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A module is faithful when its annihilator of a module is zero. Equivalently, the action ring homomorphism $R\to {\mathrm{End}}_{\mathbb{Z}}(M)$ to the endomorphism ring of its additive abelian group is injective: every nonzero scalar acts nontrivially somewhere on $M$. A primitive ring has a faithful simple module.

Over a commutative unital ring, an ideal maximal among annihilators of nonzero elements of a module is a prime ideal. Each such ideal is the annihilator of a module given by the cyclic submodule generated by that element. If $P=\mathrm{Ann}(m)$ and $ab\in P$ with $b\notin P$, then $bm\ne 0$ and $P\subseteq \mathrm{Ann}(bm)$. Maximality makes these annihilators equal, so $a\in P$. A Noetherian ring ensures that a nonzero module has a maximal element annihilator by the ascending chain condition on ideals.

The algebraic set defined by the annihilator of a module of the associated graded module for a good filtration for the order filtration is the order characteristic variety. It lies in affine space with coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$ and is invariant under scaling the ${\xi }_i$. Its dimension is the dimension of a Weyl algebra module for the order filtration. The order filtration of a Weyl algebra assigns degrees zero to the $x_i$ and one to the ${\xi }_i$; a characteristic variety formed with the Bernstein filtration need not be the same algebraic set.

Let $N$ be the nilradical of $R$. By b, the reduced locus of a ring isThe Noetherian ring hypothesis makes the ideal $N$ finitely generated; write $N=(n_1,\dots ,n_t)$. Fix $\mathfrak{p}\in \mathcal{R}$. For every $i$, the vanishing criterion in a module localization gives $s_i\notin \mathfrak{p}$ with $s_i{n}_i=0$. The product $f=s_1\cdots s_t$ lies outside the prime ideal $\mathfrak{p}$ and satisfies $fN=0$. When $N=0$, take $f=1$.

For every prime ideal $\mathfrak{q}$ in the distinguished open setthe element $f/1$ is a unit in $R_{\mathfrak{q}}$. The equality $fN=0$ therefore implies $N{R}_{\mathfrak{q}}=0$, so $R_{\mathfrak{q}}$ is reduced. Thus $\mathfrak{p}\in D(f)\subseteq \mathcal{R}$, proving that $\mathcal{R}$ is a Zariski-open subset of the spectrum of a ring.

More precisely, let $A={\mathrm{Ann}}_R(N)$, the annihilator of a module $N$. The argument shows that $N{R}_{\mathfrak{p}}=0$ if and only if some $f\in A$ lies outside $\mathfrak{p}$. HenceEquivalently, the complement is the support of a module $N$. This is vanishing of a finitely generated module near a prime applied to $N$: its finite generation lets the separately chosen denominators be combined into one element annihilating all of $N$.

A primitive ideal is the kernel of an algebraically irreducible unital representation, equivalently the annihilator of a module for a nonzero simple module. Irreducibility here excludes all proper nonzero invariant linear subspaces, not just closed ones. In a Banach algebra the simple modules can be represented as $A/L$ with $L$ a maximal left ideal. Such an ideal is closed: if its closure were all of $A$, it would contain an element close enough to $1$ to be invertible by the Neumann series, contradicting properness. Maximality then makes its closure itself. Thus the quotient carries a Banach-space norm and a bounded left action.

Let $P$ be a maximal proper two-sided ideal. The quotient ring $R=A/P$ is nonzero, unital and simple as a ring. By the Zorn lemma, choose a maximal left ideal $L$ of $R$. The simple module $R/L$ has an annihilator which is a proper two-sided ideal of $R$. Simplicity forces that annihilator to be zero. Its pullback to $A$ is exactly $P$, proving every maximal proper two-sided ideal is primitive. If one wants the Banach representation explicitly, realize the same module as the quotient of $A$ by the pullback maximal left ideal, which is closed by the preceding argument.

Assume $R$ is a nonzero unital ring and all its modules are unital. Order the proper left ideals by inclusion. A chain has an upper bound given by its union, which is still a proper left ideal: if it contained $1$, one member of the chain would contain $1$ and be all of $R$. The family is nonempty because it contains $0$. The Zorn lemma gives a maximal proper left ideal $L$. The nonzero quotient $R/L$ has no proper nonzero submodules, because its submodules correspond to left ideals between $L$ and $R$. Thus $R/L$ is a simple left module. The nonzero-ring qualification is necessary: the zero ring has no nonzero unital simple module.

Define the Jacobson radical $J(R)$ as the intersection of all maximal left ideals. It is equivalently the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module. To verify this equivalence, for a nonzero vector $v$ in a simple module $S$, the surjection $R\to S$, $r\mapsto rv$, has a maximal left-ideal kernel. Thus $x\in J(R)$ implies $xv=0$ for every $v$ in every simple module. Conversely, if $x$ annihilates all simple modules, it annihilates $1+L$ in every $R/L$, so $x\in L$. An annihilator of a module is a two-sided ideal; this description therefore proves that $J(R)$ is a two-sided ideal.

For the forward implication of the unit criterion for the Jacobson radical, first let $y\in J(R)$. If $R(1-y)$ were a proper left ideal, it would lie in a maximal one $L$. Both $y$ and $1-y$ would belong to $L$, an impossibility. Hence $c(1-y)=1$ for some $c\in R$. A left inverse alone does not yet establish that $1-y$ is a unit in a general noncommutative ring. However, $c=1+cy$, with $cy\in J(R)$, so the same argument gives $dc=1$ for some $d$. Multiplying $c(1-y)=1$ by $d$ on the left gives $d=1-y$. Therefore $(1-y)c=dc=1$ as well. This proves that $1-y$ is a unit.

For $x\in J(R)$, its two-sided-ideal property gives $axb\in J(R)$ for all $a,b$, so $1-axb$ is a unit. Conversely, if $x\notin J(R)$, choose a maximal left ideal $L$ with $x\notin L$. Then $L+Rx=R$, so $l+ax=1$ for some $l\in L$, $a\in R$. Hence $1-ax=l\in L$ is not a unit: if it were invertible, multiplying it on the left by its inverse would put $1$ in $L$. Taking $b=1$ contradicts the given unit property. We have provedFinally let $R$ be commutative and $I$ an ideal. The set $S=1+I$ contains $1$, andso it is a multiplicative subset. In the localization of a module $I_S\subseteq R_S$, any element is $i/s$ with $i\in I$ and $s\in S$. For any $r/t\in R_S$,Both $ts$ and $ts-ri$ lie in $S$, since $ts\in 1+I$ and $ri\in I$. The fraction is consequently a unit in $R_S$. In a commutative ring, the product of any two coefficients is again a single coefficient, so the proved unit criterion for the Jacobson radical applies to all expressions $1-a(i/s)b$. ThusThis proves localization away from one plus an ideal lies in the Jacobson radical. If $I=R$, then $0\in S$, $R_S=0$, and the inclusion holds trivially.

No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.

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.

A primitive ideal of a unital associative algebra is the annihilator of a module for a nonzero simple module. Every maximal proper two-sided ideal is primitive: its quotient is a simple ring, and any simple module over that quotient has zero annihilator there. In a unital Banach algebra, a simple module can be realized as $A/L$ for a maximal left ideal; this ideal is closed, making the representation a bounded algebraically irreducible action on a Banach space.

A primitive ideal of a unital associative algebra is the annihilator of a module for a nonzero simple module. Every maximal proper two-sided ideal is primitive: its quotient is a simple ring, and any simple module over that quotient has zero annihilator there. In a unital Banach algebra, a simple module can be realized as $A/L$ for a maximal left ideal; this ideal is closed, making the representation a bounded algebraically irreducible action on a Banach space.

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.