Show that $R/M$ is a division ring.

Question

Let $R$ be a ring with unity. If the set of all non-invertible elements form an ideal $M$, show that $R/M$ is a division ring.

Also show that in this case for each $r\in R$, either $r$ or $1-r$ is invertible.

I tried as follows: Let $a+M\in R/M$ we must prove that there is exist $x+M\in R/M$ such that $(a+M)(x+M)=1+M$ so $ax+M=1+M$ so $ax=1$ so $a$ is invertible, but $R$ does not necessarily have units, what can I do?

Answer 1

Let $a+M(\neq M)\in R/M\implies a\notin M\implies a$ is an invertible element $\implies \exists b\in R$ such that $ab=ba=1$.

So $(a+M)(b+M)=(b+M)(a+M)=1+M$

Second Part:

Suppose neither $r$ nor $1-r$ is invertible. Then $r,1-r\in M\implies 1\in M$ which is false.

Answer 2

Let $p:R\rightarrow R/M$ be the projection. Suppose that $p(a)\neq 0$ this implies that $a$ is not in $M$, thus $a$ is invertible, there exists $b,c$ such that $ab=1, ca=1$, this implies that $p(a)p(b)=1=p(c)p(a)$

Answer 3

As I wrote in my comment, $R$ has to be commutative. So assuming this I going to prove that $R/M$ is a simple ring (which is equivalent to be a field, why?).

Lemma: $M$ is a maximal ideal of $R$.

Proof: Suppose that $M$ is not maximal, then there is a proper ideal $I$ of $R$ such that $M\subset I$. So there exists $r\in I\setminus M$. If $r$ is a unit then $I=R$, contradiction, so $r$ is a non-unit and thus $r\in M$, contradiction again. Therefore $M$ is maximal.

Now, we apply the correspondence theorem for ideals which states that

Any ideal of $R/M$ must be of the form $J/M$ for some ideal $J$ of $R$ such that $J\supseteq M$.

But since $M$ is maximal, $J=M$ or $J=R$, then all the ideals of $R$ are $\{0\}$ and $R/M$. Hence, $R/M$ is simple.

Finally, if both $r$ and $1-r$ are non-units, then $r+(1-r)=1\in M$, so $M=R$, contradicting that $M$ is maximal. This means that either $r$ or $1−r$ is invertible.