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.

Lattice theorem sez: no ideals between $0$ and $A/M\Leftrightarrow$ no ideals between $M$ and $A$. — anon, Feb 22 '14 at 10:42
@anon what's the mean of no ideals between M and A?they have no same ideal?they both don't have any ideal? — amateur, Feb 22 '14 at 10:49
Is English your first language? If $I<J<K$ (strict inequalities) then $J$ is in between $I$ and $K$. An ideal is maximal precisely when it is proper and no ideal sits between it and the whole ring. — anon, Feb 22 '14 at 10:53

score 2 · Accepted Answer · answered Feb 22 '14 at 14:45

2

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

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

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

answered Feb 22 '14 at 14:45

DonAntonio

214,715

I want to say why too. – amateur Feb 22 '14 at 22:03
$$ab+\mathfrak m=1+\mathfrak m\implies ab-1\in\mathfrak m+\langle a\rangle\implies 1\in\mathfrak m+\langle a\rangle\implies \mathfrak m+\langle a\rangle=A $$ – DonAntonio Feb 23 '14 at 00:26
1

ab+m=1+m =>ab-1∈m+,why?ab+m=1+m=>ab-1+m=m=>ab-1∈m.why there is a ?@DonAntonio – amateur Feb 23 '14 at 00:33
First "why?": because that's the definition of the equivalence relation that gives the (left or right) cosets in quotient rings (or quotient groups): $;a+I=b+I\iff a-b\in A;$ . Second "why?": that $;\langle a\rangle =aA=$ the principal ideal generated by $;a;$ . – DonAntonio Feb 23 '14 at 04:42

score 0 · Answer 2 · answered May 25 '23 at 15:31

Here is a Brut force method.

In a commutative ring with unity, $M \subset R$ is a maximal ideal if and only if $R/M$ is a field.

$(\Rightarrow)$ Suppose $M \subset R$ is maximal. Then $R/M$ is a commutative ring with $1$ and since $M \neq R$, $R/M \neq 0$ and let $a+M \in R/M$. I claim there exists a multiplicative inverse for this element. Consider the ideal

$$I=\{ar+m : \text{for some $r \in R,m \in M$}\}$$

You can check that $I \subset R$ is an ideal and since $a \in I$ and $M$ is maximal, we have that $I=R$ is the entire ring. Thus $1 \in I$ and we have

$$1=ar+m$$

which holds if and only if

$$1+M=ar+M=(a+M)(r+M)$$

thus $r+M \in R/M$ is the multiplicative inverse for $a+M$ and $R/M$ is a field.

($\Leftarrow$) Suppose $R/M$ is a field and suppose towards a contradiction that there exists an ideal $I \subset R$ with

$$M \subsetneq I \subsetneq R.$$

Let $a \in I \setminus M$ then $a+M \in R/M$ has a multiplicatice inverse as $R/M$ is a field, call it $r+M$, but then

$$1=(a+M)(r+M)=ar+M$$

thus

$$ar=1+m; \space \text{some $m \in M$}$$

But by properties of ideals we have

$$1 = ar-m \in I$$

forcing $I=R$ to be the whole of the ring as it has $1$ and $M \subset R$ is maximal.

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

2 Answers2