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 $F$?

3$\begingroup$ No. Consider finite fields. $\endgroup$ – Gene S. Kopp Jul 28 '13 at 7:39

1$\begingroup$ sorry I forgot the infinite condition $\endgroup$ – user37834 Jul 28 '13 at 8:26

$\begingroup$ If $Char\ F=0$ then $[F:R]<\infty$ forces $R$ to be equal to $F$. $\endgroup$ – Name Jul 28 '13 at 10:52

1$\begingroup$ @silvi If $Char\ F=0$ and $[F:R]=n$ then for every $x\in F$ we have $x=n(\frac1n x)\in R$ by Lagrange's theorem. $\endgroup$ – Name Jul 28 '13 at 12:32
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 $R \cdot \theta^{j1}/R \cdot \theta^{j} \cong R\cdot \theta^{1}/R$. 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 nontrivial and hence infinite, which would be a contradiction.
Related result (W. R. Scott, “On the multiplicative group of a division ring” Proc. Amer. Math.Soc. 8 (1957) no. 2, 303–305): given a division ring $F$ (commutative or otherwise) and a division ring $R \subsetneqq F$, we have $$[F^{\ast} : R^{\ast}] = F$$ where $R^{\ast}$ and $F^{\ast}$ denote the multiplicative groups of units of the respective division rings. (This result enabled Scott to sharpen a 1956 result of I. N. Herstein that any noncentral element of a division ring has infinitely many conjugates—itself of interest in showing that a polynomial $f$ over a division ring with central coefficients either has at most $\deg f$ right roots or else has infinitely many right roots.)