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A semiprime ring has a classical right ring of quotients that is a semisimple ring and Artinian exactly when it is a right Goldie ring. In particular, the theorem applies to a semiprime ring that is a right Noetherian ring. It extends the fraction-field construction from commutative domains to rings whose quotient can be a finite product of matrix rings over division rings.

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.

A nonzero two-sided ideal of the classical right ring of quotients contains a nonzero element of the original ring: multiply a nonzero fraction by its regular right denominator. If the original ring is prime, the contractions of any two such ideals have nonzero product, so the quotient ring is prime too. The Goldie theorem makes it semisimple Artinian, and the Artin–Wedderburn theorem expresses it as a finite product of matrix rings over division rings. Primeness excludes more than one factor.

Suppose $R$ is a right Noetherian ring and every essential right ideal contains a regular element of a ring. Preimages of essential regular principal right ideals under left multiplication are essential, so they contain regular elements and prove the right Ore condition. In the classical right ring of quotients $Q$, every right ideal is generated by its contraction to $R$, making $Q$ right Noetherian. An essential right ideal of $Q$ contracts to an essential right ideal of $R$, hence contains a regular element and is all of $Q$. A maximal disjoint complement to any right ideal has essential sum with it, so every right ideal is complemented. The right regular module is therefore a finite direct sum of simple modules, making $Q$ semisimple Artinian.