Let $F$ be an infinite field and $R$ a subring of $F$. suppose that $[F:R] < \infty$ (Index of $R$ in $F$ as a subgroup is finite). Does this force $R$ to be equal to $F$?



Subgroups of finite index of fields
Suppose $[F : R] = n$ is finite. I first claim that the integral domain $R$ is a subfield of $F$. For if $0 \neq \theta \in R$, then we have inclusions of $R$-modules 


$$R \subset R \cdot \theta^{-1} \subset R \cdot \theta^{-2} \subset \ldots$$ 


where in each case $R \cdot \theta^{-j-1}/R \cdot \theta^{-j} \cong R\cdot \theta^{-1}/R$. If $[R \cdot \theta^{-1} : R] = k$, then $[R \cdot \theta^{-j} : R] = k^j$, and for sufficiently high $j$ we have $k^j \gt n$ (contradiction) unless $k = 1$. 


So now $F$ is an extension of the field $R$. If $R$ is infinite and $[F : R] \gt 1$, then the $R$-vector space $F/R$ is non-trivial and hence infinite, which would be a contradiction. 


