Theorem. Let $R$ be a ring such that $R$ is a subring of a division ring $D$. If for all $d(\ne 0)\in D$ either $d\in R$ or $d^{-1}\in R$ then $R$ is a local ring.

My proof. It suffices to prove that $R\setminus U(R)$ is an ideal of $R$ where $U(R)$ denotes the set of all units of $R$. For this we first show that if $x\in R\setminus U(R)$ then $Rx\subseteq R\setminus U(R)$. So let $x\in R\setminus U(R)$. Clearly then $x\ne 0$. Suppose that there exists $r\in R$ such that $rx\in U(R)$. Consequently, since $D$ is a division ring, we have $x\in U(R)$, a contradiction. So $Rx\subseteq R\setminus U(R)$ for all $x\in R\setminus U(R)$.

Let $x,y\in R\setminus U(R)$. Let $x^{-1},y^{-1}$ denote the inverses of $x$ and $y$ respectively in $D$ (clearly both $x$ and $y$ are non-zero so that they are invertible in $D$). Then note that by hypothesis $xy^{-1}\in R$ or $yx^{-1}\in R$. Since $1\in R$ it follows that either $1+xy^{-1}\in R$ or $1+yx^{-1}\in R$. In the former case, $$(1+xy^{-1})y\in R\setminus U(R)\Rightarrow y+x\in R\setminus U(R)$$ and in the later

$$(1+yx^{-1})x\in R\setminus U(R)\Rightarrow x+y\in R\setminus U(R)$$

So we are done.

Is this proof ok?

Anyone? any feedback?