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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.