show/hide this revision's text 4 correction

I think the answer is no, but I didn't get anywhere. [Edit: I used to think the answer was no, but Simon Thomas convinced me otherwise.] Here is the condensed version of what I posted earlier, which seems to put serious constraints on $S$.

Let $R$ be any field and let $S$ be an additive subgroup of $R$ which is closed under multiplication. If $S$ has index less than $|S|$ as an additive subgroup of $R$, then every element of $R$ is of the form $a/b$ for $a, b \in S$. To see this, pick $r \in R$ and consider the multiples $ur$ for $u \in S$. Since $S$ has index less than $|S|$, there must be $u \neq v$ such that $a = ur - vr \in S$ then $r = a/b$ where $b = u - v$.
show/hide this revision's text 3 condensed answer

I think the answer is no, but I didn't get anywhere. Here is a partial negative answerthe condensed version of what I posted earlier, maybe somebody else can finish it off...which seems to put serious constraints on $S$.

Let $R$ be any field . Suppose that and let $S$ is a proper be an additive subgroup of $R$ which happens to be is closed under multiplication. Pick $r \in R \setminus S$. Let $x,y \in S$ be distinct and consider the difference $d = (r-x)^{-1} - (r-y)^{-1}$. If $d \in S$ then $r$ satisfies the quadratic equation $dr^2-d(x+y)r+(dxy-y+x) = 0$ with coefficients in $S$. So if $r$ does not satisfy a quadratic equation with coefficients in $S$, then the reals $(r-x)^{-1}$ for $x \in S$ must all belong to different additive cosets of $S$ in $R$. Hence the index of $S$ is at least $|S|$.

Actually, if $S$ has index less than $|S|$, |S|$ as an additive subgroup of $R$, then every element of $R$ is of the form $a/b$ for $a, b \in S$. To see this, pick $r \in R$ and consider the multiples $ur$ for $u \in S$. Since $S$ has index less than $|S|$, there must be $u \neq v$ such that $a = ur - vr \in S$ then $r = a/b$ where $b = u - v$.
show/hide this revision's text 2 addendum

I think the answer is no. Here is a partial negative answer, maybe somebody else can finish it off...

Let $R$ be any field. Suppose that $S$ is a proper additive subgroup of $R$ which happens to be closed under multiplication. Pick $r \in R \setminus S$. Let $x,y \in S$ be distinct and consider the difference $d = (r-x)^{-1} - (r-y)^{-1}$. If $d \in S$ then $r$ satisfies the quadratic equation $dr^2-d(x+y)r+(dxy-y+x) = 0$ with coefficients in $S$. So if $r$ does not satisfy a quadratic equation with coefficients in $S$, then the reals $(r-x)^{-1}$ for $x \in S$ must all belong to different additive cosets of $S$ in $R$. Hence the index of $S$ is at least $|S|$.

Actually, if $S$ has index less than $|S|$, then every element of $R$ is of the form $a/b$ for $a, b \in S$. To see this, pick $r \in R$ and consider the multiples $ur$ for $u \in S$. Since $S$ has index less than $|S|$, there must be $u \neq v$ such that $a = ur - vr \in S$ then $r = a/b$ where $b = u - v$.
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