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For $a\in R$, write $m_a$ for multiplication by $a$ in ${\mathrm{End}}_k(R)$. The algebraic differential operator ring is defined by the filtrationThe order of an algebraic differential operator is the least $j$ for which $\theta \in D_k^j(R)$; the zero operator may be assigned order $-\infty$. In degree zero, commuting with all multiplication maps means $\theta (a)=a\theta (1)$, so $D_k^0(R)=\{m_a:a\in R\}\cong R$. The commutator identity $[UV,m_a]=U[V,m_a]+[U,m_a]V$ proves by induction that $D_k^r{D}_k^s\subseteq D_k^{r+s}$, so the union is a ring under addition and composition.

The derivations of an algebra formEvery such derivation has $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, so has differential order at most one. Conversely, if $\theta \in D_k^1(R)$, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]$ is a multiplication map. Evaluating at $1$ shows it is $m_{\delta (a)}$; evaluating at $b$ gives the Leibniz rule. Hence the first-order algebraic differential operator decomposition isIt is a direct sum as $k$-vector spaces: a multiplication map that is a derivation vanishes at $1$, and is zero.

The complex Weyl algebra has generators $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ and relationsTake $n\ge 1$. If a nonzero unital module $M$ had finite dimension of a vector space over $\mathbb{C}$ $m$, its representing matrices would satisfy $[{\partial }_1,x_1]=I_m$. Taking the matrix trace gives $0=m$, impossible in characteristic zero. The Weyl algebra has no nonzero finite-dimensional modules. For $n=0$ the algebra is $\mathbb{C}$ and this assertion has the obvious exception.

For a nonzero finitely generated Weyl algebra module, use the Bernstein filtrationThe ordered monomials form a vector space basis, soTo justify the ordered basis, the relations move all ${\partial }_i$ to the right and give spanning. Independence follows in the action on $\mathbb{C}[x_1,\dots ,x_n]$: apply a putative zero operator ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x){\partial }^{\mathbf{b}}$ coefficientwise to the formal exponential $e^{t\cdot x}$, obtaining ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x)t^{\mathbf{b}}=0$. Thus every coefficient vanishes.

Choose a finite-dimensional nonzero generating subspace $M_0\subseteq M$ and set $M_j=F_j{A}_n{M}_0$. The associated graded module is finitely generated over the polynomial associated graded ring, so the Hilbert-Serre theorem implies ${\dim }_{\mathbb{C}}{M}_j$ is eventually a polynomial. DefineThis Bernstein growth dimension of a Weyl algebra module equals the Gelfand–Kirillov dimension of a module, not its ordinary vector-space dimension. If a different finite generating subspace is used, each lies in some fixed filtration step of the other. The corresponding $M_j$ are sandwiched between shifted filtrations, so the degree is unchanged.

We prove Bernstein inequality for Weyl algebra modules using an injectivity estimate. The faithful finite-step action of a Weyl algebra claim is that the mapis injective for every $j$. Induct on $j$. For $j=0$, a scalar annihilating $M_0\ne 0$ is zero. Suppose $a\in F_j{A}_n$ annihilates $M_j$. For any generator $z=x_i$ or ${\partial }_i$, the commutator $[a,z]$ belongs to $F_{j-1}{A}_n$ and annihilates $M_{j-1}$: both $a(zv)$ and $z(av)$ vanish. The induction hypothesis gives $[a,z]=0$ for every generator. In the ordered basis,In characteristic zero these identities force $a$ to be a scalar. Since it annihilates $M_0$, it is zero. This proves injectivity.

ConsequentlyThe left side grows as a positive constant times $j^{2n}$; the right side has growth degree $2d(M)$. Comparing powers provesThe polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with the usual multiplication and differentiation actions, has $d(M)=n$, showing the bound is sharp.

Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.

Put $R=k[x]$. The relation $yr=ry+r^{\prime }$ gives $R+Ry=R+yR$, so part (a) makes the Weyl algebra $A_1(k)$ left Noetherian. Applying the same argument to its opposite ring makes it right Noetherian.

Assume $\mathrm{char}k=0$ and let $0\ne I◃A_1(k)$. Using the PBW basis $x^i{y}^j$, choose an element of $I$ of least positive $y$-degree. Commutation with $x$ differentiates in $y$, so minimality leaves a nonzero polynomial in $x$. Repeated commutation with $y$ differentiates that polynomial and eventually gives a nonzero scalar. Hence $1\in I$, proving simplicity. In characteristic $p>0$, both $x^p$ and $y^p$ are central, and the proper ideal $(x^p)$ proves that $A_1(k)$ is not simple.

For a nonzero element of a Weyl algebra over a characteristic-zero field, choose a monomial $x^{\alpha }{D}^{\beta }$ of maximal total degree with coefficient $c_{\alpha \beta }\ne 0$. The displayed iterated commutators kill every lower-degree monomial. Another monomial of the same degree survives only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. Hence the result is a nonzero scalar. If the element has filtered degree at most $j$, expansion of these commutators gives a sum of terms $uav$ with both $u$ and $v$ of degree at most $j$. This proves faithful finite-step action of a Weyl algebra over any characteristic-zero field.

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.

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