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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.

Use generators $x,D$ with $Dx-xD=1$. The ordered monomials $x^i{D}^j$ form a basis: construct the Weyl algebra as the differential Ore extension of $\mathbb{C}[x]$ with $Df=fD+f^{\prime }$, which has unique ordered polynomial expressions. Giving both generators degree one produces the Bernstein filtration, whose associated graded ring is $\mathbb{C}[x,\xi ]$. If $a,b\ne 0$, their nonzero leading symbols have nonzero product in this polynomial integral domain. Thus $ab$ has nonzero leading symbol and cannot vanish. Hence $A_1(\mathbb{C})$ has no nontrivial zero divisors.

Here is the left-handed Ore theorem. Let $S$ be a multiplicative set of regular elements of a ring, containing $1$ and excluding $0$. There is an injective Ore localization whose elements are left fractions $s^{-1}a$ if and only ifThis is the left Ore condition. For a noncommutative domain, taking $S=A\setminus \{0\}$ gives a division ring exactly when this condition holds. With zero-divisor denominators a further left reversibility condition is needed; the regular-denominator statement is the one used here.

Necessity follows by writing $a{s}^{-1}=t^{-1}b$ in the quotient and multiplying by $t$ and $s$, giving $ta=bs$ in the embedded original ring. For a construction proving sufficiency, use the left Ore fraction construction: pairs $(s,a)$ represent $s^{-1}a$, and $(s,a)$ and $(t,b)$ are equivalent whenfor suitable $u,v\in A$. Reflexivity and symmetry are immediate. For transitivity, combine $us=vt=w$ and $u^{\prime }t=v^{\prime }q=w^{\prime }$ by a common left multiple $hw=k{w}^{\prime }\in S$. Then $(hv-k{u}^{\prime })t=0$, so regularity gives $hv=k{u}^{\prime }$, which combines the numerator equalities and proves transitivity.

Addition uses the common denominator $w$, with numerator $ua+vb$. To multiply $s^{-1}a$ by $t^{-1}b$, choose $v\in S,c\in A$ with $va=ct$; setThe Ore condition supplies a common refinement whenever a representative or an Ore relation is changed, and regularity cancels the shared denominator, proving independence of choices. The same refinements verify the ring axioms; the formulas are simply addition and multiplication after clearing left denominators. The map $a\mapsto (1,a)$ is injective, since equivalence to zero would give $ua=0$ with $u\in S$. Each $s\in S$ becomes invertible. A homomorphism inverting $S$ extends uniquely by $(s,a)\mapsto f(s{)}^{-1}f(a)$, giving the universal property. For all nonzero denominators in a noncommutative domain, a nonzero fraction has inverse $(s^{-1}a{)}^{-1}=a^{-1}s$, so the quotient is a division ring. This is the requested proof sketch, with the fraction equivalence and multiplication made explicit.

It remains to check the left Ore condition for $A_1(\mathbb{C})$. Its graded polynomial ring is Noetherian by the Hilbert basis theorem. Lifting homogeneous generators of a graded left ideal and successively cancelling leading symbols proves that the filtered algebra is a left Noetherian ring: each subtraction lowers the nonnegative degree, so it terminates.

In any left Noetherian noncommutative domain, nonzero principal left ideals intersect. Otherwise $Ab\cap Aa=0$ with $a,b\ne 0$ would makean infinite direct sum of nonzero left ideals. To verify directness, reduce a finite relation modulo $Aa$ to make its first summand zero, then cancel $a$ on the right and repeat. Its increasing finite partial sums contradict the ascending chain condition. A nonzero element of $Aa\cap As$ therefore has the form $ta=bs\ne 0$, with $t\ne 0$, proving the required condition. Consequently $A_1(\mathbb{C})$ has a classical ring of left quotients, and it is a division ring.

A right Ore localization at $S$ is a unital homomorphism $\iota :R\to Q$ such that every $\iota (s)$, $s\in S$, is a unit, every element of $Q$ is $\iota (r)\iota (s{)}^{-1}$, andEquivalently, the usual fraction construction has this zero criterion and is universal among homomorphisms that invert $S$. Injectivity is required only when the denominators are regular elements of a ring. If $0\in S$, every homomorphism inverting $S$ has zero target; this is the degenerate zero-ring localization. For a nonzero localization we require $0\notin S$.

The necessary and sufficient conditions are that $S$ be a right denominator set:

For necessity, write the element $\iota (s{)}^{-1}\iota (r)$ as $\iota (b)\iota (t{)}^{-1}$. Multiplying by $\iota (s)$ on the left and $\iota (t)$ on the right gives $\iota (rt-sb)=0$. The zero criterion supplies $u\in S$ with $(rt-sb)u=0$. Hence $r(tu)=s(bu)$, the right Ore condition in $R$, not merely an equality of images in $Q$. Also $sr=0$ makes $\iota (r)=0$ because $\iota (s)$ is invertible, and the zero criterion gives $rt=0$. This proves both necessities.

The sufficiency is the Ore theorem for denominator sets. Concretely, right fractions can be represented by $(r,s)$; a common-denominator equality has $su=tv\in S$ and $ru=r^{\prime }v$ for suitable $u,v\in R$. The two conditions make the induced fraction equivalence transitive and make addition and multiplication well defined; the zero fraction is exactly the kernel displayed above. The requested direction of the theorem was necessity, for which the complete argument has just been given.

For a right Noetherian domain, any two nonzero principal right ideals intersect. To see this without using the desired localization, suppose $bR\cap aR=0$ with $a,b\ne 0$. The sumis direct: reduce a finite relation modulo $aR$ to get its first summand zero, then cancel $a$ in the noncommutative domain and repeat. Every summand is nonzero, so the finite partial sums form a strictly ascending chain of right ideals, contradicting the ascending chain condition. A nonzero intersection $rR\cap sR$ thus provides $rt=sb\ne 0$, with $t\ne 0$. This is the right Ore condition for $S=R\setminus \{0\}$. Reversibility is automatic in a noncommutative domain, and the zero criterion makes the map injective. The localization is a division ring, since a nonzero $r{s}^{-1}$ has inverse $s{r}^{-1}$.

The Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. A right Noetherian ring satisfies the annihilator condition because right annihilators are right ideals. Its right regular module has finite uniform dimension: an infinite direct sum of nonzero submodules would give a strictly ascending chain of finite partial sums. A prime ring is semiprime, so the theorem applies to a prime right Noetherian ring $R$.

To obtain a single matrix-ring factor, not just a finite product, prove that its quotient $Q$ is prime. A nonzero two-sided ideal $I\subseteq Q$ contains a nonzero fraction $r{s}^{-1}$ and hence the nonzero element $r=(r{s}^{-1})s\in I\cap R$. The same holds for any other nonzero ideal $K$. Since $R$ is prime, $(I\cap R)(K\cap R)\ne 0$, so $IK\ne 0$. Thus $Q$ is prime. The Artin–Wedderburn theorem gives a finite product of matrix rings over division rings; primeness excludes two different nonzero factors with zero product. This proves the prime classical quotient of a prime right Goldie ring:For the requested counterexample to Noetherianity, take $R=k[x_1,x_2,\dots ]$. It is a commutative integral domain and has its field of fractions as its classical right ring of quotients. But $(x_1)⊊(x_1,x_2)⊊\cdots$ is a strictly ascending ideal chain: evaluation at $x_1=\cdots =x_m=0$ shows $x_{m+1}$ is not in the preceding ideal. Therefore existence of a classical quotient does not imply right Noetherianity.

The mod-$p$ Iwasawa algebra is the completed group algebrawhere $U$ runs through open normal subgroups and the transition maps come from quotient maps. This is the mod-$p$ algebra, distinguished from the integral Iwasawa algebra ${\mathbb{Z}}_p[[K]]$.

Use the stated complete separated negative filtered ring structure, writing $F_{-n}={\mathfrak{m}}^n$ and $\nu (x)=\max \{n:x\in F_{-n}\}$ for a nonzero element. In the usual odd-prime congruence-group situation,The dimension-three assertion records the three independent entries of a trace-zero matrix; the argument below only needs the given graded polynomial-algebra property. Completeness and separatedness are with respect to this filtration and follow from the inverse-limit topology for the standard filtration. If $x,y\ne 0$, their leading symbols are nonzero, so their product in the graded domain is nonzero. Therefore $\nu (xy)=\nu (x)+\nu (y)$ and $xy\ne 0$. This proves the domain assertion by filtered-graded transfer for complete rings.

For a left ideal $I$, its leading symbols form a homogeneous left ideal $\mathrm{gr}I$ of the polynomial ring. The Hilbert basis theorem gives finitely many homogeneous generators; choose lifts $a_1,\dots ,a_s\in I$. Given $x\in I$, express its leading symbol using these generators, lift the homogeneous coefficients, and subtract the resulting $\sum r_j{a}_j$. The remainder is still in $I$ and has strictly larger valuation. Repeat. At a stage where the remainder has valuation $n$, the coefficient contribution for $a_j$ has valuation at least $n-\nu (a_j)$. Thus each accumulated coefficient is a Cauchy series, converging to an element of ${\Omega }_K$. The remainder tends to zero, and separatedness gives $x={\sum }_j{r}_j{a}_j$. Hence $I$ is algebraically generated by these finitely many lifts; no prior assumption that $I$ is closed was used. This proves $\boxed{{\Omega }_K\text{ is left Noetherian and a domain}}$ under the stated filtration hypothesis. The same proof for right ideals also gives right Noetherianity.

For the classical fractions, a left Noetherian domain satisfies the left Ore condition. Here is the common-multiple argument. If two nonzero principal left ideals $Ra,Rb$ had zero intersection, the ideals $Ra{b}^i$, $i\ge 0$, would form a direct sum: from a finite relation, its first term lies in $Ra\cap Rb$, and then cancellation of the nonzero $b$ proves the assertion inductively. Their partial sums give an infinite ascending chain, impossible in a left Noetherian ring. Thus $Ra\cap Rb\ne 0$, giving common left multiples for every two nonzero denominators. The right-handed argument also applies. Ore localization at all nonzero elements therefore produces the classical division ring of fractions, with expressions $s^{-1}a$ or $b{s}^{-1}$; every nonzero fraction is invertible.

There is a genuine prime restriction behind the auxiliary hypothesis. For $p=2$, the matrix $z=-I$ is a nonidentity element of $K$ of order two. In ${\mathbb{F}}_2[[K]]$, the nonzero element $z-1$ satisfies $(z-1{)}^2=z^2-2z+1=0$. It is nonzero already in a finite quotient separating $z$ from one. Thus the actual $p=2$ algebra is not a domain, and cannot have the supplied separated graded-domain filtration. The intended conclusion applies for odd $p$; at $p=2$ it instead applies to a suitable deeper torsion-free congruence subgroup, such as the kernel modulo four.

Use $q\in k^{\times }$ and $YX=qXY$ for the quantum plane. Its monomials $X^i{Y}^j$, $i,j\ge 0$, form a basis, with multiplicationOne way to verify the basis assertion without assuming it is to define this multiplication on the vector space with the displayed formal basis. This defines an associative algebra: for a third monomial $X^w{Y}^z$, the two products of three monomials have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the lexicographic order. The largest monomials of two nonzero finite sums give the uniquely largest monomial of their product, with coefficient $cd{q}^{ju}\ne 0$. Therefore the quantum plane is a domain. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.

A uniform module is a nonzero module in which any two nonzero submodules have nonzero intersection. Suppose the right regular module of a right Noetherian domain $A$ were not a uniform module. Choose nonzero $a,b$ from two right ideals with zero intersection. Then $aA\cap bA=0$. The right idealsform a direct sum. For if ${\sum }_{i=0}^m{b}^{i}a{c}_i=0$, then $a{c}_0\in aA\cap bA$ is zero, so $c_0=0$ by the noncommutative domain property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of right ideals. This contradicts the ascending chain condition of a right Noetherian ring. Hence $A$ is a uniform right module.

It follows that $aA\cap sA\ne 0$ whenever $a,s\ne 0$: there are nonzero $u,v$ with $au=sv$. This is the right Ore condition for the multiplicative set $S=A\setminus \{0\}$; zero numerators cause no difficulty. The Ore localization theorem therefore constructs the ring of right fractionsThe map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a noncommutative domain. To make the denominator convention concrete, if $su=tv\ne 0$ thenFor multiplication, choose $bu=sc$ with $u\ne 0$; thenCommon right multiples make these operations independent of the chosen representatives. Every nonzero $a{s}^{-1}$ has inverse $s{a}^{-1}$, so $Q$ is a division ring. The original field $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a division algebra over $k$. No commutative fraction field construction is being assumed.

A multiplicative set $S$ containing $1$ and not $0$ is a right denominator set if it satisfies the right Ore condition and right reversibility: $sr=0$, with $s\in S$, implies $rt=0$ for some $t\in S$. Its Ore localization consists of right fractions $r{s}^{-1}$ and has kernel $\{r:rt=0\text{ for some }t\in S\}$. This condition permits zero-divisor denominators; it is automatic for a set of regular elements of a ring. When all denominators are regular, the localization map is injective.

A noncommutative domain whose right ideals satisfy the ascending chain condition is a right Noetherian domain. Its right regular module is a uniform module, so any two nonzero principal right ideals intersect. Thus its nonzero elements satisfy the right Ore condition and it embeds in a division ring by Ore localization.

A multiplicative subset $S$ of a ring is a right Ore set if $1\in S$, $0\notin S$ and it satisfies the right Ore condition. In a noncommutative domain, this condition lets nonzero denominators be used in Ore localization.