Noncommutative algebra

Noncommutative algebra studies rings whose multiplication need not commute and their one-sided ideals and modules.

A ring is semiprime when it has no nonzero nilpotent ideals. Equivalently, a two-sided ideal with square zero must be zero. This is weaker than having no nonzero nilpotent elements: a matrix ring over a field is semiprime and can have nilpotent matrices.

A ring is right Goldie if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. The left-sided definition reverses the multiplication. These two definitions need not be equivalent.

A right Goldie ring satisfies the right-sided conditions in the definition of a Goldie ring. Every right Noetherian ring does: its right annihilators are right ideals, and its right regular module has finite uniform dimension.

A semiprime ring has a classical right ring of quotients that is a semisimple ring and Artinian exactly when it is a right Goldie ring. In particular, the theorem applies to a semiprime ring that is a right Noetherian ring. It extends the fraction-field construction from commutative domains to rings whose quotient can be a finite product of matrix rings over division rings.

A nonzero two-sided ideal of the classical right ring of quotients contains a nonzero element of the original ring: multiply a nonzero fraction by its regular right denominator. If the original ring is prime, the contractions of any two such ideals have nonzero product, so the quotient ring is prime too. The Goldie theorem makes it semisimple Artinian, and the Artin–Wedderburn theorem expresses it as a finite product of matrix rings over division rings. Primeness excludes more than one factor.

Suppose $R$ is a right Noetherian ring and every essential right ideal contains a regular element of a ring. Preimages of essential regular principal right ideals under left multiplication are essential, so they contain regular elements and prove the right Ore condition. In the classical right ring of quotients $Q$, every right ideal is generated by its contraction to $R$, making $Q$ right Noetherian. An essential right ideal of $Q$ contracts to an essential right ideal of $R$, hence contains a regular element and is all of $Q$. A maximal disjoint complement to any right ideal has essential sum with it, so every right ideal is complemented. The right regular module is therefore a finite direct sum of simple modules, making $Q$ semisimple Artinian.

A regular element of a ring is a two-sided non-zero-divisor. The two cancellation conditions must both hold in a noncommutative ring. Products of regular elements remain regular.

A ring is semilocal when its quotient by the Jacobson radical is a semisimple ring that is Artinian. Equivalently, that quotient is a right Artinian ring: it has zero Jacobson radical, and the Artinian structure theorem makes it semisimple. A commutative ring is semilocal exactly when it has finitely many maximal ideals.

For a finitely generated module $V$ on the right over a semilocal ring, the displayed condition makes the induced module endomorphism on $V/VJ$ injective. This quotient is an Artinian module, so Artinian modules are co-Hopfian makes that map surjective. The remaining quotient module $V/\alpha (V)$ equals its product with $J$, and Nakayama lemma makes it zero.

For a ring $D$, an automorphism $\sigma$ and central $t$, adjoin $X,Y$ with $Xd=\sigma (d)X$, $Yd={\sigma }^{-1}(d)Y$, $YX=t$ and $XY=\sigma (t)$. This construction includes Weyl algebras and rank-one universal enveloping algebras.

For a commutative Noetherian base $D$ of finite dimension $d$, the noncommutative Krull dimension of $D(\sigma ,t)$ equals $d$ unless a height-$d$ maximal ideal is periodic under $\sigma$ or contains infinitely many translates ${\sigma }^n(t)$. This is the commutative-base specialization of Theorem 5.3 in Krull dimension of Generalized Weyl Algebras with non-commutative coefficients.

Set the dimension of the zero module to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Inductively, dimension at most $d$ means that in every descending chain the successive quotients eventually have dimension less than $d$. The least such ordinal is the Krull dimension; the ordinal definition also covers dimensions beyond finite integers. The left dimension of a ring is that of its left regular module. In noncommutative rings this is not defined simply as the length of a chain of two-sided prime ideals.

A noncommutative domain is a nonzero unital ring without zero divisors; commutativity is not required, but commutative integral domains are included in this usage. Left and right cancellation hold for nonzero elements. A right Noetherian domain embeds in a division ring by the right Ore condition.

For a field $k$ and $q\in k^{\times }$, the quantum plane is $k_q[X,Y]=k⟨X,Y⟩/(YX-qXY)$. Its ordered monomials $X^i{Y}^j$, $i,j\ge 0$, are a basis, and their multiplication is $X^i{Y}^j{X}^u{Y}^v=q^{ju}{X}^{i+u}{Y}^{j+v}$. Leading monomials show it is a noncommutative domain. Some sources interchange $q$ with $q^{-1}$ in the defining relation.

For $yx=qxy$, the quantum enveloping algebra of sl2 acts by $K(x^r{y}^s)=q^{r-s}{x}^r{y}^s$, $E(x^r{y}^s)=[s{]}_q{x}^{r+1}{y}^{s-1}$ and $F(x^r{y}^s)=[r{]}_q{x}^{r-1}{y}^{s+1}$. Negative exponents here have zero coefficients and are omitted. The quantum integer identity $[r][s+1]-[s][r+1]=[r-s]$ verifies the commutator. These operations preserve total degree and make the quantum plane a module algebra.

At a primitive cube root of unity, the cubic homogeneous module has basis $x^3,x^{2}y,x{y}^2,y^3$. Its only simple modules occurring as submodules are the one-dimensional subspaces of $⟨x^3,y^3⟩$, on which $K=1$ and $E=F=0$. Its quotient by this two-dimensional submodule is simple, but it has no simple complement: any lift of the highest $K$ weight space contains $x^{2}y$, whose $E$ image is $x^3\ne 0$. Thus the cubic module is not a semisimple module.

For $q\ne 0$, define a vector space with basis $u^i{v}^j$, $i\in \mathbb{Z}$, $j\ge 0$, and product $(u^i{v}^j)(u^k{v}^{\ell })=q^{-jk}{u}^{i+k}{v}^{j+\ell }$. Bilinearity of the exponent verifies associativity. This constructs the algebra and proves the basis independently of its presentation. It localizes the quantum plane at powers of $u$ and gives a quotient witness for noncommutativity and noncocommutativity of the coordinate Hopf algebra of quantum SL2.

The two-dimensional quantum torus is the algebra with invertible generators $X,Y$ and relation $YX=qXY$, for $q\in k^{\times }$. Its basis consists of $X^i{Y}^j$ with $i,j\in \mathbb{Z}$, with the same multiplication rule as the quantum plane. It is a right Noetherian ring: successive adjunction of $X^{-1}$, $Y$, and $Y^{-1}$ satisfies the hypotheses of the noncommutative Hilbert basis theorem.

A ring is right Artinian if its right ideals satisfy the descending chain condition, equivalently its right regular module is an Artinian module. Left and right chain conditions should be distinguished for a general ring.

Let $K/k$ be an infinite-dimensional field extension. For the triangular ring in the display, the right regular module has length three: $e_{1}T$ has its off-diagonal simple submodule and simple quotient, and $e_{2}T$ is simple. Thus $T$ is right Artinian. Its off-diagonal ideal is, on the left, the infinite-dimensional $k$-vector space $K$. Every $k$-subspace determines a left ideal, so an ascending sequence of finite-dimensional subspaces violates left Noetherianity. For example take $K=k(z)$ and the spans of $1,z,\dots ,z^m$.

The finite intersections of maximal right ideals of a right Artinian ring have a minimal member by the descending chain condition. Intersecting it with any other maximal right ideal changes nothing, so it equals the Jacobson radical. If that radical is zero, the diagonal map from the regular module into a finite direct sum of its simple quotients is injective. A submodule of a finite direct sum of simple modules is semisimple and has finite length. This proves semisimplicity of the regular module without assuming Noetherianity first.

A right Artinian ring is a right Noetherian ring. Its right regular module therefore has finite composition length. Its Jacobson radical is a nilpotent ideal, and its quotient by that Jacobson radical is a semisimple ring. In particular every finitely generated right module over such a ring has finite composition length.

In a right Artinian ring, the powers of its Jacobson radical stabilize at an ideal $I$ satisfying $I^2=I$. If $I\ne 0$, choose a minimal right ideal $K$ with $KI\ne 0$. Since $(KI)I=KI$, minimality gives $KI=K$. Choosing $k\in K$ with $kI\ne 0$ gives $K=kR$ by the same minimality. Consequently $k=kj$ for some $j\in J(R)$. The unit criterion for the Jacobson radical makes $1-j$ invertible, forcing $k=0$, a contradiction. This proof avoids incorrectly applying the finitely generated form of Nakayama lemma to an arbitrary Artinian module.

A ring is left Noetherian when every ascending chain of left ideals stabilizes, equivalently when every left ideal is finitely generated.

A ring is right Noetherian when every ascending chain of right ideals stabilizes.

Suppose $A\subseteq B$ are unital rings and $B$ is a finitely generated module over $A$ on the right. If $A$ is a right Noetherian ring, then $B_A$ is a Noetherian module. Every right ideal of $B$ is an $A$-submodule, hence has finitely many generators over $A$; these also generate it over $B$. Thus $B$ is a right Noetherian ring. The same argument works on the left under the corresponding hypotheses. Finite-index group ring extensions give an important example, using coset representatives as a finite module basis.

A noncommutative domain whose right ideals satisfy the ascending chain condition is a right Noetherian domain. Its right regular module is a uniform module, so any two nonzero principal right ideals intersect. Thus its nonzero elements satisfy the right Ore condition and it embeds in a division ring by Ore localization.

If an algebra $B$ is generated by a subalgebra $A$ that is a right Noetherian ring and one element $x$, and $A+xA=A+Ax$, then $B$ is a right Noetherian ring. The proof uses $F_n={\sum }_{i\le n}A{x}^i={\sum }_{i\le n}{x}^{i}A$ and, for each right ideal $I$, the ascending right ideals $L_n=\{a:a{x}^n\in I+F_{n-1}\}$. After these stabilize, finitely many lifts of generators for $L_0,\dots ,L_N$ generate $I$ by induction on degree. Neither unique normal forms nor an automorphism moving coefficients is required.

For a right ideal $I\subseteq R[t]$, let $L_d$ be the set of degree-$d$ coefficients of its elements of degree at most $d$. These are right ideals of $R$, and multiplication by the central variable makes $L_d\subseteq L_{d+1}$. They stabilize, and every $L_d$ before stabilization is finitely generated. Choose polynomial lifts of those generators. Subtracting a suitable right linear combination of the lifts, shifted by powers of $t$, cancels the leading coefficient of any element of $I$. Induction on degree proves that the finitely many chosen lifts generate $I$.

// Destination: noncommutative-algebra.bigb

For an automorphism $\sigma$ of a left Noetherian ring $A$, the skew polynomial ring of an automorphism $A[t;\sigma ]$ is left Noetherian. For a left ideal $L$, the normalized leading-coefficient ideals $I_n=\{{\sigma }^{-n}(a_n):{\sum }_{j=0}^n{a}_j{t}^j\in L\}$ form an ascending chain of left ideals of $A$. Choose finite generating lifts through the stabilization degree. Every element of $L$ has its leading coefficient cancelled by a suitable left multiple of one of these lifts, and induction on degree proves finite generation. Passing to the skew Laurent polynomial ring preserves the property by clearing negative powers. The right-sided statement follows by applying the argument to the opposite ring.

The first Weyl algebra is $A_1(k)=k⟨x,y⟩/(yx-xy-1)$, the algebra of polynomial-coefficient differential operators in one variable.

For the relative differential operators over $\mathbb{C}$ on a polynomial ring, coordinates have order zero and partial derivatives have order one. The associated graded ring is $\mathbb{C}[x,\xi ]$ with degrees zero for $x_i$ and one for ${\xi }_i$. This differs from the Bernstein filtration, in which both sets of generators have degree one and every filtration piece is finite-dimensional.

In a nonzero two-sided ideal choose an element of least Bernstein filtration degree. Commuting it with any position or differentiation generator lowers degree, so all such commutators vanish. The ordered monomial basis of a Weyl algebra and characteristic zero then force it to be a nonzero scalar. The ideal is the whole ring. The argument fails in positive characteristic, when powers of the generators can be central.

The defining commutators $[{\partial }_i,x_j]={\delta }_{ij}$ reorder every word into the displayed monomials. Over a characteristic-zero field, the polynomial representation proves independence: iterated commutation with the coordinate multiplications extracts a highest derivative coefficient, multiplied by the nonzero factorial $\beta !$. Hence the representation is faithful and every polynomial-coefficient operator has a unique ordered expression.

In characteristic $p>0$, multiplication by $x_i$ and formal differentiation by $x_i$ act on the displayed vector space, because differentiation preserves the ideal generated by the $p$th powers. They satisfy the Weyl algebra relations and yield a nonzero module of dimension $p^n$. The central elements $x_i^p$ and ${\partial }_i^p$ also show why the characteristic-zero simplicity argument cannot apply. In particular the characteristic-zero Bernstein inequality for Weyl algebra modules fails for this finite-dimensional module.

// Destination: module-theory.bigb

For the complex Weyl algebra $A_n$ with $n\ge 1$, a unital finite-dimensional module would give matrices satisfying $[{\partial }_1,x_1]=I$. Their matrix trace gives $0={\dim }_{\mathbb{C}}M$, so the module is zero. Characteristic zero and $n\ge 1$ are essential.

Give both position and differentiation generators of a Weyl algebra degree one. The ordered monomial basis gives $\dim F_j{A}_n=\left(\genfrac{}{}{0ex}{}{j+2n}{2n}\right)$ and ${\mathrm{gr}}_F{A}_n$ equal to a polynomial ring in $2n$ variables. Commuting a filtered element with any generator lowers its degree by at least one.

A left primitive ring has a faithful simple left module.

If $V$ is a simple left $R$-module and $D={\mathrm{End}}_R(V{)}^{\mathrm{op}}$, then every prescribed map on a finite $D$-linearly independent subset of $V$ is induced by an element of $R$.

Morita-equivalent rings have equivalent module categories. In particular, $R$ and ${\mathrm{Mat}}_n(R)$ are Morita equivalent.

The Jacobson radical is the intersection of all maximal left ideals, equivalently the intersection of the annihilators of all simple left modules.

The unit criterion for the Jacobson radical implies that every radical element is a quasinilpotent element, because $1-x/\lambda$ is invertible for every nonzero $\lambda$. Conversely, if a two-sided ideal consists of quasinilpotent elements, then $yx$ is quasinilpotent for every member $x$ and every $y\in A$. Hence $1+yx$ is invertible and $x$ lies in the Jacobson radical. The maximality statement concerns ideals contained in the quasinilpotent set, not the whole set itself.

The intersection of maximal left ideals is the intersection of annihilators of simple modules on the left, hence is a two-sided ideal. If $j$ belongs to it, $1-rj$ has a left inverse for every $r$. Its left inverse also differs from $1$ by a radical element and has a left inverse; this makes the original inverse two-sided. The identity $(1-ba{)}^{-1}=1+b(1-ab{)}^{-1}a$ then shows that $1-jr$ is invertible too. If $j$ were outside a maximal right ideal $I$, one would have $1-jr\in I$ for some $r$, a contradiction. Applying the same argument to the opposite ring proves equality.

The Jacobson radical is the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module, so it is a two-sided ideal. If $y\in J(R)$, the left ideal $R(1-y)$ cannot be proper, since a containing maximal left ideal would also contain $y$ and hence $1$. Thus $c(1-y)=1$ for some $c$. Now $c=1+cy$ also differs from $1$ by a radical element and has a left inverse $d$. Multiplying $c(1-y)=1$ by $d$ gives $d=1-y$, and hence $(1-y)c=1$. This proves the unit assertion. Conversely, if $x$ is outside a maximal left ideal $L$, then $ax+l=1$ for some $a\in R$, $l\in L$. Consequently $1-ax\in L$ cannot be a unit, proving the reverse direction.

If $I$ is a nilpotent ideal and $J(A/I)=0$, then $I=J(A)$. Every $1-ax$, $x\in I$, has a finite geometric-series inverse, giving $I\subseteq J(A)$. The radical's image lies in $J(A/I)=0$, giving the reverse inclusion. A semisimple algebra as quotient supplies the required zero radical.

For an integral extension $A\subseteq B$ of commutative rings,Contraction sends maximal ideals of $B$ to maximal ideals of $A$, while the Lying-over theorem puts a maximal ideal of $B$ over every maximal ideal of $A$.

A commutative ring $R$ is Jacobson when every prime ideal is the intersection of the maximal ideals containing it. Equivalently, for every ideal $I$, its radical is the intersection of the maximal ideals containing $I$.

For a finite-type integer algebra, a nonnilpotent element $r$ gives a nonzero localization of a ring $R[1/r]$. A maximal ideal there has finite residue field by the finite-field theorem for finitely generated integer algebras. The image of $R$ in that field is a finite integral domain, hence a field. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every maximal ideal, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are Jacobson rings. Merely contracting a localized maximal ideal without the finite-image argument would not prove maximality.

Every ring integral over a Jacobson ring is Jacobson. After quotienting by a prime, an integral equation for a nonzero element has nonzero constant term; choose a maximal ideal of the base avoiding that term and apply the Lying-over theorem.

An injective module has the extension property for homomorphisms defined on submodules.

The divisible group $D=\mathbb{Q}/\mathbb{Z}$ is injective: every homomorphism $n\mathbb{Z}\to D$ extends to $\mathbb{Z}$ by choosing an $n$th division of its value at $n$, and the Baer criterion applies. It is a cogenerator because any nonzero element generates a cyclic subgroup admitting a character nonzero on that element, and injectivity extends that character to the whole group. Consequently a map $u:T\to U$ of abelian groups is injective exactly when restriction $U^{\ast }\to T^{\ast }$ is surjective: injectivity gives extension of characters, while a nonzero kernel element is detected by a character that cannot extend along $u$.

For a Noetherian ring, injectivity of a module can be tested at all prime localizations. By the Baer criterion, test ${\mathrm{Ext}}_R^1(R/J,M)$ for every ideal $J$. Localization of Ext over a Noetherian ring identifies its localization with the corresponding ideal test over $R_P$. Every ideal of $R_P$ is extended from its contraction, and localization detects zero elements applies to the resulting Ext module even when it is not finitely generated. These observations prove both directions.

This criterion characterizes an injective module using extension from ideals. A left $R$-module is injective exactly when every homomorphism from a left ideal of $R$ extends to a homomorphism from $R$.

A ring is left Noetherian exactly when every direct sum of injective left modules is injective.

Over an integral domain, a module $M$ is divisible when $rM=M$ for every nonzero $r$. A divisible group is precisely a divisible $\mathbb{Z}$-module, and over a principal ideal domain divisibility is equivalent to injectivity.

The injective hull of a module $M$ is a minimal injective extension of $M$ in which $M$ is an essential submodule.

For a domain satisfying the left Ore condition, its division ring of fractions $Q$ is an injective essential extension of the left regular module. For a homomorphism $f:I\to Q$ on a nonzero left ideal, choose $a\in I\setminus \{0\}$. A common left multiple $sx=ta$ gives $sf(x)=tf(a)$, hence $f(x)=x{a}^{-1}f(a)$; right multiplication by $a^{-1}f(a)$ extends it. The Baer criterion proves injectivity. Clearing a left denominator proves essentiality. The same calculation gives ${\mathrm{End}}_R({}_{R}Q)\cong Q^{\mathrm{op}}$, with the opposite ring required because right multiplications compose in reversed order.

The injective hull of a uniform module is uniform. An injective endomorphism of a uniform injective module splits its image, so its complementary summand must vanish and it is an automorphism. Consequently all nonunits are precisely the noninjective endomorphisms. Their nonzero kernels are essential; intersection of two such kernels is nonzero, making their sum noninjective. Composition on either side preserves noninjectivity, unless the other map is a unit, which merely transports the kernel. Thus the nonunits form the unique maximal ideal, and the quotient is a division ring.

The Ore condition is a common-multiple condition that permits fractions to be formed in a noncommutative ring.

A multiplicative set $S$ of regular elements of a ring admits a right ring of fractions exactly when it satisfies the right Ore condition. The resulting localization embeds the original ring, inverts $S$, and expresses every element as $a{s}^{-1}$. For denominators that can be zero divisors, an additional denominator reversibility condition is required.

If two nonzero principal right ideals in a right Noetherian domain had zero intersection, the sums $bR\oplus abR\oplus \cdots \oplus a^{m}bR$ would be strictly increasing. Directness follows by splitting the lowest term into $bR$ and the other terms into $aR$, then cancelling $a$ and repeating. This contradicts the ascending chain condition. A nonzero common element $au=bv$ gives the right Ore condition. The left version follows from the opposite ring. Consequently a noncommutative domain Noetherian on both sides has a single division ring of fractions with both left and right fraction presentations.

For a multiplicative set $S$ of regular elements satisfying the left Ore condition, represent a fraction by $(s,a)$. Pairs $(s,a)$ and $(t,b)$ agree when there are $u,v$ with $us=vt\in S$ and $ua=vb$. Addition uses this common denominator. For multiplication choose $v\in S,c\in R$ with $va=ct$, then $(s^{-1}a)(t^{-1}b)=(vs{)}^{-1}cb$. The Ore condition gives common multiples for successive operations; regularity supplies cancellation. In a domain with $S$ all nonzero elements, every nonzero fraction has inverse $a^{-1}s$.

// Destination: associative-algebra.bigb

A multiplicative subset $S\subseteq A$ with $1\in S$ and $0\notin S$ of a ring $A$ satisfies the right Ore condition if for every $a\in A$ and $s\in S$ there are $u\in S$ and $b\in A$ such that $au=sb$. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a ring of right fractions $a{s}^{-1}$; every nonzero fraction is invertible. The usual additional denominator reversibility condition is automatic when the elements of $S$ are nonzero in a noncommutative domain.

A multiplicative set $S$ containing $1$ and not $0$ is a right denominator set if it satisfies the right Ore condition and right reversibility: $sr=0$, with $s\in S$, implies $rt=0$ for some $t\in S$. Its Ore localization consists of right fractions $r{s}^{-1}$ and has kernel $\{r:rt=0\text{ for some }t\in S\}$. This condition permits zero-divisor denominators; it is automatic for a set of regular elements of a ring. When all denominators are regular, the localization map is injective.

The classical right ring of quotients inverts all regular elements of a ring. Existence requires the right Ore condition for these denominators. The original ring embeds, every regular element becomes a unit, and every fraction has a right denominator. The right Ore condition also permits finitely many fractions to be placed over a common right denominator.

A multiplicative subset $S$ of a ring is a right Ore set if $1\in S$, $0\notin S$ and it satisfies the right Ore condition. In a noncommutative domain, this condition lets nonzero denominators be used in Ore localization.

Ore localization extends localization to suitable multiplicative subsets of noncommutative rings.

This is the localization of a ring at all its regular elements of a ring, with left fractions. It exists precisely when those denominators satisfy the left Ore condition. If $R$ embeds in a ring $Q$, all regular elements become units in $Q$, and every element of $Q$ is such a fraction, the Ore condition follows by writing $r{c}^{-1}=s^{-1}b$, which gives $sr=bc$.

Let $A$ be a domain satisfying both Ore conditions, with quotient division ring $K$. A matrix over $A$ is regular exactly when it is invertible over $K$: a kernel vector of a singular matrix can have its denominators cleared to give a nonzero annihilator over $A$. An upper triangular matrix is regular exactly when all diagonal entries are nonzero. Each finite list of entries of a matrix over $K$ has a common left denominator $a\in A\setminus \{0\}$. Thus it is a fraction with denominator $a{I}_r$, proving the displayed classical localizations. The triangular localization retains its strictly upper triangular nilpotent ideal; it is not a semisimple matrix ring.

A multiplicative subset $S$ is left Ore when, for every $s\in S$ and $r\in R$, there are $s^{\prime }\in S$ and $r^{\prime }\in R$ such that $s^{\prime }r=r^{\prime }s$.

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.