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A central simple algebra over a field $k$ is a finite-dimensional associative $k$-algebra that is a simple ring and whose center is exactly $k$.

A regular element of a ring is an element $c$ for which both multiplication maps are injective:It is thus neither a left nor a right zero divisor; it need not be a unit. Here “ideals” means two-sided ideals. Put $A=A_n(\mathbb{C})$, and let $T_2(A)$ denote the specified upper triangular ring.

We first establish the additional facts about the Weyl algebra needed below. Its generators satisfyReordering gives a spanning set $x^{\alpha }{\partial }^{\beta }$. These monomials are independent. Indeed, represent the generators by multiplication and differentiation on $\mathbb{C}[x_1,\dots ,x_n]$. In a putative zero operator $P={\sum }_{\beta }{f}_{\beta }(x){\partial }^{\beta }$, choose $\beta$ of maximal total derivative degree. Iterated commutation with multiplication by the coordinates, ${\beta }_i$ times in coordinate $i$, gives multiplication by $\beta !f_{\beta }(x)$. All other derivative terms of that total degree disappear unless they have exactly the same multiindex, and lower-degree terms disappear. Thus $f_{\beta }=0$, and descending induction eliminates all coefficients. This proves the ordered monomial basis of a Weyl algebra and faithfulness of its polynomial action.

The Weyl algebra is a simple ring. In a nonzero two-sided ideal choose a nonzero element of least Bernstein filtration degree, where both $x_i$ and ${\partial }_i$ have degree one. Commuting it with any generator stays in the ideal and lowers that degree. Minimality therefore makes all these commutators zero. Commutation with every $x_i$ forces all derivative coefficients with $\beta \ne 0$ to vanish; commutation with every ${\partial }_i$ then forces all partial derivatives of the remaining polynomial to vanish. In characteristic zero the element is a nonzero scalar. The ideal consequently contains $1$. This proves simplicity of a Weyl algebra in characteristic zero. Its Jacobson radical is zero, since the radical is a proper two-sided ideal of a nonzero unital simple ring.

Use the two diagonal idempotents to isolate the entries of any ideal of $T_2(A)$. It must have the formwhere all three entries are two-sided ideals of $A$ and $I_{11}A+A{I}_{22}\subseteq I_{12}$. Conversely this condition makes the indicated set an ideal. Simplicity says that each $I_{ij}$ is $0$ or $A$, and either nonzero diagonal entry forces $I_{12}=A$. Hence the five ideals of $R_1$ areFor a two-sided ideal of $M_2(A)$, multiplication on both sides by elementary matrices isolates any entry and moves it to any matrix position. The entries obtained form one two-sided ideal $I$ of $A$, and the original ideal is $M_2(I)$. Thus the only ideals of $R_2$ are $0$ and $R_2$.

Since $N^2=0$, $N$ lies in the Jacobson radical of $R_1$: for $z\in N$ and $r\in R_1$, $zr\in N$ and $1-zr$ has inverse $1+zr$. On the other hand, $R_1/N\cong A\times A$ has zero radical, so the image of the radical in this quotient is zero. The elementary unit criterion for the Jacobson radical justifies both assertions. ThereforeThe second equality follows from simplicity of $R_2$.

We now construct the denominators used to describe regular matrices. A left Noetherian noncommutative domain satisfies the left Ore condition. For if $Aa\cap Ab=0$ for two nonzero elements, the left ideals $Ab,Aba,Ab{a}^2,\dots$ have direct sum: in a relation the first summand lies in $Ab$, all the others in $Aa$, so it vanishes; cancellation of the final $a$ repeats the argument. Their finite partial sums form a strictly ascending chain, contradicting left Noetherianity. Thus $Aa\cap Ab\ne 0$, giving common nonzero left multiples and the required Ore equations. The analogous argument with right ideals gives the right Ore condition under right Noetherianity. Both hypotheses for $A$ are allowed in the question.

The Ore localization therefore embeds $A$ in a division ring $Q$. Briefly, its elements are left fractions $a^{-1}b$, with nonzero $a$, and two fractions are identified after passing to a common left denominator. Common multiples supply addition and multiplication, while cancellation in the domain makes the embedding injective. A nonzero fraction has an inverse, obtained by interchanging its numerator and denominator in the appropriate Ore fraction calculation. The right fraction construction gives the same division ring, so finite lists have common left and common right denominators.

A triangular matrix $c=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ with $r,t\ne 0$ is invertible over $T_2(Q)$, with inverseIt is consequently regular over $T_2(A)$. If $r=0$, then $c{E}_{11}=0$; if $t=0$, then $E_{22}c=0$. Hence

A full matrix is regular exactly when it is invertible over $Q$:One implication follows immediately from invertibility. Conversely, a singular matrix over a division ring has a nonzero right kernel column and a nonzero left kernel row, by row reduction. A common right denominator turns the column into a nonzero column over $A$, providing a right annihilator matrix over $A$. A common left denominator does the same for the row and gives a left annihilator. Thus a singular matrix is not regular. Explicitly, for $c=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ with $r\ne 0$, row elimination says that regularity is equivalent to $u-t{r}^{-1}s\ne 0$ in $Q$. If $r=0$, it is equivalent to both $s\ne 0$ and $t\ne 0$. A commutative determinant formula must not be substituted here.

Finally both classical left rings of quotients exist, andTo prove this directly, clear a common left denominator $a\ne 0$ from all the finitely many entries of a matrix over $Q$. It becomes $(a{I}_2{)}^{-1}b$ with $b$ in the original triangular or full ring and $a{I}_2$ regular. Every original regular element becomes a unit in the corresponding ring over $Q$, by the descriptions above. For $c$ regular and $r$ in the original ring, write $r{c}^{-1}=d^{-1}b$ in this way; then $dr=bc$, the left Ore condition. Since the denominators are regular, the additional annihilator condition for localization is automatic. This proves the matrix and triangular localization over an Ore domain assertions. In particular, the nonzero nilpotent radical of the triangular ring does not prevent its classical localization; that localization retains the triangular radical.

For $n\ge 1$, the Weyl algebra $A=A_n(\mathbb{C})$ is generated by $x_1,\dots ,x_n,D_1,\dots ,D_n$ with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. Iterating the differential Ore extension construction gives the ordered basis $x^{\alpha }{D}^{\beta }$. In the Bernstein filtration, both kinds of generator have degree one, andLeading symbols show that $A$ is a noncommutative domain. The Hilbert basis theorem and homogeneous-generator lifting by decreasing degree make it both left and right Noetherian.

It is also a simple ring. In a nonzero two-sided ideal, choose a nonzero element and a monomial $x^{{\alpha }_0}{D}^{{\beta }_0}$ of maximal total degree with nonzero coefficient $c$. Applying $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ eliminates every lower-degree term and every different term of that maximal degree. The result is the nonzero scalar $(-1{)}^{|{\beta }_0|}{\alpha }_0!{\beta }_0!c$. Thus the ideal contains $1$. Here the scalar extraction by iterated Weyl commutators uses characteristic zero. Despite simplicity, $A$ has no nonzero finite-dimensional modules: in such a representation $[D_1,x_1]=I$ would have trace $0=\dim M$, impossible over $\mathbb{C}$. Its interesting module theory is necessarily infinite-dimensional.

For a finitely generated left module $M$, choose a finite-dimensional generating subspace $M_0$ and set $M_j=F_{j}A{M}_0$. This is a good filtration of a module. Its associated graded module is finite over the graded polynomial ring, so its cumulative Hilbert function is eventually a polynomial,with positive integer multiplicity $e(M)$. These are the Bernstein growth dimension of a Weyl algebra module and its multiplicity of a filtered module. Two good filtrations bound each other after fixed shifts: lift finite homogeneous generators in one filtration, bound their degrees in the other, and use compatibility with $F_{j}A$; then reverse the roles. The growth degree and leading coefficient therefore do not depend on the good filtration of a module. The degree is also the Gelfand–Kirillov dimension of a module and the dimension of the characteristic variety of a Weyl algebra module, the support of $\mathrm{gr}M$. Since $M$ is a quotient of a finite sum of copies of $A$, $d(M)\le 2n$.

The essential lower bound is the Bernstein inequality for Weyl algebra modules:Here is a full proof. Multiplication gives a linear map $F_{j}A\to {\mathrm{Hom}}_{\mathbb{C}}(M_j,M_{2j})$. Prove it injective by induction on $j$. At $j=0$ a scalar killing $M_0\ne 0$ must vanish. If $a\in F_{j}A$ kills $M_j$, then for every generator $z$, its commutator $[a,z]\in F_{j-1}A$ kills $M_{j-1}$: both $azm$ and $zam$ are zero for $m\in M_{j-1}$. Induction gives $[a,z]=0$ for all generators. In the ordered basis, $[D_i,a]$ differentiates the $x_i$ exponents and $[x_i,a]$ differentiates the $D_i$ exponents with a minus sign. Vanishing of all these commutators in characteristic zero forces $a$ to be a scalar. It then vanishes because it kills $M_0$. This proves the faithful finite-step action of a Weyl algebra and henceComparison of degrees gives $2n\le 2d(M)$, proving the inequality. Equally, without invoking eventual polynomiality, this estimate directly bounds the Gelfand–Kirillov dimension of a module below by $n$. Right modules satisfy the same inequality through the filtration-preserving isomorphism $A\cong A^{\mathrm{op}}$, $x_i\mapsto x_i$, $D_i\mapsto -D_i$.

A nonzero finitely generated module with $d(M)=n$ is a holonomic Weyl algebra module; the zero module is conventionally included. This minimal-growth class is stable under submodules, quotient modules and extensions. For $0\to N\to M\to Q\to 0$, take the induced and quotient filtrations. Their graded sequence is exact; Noetherianity of $\mathrm{gr}A$ makes the induced filtrations good. Thus the Hilbert functions add andif both nonzero terms have the top dimension, their multiplicities add. For a holonomic Weyl algebra module, every nonzero subquotient has dimension $n$, by the upper bound inherited from $M$ and Bernstein's lower bound. Each nonzero factor in a strict chain therefore consumes positive integral multiplicity. This proves the multiplicity bounds the length of a holonomic Weyl module:In particular a holonomic module is both Noetherian and Artinian and has finite length, although its vector-space dimension is infinite. This distinction separates finite-dimensional representation theory from finite-length Weyl module theory.

The polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with $x_i$ acting by multiplication and $D_i$ by differentiation, is generated by $1$. Its $M_j$ consists of polynomials of degree at most $j$, with dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$, so $d=n$ and $e=1$. It is simple: differentiating a maximal-degree monomial of a nonzero polynomial isolates a nonzero constant, and multiplication by polynomials then generates the whole module. A second basic example is the delta moduleUsing the reverse ordered basis $D^{\beta }{x}^{\alpha }$, its vector space is $\mathbb{C}[D_1,\dots ,D_n]$; $D_i$ acts by multiplication and $x_i$ by negative differentiation. The same argument proves simplicity and holonomicity. These two examples are exchanged by the automorphism $x_i\mapsto D_i$, $D_i\mapsto -x_i$. Free modules, in contrast, have growth dimension $2n$ and are not holonomic. Thus the central organizing invariants are good filtration of a module, growth dimension, characteristic support and multiplicity, with Bernstein's estimate explaining why dimension $n$ is the natural smallest nonzero size.

This part proves $(a)\Rightarrow (b)$. If $M\ne 0$ were a finite-dimensional unital module, its action would give a homomorphismThe kernel is a two-sided ideal and is proper because the identity acts as the identity on $M$. A simple ring therefore makes the homomorphism injective. This is impossible: $A_n(k)$ has the infinitely many linearly independent ordered monomials established in part (a), whereas the endomorphism algebra has dimension $({\dim }_{k}M{)}^2$. Thus simplicity excludes every nonzero finite-dimensional module.

There is also a direct characteristic-zero check: the relation $[D_1,x_1]=1$ acts as a matrix identity. Taking its trace gives $0={\dim }_{k}M$ in $k$. In characteristic zero this forces $M=0$. This agrees with the Weyl algebra has no nonzero finite-dimensional modules criterion, but the injectivity argument proves the requested implication without presupposing characteristic zero.