= Solution
{c}
Put $B=R/I$. By the definition of a <left-primitive ideal>, there is a <faithful module> $V$ that is a <simple module> on the left over $B$. The assumed relation passes to the <quotient ring> $B$, since any pair of elements of $B$ can be lifted to $R$.
By <Schur's lemma>, the <endomorphism ring> $\operatorname{End}_B(V)$ is a <division ring>: the <kernel> and <image> of a nonzero <module endomorphism> are <submodules>, so simplicity makes that <module endomorphism> both <injective> and <surjective>. Set
$$
\Delta=\operatorname{End}_B(V)^{\mathrm{op}}.
$$
Evaluation makes $V$ a right <vector space over a division ring> $\Delta$, and the $B$-action is $\Delta$-linear. The <opposite ring> in the definition ensures the correct order for this right action.
The <Jacobson density theorem> says that, for any finite $\Delta$-<linearly independent> family $v_1,\ldots,v_m$ and any $w_1,\ldots,w_m\in V$, some $b\in B$ satisfies $bv_i=w_i$ for every $i$.
Suppose that $V$ contains two $\Delta$-<linearly independent> vectors $v,w$. Applying the <Jacobson density theorem> twice, choose $a,b\in B$ satisfying
$$
av=0,\qquad aw=v,\qquad bv=w,\qquad bw=0.
$$
The inherited relation gives
$$
ba=\sum_{k=1}^n c_ka^kb
$$
with integers $c_k$. Acting on $w$, the left side gives $baw=bv=w$, but every term on the right gives $a^kbw=0$. This contradicts $w\ne0$. Thus $\dim_\Delta V=1$.
The <ring homomorphism>
$$
B\longrightarrow\operatorname{End}_\Delta(V)
$$
given by the action is <injective> because $V$ is a <faithful module>. It is <surjective> because the <Jacobson density theorem>, applied to a single nonzero <basis vector>, realizes every $\Delta$-linear <endomorphism> of this one-dimensional <vector space over a division ring>. Every nonzero such <endomorphism> is invertible. Hence
$$
\boxed{R/I\cong\operatorname{End}_\Delta(V)\text{ is a division ring}.}
$$
The use of the <Jacobson density theorem> here only prescribes actions on finitely many vectors; it does not assume that an arbitrary <primitive ring> is already a full <endomorphism ring>.
Back to article page
