Non-commutative rings course of the University of Oxford (C2.5)

Source: original examination PDF, bundle pages 343–345 inclusive; paper pages 1–3.

A left-primitive ideal is a two-sided ideal of the formfor a nonzero simple module $V$ on the left. Equivalently, the quotient ring $R/P$ has a faithful module that is a simple module, and is therefore a primitive ring.

To prove that $P$ is a prime ideal of a noncommutative ring, let $A,B$ be two-sided ideals with $AB\subseteq P$. Since $B$ is two-sided, $BV$ is a submodule of $V$: for $r\in R$, $b\in B$ and $v\in V$, one has $r(bv)=(rb)v\in BV$. If $B⊈P$, then $BV\ne 0$, so simplicity gives $BV=V$. Consequentlyand $A\subseteq P$. Thus $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. Also $P\ne R$, since $1$ acts as the identity element on the nonzero module $V$. These are precisely the required prime ideal of a noncommutative ring conditions.

The Jacobson radical isHere the second intersection ranges over isomorphism classes of nonzero simple modules. For the equality, every nonzero $v$ in a simple module gives a surjective module homomorphism $R\to V$, $r\mapsto rv$, whose kernel is a maximal left ideal. Conversely, every maximal left ideal $L$ gives the simple module $R/L$. This description by annihilators shows that $J(R)$ is a two-sided ideal.

Let $j\in J(R)$ and put $a=1+j$. If the left ideal $Ra$ were proper, it would lie in a maximal left ideal $L$. Both $a$ and $j$ would belong to $L$, giving $1\in L$, a contradiction. Therefore there is $u\in R$ with $ua=1$.

A left inverse alone need not be a two-sided inverse in a general ring, so one more step is necessary. The equality $u(1+j)=1$ gives $u=1-uj\in 1+J(R)$. Applying the same argument to $u$ supplies $v\in R$ with $vu=1$. Thenand hence $au=vu=1$. ThusFor the zero ring the conclusion holds directly; the maximal left ideal argument concerns the nonzero case.

Let $L$ be any maximal left ideal. If $I⊈L$, maximality gives $L+I=R$, because the two-sided ideal $I$ is in particular a left ideal. Thus $1=\ell +i$ for some $\ell \in L$ and $i\in I$. Butis a unit by hypothesis. A proper left ideal cannot contain a unit, since left multiplication by its inverse would put $1$ in the left ideal. This contradiction proves $I\subseteq L$ for every maximal left ideal $L$, and thereforeTogether with ii, this is the ideal form of the unit criterion for the Jacobson radical.

Assume $A\ne 0$, as is implicit in the claimed conclusion. Suppose that a surjective ring homomorphism $f:R\to M_2(A)$ exists. Choose $r_1,r_2\in R$ mapping to the matrix units $E_{12}$ and $E_{21}$. Applying $f$ to the assumed relation gives integers $c_1,\dots ,c_n$ such thatBut $E_{12}^2=0$ and $E_{12}{E}_{21}=E_{11}$, so the right side is $c_1{E}_{11}$. Its $(2,2)$ entry is zero, whereas the $(2,2)$ entry of $E_{22}$ is $1_A\ne 0$. This contradiction proves that no such surjective ring homomorphism exists.

If the convention for rings allows $A$ to be the zero ring, that case must be excluded from the assertion: $M_2(0)=0$, and the zero ring homomorphism is surjective. This is a convention issue in the printed question.

Put $B=R/I$. By the definition of a left-primitive ideal, there is a faithful module $V$ that is a simple module on the left over $B$. The assumed relation passes to the quotient ring $B$, since any pair of elements of $B$ can be lifted to $R$.

By Schur's lemma, the endomorphism ring ${\mathrm{End}}_B(V)$ is a division ring: the kernel and image of a nonzero module endomorphism are submodules, so simplicity makes that module endomorphism both injective and surjective. SetEvaluation makes $V$ a right vector space over a division ring $\Delta$, and the $B$-action is $\Delta$-linear. The opposite ring in the definition ensures the correct order for this right action.

The Jacobson density theorem says that, for any finite $\Delta$-linearly independent family $v_1,\dots ,v_m$ and any $w_1,\dots ,w_m\in V$, some $b\in B$ satisfies $b{v}_i=w_i$ for every $i$.

Suppose that $V$ contains two $\Delta$-linearly independent vectors $v,w$. Applying the Jacobson density theorem twice, choose $a,b\in B$ satisfyingThe inherited relation giveswith integers $c_k$. Acting on $w$, the left side gives $baw=bv=w$, but every term on the right gives $a^{k}bw=0$. This contradicts $w\ne 0$. Thus ${\dim }_{\Delta }V=1$.

The ring homomorphismgiven by the action is injective because $V$ is a faithful module. It is surjective because the Jacobson density theorem, applied to a single nonzero basis vector, realizes every $\Delta$-linear endomorphism of this one-dimensional vector space over a division ring. Every nonzero such endomorphism is invertible. HenceThe use of the Jacobson density theorem here only prescribes actions on finitely many vectors; it does not assume that an arbitrary primitive ring is already a full endomorphism ring.

A multiplicative subset $S$ is a left Ore set if for every $s\in S$ and $r\in R$ there are $s^{\prime }\in S$ and $r^{\prime }\in R$ withEquivalently, the subsets $Sr$ and $Rs$ intersect for each such pair. This is the common-left-multiple condition needed for Ore localization with denominators on the left. In a noncommutative domain, no extra condition concerning annihilation by denominators is needed, since all members of $S$ are regular elements of a ring.

We first prove that any two nonzero principal left ideals $Ra$ and $Rb$ intersect nontrivially. Suppose instead that $Ra\cap Rb=0$. The nonzero left idealsthen form a direct sum. Indeed, take a finite relationIt impliesAs $R$ is a noncommutative domain and $a\ne 0$, this gives $u_0=0$. Right cancellation of the nonzero element $b$ reduces the remaining relation to one of the same form with one fewer term. Repetition gives $u_1=\cdots =u_m=0$, proving the asserted direct sum.

Consequently its finite partial sums give a strictly ascending chain of left ideals, contradicting the ascending chain condition in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$ whenever $a,b\ne 0$. This is the left Noetherian domain Ore lemma.

Now take $s\in S=R\setminus \{0\}$ and $r\in R$. If $r=0$, choose $s^{\prime }=1$ and $r^{\prime }=0$. Otherwise chooseSince $z\ne 0$, the coefficient $s^{\prime }$ is nonzero and hence belongs to $S$. This proves the left Ore set condition in all cases.

Use $s=x_1$ and $r=x_2$. A putative left Ore set equality would beThe left side is a single word ending in $x_2$, with coefficient $1$. The right side is a finite linear combination of words ending in $x_1$. Distinct words are linearly independent in the free associative algebra, so equality is impossible. ThereforeThe last letters matter because the denominators in the left Ore set equation occur on the right of the common left multiple.

View $\hat{R}$ as a noncommutative formal power series ring, and let $\epsilon :\hat{R}\to \mathbb{Q}$ take the coefficient of the empty word. This is a ring homomorphism, since the empty word has only the factorization $1\cdot 1$. In particular the two-sided ideal $I$ generated by $x_1,x_2$ lies in $\ker \epsilon$.

In fact $I=\ker \epsilon$. Every noncommutative formal power series with zero constant term can be partitioned according to the first letter of its words and writtenBoth summands belong to the two-sided ideal $I$.

For $a=1+h\in 1+I$, define the formal geometric seriesThis defines an element of $\hat{R}$: since $h$ has zero constant term, every word in $h^k$ has length at least $k$. Thus the coefficient of a fixed word of length $d$ receives contributions only from $0\le k\le d$, each a finite sum.

For each finite partial sum $g_N={\sum }_{k=0}^N(-h{)}^k$, associativity gives both identitiesIn every fixed word degree the last term vanishes once $N$ is large enough. It follows coefficient by coefficient that $ag=ga=1$. HenceThe argument is formal and uses no analytic convergence. More generally, every noncommutative formal power series with a nonzero rational constant term is a unit, after factoring out that scalar. This is the unit criterion for noncommutative formal power series.

Take $D=\hat{R}$. We first check that this noncommutative formal power series ring is a noncommutative domain. For nonzero $F,G\in D$, let $p,q$ be the smallest lengths of words with nonzero coefficients in $F,G$, respectively. Their degree-$p$ and degree-$q$ parts $F_p,G_q$ are nonzero homogeneous polynomials in the free associative algebra.

Choose words $u,v$ with nonzero coefficients in $F_p,G_q$. In $F_p{G}_q$, the coefficient of $uv$ is the product of those two coefficients: a word of length $p+q$ has a unique splitting into a prefix of length $p$ and a suffix of length $q$. This product is nonzero in $\mathbb{Q}$. The degree-$p+q$ part of $FG$ is exactly $F_p{G}_q$, since all lower parts of $F$ and $G$ vanish. Thus $FG\ne 0$, proving that $D$ is a noncommutative domain.

By the defining property of the free associative algebra, there is a ring homomorphismThe same substitution defines a ring automorphism of $R$, with inverse $x_1\mapsto x_1-1$, $x_2\mapsto x_2-2$. Composing this ring automorphism with the natural inclusion $R\hookrightarrow D$ proves that $f$ is injective.

Finally,Their constant terms are $3,-1,1,2$, respectively, all nonzero. By the unit criterion for noncommutative formal power series proved in c, all four elements are units. Therefore $D$ and $f$ have every required property. The embedding changes the images of the generators, so the failure of the original monomial left Ore set in b presents no obstruction.

Use the Weyl algebra convention $[y_i,x_j]={\delta }_{ij}$, with $[x_i,x_j]=[y_i,y_j]=0$, where the brackets are commutators. Its order filtration of a Weyl algebra isThus $x_i$ has degree zero and $y_i$ degree one. By the ordered monomial basis of a Weyl algebra, the associated graded ring iswhere ${\xi }_i$ is the symbol of $y_i$.

The Rees ring of a filtered algebra is the graded ringwhere $t$ is a central indeterminate. Here it is generated by $x_i$, $Y_i=y_{i}t$, and $t$, with degrees $0,1,1$, respectively, and relationsIt interpolates between the associated graded ring and the original filtered ring:

A good filtration for the order filtration on $M$ is an increasing exhaustive family of complex vector subspaces $(M_j{)}_{j\in \mathbb{Z}}$, zero for sufficiently negative $j$, with $F_{i}R{M}_j\subseteq M_{i+j}$, such that the associated graded moduleis a finitely generated module over $A$. Equivalently, there are finitely many $m_1,\dots ,m_s$ and integer shifts $d_1,\dots ,d_s$ such that $M_j={\sum }_{\ell =1}^s(F_{j-d_{\ell }}R)m_{\ell }$. In particular, choose any finite generating family $m_1,\dots ,m_s$ of $M$ and putThis is a good filtration for the order filtration.

For $M\ne 0$, its dimension of a Weyl algebra module for the order filtration isHere support of a module, annihilator of a module, and Krull dimension are taken over the commutative polynomial ring $A$. The corresponding algebraic set is the order characteristic variety of a Weyl algebra module. This dimension is independent of the chosen good filtration for the order filtration; one may set $d(0)=-\infty$.

The pieces $F_{j}R$ generally have infinite complex dimension, since they contain all polynomials in the $x_i$. Thus a polynomial-growth definition based directly on ${\dim }_{\mathbb{C}}{M}_j$ would be inappropriate for this order filtration of a Weyl algebra.

For the submodule $N\subseteq M$, define its subspace filtration and the quotient filtration on the quotient module $Q=M/N$ byThey are compatible filtrations of a module, since multiplication by $F_{i}R$ sends $N_j$ into $N_{i+j}$ and $Q_j$ into $Q_{i+j}$.

In each degree, inclusion and the quotient map induceThe first map is injective: if $n\in N_j$ belongs to $M_{j-1}$, then $n\in N\cap M_{j-1}=N_{j-1}$. The second map is surjective by the definition of $Q_j$.

For exactness in the middle, a class $m+M_{j-1}$ lies in $\ker {\pi }_j$ precisely when $m\in M_j$ and $m+N\in Q_{j-1}$. This means $m=m^{\prime }+n$ for $m^{\prime }\in M_{j-1}$ and $n\in N$. Since $m,m^{\prime }\in M_j$, one also has $n\in N_j$, andConversely, every class from $N_j$ maps to zero in $Q_j/Q_{j-1}$. Taking the direct sum over all degrees gives the short exact sequence of graded modulesThe maps respect the $\mathrm{gr}R$-action because they come from module homomorphisms compatible with the filtrations of a module. Consequentlyas $\mathrm{gr}R$-modules. This proves exactness of associated graded modules for induced filtrations.

Furthermore $A=\mathrm{gr}R$ is a Noetherian ring, and $\mathrm{gr}M$ is a finitely generated module. Its submodule $\mathrm{gr}N$ and its quotient module $\mathrm{gr}(M/N)$ are therefore finitely generated modules. Hence the induced subspace filtration and quotient filtration are both good filtrations for the order filtration.

Use the quotient filtration induced from the order filtration of a Weyl algebra. The key step is to identify the entire associated graded module of the left ideal $I$, rather than merely take symbols of its displayed generators.

For every member $x^{\alpha }{y}^{\beta }$ of the ordered monomial basis of a Weyl algebra, right multiplication by the pure $y$-monomials givesNo commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex linear span of the basis monomials satisfying eitherIntersecting this linear span with each filtration piece and then taking symbols therefore yields the monomial idealBy b,

To determine its Krull dimension, compute the radical of an ideal:Indeed, $K$ lies in the displayed intersection, an intersection of prime ideals. Conversely, $({\xi }_1{\xi }_2{)}^3\in K$ and $({\xi }_1{\xi }_3{)}^3\in K$, so both generators of that intersection lie in $\sqrt{K}$. The equality of the intersection with $({\xi }_1{\xi }_2,{\xi }_1{\xi }_3)$ follows by factoring out ${\xi }_1$ and reducing modulo $({\xi }_2,{\xi }_3)$.

The order characteristic variety of a Weyl algebra module in this case has two irreducible components:They are coordinate affine spaces of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$. The Krull dimension is the larger of these component dimensions. ThusIn particular, two displayed generators do not force a drop of two in dimension: their common factor creates the component ${\xi }_1=0$ of dimension $2n-1$.

In fact $J=R$, so the quotient module is zero. This requires an argument using left ideals: a left ideal is not generally closed under multiplication on the right.

Write $x=x_1$, $y=y_1$, so that $yx-xy=1$. The commutator product rule givesBoth $y^2{x}^2$ and $x^2{y}^2$ lie in $J=R{x}^2+R{y}^2$, so their differencebelongs to $J$. Multiplying on the left and subtracting another known element of the left ideal givesBecause the scalar $6$ is invertible in $\mathbb{C}$, it follows that $y\in J$, and hence $xy\in J$. Therefore $h-4xy=2\in J$, giving $1\in J$. We concludeThus the answer is yes for every $n\ge 1$. All ideal operations in the proof use left multiplication. In particular, simply replacing $J$ by the commutative ideal $(x_1^2,{\xi }_1^2)$ of the associated graded ring would miss the lower-order consequences that force $J=R$.