= Solution
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 $as^{-1}$.
<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 sum
$$
B+cB+c^2B+\cdots
$$
would be direct. Indeed, a finite relation $b_0+cb_1+\cdots+c^mb_m=0$ gives $b_0\in B\cap cR$, hence $b_0=0$; cancellation of $c$ repeats the argument. Each $c^jB$ 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 $c\in S$, put
$$
E=\{x\in R:rx\in cR\}.
$$
This 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 $d\in S$, giving
$$
rd=cb\qquad\text{for some }b\in R.
$$
This 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 $Q$,
$$
J=(J\cap R)Q.
$$
Indeed, if $x=as^{-1}\in J$, 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 $s\in S$ with
$$
0\ne xs\in B\cap(J\cap R).
$$
Thus $J\cap R$ 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 \b[semisimple Artinian ring]. This is <semisimplicity of a classical quotient from regular elements in essential ideals>.
Back to article page
