If $K$ is a field of characteristic $0$, $A$ is a subring of $K$ maximal subring of $K$ which doesn't contain $\frac{1}{2}$, and $F$ is the field of fractions of $A$, then I have proved that $K$ is algebraic over $F$ and that $A$ is integrally closed in $K$. I'm supposed to use these two facts to conclude that $F=K$, but I'm not sure how to do that.
I tried assuming that $\alpha$ was an element of $K-F$ and showing that $\alpha$ would have to be integral over $A$. Since $K$ is algebraic over $F$, it satisfies a monic polynomial with coefficients in $F$, but not necessarily with coefficients in $A$. This is where I get stuck, because I don't see any reason why those coefficients would be in $A$.
Suppose $\alpha\in K$. Then there is some nonzero polynomial $p(x)\in F[x]$ such that $p(\alpha)=0$. Clearing denominators, we get a polynomial $q(x)\in A[x]$ such that $q(\alpha)=0$. Now if $q(x)=cx^n+dx^{n-1}+ex^{n-2}+\dots$, we see that $$c^{n-1}q(x)=(cx)^n+d(cx)^{n-1}+ce(cx)^{n-2}+\dots=r(cx)$$ for some monic polynomial $r(x)\in K[x]$. Thus $r(c\alpha)=0$, so $c\alpha$ is integral over $A$. Since $A$ is integrally closed in $K$, $c\alpha\in A$. Thus $\alpha=c\alpha/c\in F$.
More generally, this shows that if $A$ is a reduced ring and $K$ is a ring containing $A$, an element $\alpha\in K$ is algebraic over $A$ iff there exists a nonzero $c\in A$ such that $c\alpha$ is integral over $A$.