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For a normal p-subgroup $P◃G$, every central idempotent $e$ of $kG$ satisfies $e={\mathrm{Br}}_P(e)\in k{C}_G(P)$. Indeed a nonfixed conjugation orbit sum maps to zero in $k(G/P)$, so $e-{\mathrm{Br}}_P(e)$ belongs to the nilpotent ideal $J(kP)kG$. Normality makes the Brauer image a central idempotent of $kG$, and two central idempotents congruent modulo a nilpotent ideal coincide. Consequently the blocks of a group algebra of $kG$ correspond to $G$-orbits of blocks of $k{C}_G(P)$.

Over a field of characteristic three, the group algebra $A=k{S}_3$ has a single block of an Artinian algebra. Put $r=(123)$, $s=(12)$ and $a=r-1$. Then $a^3=0$, $sas=-a+a^2$, and $J(A)=aA$: the latter is a nilpotent ideal, with semisimple ring quotient $k{C}_2$. The center of an associative algebra is $k1\oplus k{a}^2\oplus k{a}^{2}s$, whose last two summands form a square-zero ideal. A central idempotent $d1+n$ satisfies $d^2=d$ and $(2d-1)n=0$, hence is zero or one. Thus no nontrivial central block decomposition exists.

Let $Z=⟨z⟩\le Z(G)$ have order $p$ and set $t=z-1$. The nilpotent ideal $tkG$ is generated by a central element. If an idempotent lifts a central idempotent of the quotient, its off-diagonal corner satisfies $eA(1-e)\subseteq teA(1-e)$ and therefore is zero. Thus central block idempotents lift centrally and uniquely. A principal indecomposable module is free over $kZ$, and the $p$ factors of its $t$-filtration are all its reduction modulo $t$. Consequently, under the natural indexing of simple modules, $C_G=p{C}_{G/Z}$.

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.

With $s=(12)$, the idempotents $e_{\pm }=(1\pm s)/2$ split the right regular module into two three-dimensional projective modules. Their tops are respectively the trivial representation and sign representation. The nilpotent ideal $J(k{S}_3)$ ensures that any nonzero direct summand has nonzero top, so the one-dimensional tops make these projective modules indecomposable. Their successive radical series of a module factors are trivial, sign, trivial and sign, trivial, sign. This decomposition concerns right modules; the block of S3 in characteristic three remains a single algebra.

For a nilpotent ideal $I$, its nilpotence index of an ideal is the least positive integer $n$ for which $I^n=0$. In $k[X]/(g^e)$, with $g$ an irreducible polynomial, the unique maximal ideal $(g)$ has index $e$. In a discrete valuation ring with $p=u{\pi }^e$, the maximal ideal of its reduction modulo $p$ has the same index. Comparing these indices under a quotient ring isomorphism recovers the ramification exponent in the Kummer-Dedekind theorem.

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.

The nilradical $N$ of a commutative Artinian ring is a nilpotent ideal. Its powers stabilize, say $N^r=N^{r+1}=I$. If $I\ne 0$, choose an ideal $J$ minimal subject to $IJ\ne 0$, and then $x\in J$ with $Ix\ne 0$. Minimality and $I^2=I$ give $Rx=Ix$, so $x=ax$ for some $a\in I$. Since $a$ is nilpotent, $1-a$ is a unit, contradicting $(1-a)x=0$ and $x\ne 0$.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

Take rings and modules to be unital. Let $J_{\ell }$ denote the intersection of the maximal left ideals. For each nonzero vector $v$ of a simple module $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a maximal left ideal. Consequently $J_{\ell }$ annihilates every simple module on the left. Conversely, annihilating $R/L$ forces an element to belong to the maximal left ideal $L$. Thuswhich shows that $J_{\ell }$ is a two-sided ideal.

For $j\in J_{\ell }$ and $r\in R$, the left ideal $R(1-rj)$ cannot be proper: a containing maximal left ideal would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_{\ell }$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a unit, not just left invertible.

The identityis verified by multiplying on either side, using $(1-ab{)}^{-1}(1-ab)=(1-ab)(1-ab{)}^{-1}=1$. It implies that $1-jr$ is a unit as well. If $j$ were outside a maximal right ideal $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper right ideal cannot contain a unit, giving a contradiction. Thus $J_{\ell }$ belongs to every maximal right ideal. Applying the same argument to the opposite ring proves the reverse inclusion. This proves the Jacobson radical is independent of handedness:For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.

If a two-sided ideal $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx{)}^m=0$. The finite geometric series ${\sum }_{i=0}^{m-1}(rx{)}^i$ inverts $1-rx$. The unit criterion for the Jacobson radical, also proved above by the maximal-ideal argument, gives $N\subseteq J(R)$. This assertion concerns a nilpotent ideal; arbitrary nilpotent elements of a noncommutative ring need not lie in its Jacobson radical.

For the formal power series ring over the p-adic integers, a series $f={\sum }_{i\ge 0}{a}_i{t}^i$ is a unit exactly when $a_0$ is a unit of ${\mathbb{Z}}_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p{\mathbb{Z}}_p$. They form the unique maximal ideal $(p,t)$, and the quotient is ${\mathbb{F}}_p$. Hence

The prime radical of a noncommutative ring is $N(R)={\bigcap }_{P}P$, with $P$ ranging over all prime ideals of a noncommutative ring. Here primeness means that for two-sided ideals $A,B$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. A minimal prime over an ideal $I$ is a prime ideal containing $I$ with no strictly smaller prime ideal containing $I$.

The ascending chain condition on left ideals also applies to two-sided ideals. We first prove by Noetherian induction that every proper $I$ contains a finite product of prime ideals, each containing $I$. If there were counterexamples, choose a maximal one. It is not prime, so there are strictly larger two-sided ideals $A,B$ with $AB\subseteq I$. Neither is a counterexample; products of prime ideals contained in $A$ and in $B$ then concatenate to a product contained in $I$, a contradiction.

Write $P_1\cdots P_t\subseteq I\subseteq P_j$. Every prime ideal $P$ over $I$ contains at least one $P_j$, by primeness. The inclusion-minimal members of the finite list $P_j$ are therefore precisely the minimal primes over an ideal $I$: any smaller prime over $I$ would contain a member of that same list. Consequently there are finitely many. Replacing each factor $P_j$ by a minimal prime contained in it giveswith each $Q_j$ minimal over $I$. Repeated factors may be necessary; in $k[\epsilon ]/({\epsilon }^m)$ the sole minimal prime $(\epsilon )$ requires its $m$th power to reach zero.

Apply this to $I=0$. Since $N(R)\subseteq Q_j$ for every $j$,Thus the prime radical is a nilpotent ideal, as in finite minimal primes and nilpotent prime radical.

Now assume $R$ is commutative. A prime ideal $P$ is an associated prime of a module $M$ when $P={\mathrm{Ann}}_R(m)$ for some $0\ne m\in M$. Every nonzero module over $R$ has an associated prime: the ascending chain condition gives a maximal annihilator of a nonzero element, and maximal annihilator of a module element is prime.

Since the given $M$ is a Noetherian module, repeatedly choosing such a cyclic submodule in the next quotient yields a finite prime filtrationThe process terminates by the ascending chain condition. In a short exact sequence $0\to U\to M\to W\to 0$, one has $\mathrm{Ass}(M)\subseteq \mathrm{Ass}(U)\cup \mathrm{Ass}(W)$. Indeed, if $P=\mathrm{Ann}(m)$ and $Rm\cap U=0$, the image of $m$ in $W$ still has annihilator $P$. Otherwise choose $0\ne am\in U$; then $a\notin P$, and primeness gives $\mathrm{Ann}(am)=P$. As $\mathrm{Ass}(R/P_i)=\{P_i\}$, induction along the prime filtration provesIn particular there are only finitely many associated primes of a module.

Finally let $P$ be a minimal prime ideal of $R$. Its localization at a prime ideal $R_P$ has exactly one prime, its maximal ideal $P{R}_P$. The preceding nilpotence result makes this maximal ideal nilpotent. Choose $0\ne y\in R_P$ annihilated by $P{R}_P$, for example a nonzero element in its last nonzero power, or $1$ if $P{R}_P=0$. Since elements outside the maximal ideal are units, ${\mathrm{Ann}}_{R_P}(y)=P{R}_P$.

Write $y=r/s$. The element $r/1$ is nonzero and also has annihilator $P{R}_P$. Generate $P$ by $p_1,\dots ,p_{\ell }$. For each $i$, choose $u_i\notin P$ with $u_i{p}_{i}r=0$, and put $u={\prod }_i{u}_i$. Then $m=ur$ remains nonzero after localization, while $Pm=0$. If $am=0$, localization forces $a/1\in P{R}_P$, and hence $a\in P$. Therefore ${\mathrm{Ann}}_R(m)=P$. This proves that minimal primes are associated primes.

Write $A=RG$. An idempotent is an element $e$ with $e^2=e$; it is central idempotent when it commutes with every element of the ring. In a commutative ring centrality is automatic. Two idempotents are orthogonal idempotents if $ef=fe=0$. A nonzero primitive central idempotent cannot be expressed as the sum of two nonzero orthogonal central idempotents; this is primitivity in $Z(A)$, not necessarily primitivity in $A$.

A block of a group algebra is an indecomposable two-sided direct-summand ideal of $A$, equivalently an ideal $Ae$ with $e$ a primitive central idempotent. Here is the equivalence. If $A=B\oplus C$ as two-sided ideals, then $BC=CB=0$. Write $1=e+f$ with $e\in B,f\in C$. Multiplication by this equality shows that $e$ is the identity of $B$, that $f$ is the identity of $C$, and that $e,f$ are orthogonal central idempotents. Conversely a central idempotent gives $A=Ae\oplus A(1-e)$. Splitting the ideal further is therefore equivalent to splitting its central idempotent further.

For uniqueness, suppose $1={\sum }_i{e}_i={\sum }_j{f}_j$ are decompositions into primitive central idempotents. For each $i$,is an orthogonal central-idempotent decomposition. Exactly one summand is nonzero, and that summand equals $e_i$. Applying the same reasoning to the corresponding $f_j$ gives $e_i=f_j$. Thus the blocks, and their identities, are unique up to ordering. Existence in the paper's coefficient rings follows from finite-dimensionality over $k$, and over $\mathcal{O}$ from completeness and idempotent lifting of the decomposition modulo its maximal ideal.

Over $k$, a principal indecomposable module is an indecomposable summand of the regular left module $A=kG$, equivalently $P=Ae$ for a primitive idempotent $e$ of $A$. Put $J=J(A)$, the Jacobson radical. The head of a module $P/JP$ is nonzero and semisimple. It is simple: projectivity gives a surjectionwhose kernel consists of maps into $JP$ and is nilpotent, since a product of $r$ such maps sends $P$ into $J^{r}P$. If the head of a module split, a nontrivial projection on it would lift by idempotent lifting to a nontrivial projection on $P$, contradicting indecomposability. Every maximal submodule contains $JP$, since $J$ annihilates simple modules. The simplicity of the head of a module now proves $J(P)=JP$ is the unique maximal submodule, and $P$ is its head of a module's projective cover.

Choose the simple modules $S_1,\dots ,S_r$, with corresponding projective covers $P_1,\dots ,P_r$. Our convention for the Cartan matrix of a group algebra iswhere brackets denote the multiplicity in a composition series.

For the final two unheaded requests, retain the additional hypothesis $G=N{C}_G(N)$. Choose a central subgroup $Z=⟨z⟩$ of $N$ of order $p$. Every element of $N$ centralizes $Z$, and so does every element of $C_G(N)$; hence $Z\le Z(G)$. Set $t=z-1$, so $t$ is central and $t^p=0$, and $A/tA\cong k(G/Z)$.

To lift a central idempotent $\bar{e}$ of this quotient, first lift it as an idempotent $e$ by idempotent refinement theorem. Since $\bar{e}$ is central, $eA(1-e)\subseteq tA$. Multiplying a representation $ea(1-e)=tb$ on the left and right by $e,1-e$ givesThe other off-diagonal corner vanishes likewise, so $e$ is central. Two central lifts of the same idempotent coincide: their mismatched products $e(1-f)$ and $f(1-e)$ are idempotents in the nilpotent kernel and therefore zero. Primitivity is preserved because orthogonal idempotent decompositions lift. Thus the central $p$-quotient gives a bijection of block idempotents.

The quotient $G/Z$ still satisfies the required hypothesis for $N/Z$, because the image of $C_G(N)$ centralizes $N/Z$. Induction on $|N|$ therefore proves $\tau$ bijects the block idempotents of $kG$ and $k(G/N)$. The centrality argument is essential: a general nilpotent algebra quotient need not give a bijection of central block idempotents.

For the Cartan matrix of a group algebra, every projective $A$-module is free over $kZ$: $A$ is free over $kZ$, a projective $A$-module is a summand of a finite free $A$-module, and finite projectives over the local ring $kZ\cong k[t]/(t^p)$ are free. If $P$ is a principal indecomposable module, $\bar{P}=P/tP$ is a principal indecomposable module over $A/tA$: an idempotent splitting of its endomorphism ring would lift and split $P$. The $p$ quotients of its filtrationare all isomorphic to $\bar{P}$ as $k(G/Z)$-modules, by multiplication by the corresponding power of $t$. All simple modules are annihilated by the nilpotent ideal $tA$. Hence $[P:S]=p[\bar{P}:S]$, with the natural indexing of simple modules, and $C_G=p{C}_{G/Z}$. Iterating over $N$ givesThis is blocks and Cartan matrices under central p-quotients. The extra hypothesis is still in force: for example $N=C_3⊴S_3$ in characteristic three does not satisfy it, and its Group algebra Cartan matrix is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ rather than $3I$.

Use the following Jacobson radical facts for a unital algebra: a nilpotent ideal is contained in the Jacobson radical; under a surjective algebra homomorphism the image of the Jacobson radical is contained in the radical of the quotient; and a finite product of fields has zero Jacobson radical. For completeness, if $I^N=0$ and $x\in I$, then $ax\in I$ for every $a\in A$, and $1-ax$ is invertible with inverse ${\sum }_{j=0}^{N-1}(ax{)}^j$. The usual unit criterion for the Jacobson radical therefore gives $I\subseteq J(A)$. The quotient assumption gives $J(A)\subseteq I$. HenceThis is the nilpotent ideal with semisimple quotient radical criterion.

For the finite quiver under consideration, let $R$ be the arrow ideal of a path algebra, spanned by paths of positive length. If $Q$ has $r$ vertices and no oriented cycle, a path cannot repeat a vertex, so $R^r=0$. Meanwhile $kQ/R\cong {\prod }_{i\in Q_0}k$, with the constant paths giving the coordinate idempotents. The criterion just proved yields $J(kQ)=R$.

The condition on cycles is necessary. A quiver with one loop has path algebra $k[t]$, whose arrow ideal is $(t)$. But $J(k[t])=0\ne (t)$: the maximal ideals $(t-a)$, $a\in k$, have intersection zero, since the algebraically closed field $k$ is infinite and a nonzero polynomial has only finitely many roots. Thus the arrow ideal need not be the Jacobson radical when oriented cycles are present.

The algebra must be finite-dimensional, and there is no universal bound on its dimension. Let $e=\epsilon m$ and write its coevaluation morphism as a finite sumFor any $v\in A$, one snake identity givesThus the finitely many $b_i$ span $A$, making it a finite-dimensional vector space. The other snake identity shows that $e$ is nondegenerate in the other variable as well, so the maps to the dual space determined by this pairing are isomorphisms. In the standard algebraic terminology this is a Frobenius algebra.

Every positive finite dimension of a vector space occurs: take $A=k^d$ with coordinatewise multiplication and $\epsilon (a_1,\dots ,a_d)={\sum }_i{a}_i$. The coordinate idempotents $e_i$ give $e(e_i,e_j)={\delta }_{ij}$ and $n(1)={\sum }_i{e}_i\otimes e_i$. If zero unital algebras are allowed, the zero object gives dimension zero too.

The pairing need not make $A$ a semisimple algebra. For example, the dual numbers $k[t]/(t^2)$ with $\epsilon (a+bt)=b$ have pairing matrixand coevaluation morphism $1\mapsto 1\otimes t+t\otimes 1$. This pairing is nondegenerate over every field, despite the nilpotent ideal $(t)$. Finite-dimensionality is the dimension conclusion; separability is not part of the given definition.

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.

The Jacobson radical is $J(A)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals of $A$; equivalently, it is the intersection of the annihilators of all simple modules. It is a two-sided ideal. A projective module has the lifting property against every surjective R-module homomorphism. A finitely generated module is a projective module precisely when it is a direct summand of a finitely generated free module. An indecomposable module is nonzero and admits no direct sum decomposition into two nonzero submodules.

To calculate the top of an indecomposable projective module, use the right Artinian ring hypothesis, namely the descending chain condition on right ideals. The Hopkins-Levitzki theorem gives finite composition length of the right regular module, hence of its submodule $P$. Also $J=J(A)$ is a nilpotent ideal, and $A/J$ is a semisimple ring. Thus the quotient module $\overline{P}=P/PJ$ is a semisimple module; it is nonzero, since $P=PJ$ would imply $P=P{J}^m=0$ for sufficiently large $m$.

Suppose $\overline{P}$ were not a simple module. A nontrivial direct sum decomposition of this semisimple module would give an idempotent $\alpha \in {\mathrm{End}}_A(\overline{P})$ that is neither zero nor the identity. Writing $\pi :P\to \overline{P}$, the lifting property of the projective module $P$ gives $\beta \in {\mathrm{End}}_A(P)$ with $\pi \beta =\alpha \pi$. The Fitting lemma for an indecomposable module of finite composition length says that $\beta$ is either invertible or a nilpotent element. Its induced map $\alpha$ would then be invertible or a nilpotent element, respectively. Neither is possible for a nontrivial idempotent. Therefore $P/PJ(A)$ is simple. This argument does not assume that an embedded projective module automatically splits off from the ambient module.

A block of an Artinian algebra is a nonzero two-sided direct summand $Ae$ determined by a primitive central idempotent $e$: $e$ cannot be written as a sum of two nonzero orthogonal central idempotents. Its identity is $e$. For a finite-dimensional associative algebra, the blocks of an Artinian algebra give its unique decomposition as a finite product of indecomposable algebras, or equivalently as a direct sum of two-sided ideals.

For the block of S3 in characteristic three, put $A=k{S}_3$, $r=(123)$, $s=(12)$ and $a=r-1$. The group algebra has basis $r^i,r^{i}s$ for $0\le i<3$. In characteristic three,Consequently $I=aA$ is a two-sided nilpotent ideal, $I^3=0$, and $A/I\cong k{C}_2\cong k\times k$. A nilpotent ideal lies in the Jacobson radical, and a quotient that is a semisimple ring forces the reverse inclusion. HenceDefine orthogonal idempotents $e_{+}=(1+s)/2$ and $e_{-}=(1-s)/2$. They sum to one, so the right regular module decomposes asThe two summands are the indecomposable projectives of S3 in characteristic three. Each direct summand is a projective module, with basis $e_{\pm },e_{\pm }r,e_{\pm }{r}^2$. Its quotient module modulo multiplication by $J(A)$ is one-dimensional: the trivial representation for $e_{+}A$, and the sign representation for $e_{-}A$. Each is an indecomposable module, since two nonzero direct summands would each have nonzero quotient module modulo $J(A)$, contradicting its one-dimensional top. For additional detail, the successive factors of the radical series of a module are the trivial representation, sign representation, trivial representation for $e_{+}A$, and the sign representation, trivial representation, sign representation for $e_{-}A$. Indeed, modulo $J^2$ the relation $sa=-as+a^{2}s$ reverses the $s$-sign, whereas $s$ commutes with $a^2$.

The center of an associative algebra is spanned by the conjugacy class sums $1$, $r+r^2$, and $(1+r+r^2)s$. Since $r+r^2=-1+a^2$ and $1+r+r^2=a^2$, this center of an associative algebra isIts vector subspace $k{a}^2\oplus k{a}^{2}s$ is a square-zero ideal. If $d1+n$ is a central idempotent, then $d^2=d$ and $(2d-1)n=0$. Thus $d=0$ or $1$ and $n=0$. There is no nontrivial central idempotent, so the whole group algebra is its single block. The two three-dimensional projective modules above are a decomposition of the regular module, not two blocks of an Artinian algebra.

A right Noetherian ring satisfies the ascending chain condition on right ideals; equivalently, every right ideal is a finitely generated module.

Here is a noncommutative Hilbert basis theorem proof adapted to the stated hypothesis. Set $F_{-1}=0$ andThe equality follows inductively from $A+xA=A+Ax$, by moving one coefficient past one $x$ at a time. These spaces form an exhaustive filtered algebra structure on $B$, with $F_n{F}_m\subseteq F_{n+m}$. This does not assert uniqueness of the displayed expressions or the existence of a coefficient-moving automorphism.

For a right ideal $I\subseteq B$, defineEach $L_n$ is a right ideal of $A$. In fact, if $a\in L_n$ and $c\in A$, write $c{x}^n=x^{n}b+u$ with $b\in A$, $u\in F_{n-1}$; then $ac{x}^n=(a{x}^n)b+au\in I+F_{n-1}$. Also $L_n\subseteq L_{n+1}$ by right multiplication by $x$. Since $A$ is a right Noetherian ring, this ascending chain stabilizes at some $L_N$.

Choose finite generators $a_{nj}$ for each $L_n$, $0\le n\le N$, and choose $f_{nj}\in I$ with $f_{nj}-a_{nj}{x}^n\in F_{n-1}$. These finitely many elements generate $I$ as a right ideal. To see this, induct on $m$ for $f\in I\cap F_m$. Write $f=a{x}^m+u$ with $u\in F_{m-1}$; then $a\in L_m$. Put $d=\min (m,N)$ and express $a={\sum }_j{a}_{dj}{c}_j$. Write $c_j{x}^d=x^d{b}_j+u_j$ with $u_j\in F_{d-1}$. The differencelies in $I\cap F_{m-1}$, so the induction applies. At $m=0$ the remainder is zero. Hence $B$ is right Noetherian.

For the quantum torus, take $q\in k^{\times }$ and the convention $YX=qXY$. Begin with the polynomial ring $k[X]$, which is Noetherian by the Hilbert basis theorem. Adjoining $X^{-1}$ preserves the hypothesis because it commutes with $k[X]$, and gives $k[X^{\pm 1}]$. Adjoin $Y$ next. The relation $Yf(X)=f(qX)Y$ and its inverse coefficient-moving relation give $A+YA=A+AY$ for $A=k[X^{\pm 1}]$. Finally adjoin $Y^{-1}$ to $k[X^{\pm 1}][Y;\sigma ]$, where $\sigma (X)=qX$. On a monomial, $Y^{-1}{X}^i{Y}^j=q^{-i}{X}^i{Y}^{j-1}$; when $j=0$ this is in $A{Y}^{-1}$, and when $j>0$ it is in $A$. The reverse inclusion follows by the same relation. The preceding argument applies at each step, proving the quantum torus is right Noetherian. Nonzero $q$ is required for this notation.

For a noncommutative ring, a prime ideal of a noncommutative ring means a proper two-sided ideal $P$ such that $UV\subseteq P$ for two-sided ideals $U,V$ implies $U\subseteq P$ or $V\subseteq P$. Equivalently, $aAb\subseteq P$ implies $a\in P$ or $b\in P$. This definition does not require $A/P$ to be a noncommutative domain.

Retain the right Noetherian ring hypothesis for the last assertion. More generally, the ascending chain condition on two-sided ideals suffices. We claim that every proper two-sided ideal $I$ contains a product of finitely many prime ideals of a noncommutative ring, each containing $I$. If not, choose a maximal counterexample $I$. It cannot be a prime ideal of a noncommutative ring. Thus there are two-sided ideals $U,V$ strictly containing $I$ with $UV\subseteq I$: add $I$ to the two witnesses for failure of the defining condition for a prime ideal of a noncommutative ring. By maximality, both $U$ and $V$ contain products of finitely many prime ideals of a noncommutative ring containing them. Concatenating these products gives a product inside $UV\subseteq I$, a contradiction.

Apply the claim to $I=0$ in a nonzero $A$, obtaining $P_1\cdots P_r=0$. Every prime ideal of a noncommutative ring $Q$ contains one of the $P_i$, by repeated application of the definition of a prime ideal of a noncommutative ring. For the prime radical of a noncommutative ring $N$, it follows thatIndeed, $N\subseteq P_i$ gives $N^r\subseteq P_1\cdots P_r=0$, and ${\bigcap }_i{P}_i\subseteq Q$ for every $Q$ gives equality of the intersections. If $A=0$, the empty intersection is the whole zero ring and the conclusion is immediate.

The final assertion is false for arbitrary algebras without the preceding chain condition. For example, in the commutative ring $k[z_1,z_2,\dots ]/(z_i^2:i\ge 1)$ the nilradical is $(z_1,z_2,\dots )$, its only prime ideal, but the product of any number of distinct $z_i$ is nonzero. Thus this nilradical is not a nilpotent ideal.

First the product is also split. Write $A_i=k{G}_i$ and $J_i=\mathrm{rad}{A}_i$. The idealis nilpotent: if $J_1^r=J_2^s=0$, every product of $r+s-1$ of its factors vanishes. The quotient isa product of full matrix algebras over $k$, because each $A_i$ is split. A nilpotent ideal lies in the Jacobson radical, and a semisimple quotient forces the reverse inclusion, so $I$ is exactly the radical. Thus $k$ is a splitting field for finite group representations of the product.

For simple modules $S_i$, Schur lemma and splitting give ${\mathrm{End}}_{A_i}(S_i)=k$. The Jacobson density theorem therefore makes the image of $A_i$ on $S_i$ the full ${\mathrm{End}}_k(S_i)$. Tensoring these maps shows that the product algebra acts on $S_1\otimes S_2$ through its full endomorphism algebra. Hence this tensor product of group representations is simple.

On restriction to $G_1$, it is a direct sum of $\dim S_2$ copies of $S_1$. If two external tensor products are isomorphic, their restrictions and the Jordan–Hölder theorem force $S_1\cong S_1^{\prime }$; restricting to $G_2$ similarly forces $S_2\cong S_2^{\prime }$. The converse follows by tensoring the isomorphisms.

Finally, the p-regular conjugacy classes of $G_1\times G_2$ are precisely pairs of such classes in the factors. The corrected result in part (b) counts as many simples for the product as pairs of simples for the two factors. Our pairwise nonisomorphic tensor products already attain that count, so they exhaust all simples. Thus $\boxed{{\mathrm{Irr}}_k(G_1\times G_2)=\{S_1\otimes S_2\}}$, uniquely indexed by pairs of simple isomorphism classes.