Proof that an ideal $M$ is maximal iff $R/M$ is a field I am referencing the proof located at http://www.maths.nuigalway.ie/MA416/section3-4.pdf, Theorem 3.4.2.
I am only looking at the right to left direction. I understand the following:
Let $a \in I, a \not\in M$.
$\rightarrow a + M$ is not the zero element of $R/M$.
$\rightarrow \exists \; b \in R \;$  s.t.  $\;(a+M)(b+M) = 1 + M$.
$\rightarrow ab-1 \in M$ *
Define $m = ab-1$.
$\rightarrow 1=ab-m$
$\rightarrow 1 \in I$ *
Except the starred steps, I do not understand how those were obtained from the information provided before them.
 A: It seems that you are trying to show that if $R/M$ is a field, then $M$ is maximal. So you assume that $I$ is another ideal of $R$ with $M \subset I$, and you want to show that $I = R$.
Note that $(a+M)(b+M) = ab + M$ by definition of multiplication on the quotient ring, so $(a+M)(b+M) = 1 + M \Leftrightarrow ab+M = 1 + M \Leftrightarrow (ab - 1) + M = M \Leftrightarrow ab - 1 \in M$.
As $a \in I$, $ab \in Ib = I$ as $I$ is an ideal. As $m = ab - 1 \in M$ and $M \subset I$, $m \in I$. As $I$ is an ideal $1 = ab - m \in I$.
A: For $ab - 1 \in M$.  The previous step was $(a + M)(b + M) = 1 + M$.  Note that $(a + M)(b + M) = ab + M$ by definition so $ab + M = 1 + M$.  This says in particular that $ab \in 1 + M$ so there is an $m \in M$ such that $ab = 1 + m$.  But then $ab - 1 = m$ for some $m \in M$.
For $1 \in I$ note that $a \in I$ and $I$ is an ideal so $ab \in I$.  Also $m \in M$ and $M$ is contained in $I$ so $m \in I$.  Thus $ab - m$ is contained in $I$, but $ab - m = 1$ so $1 \in I$.
A: What you wrote down is the right-to-left, not the left-to-right: if $M$ is an ideal in the commutative ring $R$ and $R/M$ is a field, then $M$ is a maximal ideal.
