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 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.

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.

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.

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.

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.

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.

No. The denominator test from part ii is satisfied here, so a stronger obstruction is needed. Put $a_i=v(\{i\})$ and $b_{ij}=v(\{i,j\})$. In a three-player simple cooperative game, the Shapley value formula gives, for distinct $i,j,k$,Subtract the corresponding expression for ${\varphi }_j$:Each coalition value is zero or one, so this difference must be an integer multiple of $1/2$. This is the obstruction from pairwise Shapley differences in a three-player simple game. The proposed first two coordinates differ by $1/6$, which is not an integer multiple of $1/2$. Therefore the vector cannot occur, even though each coordinate separately passes the $1/6$ denominator test.

Yes. Give players $1,2,3$ weights $1,1,2$, respectively, and let the quota be three. This weighted voting game is a simple cooperative game whose minimal winning coalitions are $\{1,3\}$ and $\{2,3\}$. In orders $123,132,213,231$, player 3 supplies the winning marginal contribution. In order $312$ player 1 does, and in order $321$ player 2 does. Averaging the six marginal contribution vectors gives exactly $(1/6,1/6,4/6)$ as the vector of Shapley values.

No. In a simple cooperative game, a player's marginal contribution along any permutation is zero or one. With three players there are six equally likely permutations, so every Shapley value is an integer multiple of $1/6$, by the Shapley value denominator in a simple game. But $6(5/12)=5/2$ and $6(3/12)=3/2$ are not integers. The vector sums to one, but fails this necessary denominator condition.

Yes. Use the simple cooperative game in which a coalition wins exactly when it contains at least two players. In every permutation, precisely the second arrival changes a losing coalition into a winning one. Each player is second in two of the six permutations, so each Shapley value is $2/6=1/3$. This constructs the required vector while verifying that singleton coalitions lose and the grand coalition wins.

The core is empty. Efficiency would require $x_1+x_2+x_3=v(N)=1$. Each two-player coalition has value $1$, so core membership would also requireAdding gives $2(x_1+x_2+x_3)\ge 3$, contradicting efficiency, which makes the left side $2$. Thus no allocation can meet all coalition demands. The obstruction uses only pair coalitions and the grand coalition; imposing individual rationality cannot remove it. The equal allocation $(1/3,1/3,1/3)$ is efficient but gives each pair only $2/3<1$, so symmetry does not supply a core allocation.

The core is the singleton $\{(1/2,1/2,1/2,1/2)\}$. Here the characteristic function of a coalitional game is $v(C)=\lfloor |C|/2\rfloor$, the two-person instance of the core of the miners game calculation.

Let $x$ be in the core. Efficiency requires $x_1+x_2+x_3+x_4=2$, and every pair coalition requires $x_i+x_j\ge 1$. Summing the six pair inequalities counts each coordinate three times, giving $3{\sum }_i{x}_i\ge 6$. Efficiency makes this an equality; since every individual slack is nonnegative, all six pair inequalities must be equalities. In particular $x_1+x_2=x_1+x_3=x_1+x_4=1$ forces $x_2=x_3=x_4$, and comparison with $x_2+x_3=1$ gives every $x_i=1/2$.

Conversely, for the uniform allocation every coalition satisfiesand its total is $2=v(N)$. It is therefore a core member, completing both existence and uniqueness. In particular, a three-player coalition receives $3/2\ge 1$; the pair constraints are the tight ones.