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

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  • $\begingroup$ Are you familiar with Galois theory? $\endgroup$ – paul blart math cop Nov 23 '20 at 7:46
  • $\begingroup$ @paulblartmathcop Yes. but I don't know how to apply that theory here $\endgroup$ – love_sodam Nov 23 '20 at 10:03
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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 $L = \mathbb Q(\sum \sqrt[q_i]{p_i} + \sum \zeta_i)$.

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  • $\begingroup$ You have interchanged $p_i$ by $q_i$ throughout the answer $\endgroup$ – Why Nov 23 '20 at 11:25
  • $\begingroup$ 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? $\endgroup$ – love_sodam Nov 23 '20 at 11:26
  • $\begingroup$ Ah I was worried about that, thanks for mentioning it! $\endgroup$ – paul blart math cop Nov 23 '20 at 11:27
  • $\begingroup$ @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 $\zeta_i^k \sqrt[q_i]{p_i}$. $\endgroup$ – paul blart math cop Nov 23 '20 at 11:34
  • $\begingroup$ 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. $\endgroup$ – paul blart math cop Nov 23 '20 at 11:35

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