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After replacing $X$ by a maximal $D$-linearly independent subset with the same span, write $X=\{x_1,\dots ,x_n\}$. The standard independence lemma, proved by induction using Schur's lemma, saysOtherwise $r{x}_n$ would depend only on $(r{x}_1,\dots ,r{x}_{n-1})$, defining an $R$-map whose coordinate maps lie in ${\mathrm{End}}_R(V)$ and forcing $x_n\in {\sum }_{i<n}{x}_{i}D$. Applying the same argument with $y$ shows that ${\bigcap }_i{\mathrm{ann}}_R(x_i)y=0$ implies $y\in XD$, proving the requested claim.

The Jacobson density theorem states that if $x_1,\dots ,x_n$ are $D$-independent and $y_1,\dots ,y_n\in V$, there is $r\in R$ with $r{x}_i=y_i$ for every $i$. Induct on $n$. First match the first $n-1$ values. The independence lemma makes $I{x}_n$, for $I={\bigcap }_{i<n}{\mathrm{ann}}_R(x_i)$, a nonzero submodule and hence all of $V$; an element of $I$ supplies the final correction.

If $R$ is primitive, choose a faithful simple module $V$. When ${\dim }_{D}V=n<\infty$, density makes $R\to {\mathrm{End}}_D(V)\cong {\mathrm{Mat}}_n(D)$ surjective and faithfulness makes it injective. If ${\dim }_{D}V$ is infinite, choose an $n$-dimensional $D$-subspace $W$ and let $R_n=\{r:rW\subseteq W\}$. Density makes restriction $R_n\to {\mathrm{End}}_D(W)\cong {\mathrm{Mat}}_n(D)$ surjective.

Solved by gpt-5.6-sol high.