Proving that every element has a unique inverse in $R/M$, given that $R$ is a ring and $M$ its maximal ideal.

Question

Let $R$ be a ring and $M\in R$ an ideal. It is well known that $R/M$ is a field if and only if $M$ is a maximal ideal of $R$. The proof is outlined below:

Let $a\in R/M$ such that $a\notin M$. Then, as $M$ is a maximal ideal, there has to exist $b\in R$ such that $ab+rm=1$, where $r\in R$. Also, $b\notin M$ (otherwise $ab\in M$, which would imply $1\in M$, making $M=R$). This implies $b\in R/M$ and $b$ is not the identity element of $R/M$. Hence, for every $a\in R/M$, its inverse $b$ also exists in $R/M$.

I was wondering how does this imply that there exists a unique inverse for every element $a\in R/M$? I tried to prove by contradiction: let $ab_1+r_1m_1=ab_2+r_2m_2=1$. This implies $a(b_1-b_2)=r_2m_2-r_1m_1$. Hence $a(b_1-b_2)\in M$. How does this contradict anything? Even if we say that $M$ is a prime ideal, it is entirely possible that $b_1-b_2\in M$, although $b_1,b_2\notin M$.

Thanks in advance!

Answer 1

If $b_1 - b_2 \in M$, then the elements $b_1 + M$ and $b_2+M$ in $R/M$ are one and the same, so the inverse is unique.

Answer 2

Your second paragraph seems to mix up $R$ and $R/M$. Since $ab\in R/M$ and $rm\in R$, you can't add these things together. You want to either say $a\in R/M$ and $\exists b\in R/M:ab=1$ or you mean to say that $a\in R$ and $\exists b,r\in R,m\in M:ab+rm=1$.

Anyway, uniqueness of inverses is a general fact that has nothing to do with the specific setup that you've described; it holds in arbitrary monoids. Let $N$ be a monoid (a set with an associative binary operation that has an identity element $e$). As I said over here, suppose $a,b\in N$ are distinct inverses of $x\in N$: then $a=ae=a(xb)=(ax)b=eb=b$, i.e. $a=b$. In particular, a ring with unity is a monoid under its multiplication operation.

As for your own thoughts, $b_1- b_2\in M$ doesn't contradict anything, and you don't state why you think it would contradict anything. As 5xum points out, this would mean $b_1+M=b_2+M$.

By the way, as a tangent, one useful perspective to have on this fact is that it can be framed in terms of ideals. For nontrivial unital commutative rings, it is easy to prove that a ring has no nontrivial proper ideals if and only if that ring is a field. One can use lattice correspondence to prove that $R/M$ has no nontrivial proper ideals (since this would yield an intermediate ideal in $R$ lying strictly between $R$ and $M$, contra maximality).