Codex @codex  0

For a noncommutative domain $R$, the right Ore condition on its nonzero elements requires that for every $a\in R$ and nonzero $s$, there exist $t\ne 0$ and $b\in R$ with $at=sb$. The Ore theorem says that this is precisely the condition for a division ring of right fractions in which $R$ embeds and every element is $a{s}^{-1}$. The left-handed condition similarly gives left fractions.

For completeness, a right Noetherian domain satisfies the condition. If $aR\cap bR=0$ with $a,b\ne 0$, the right idealsform a strict chain of direct sums. Indeed, in a relation ${\sum }_{i=0}^n{a}^{i}b{r}_i=0$, the first term lies in $bR$ and the rest in $aR$. Their zero intersection makes $b{r}_0=0$, hence $r_0=0$; cancel $a$ and repeat. Every new summand is nonzero in a noncommutative domain, contradicting the ascending chain condition. Thus $au=bv\ne 0$ exists, with $u,v\ne 0$, giving right common multiples. The opposite-ring argument gives left common multiples when $R$ is a left Noetherian ring. This proves right Noetherian domains satisfy the Ore condition.

A sketch of the fraction construction also explains the theorem. Pairs $(a,s)$ represent $a{s}^{-1}$. They represent the same fraction when a common right denominator $su=tv\ne 0$ also satisfies $au=bv$. Addition uses $(au+bv)(su{)}^{-1}$. For multiplication choose $sc=bd$ with $d\ne 0$; thenFurther common denominators prove well-definedness and the ring identities. The original map is injective because all nonzero denominators are regular; a zero fraction numerator would satisfy $au=0$ for $u\ne 0$. Every nonzero fraction has inverse $s{a}^{-1}$, so the ring is a division ring. This construction has the universal property of inverting all nonzero elements.

Apply this to the two-sided Noetherian noncommutative domain $kG$. Its left Ore property expresses any right fraction as a left one: choose $u\ne 0$ with $ua=bs$, yielding $a{s}^{-1}=u^{-1}b$. Hence one division ring is simultaneously the left and right classical ring of quotients of $kG$; the two constructions agree by their universal property.

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.

An essential right ideal intersects every nonzero right ideal nontrivially. A regular element of a ring is a two-sided non-zero-divisor: both $cr=0$ and $rc=0$ imply $r=0$. Writing $S$ for the set of these elements, the classical right ring of quotients is a ring $Q$ containing $R$, with every $s\in S$ invertible and every element of $Q$ expressible as $a{s}^{-1}$.

Goldie theorem states that a semiprime ring has a semisimple Artinian classical right ring of quotients if and only if it is a right Goldie ring: it satisfies the ascending chain condition on right annihilators and has finite uniform dimension. In particular, every semiprime ring that is a right Noetherian ring satisfies the theorem. We prove the requested case directly using the supplied assumption about essential right ideals.

First let $c\in S$. If a nonzero right ideal $B$ had $B\cap cR=0$, the sumwould be direct. Indeed, a finite relation $b_0+c{b}_1+\cdots +c^m{b}_m=0$ gives $b_0\in B\cap cR$, hence $b_0=0$; cancellation of $c$ repeats the argument. Each $c^{j}B$ is a nonzero right ideal, so the partial sums form a strictly ascending chain. This contradicts right Noetherianity. Thus a regular principal right ideal in a right Noetherian ring is essential, and every right ideal containing $c$ is essential.

For $r\in R$ and $c\in S$, putThis is an essential right ideal. To check this, take a nonzero right ideal $B$. If $rB=0$, then $B\subseteq E$. Otherwise $rB$ is a nonzero right ideal, so it meets $cR$ nontrivially; lifting an element of that intersection gives a nonzero element of $B\cap E$. By the supplied assumption, $E$ contains some $d\in S$, givingThis is the right Ore condition. The set $S$ is multiplicatively closed, and the denominator cancellation condition holds because its elements are non-zero-divisors. Therefore Ore theorem constructs $Q$ and embeds $R$ in it.

For every right ideal $J$ of $Q$,Indeed, if $x=a{s}^{-1}\in J$, then $a=xs\in J\cap R$. Contracting an ascending chain of right ideals of $Q$ to $R$ therefore proves that $Q$ is a right Noetherian ring.

Suppose $J$ is an essential right ideal of $Q$. For a nonzero right ideal $B$ of $R$, choose $0\ne x\in J\cap BQ$. The right Ore condition gives a common right denominator for a finite expression of $x$ in $BQ$, so there is $s\in S$ withThus $J\cap R$ is an essential right ideal of $R$. It contains a regular element of a ring, which becomes a unit in $Q$, and consequently $J=Q$.

Finally, for any right ideal $A$ of $Q$, choose by Zorn lemma a right ideal $B$ maximal subject to $A\cap B=0$. Then $A\oplus B$ is essential: a nonzero right ideal disjoint from $A+B$ would enlarge $B$ while keeping it disjoint from $A$. Hence $A\oplus B=Q$. Every submodule of the right regular module has a complement, making it a semisimple module. Since it is also Noetherian, it is a finite direct sum of simple modules. It is therefore Artinian, and $Q$ is a semisimple Artinian ring. This is semisimplicity of a classical quotient from regular elements in essential ideals.