How to show that $h$ is onto? Let $X\subseteq\mathbb C$ and $\mathbb Q(X)$ be the intersection of all subfields of $\mathbb C$ that contain $X$. I want to show that $\mathbb Q(X)$ consists of all rational expressions $\frac{p(\alpha_1,\cdots,\alpha_n)}{q(\beta_1,\cdots,\beta_n)}$ for all $n$ where $p,q\in\mathbb Q[t_1,\cdots,t_n]$, the $\alpha_j$'s and $\beta$'s belong to $X$, and $q(\beta_1,\cdots,\beta_n)\neq0$. 
Let $R$ be the set of all $p(\alpha_1,\cdots,\alpha_n)$ for all $n\in\mathbb N$ and $K$ be the set of all rational expressions $\frac{p(\alpha_1,\cdots,\alpha_n)}{q(\beta_1,\cdots,\beta_n)}$ for all $n\in\mathbb N$, where $p,q$ and $\alpha_j,\beta_j$'s are as above.
Let $f:R\to\mathbb Q(X)$ be the inclusion map and $g:R\to K$ be the map defined by $g(p)=\frac p1$. Then $g$ is an one-to-one rings homomorphism and $g(1)=1$. Let $h:K\to\mathbb Q(X)$ be the map defined by 
$$h\left(\frac pq\right)=f(p)(f(q))^{-1}.$$
Then $h$ is a rings homomorphism and $g\circ h=f$. Since $K$ is a field, it follows that $h$ is one-to-one. Thus if we prove that $h$ is onto, then isomorphic image $h(K)$ is a subfield of $\mathbb Q(X)$ which shows that $\mathbb Q(X)=K$. I've tried to show that $h$ is onto but couldn't arrive to any result. Could you please give me some hint?
 A: I think it may be easier to show $K=\mathbb{Q}(X)$ directly, i.e., by proving $LHS\subset RHS$ and $RHS\subset LHS$. 
$K\subset \mathbb{Q}(X)$: let $p(\alpha_{1},\cdots,\alpha_{n})/q(\beta_{1},\cdots,\beta_{n})\in K$. Since $\alpha_{i},\beta_{i}\in X$ and $\mathbb{Q}(X)$ is a field, $p(\alpha_{1},\cdots,\alpha_{n})/q(\beta_{1},\cdots,\beta_{n})\in \mathbb{Q}(X)$.
$\mathbb{Q}(X)\subset K$: we first show that $K$ is a subfield of $\mathbb{Q}(X)$. Check that $K$ is non-empty, closed under addition, multiplication, and taking inverse of non-zero element. Then $K$ is a subfield of $\mathbb{C}$ containing $X,\mathbb{Q}$. So $K$ contains $\mathbb{Q}(X)$. 
A: Since $R\subseteq K$ and $h(R)=R$, it follows that
$$X\subseteq R=h(R)\subseteq h(K)\subseteq\mathbb Q(X)$$
Now, since $h(K)$ is a field containing $X$ and $\mathbb Q(X)$ is the smallest subfield of $\mathbb C$ that contains $X$, so $h(K)=\mathbb Q(X)$. That is, $h$ is onto which implies that $h$ is an isomorphism and $\mathbb Q(X)\cong K$.
So finally, by solving this question, I've finished the wintersemester 2017/2018!
