I am reading Rotman's Advanced Modern Algebra. In the proof of Corollary C-2.47, the author use the fact $kG\cong M_{n_1}(k)\times \cdots \times M_{n_r}(k)$ as vector spaces over $k$. But he only proof it as rings in Theorem C-2.35 and Corollary C-2.44.
I let $\theta$ be the ring isomorphism $kG\cong M_{n_1}(k)\times \cdots \times M_{n_r}(k)$ and try to prove that $\theta$ is a linear transformation. That is, $\theta(\lambda \cdot \sum \lambda_g g)=\lambda\cdot\theta(\sum \lambda_g g)$. But I get stuck.
Question 1. Is my direction right and how do I prove it?
Question 2. Where can I find the proof $kG\cong M_{n_1}(k)\times \cdots \times M_{n_r}(k)$ as vector spaces over $k$?
Corollary C-2.47. If $G$ is a finite group and $k$ is an algebraically closed field whose characteristic does not divide $|G|$, then $$|G|=n_1^2+n_2^2+\cdots+n_r^2,$$ where the $i$th simple component $B_i$ of $kG$ consists of $n_i\times n_i$ matrices.
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Theorem C-2.35 (Wedderburn-Artin I). A ring $R$ is left semisimple if and only if $R$ is isomorphic to a direct product of matrix rings over division rings.
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Corollary C-2.44 (Molien). If $G$ is a finite group and $k$ is an algebraically closed field whose characteristic does no divide $|G|$, then $$kG\cong M_{n_1}{(k)}\times \cdots \times M_{n_r}{(k)}.$$
Proof of Corollary C-2.44: By Maschke's Theorem, $kG$ is a semisimple ring, and its simple component are isomorphic to matrix rings of the form $M_{n}{D}$, where $D$ arises as $End_{kG}{(L)}^{\text{op}}$ for some minimal left ideal $L$ in $kG$. Therefore, it suffices to show that $End_{kG}{(L)}^{\text{op}}=D=k$. Now $End_{kG}{(L)}^{\text{op}}\subseteq End_{k}{(L)}^{\text{op}}$, which is finite-dimensional over $k$ because $L$ is; hence, $D=End_{kG}{(L)}^{\text{op}}$ is finite-dimensional over $k$. Each $f\in End_{kG}{(L)}$ is a $kG$-map, hence is a $k$-map; that is, $f(au)=af(u)$ for all $a\in k$ and $u\in L$. Therefore, the map $\varphi_a:L\to L$, given by $u\mapsto au$, commutes with $f$; that is, $k$ (identified with all $\varphi_a$) is contained in $Z(D)$, the center of $D$. If $\delta\in D$, then $\delta$ commutes with every element in $k$, and so $k(\delta)$, the subdivision ring generated by $k$ and $\delta$, is a (commutative) field. As $D$ is finite-dimensional over $k$, so is $k(\delta)$; that is, $k(\delta)$ is a finite extension of the field $k$, and so $\delta$ is algebraic over $k$, by Proposition A-3.84 in Part 1. But $k$ is algebraically closed, so that $\delta\in k$ and $D=k$.
Since nobody has given a complete answer to this question, I am expanding on what I wrote. Note that $k[G]$ is a $k$-algebra with center containing $k$. Suppose that $D$ is a ring direct summand of $k[G]$, I claim that $D$ is a $k$-subalgebra of $k[G]$.
To prove this, we note that $D$ is a two-sided ideal of $k[G]$. Therefore, for each $x\in k\subseteq k[G]$ and $t\in D$, we have $x\cdot t \in D$ and the fact that $k$ lies within the center of $k[G]$ implies that $D$ is an associative $k$-algebra under the multiplication induced by that of $k[G]$. That is, in any decomposition $k[G]=D_1\oplus D_2\oplus \ldots \oplus D_n$ of $k[G]$ into a ring direct sum of simple rings $D_i$, each $D_i$ is a $k$-algebra with the same $k$-algebra structure inherited from $k[G]$.