Suppose $R$ is a right Noetherian ring containing the Wedderburn ring $S$ and write $S=\cdot \sum_{i=1}^{k} S_{i}$, a direct sum of simple rings. If $f_{i}$ is a primitive idempotent of $S_{i}$, prove that
$$
\text { u. } \operatorname{dim}_{R} R=\sum_{i=1}^{k}\left(\mathrm{u} \cdot \operatorname{dim}_{S} S_{i}\right)\left(\mathrm{u} \cdot \operatorname{dim}_{R} f_{i} R\right)
$$
To this end, first observe that $S_{i} \cong \mathrm{M}_{n_{i}}\left(D_{i}\right)$ has a family of orthogonal primitive idempotents $\left\{f_{i, 1}, f_{i, 2}, \ldots, f_{i, n_{i}}\right\}$ with $n_{i}=$ u.dim $_{S} S_{i}$ and $f_{i, j} S_{S} \cong f_{i} S_{S} .$ Then note that $\left\{f_{i, j}\right\}$ is an orthogonal decomposition of 1 in $R$ and hence that $R=\cdot \sum_{i, j} f_{i, j} R$. Finally, use the preceding exercise to conclude that $f_{i, j} R_{R} \cong f_{i} R_{R}$. This is a special case of the additivity principle.