Proof that an ideal $M$ is maximal iff $R/M$ is a field

Question

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.

Answer 1

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 $1 \in I$.

Answer 2

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 $1 = ab - m \in I$.