Prime ideal of a noncommutative ring

A proper two-sided ideal $P$ of a noncommutative ring is prime when $AB\subseteq P$ for two-sided ideals $A,B$ implies $A\subseteq P$ or $B\subseteq P$.

A prime ring is a nonzero ring whose zero ideal is a prime ideal of a noncommutative ring. Equivalently, the product of any two nonzero two-sided ideals is nonzero. This condition is weaker than absence of zero divisors: a full matrix ring over a division ring is prime even when its size is greater than one. For a commutative ring, primeness is exactly being an integral domain.

The prime radical is the intersection of all prime ideals of a noncommutative ring. For a nonzero ring satisfying the ascending chain condition on two-sided ideals, there exist prime ideals of a noncommutative ring $P_1,\dots ,P_r$ with $P_1\cdots P_r=0$. Indeed, a maximal counterexample among two-sided ideals could not itself be a prime ideal of a noncommutative ring, and two larger witness ideals would contradict its maximality. Every prime ideal of a noncommutative ring contains one $P_i$, so the intersection equals ${\bigcap }_i{P}_i$, and its $r$th power is zero. This nilpotence assertion fails without a chain condition.

In a left Noetherian ring, Noetherian induction shows that every proper two-sided ideal contains a finite product of prime ideals of a noncommutative ring that contain it. A maximal counterexample would be nonprime and contain a product of two strictly larger ideals; their prime products give the contradiction. Every prime over the original ideal contains a factor, so the inclusion-minimal factors give all its minimal primes over an ideal. Replacing each factor by a minimal prime beneath it gives a product of minimal primes inside the original ideal, with repetitions allowed. Applying this to zero makes a power of the prime radical zero.