How to prove m is maximal iff A/m is a field?

Question

m is a maximal ideal of a commutative ring A. then m is maximal iff A/m is a field.

Use Lattice theorem we get there is a bijection between m and an ideal of A/M.

A/M is a field =>the only odeals in A are 0 and (1).

Then I have no clue.

I have search this question and find How to directly prove that $M$ is maximal ideal of $A$ iff $A/M$ is a field?. But I still don't get it.

Answer 1

A unitary commutative ring is a field iff every non-zero element has a multiplicative inverse.

In $\;A/\frak m\;$ , an element $\;a+\frak m\;$ is not zero iff $\;a\notin\frak m\;$ , so we have

$$A/\mathfrak m\;\;\text{is a field}\;\iff\;\forall\,a\in A\setminus\mathfrak m\;\;\exists\,b\in A\;\;s.t.\;\;(a+\mathfrak m)(b+\mathfrak m):=ab+\mathfrak{m}=1+\frak m\iff$$

$$\iff \forall a\notin\mathfrak m\;,\;\;\mathfrak m+\langle a\rangle=A\;\text{(why?!)}\iff\;\mathfrak m\;\;\text{is a maximal ideal}$$