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A well known result in Commutative Algebra says: for a commutative ring $R$ with $1$, $R/I$ is an Integral Domain if and only if $I$ is a Prime Ideal of $R$. Can this result be generalised for non commutative rings?

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If $R$ is a noncommutative ring, there are two different notions of prime ideals, both of which are equivalent to the usual definition in case $R$ is commutative.

A proper ideal $P$ of $R$ is a prime ideal if the following equivalent conditions hold.

  1. For ideals $I$, $J$ of $R$, if $IJ \subset P$ then $I \subset P$ or $J \subset P$.
  2. For left ideals $I$, $J$ of $R$, if $IJ \subset P$ then $I \subset P$ or $J \subset P$.
  3. For right ideals $I$, $J$ of $R$, if $IJ \subset P$ then $I \subset P$ or $J \subset P$.
  4. For all $a$, $b \in R$, if $aRb \in P$, then $a \in P$ or $b \in P$.

A proper ideal $P$ of $R$ is a completely prime ideal if, for all $a$, $b \in R$, $ab \in P$ implies $a \in P$ or $b \in P$.

It follows that an ideal $P$ of $R$ is a completely prime ideal if and only if $R/P$ is a domain (in the sense that $R/P$ is a nonzero ring which has no zero-divisors except for $0$ itself; it need not be commutative).

A ring in which $0$ is a prime ideal is called a prime ring. Thus an ideal $P$ of $R$ is prime if and only if $R/P$ is a prime ring.

Every completely prime ideal is a prime ideal. The converse is true for commutative rings, but not in general for noncommutative rings. Every maximal ideal is a prime ideal, but not necessarily completely prime. For instance, this is the case in any simple ring which is not a domain.

So, if $K$ is a field, then $M_n(K)$ is a prime ring (the unique proper ideal, $0$, is a prime ideal). However $M_n(K)$ is not completely prime for $n > 1$ since there are non-trivial zero-divisors. In the matrix ring $M_n(\mathbb Z)$ ($n>1$) every prime number generates a prime but not completely prime ideal, since $M_n(\mathbb Z)/pM_n(\mathbb Z) \cong M_n(\mathbb Z/p \mathbb Z)$ is a prime ring which is not a domain.

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