= Solution
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 $\iota(r)\iota(s)^{-1}$, and
$$
\ker\iota=\{r\in R:rt=0\text{ for some }t\in S\}.
$$
Equivalently, 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 $r\in R$ and $s\in S$, there are $t\in S$ and $b\in R$ with $rt=sb$; this is the <right Ore condition>.
* If $sr=0$ with $s\in S$, there is $t\in S$ with $rt=0$; this is right reversibility.
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 $ru=r'v$ 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 $a,b\ne0$. The sum
$$
bR+abR+a^2bR+\cdots
$$
is 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 $rt=sb\ne0$, with $t\ne0$. 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 $rs^{-1}$ has inverse $sr^{-1}$.
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 $rs^{-1}$ and hence the nonzero element $r=(rs^{-1})s\in I\cap R$. The same holds for any other nonzero <ideal> $K$. Since $R$ is prime, $(I\cap R)(K\cap R)\ne0$, so $IK\ne0$. 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>:
$$
\boxed{Q_{\mathrm{cl}}^r(R)\cong M_n(D)\quad\text{for a division ring }D.}
$$
For the requested counterexample to Noetherianity, take $R=k[x_1,x_2,\ldots]$. It is a commutative <integral domain> and has its <field of fractions> as its <classical right ring of quotients>. But $(x_1)\subsetneq(x_1,x_2)\subsetneq\cdots$ 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 \b[not] imply right Noetherianity.
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