Prove $\Bbb{Q}(M)=\Bbb{Q}(N)$ where $M=\{ \sqrt{r} | r \in \Bbb{Q}\} \subseteq \Bbb{C},N=\{ \sqrt{1-a^2} | a \in \mathbb{Q}\} \subseteq \Bbb{C}$

Question

Prove $\mathbb{Q}(M)=\mathbb{Q}(N)$ where $M=\{ \sqrt{r} | r \in \mathbb{Q}\} \subseteq \mathbb{C}$ and $ N=\{ \sqrt{1-a^2} | a \in \mathbb{Q}\} \subseteq \mathbb{C}$.

What I have found out so far:

• $\mathbb{Q}(M)$ is the field we get when we adjunct all roots of primes to $\mathbb{Q}$
• Adjuncting finitely many roots of primes is the same as adjuncting their sum to $\mathbb{Q}$
• Elements from $\mathbb{Q}(M)$ have the form $q+\sum^\infty_{i_0}c_i \sqrt{r_i}$ where $q,c_i, r_i \in \mathbb{Q}$
• Elements from $\mathbb{Q}(N)$ have the form $p+\sum^\infty_{i_0}d_i \sqrt{1-a_i^2}$ where $p,d_i, a_i \in \mathbb{Q}$

I would like to ask for ideas and hints, not solutions.

Answer 1

clearly $\mathbb Q (N) \subset \mathbb Q(M)$. Now take a prime $p$ and observe $\sqrt p=\sqrt{(\frac{p+1}{2})^2-(\frac{p-1}{2})^2}=\frac{p+1}{2}\sqrt{1-(\frac{p-1}{p+1})^2}\in \mathbb Q(N)$. Thus $M\subset \mathbb Q(N)$ and hence $\mathbb Q(M)=\mathbb Q(N)$