The relation between integral domains and their field of fractions. Given a integral domain , we can construct its field of fractions.But if given a field $K$ ,how can know what are the intergal domains such that their field of fractions are $K$? Can we constuct them? For example , for the complex field $\mathbb{C}$ ,do  there exist intergal domains such that their field of fractions are $\mathbb{C}$? If there exist ,how many kind of them up to isomorphism?
 A: Let $\mathcal{E}$ be the set of subrings $\mathcal{O}$ of $\mathbb{C}$ which do not contain $1/2$. The set $\mathcal{E}$ is not empty, as it contains $\mathbb{Z}$. If $(\mathcal{O}_i)_{i\in I}\subset \mathcal{E}$ is a totally ordered (by inclusion) subset of $\mathcal{E}$, then $\cup_{i\in I} \mathcal{O}_i$ is a ring, not containing $1/2$, so is in $\mathcal{E}$ and an upper bound for all the $\mathcal{O}_i$ . 
So Zorn's Lemma applies: there exist a maximal subring $\mathcal{O}\subset \mathbb{C}$, not containing $1/2$.
Let $k$ be the field of fraction of $\mathcal{O}$. We wish to prove that $k=\mathbb{C}$. 
Otherwise, there would exist $x\in \mathbb{C}-k$. Since $\mathcal{O}[x]\neq \mathcal{O}$, maximality of $\mathcal{O}$ means that $1/2\in \mathcal{O}[x]$. so there exists $n\geq 1$ and $a_i \in \mathcal{O}$, such that
$$\frac12=a_0+a_1 x + ...+ a_n x^n.$$
So $x$ is algebraic over $k$, of degree $d\geq 2$ since $x\notin k$. 
We would like to replace $x$ by a multiple $y=bx$, $b\in \mathcal{O}-\{0\}$, so that the minimal polynomial of $y$ has coefficients in $\mathcal{O}$. This is done as follows.
There exists $b\in \mathcal{O}-\{0\}$ such that the minimal polynomial of $bx$ in $k[X]$ has coefficients in $\mathcal{O}$; indeed, if $P_x\in k[X]$ is the minimal polynomial of $x$,
$$P_x=X^d+\sum_{i=0}^{d-1} \frac{p_i}{q_i}X^i,$$
where $p_i,q_i \in \mathcal{O}$, $q_i\neq 0$, then $b=q_0...q_{d-1}$ does the trick, and $k[bx]=k[x]$. So, putting now $y=bx$, we still have $\mathcal{O}[y]\neq \mathcal{O}$, so by maximality $1/2 \in \mathcal{O}[y]$. So there exist $m>0$ and coefficients $c_i \in \mathcal{O}$ such that
$$\frac12=c_0+c_1 y + ...+ c_m y^m=Q(y).$$
We may assume that $m<d$: to see this, since the minimal polynomial $P_y$ of $y$ over $k$ is in $\mathcal{O}[X]$, and is unitary, we may use an euclidean division in $\mathcal{O}[X]$, $Q=LP_y+R$, where $R \in \mathcal{O}[X]$ has degree $<d$, and replace $Q$ by $R$ in the preceding equality.  Therefore, $y$ is a zero of the following polynomial
$$2c_mX^m+...+2c_1X+(2c_0-1).$$
Since $2c_0-1\neq 0$, the polynomial is nonzero, but $y$ is algebraic over $k$ of degree $d>m$. This is a contradiction, showing that no such $x$ exists.
(The proof for $\mathbb{R}$ is the same.)
