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A left-primitive ideal is a two-sided ideal of the formfor a nonzero simple module $V$ on the left. Equivalently, the quotient ring $R/P$ has a faithful module that is a simple module, and is therefore a primitive ring.

To prove that $P$ is a prime ideal of a noncommutative ring, let $A,B$ be two-sided ideals with $AB\subseteq P$. Since $B$ is two-sided, $BV$ is a submodule of $V$: for $r\in R$, $b\in B$ and $v\in V$, one has $r(bv)=(rb)v\in BV$. If $B⊈P$, then $BV\ne 0$, so simplicity gives $BV=V$. Consequentlyand $A\subseteq P$. Thus $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. Also $P\ne R$, since $1$ acts as the identity element on the nonzero module $V$. These are precisely the required prime ideal of a noncommutative ring conditions.

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 ${\mathrm{End}}_B(V)$ 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. SetEvaluation 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 $v_1,\dots ,v_m$ and any $w_1,\dots ,w_m\in V$, some $b\in B$ satisfies $b{v}_i=w_i$ for every $i$.

Suppose that $V$ contains two $\Delta$-linearly independent vectors $v,w$. Applying the Jacobson density theorem twice, choose $a,b\in B$ satisfyingThe inherited relation giveswith integers $c_k$. Acting on $w$, the left side gives $baw=bv=w$, but every term on the right gives $a^{k}bw=0$. This contradicts $w\ne 0$. Thus ${\dim }_{\Delta }V=1$.

The ring homomorphismgiven 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. HenceThe 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.