A sufficient condition for a unitary ring to be a local ring

Question

Theorem. Let $R$ be a unitary 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.

Question

Is my argument correct?

Answer 1

There is a small gap like "Let $x,y\in R\setminus U(R)$... clearly $x,y$ are nonzero..." but of course, they could be zero, and may not have inverses. But of course that's rectified easily enough since in any such case $x+y$ is obviously a nonunit of $R$.

Otherwise the idea is sound.

But look at this:

Then note that by hypothesis $xy^{-1}\in R$ or $yx^{-1}\in R$.

This observation is actually the key to a very short solution. Let $x,y$ actually be arbitrary nonzero elements of $R$. Then (by what you said) either $Rx\subseteq Ry$ or $Ry\subseteq Rx$.

In other words, the principal left ideals are totally ordered.

Then it is a simple exercise to show the entire collection of left ideals is totally ordered.

Thus any maximal left ideal would have to be unique, and of course maximal left ideals always exist in a ring with identity.

So in fact, beyond being local the left ideals actually form a chain, a much stronger property.