# Characterization of maximal ideals in an arbitrary ring.

We know that for a commutative ring $R$ with identity $1_R$,

(a) An ideal $P$ is prime iff $R/P$ is an integral domain

(b) An ideal $M$ is maximal iff $R/M$ is a field.

The way i proved the statement (a) doesn't need the existence of $1_R$. But then the way i proved (b) uses the existence of $1_R$. Also for the proof of "Every maximal ideal is prime" I used (a),(b)

(Note that if we take $R=x\mathbb{R}[x],I=x^2\mathbb{R}[x]$ then $I$ is a maximal ideal of $R$ which is not prime. Indeed, assume $I\subsetneq M$ then we can show $\exists ax\in M\text{ for some }a\in\mathbb{R}^*$. Thus $M=R$. But $x.x=x^2\in I$ but $x\not\in I$ which shows that $I$ is not prime). So I used the existence of $1_R$.

When I proved the existence of "Maximal ideal", i used Zorn's lemma along with the existence of $1_R$.

My question: I can formulate the characterization of prime ideal in an arbitrary ring as

"An ideal $P$ is prime iff $R/P$ has no zero-divisor."

(I hope it is correct! If not, please tell me where i am wrong.)

So, the question is how to characterize the maximal ideal of a ring $R$ if

(i) $R$ is commutative but has no unity (For this part I assume that a maximal ideal $M$ is given. As in an arbitrary ring maximal ideals may not exist, I'm assuming the existence and want to conclude about the ring $R/M$.)

(ii) $R$ is not commutative but has unity (I think $R/P$ is division ring, is necessary and sufficient. If I'm wrong, please correct my statement)

(iii) $R$ is non-commutative and without unity.

"An ideal $$P$$ is prime iff $$R/P$$ has no zero-divisor."

Well, if you are saying a domain is a ring without zero divisors (i.e. zero is not a zero divisor) then you have to do something about the case $$R=P$$. You could, for example, just require that $$P$$ is a proper ideal of $$R$$. Otherwise, yes, the definition of a prime ideal in a commutative ring is equivalent to this.

It is not, however, the most useful defintion of prime ideal in noncommutative rings. The useful definition for noncommutative rings is this: a proper ideal $$P$$ of $$R$$ is said to be prime if, for any two ideals $$A,B$$ of $$R$$, $$AB\subseteq P$$ implies $$A\subseteq P$$ or $$B\subseteq P$$.

This is equivalent to $$R/P$$ being a prime ring, which is the noncommutative analogue of an integral domain.

An ideal in a noncommutative ring which satisfies the commutative definition of a prime ideal is called completely prime. It is equivalent to $$R/P$$ having no zero divisors. So my preferred way to state what you are trying to prove is: a proper ideal $$P$$ is completely prime iff $$R/P$$ has no zero divisors.

There is a ton of information on this already in the wiki article, which you should probably consult.

So, the question is how to characterize the maximal ideal of a ring $$R$$ if

(i) $$R$$ is commutative but has no unity (For this part I assume that a maximal ideal $$M$$ is given. As in an arbitrary ring maximal ideals may not exist, I'm assuming the existence and want to conclude about the ring $$R/M$$.)

(ii) $$R$$ is not commutative but has unity (I think $$R/P$$ is division ring, is necessary and sufficient. If I'm wrong, please correct my statement)

(iii) $$R$$ is non-commutative and without unity.

The answer to all these questions is uniformly this: a proper ideal $$M$$ is a maximal ideal if $$R/M$$ is a simple ring, meaning that it has exactly two ideals. It does not depend on identity or commutativity at all. As you've already noticed, it's possible for a maximal ideal of a commutative rng to fail to be prime.

Good examples to keep in mind include the ring of square matrices $$M_n(\mathbb R)$$ and $$2\mathbb Z/4\mathbb Z$$.

## Information

In his analysis of noncommutative rings without identity, Jacobson especially tackled maximal left-, right-, and two-sided ideals. He distinguished modular, (or sometimes called regular) maximal right ideals from ones that aren't modular. Modular maximal ideals are relatively well behaved. In the case of commutative rngs, a modular maximal ideal $$M$$ is just one for which $$R/M$$ has an identity, and is therefore a field.