Proof verification: If $M$ is maximal, $R/M$ is a field. Let $M$ be maximal, we want to show that $R/M$ is a field. To do this we wish to show that for all $a\in R$, where $a+M\not\equiv 0+M$,  $a+M \in R/M$ is a unit.
So we have already excluded $a\in M$, so as to avoid $a+M\equiv 0+M$.
1. Is it valid to exclude these elements? Surely these cannot be invertible in $R/M$
Now we have two cases:
Case 1: $a$ is a unit in $R$, then $\exists a^{-1}\in R$ such that $(a+M)(a^{-1}+M)=1+M$ and $a+M\in R/M$ is invertible.
Case 2: $a$ is not a unit in $R$, and thus $\langle a\rangle \subsetneq R$. Then $M\subseteq\langle a\rangle \cup M$, but $a\not\in M$ and $M$ is maximal, contradition, then $a\in M$ or $a$ is invertible in $R$.
So we have that either $a\in M$ or $a+M$ is invertible in $R/M$. Then all nonzero elements of $R/M$ are invertible, and $R/M$ is therefore a field.
2. Is my argument valid?

Let $a\in R$ where $a\not\in M$.  Then $(a)+M$ is an ideal properly containing $M$.  Thus $(a)+M=R$.  So $\exists$ $b\in R$ and $m\in M$ such that $1=m+ab$.  Then $\overline b$ is the multiplicative inverse of $\overline a$.  Thus every non-zero element in $R/M$ has an inverse.
