Prime Rings

Introduction

This article is the second rung of the non-commutative chain of Rings and Fields, directly above Semiprime Rings, and it stands after the commutative half of the category, every rung of which is above it. Throughout, $A$ is a ring with $1 \neq 0$ not assumed commutative, and ideals are two-sided unless the contrary is said; the commutative theory of prime ideals is above, in Commutative Rings, and the default of the corpus is the commutative ring, not the ring of this article.

A ring is prime when the product of two nonzero ideals is never zero. The condition is the ideal-theoretic weakening of the absence of zero divisors, and the article is organised around the exact sense in which it is a weakening: a ring with no zero divisors is prime, a matrix ring is prime with plenty of zero divisors, and a prime ring need not be either. Prime ideals are the ideals whose quotients are prime rings, so that the commutative dictionary — prime ideal, quotient is a domain — is replaced by the two-sided one, and the article records where the old dictionary fails. The semiprime rings of the rung above are shown to contain the prime ones properly; the rings with no zero divisors, in the commutative and in the general case, are Integral Domains, above this article in this category, and Non-Commutative Domains, below it.

Prime Rings

Definition. A ring $A$ is prime if $IJ \neq 0$ for all nonzero two-sided ideals $I, J \trianglelefteq A$.

Equivalently, $A$ is prime exactly when its zero ideal is prime in the sense of the section on prime ideals below.

Theorem. $A$ is prime if and only if for all nonzero $a, b \in A$ there is $x \in A$ with $axb \neq 0$.

Proof. Suppose first that $A$ is prime and let $a, b \neq 0$. The two-sided ideals $AaA$ and $AbA$ are nonzero, so their product is nonzero; and

$$ (AaA)(AbA) \subseteq A(aAb)A , $$

so $aAb \neq 0$ and there is $x$ with $axb \neq 0$. Conversely, suppose the displayed property holds, let $I, J$ be nonzero ideals and choose $a \in I$, $b \in J$ nonzero. Then $axb \in IJ$ for every $x \in A$, and $axb \neq 0$ for some $x$, so $IJ \neq 0$.

Corollary. A ring in which a product of nonzero elements is nonzero is prime. In particular every ring with no zero divisors is prime, in the commutative and in the general case; the classes themselves are Integral Domains, above this article in this category, and Non-Commutative Domains, below it.

Proof. If $ab \neq 0$ for all nonzero $a$ and $b$, take $x = 1$ in the criterion.

Theorem. Every prime ring is semiprime.

Proof. If $A$ is prime and $I \neq 0$ satisfies $I^2 = 0$, then the product of the two nonzero ideals $I$ and $I$ is zero, which contradicts primeness.

The converse fails, and the standard witness is a product of two rings.

Proposition. A direct product $A_1 \times A_2$ of nonzero rings is not prime, while it is semiprime exactly when both factors are semiprime.

Proof. The ideals $I = A_1 \times (0)$ and $J = (0) \times A_2$ are nonzero and $IJ = 0$, so the product is not prime. An ideal of a product has square zero exactly when both of its projections do, by the characterisation of the semiprime rings in Semiprime Rings, above.

Example. The ring $k \times k$ is semiprime and not prime, and the matrix ring $M_n(F)$ is prime and contains the nonzero nilpotent element $E_{12}$ for $n \geq 2$, so a prime ring need not be reduced; the two conditions of the pair are ordered as

$$ \text{no zero divisors} \implies \text{prime} \implies \text{semiprime} , $$

and each implication is strict in the non-commutative case, the first witnessed by $M_n(F)$ and the second by $k \times k$. The class of reduced rings, those with no nonzero nilpotent element, lies below semiprime and is not comparable with prime, as the table below records.

Prime Ideals

Definition. A proper two-sided ideal $P \subsetneq A$ is prime if $IJ \subseteq P$ implies $I \subseteq P$ or $J \subseteq P$ for all two-sided ideals $I, J$, as in Rings, above. The set of prime ideals of $A$ is written $\operatorname{Spec} A$; in the commutative case it is the spectrum of Commutative Rings, above, considered there as a partially ordered set, and no further structure on it is used in this chain.

Theorem. Let $P \subsetneq A$ be two-sided. Then $P$ is prime if and only if the quotient ring $A/P$ is prime.

Proof. By the correspondence theorem of Rings, above, the two-sided ideals of $A/P$ are exactly the $\pi(I)$ for ideals $I$ of $A$ containing $P$, and $\pi(I)\pi(J) = \pi(IJ)$. Hence $A/P$ has nonzero ideals with zero product exactly when there are ideals $I, J \not\subseteq P$ with $IJ \subseteq P$, that is, exactly when $P$ is not prime.

Theorem. A proper two-sided ideal $P$ is prime if and only if for all $a, b \notin P$ there is $x \in A$ with $axb \notin P$.

Proof. Apply the criterion for a prime ring to the quotient $A/P$: the elements outside $P$ are exactly the nonzero elements of the quotient.

Corollary. The zero ideal $(0)$ is prime if and only if $A$ is prime; and a ring with no zero divisors has $(0)$ prime.

Proof. The first statement is the definition with $P = (0)$; the second follows from the corollary of the criterion above.

Remark. The commutative case is the familiar one: if $A$ is commutative, $P$ is prime exactly when $ab \in P$ implies $a \in P$ or $b \in P$, and $A/P$ is then a domain, in the terminology of Integral Domains, above this article in this category. In the general case the second half of that statement is false, and the next section shows how.

Proposition. Let $I \trianglelefteq A$. Then the prime ideals of $A/I$ are exactly the ideals $P/I$ with $P$ a prime ideal of $A$ containing $I$.

Proof. The correspondence theorem of Rings, above, gives the bijection between the ideals of $A/I$ and the ideals of $A$ containing $I$, and it preserves products, so $P$ is prime exactly when $P/I$ is. The case $I = P$ gives the criterion for a ring to be prime in terms of its zero ideal.

Minimal Primes

Definition. A prime ideal $P$ is minimal over an ideal $I$ if $P \supseteq I$ and no prime ideal strictly between $I$ and $P$ exists; $P$ is a minimal prime of $A$ if it is minimal over $(0)$.

Theorem. Every prime ideal contains a minimal prime ideal, and the intersection of the minimal primes of $A$ is $\operatorname{Nil}_*(A)$.

Proof. Let $P_{\lambda}$ be a chain of prime ideals under inclusion and let $I, J$ be two-sided ideals with $IJ \subseteq \bigcap_{\lambda} P_{\lambda}$, so that for every $\lambda$ we have $I \subseteq P_{\lambda}$ or $J \subseteq P_{\lambda}$. Suppose neither $I$ nor $J$ is contained in the intersection, and choose $\mu$ with $I \not\subseteq P_{\mu}$ and $\nu$ with $J \not\subseteq P_{\nu}$; then $J \subseteq P_{\mu}$ and $I \subseteq P_{\nu}$. Since the family is a chain, one of $P_{\mu}, P_{\nu}$ contains the other. If $P_{\mu} \subseteq P_{\nu}$ then $J \subseteq P_{\mu} \subseteq P_{\nu}$, contradicting $J \not\subseteq P_{\nu}$; if $P_{\nu} \subseteq P_{\mu}$ then $I \subseteq P_{\nu} \subseteq P_{\mu}$, contradicting $I \not\subseteq P_{\mu}$. Hence the intersection of a chain of prime ideals is prime.

Zorn's lemma applied to the prime ideals contained in a given prime ideal, ordered by reverse inclusion, now produces one minimal among them: a chain in this order is a descending chain of primes, its intersection is prime by the argument just given, and it lies below the given prime. Hence every prime ideal contains a minimal prime. The intersection of the minimal primes is contained in every prime, hence in $\operatorname{Nil}_*(A)$; and it contains $\operatorname{Nil}_*(A)$, which lies in every prime ideal. The two intersections therefore coincide.

Corollary. $\operatorname{Nil}_*(A)$ is the intersection of the minimal prime ideals, and $A$ is semiprime exactly when its minimal primes intersect in zero.

Proof. The first statement is the theorem; the second is the criterion $\operatorname{Nil}_*(A) = 0$ of Semiprime Rings, above.

The Failure of the Commutative Correspondence

Theorem. If $A/P$ has no zero divisors then $P$ is prime; the converse fails, and the failure is total: there are prime ideals whose quotients have as many zero divisors as a matrix ring.

Proof. If $A/P$ has no zero divisors then it is prime, by the corollary above, hence $P$ is prime. For the failure, take $A = M_n(F)$ with $n \geq 2$ and $P = (0)$: the ring is prime, by the example below, while $E_{12} \neq 0$ and $E_{12}^2 = 0$ show that it has zero divisors.

Example ($M_n(F)$). The only two-sided ideals of $M_n(F)$ are $(0)$ and the whole ring, by Rings, above. Hence the product of two nonzero ideals is the whole ring, which is nonzero, so $M_n(F)$ is prime for every $n \geq 1$, and for $n = 1$ it is the field $F$. For $n \geq 2$ it has zero divisors and nonzero nilpotent elements, so it is prime, semiprime and not reduced; and $(0)$ is its only proper prime ideal, so $\operatorname{Spec} M_n(F)$ has a single point.

Example. The ring of integers $\mathbb{Z}$ is prime, and every nonzero prime ideal of $\mathbb{Z}$ has a quotient with no zero divisors; the matrix example above is the standard demonstration that this behaviour is special to the commutative case.

Theorem. Let $S$ be a ring and $n \geq 1$. Then the matrix ring $M_n(S)$ is prime if and only if $S$ is prime.

Proof. Every two-sided ideal of $M_n(S)$ has the form $M_n(I)$ for a two-sided ideal $I \trianglelefteq S$, and $M_n(I)M_n(J) = M_n(IJ)$: the containment $M_n(I)M_n(J) \subseteq M_n(IJ)$ is immediate from the definition of the matrix product, the entries of a product of a matrix over $I$ and a matrix over $J$ being sums of products $uv$ with $u \in I$ and $v \in J$, and the reverse containment follows from $E_{i1} M E_{1j}$ picking out the $(i,j)$ entry of a matrix $M$. Hence $M_n(I)M_n(J) = 0$ if and only if $IJ = 0$, so $M_n(S)$ has two nonzero ideals with zero product exactly when $S$ does.

Corollary. $M_n(S)$ is semiprime if and only if $S$ is semiprime, and $M_n(S)$ has no zero divisors only when $n = 1$.

Proof. The first statement repeats the proof above with $I = J$; the second is the computation $E_{12}E_{12} = 0$ for $n \geq 2$.

Remark. Two further differences from the commutative case are worth recording. First, in the commutative case a proper ideal is maximal exactly when its quotient is a field, so that maximal ideals are prime; in the general case a maximal two-sided ideal is prime, by Rings, above, but its quotient only has no nonzero proper two-sided ideals, and $M_n(F)$ is the quotient of itself by $(0)$ and shows that this is weaker than being a field. Second, the intersection of the prime ideals is the lower nilradical of Semiprime Rings, above, whose computation in the non-commutative case uses the strong nilpotence of that article and not the elementwise nilpotence of the commutative case.

The Centre of a Prime Ring

Proposition. The centre $Z(A)$ of a prime ring is a commutative ring with no zero divisors.

Proof. Let $z, w \in Z(A)$ with $zw = 0$. The ideals $AzA$ and $AwA$ are two-sided, because $z$ and $w$ are central, and

$$ (AzA)(AwA) \subseteq A(zAw)A \subseteq AzwA = 0 . $$

Since $A$ is prime, one of the two ideals is zero, and an ideal generated by a central element is zero only when that element is zero: hence $z = 0$ or $w = 0$.

Corollary. If $A$ is prime then $Z(A)$ is an integral domain, and $A$ is an algebra over the field $Z(A)$ when $Z(A)$ is a field. The centre of a prime ring and the structure of a division ring over its centre are taken up in Division Rings, below this article in this category.

Proof. The centre is a commutative ring with no zero divisors by the proposition, and it is a field exactly when every nonzero central element is invertible in $A$; the algebra structure is the multiplication by central elements.

Example. For $A = M_n(F)$ the centre is the ring of scalar matrices, isomorphic to $F$, and it is a field. For a commutative domain $R$, the centre is $R$ itself.

Examples and Non-Examples

Example (rings with no zero divisors are prime). Let $A$ be a ring in which $ab \neq 0$ whenever $a, b \neq 0$. If $I$ and $J$ are nonzero ideals and $a \in I$, $b \in J$ are nonzero, then $ab \in IJ$ is nonzero, so $IJ \neq 0$ and $A$ is prime. The quaternion division ring $\mathbb{H}$, the free algebra $k\langle x_1, x_2\rangle$, the Weyl algebra $A_1(k)$ and the group ring of an ordered group — conjecturally, of any torsion-free group, by Kaplansky's zero divisor conjecture — are all of this kind, and are treated in Non-Commutative Domains and Ore Domains and Division Rings of Fractions, below this article in this category.

Example (prime, not with no zero divisors). $M_n(F)$ for $n \geq 2$, as above: prime, and its zero divisors are all the non-invertible matrices.

Example (semiprime, not prime). $k \times k$, and more generally any product of two nonzero rings; also the ring $k[x,y]/(xy)$, whose two nonzero ideals $(x)$ and $(y)$ have zero product. This is the example that separates the two rungs of the chain.

The Chain of Classes

The four conditions that meet at this rung are tabulated below, with a ring satisfying each and a ring failing it. The first three are defined for a general ring, and the fourth, reducedness, is the commutative rung of Reduced Rings and the Nilradical, above.

Condition	Definition	Example	Non-example
no zero divisors	$ab = 0$ forces $a = 0$ or $b = 0$; the class is Non-Commutative Domains, below this article, and Integral Domains, above it	$\mathbb{H}$	$M_n(F)$, $n \geq 2$: $E_{12}^2 = 0$
prime	$IJ \neq 0$ for all nonzero two-sided ideals $I, J$	$M_n(F)$, any $n \geq 1$	$k \times k$: $(k \times 0)(0 \times k) = 0$
semiprime	no nonzero nilpotent two-sided ideal	$k \times k$	$k[x]/(x^2)$: $(x)$ has square zero
reduced	no nonzero nilpotent element	$k[x,y]/(xy)$	$\mathbb{Z}/4\mathbb{Z}$: $2 \neq 0$ and $2^2 = 0$

The implications $\text{no zero divisors} \implies \text{prime} \implies \text{semiprime}$ hold, and each is strict, by the first two rows. Reducedness implies semiprimality, since a nilpotent ideal consists of nilpotent elements, and the example $M_2(F)$ of the second row shows that the converse fails outside the commutative case; in the commutative case reducedness and semiprimality coincide, by Semiprime Rings, above. Reducedness and primality are incomparable, the ring $k \times k$ being reduced and not prime, and $M_n(F)$ for $n \geq 2$ prime and not reduced.

Remark. The implication from prime to semiprime has no counterexample: a prime ring that is not semiprime would be a product of two nonzero ideals equal to zero, which is exactly what primality forbids. The separating example is therefore the semiprime ring $k \times k$, which is not prime.

Example (an ideal prime in the two-sided sense and not in the elementwise one). Let $A = M_2(F)$ and let $P = (0)$. The elementwise condition "$ab \in P$ implies $a \in P$ or $b \in P$" fails, since $E_{12} \neq 0$ and $E_{12}^2 = 0$; the ideal $P$ is nevertheless prime in the two-sided sense, by the example above. Hence the elementwise definition of primality from the commutative case is strictly stronger than the two-sided definition, and it is the two-sided definition that survives without commutativity.

Summary

A ring is prime when the product of two nonzero two-sided ideals is nonzero, equivalently when for all nonzero $a, b$ there is $x$ with $axb \neq 0$. A ring without zero divisors is prime, a matrix ring $M_n(F)$ with $n \geq 2$ is prime and has zero divisors, and a product of two nonzero rings is semiprime and not prime; so the chain of classes is: no zero divisors, then prime, then semiprime, with both implications strict in the non-commutative case. A two-sided ideal is prime exactly when its quotient ring is prime, and exactly when for all $a, b$ outside it there is $x$ with $axb$ outside it; the commutative correspondence between prime ideals and domains holds in one direction only, $M_n(F)$ by its zero ideal being the counterexample. The centre of a prime ring has no zero divisors, so it is a domain in the commutative sense when it is nonzero.

Summary of Notation

Symbol	Meaning
$A$	A ring with $1 \neq 0$, not assumed commutative; the base of this article
prime ring	A ring with $IJ \neq 0$ for all nonzero two-sided ideals $I, J$
prime ideal	A proper two-sided ideal with $IJ \subseteq P \Rightarrow I \subseteq P$ or $J \subseteq P$
semiprime	Having no nonzero nilpotent two-sided ideal, of Semiprime Rings
reduced	Having no nonzero nilpotent element, of Reduced Rings and the Nilradical
$\operatorname{Spec} A$	The set of prime ideals of $A$; in the commutative case the poset of Commutative Rings
$Z(A)$	The centre of $A$
$aAb$, $axb$	The two-sided product test for primeness
$M_n(F)$, $E_{ij}$	The matrix ring and its matrix units; $M_n(F)$ is prime for every $n \geq 1$
$k \times k$	A product of two nonzero rings: semiprime and not prime
$k[x,y]/(xy)$	A commutative semiprime ring that is not prime
$\mathbb{Z}$	The model prime ring whose quotients by nonzero primes have no zero divisors

Prime Rings

Introduction

Prime Rings

Prime Ideals

Minimal Primes

The Failure of the Commutative Correspondence

The Centre of a Prime Ring

Examples and Non-Examples

The Chain of Classes

Summary

Summary of Notation

Further Reading