sqrt[q_i]{p_i}\\ \zeta_i \mapsto \zeta_i \end{cases} \\ \tau_i: \begin{cases} \sqrt[q_i]{p_i} \mapsto \sqrt[q_i]{p_i}\\ \zeta_i \mapsto \zeta_i^{k_i} \end{cases}$$

where $$k_i$$ and $$q_i$$ are relatively prime. In fact, by doing some conjugation computations you can show that every element of $$G_i$$ is of the form $$\sigma_i^{a_i} \tau_i^{b_i}$$ and that this expression is unique for $$a_i$$ modulo $$q_i$$ and $$b_i$$ modulo $$\phi(q_i) = q_i - 1$$. Essentially, we are realizing $$G_i$$ as a semidirect product of $$\langle \sigma_i \rangle$$ and $$\langle \tau_i \rangle$$.

What then is the fixed field of $$K$$? Well if we let $$\phi \in G(L/\mathbb Q)$$ then $$\phi$$ is determined by its restrictions $$\phi|_{L_i} \in G_i$$ as $$L = L_1 \cdots L_n$$. Each $$\phi|_{L_i} \in G_i$$, so it can be written uniquely as $$\sigma_i^{a_i} \tau_i^{b_i}$$. With this representation, we can see that $$\phi$$ fixes $$K_i$$ iff $$\phi|_{L_i}$$ is purely a power of $$\tau_i$$. Thus, $$\phi \in G(L/K)$$ iff $$\phi|_{L_i}$$ is a power of $$\tau_i$$ for all $$i$$.

We are trying to show that $$\mathbb Q(\alpha) = K$$. By Galois theory, this is the same as showing that $$G(L/K) = G(L/\mathbb Q(\alpha))$$. All the work we did above allowed us to figure out what $$G(L/K)$$ is. Furthermore, as $$\mathbb Q(\alpha) \subseteq K$$, we have $$G(L/K) \subseteq G(L/\mathbb Q(\alpha))$$. Take now some $$\phi \in G(L/\mathbb Q(\alpha))$$. Then by definition, $$\phi(\alpha) = \alpha$$. We can rewrite this as $$\sum \zeta_i^{m_i}\sqrt[q_i]{p_i} = \sum \sqrt[q_i]{p_i}$$ for some $$m_i \in \mathbb Z$$. Hence, $$\sum \sqrt[q_i]{p_i} = \sum Re(\zeta_i^{m_i} \sqrt[q_i]{p_i}) = \sum \sqrt[q_i]{p_i} Re(\zeta_i^{m_i})$$. For this to happen, each $$\zeta_i^{m_i} = 1$$. In other words, $$\phi|_{L_i}$$ contains no powers of $$\sigma_i$$ and therefore fixes $$K_i$$. Hence, $$\phi$$ fixes $$K$$ and $$G(L/K) = G(L/\mathbb Q(\alpha))$$.

Bonus: Using the same ideas as above, you can show that $$L = \mathbb Q(\sum \sqrt[q_i]{p_i} + \sum \zeta_i)$$.

• You have interchanged $p_i$ by $q_i$ throughout the answer – Why Nov 23 '20 at 11:25
• Thank you for your effort. I have a question. in the last paragraph, you said 'We can write this as $\sum\zeta_{i}^{k_i}\sqrt[q_i]{p_i} = \sum\sqrt[q_i]{p_i}$'. Why is that? – love_sodam Nov 23 '20 at 11:26
• Ah I was worried about that, thanks for mentioning it! – paul blart math cop Nov 23 '20 at 11:27
• @love_sodam The LHS is $\phi(\alpha) = \sum \phi(\sqrt[q_i] {p_i})$ and the RHS is $\alpha$. As discussed in my answer, $\phi|_{L_i}$ is a product of $\sigma_i$'s and $\tau_i$'s. Thus, $\phi$ must send $\sqrt[q_i]{p_i}$ to some $\zeta_i^k \sqrt[q_i]{p_i}$. – paul blart math cop Nov 23 '20 at 11:34
• Ah I see, I overloaded $k_i$. At the end, it should be an arbitrary integer, not the specific $k_i$ chosen in the definition of $\tau_i$. I'll edit this in. – paul blart math cop Nov 23 '20 at 11:35