# Subgroups of finite index of fields

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 $F$?

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No. Consider finite fields. – Gene S. Kopp Jul 28 '13 at 7:39
sorry I forgot the infinite condition – user37834 Jul 28 '13 at 8:26
If $Char\ F=0$ then $[F:R]<\infty$ forces $R$ to be equal to $F$. – Name Jul 28 '13 at 10:52
@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. – 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 $R \cdot \theta^{-j-1}/R \cdot \theta^{-j} \cong R\cdot \theta^{-1}/R$. 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.

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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.)

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