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Q. We know that if $R$ is commutative ring with unity, then

(*) $M$ is maximal ideal $\Longleftrightarrow$ $R/M$ is field.

(**) $M$ is prime ideal $\Longleftrightarrow$ $R/M$ is integral domain.

Suppose we drop one of these two conditions - "$R$ is commutative" or "$R$ contains unity", or both the conditions, does any implication in statement (*) and (**) remains true?

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  • $\begingroup$ go through the proofs of the results .you will understand it $\endgroup$ – Learnmore Jan 5 '15 at 13:38
  • $\begingroup$ Even $R$ is not comutative, $R/I$ may be commutative (I think this can happen; I don't know example). $\endgroup$ – Groups Jan 5 '15 at 13:41
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    $\begingroup$ If $R = M_2(\mathbb{C})$, the non-commutative ring of $2\times 2$ matrices over $\mathbb{C}$, then $\{0\}$ is a maximal ideal. $\endgroup$ – Prahlad Vaidyanathan Jan 5 '15 at 13:51
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If $R$ is a not necessarily commutative ring with unity then:

$M$ is a maximal ideal $\iff R/M$ is a simple ring, i.e. it has no non-zero two-sided ideals.

There are many examples of noncommutative rings that are simple but not division rings (a division ring is a noncommutative ring in which every element has a two-sided inverse, so the noncommutative analogue of a field) - $M_2(\mathbb{C})$, as pointed out above, is one example.

For prime ideals, the definition of "prime" is generally slightly different for noncommutative rings: $P$ is prime iff for any two (left / right / two-sided) ideals $A$ and $B$ then $AB \subset P$ implies $A \subset P$ or $B \subset P$. This contrasts with the usual commutative algebra definition that $P$ is prime iff for any two elements $a$ and $b$ then $ab \in P$ implies $a \in P$ or $b \in P$. The latter definition is usually called "completely prime" (and does imply prime, but not vice versa; however if $R$ is commutative then the two definitions of "prime" do coincide).

With that in mind, $M$ is completely prime iff $R/M$ has no zero-divisors, i.e. is a domain (I would say the term "integral domain" implied commutativity, and that "domain" is the more general term, but that is debatable). But there are prime ideals $M$ such that $R/M$ does have zero-divisors. As Konstantin points out below, the $0$ ideal in $M_2(\mathbb{C})$ is again an example.

For commutative rings without unity, $4\mathbb{Z}$ is a maximal ideal in $2\mathbb{Z}$, but $2\mathbb{Z}/4\mathbb{Z}$ is not a field (as it cannot be, since it doesn't have a $1$, so how can elements have multiplicative inverses?).

However "$M$ completely prime $\iff R/M$ has no zero divisors" does still hold.

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  • $\begingroup$ The zero ideal $M$ in the matrix ring $R := M_2(\mathbb{C})$ is prime, but $R / M \cong R$ has zero divisors. In your last sentence, "prime" should be replaced with "completely prime". $\endgroup$ – Konstantin Ardakov Jan 5 '15 at 14:45
  • $\begingroup$ That's helpful further up, thanks :) I've edited the last sentence for clarity, too, though it was only meant to refer to commutative rngs - do we not have prime=completely prime for commutative rings-without-unity? $\endgroup$ – Christopher Jan 5 '15 at 14:56
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    $\begingroup$ @user73985 I don't think I've ever seen prime ideals discussed in rings without identity in print. $\endgroup$ – rschwieb Jan 5 '15 at 23:34

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