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A ring is right Goldie if it has finite right uniform dimension and satisfies the ascending chain condition on right annihilators. The left-sided definition reverses the multiplication. These two definitions need not be equivalent.

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 right Goldie ring satisfies the right-sided conditions in the definition of a Goldie ring. Every right Noetherian ring does: its right annihilators are right ideals, and its right regular module has finite uniform dimension.