Question about a proof that simple algebras are semi-simple

Question

In the book Clifford Algebras: An Introduction by D. J. H. Garling, Theorem 2.7.2 states that a finite-dimensional simple unital algebra $A$ is semi-simple. My question is specifically about the proof that Garling gives for this statement, and I am not looking for an alternative proof.

Proof:

1. Suppose that $J$ is a non-zero left ideal of minimal dimension. Then $J$ is a simple left A-module. (Understood)
2. Let $$JA=\left\{\sum_{i=1}^kj_ia_i:k\in\mathbb{N},j_i\in J,a_i\in A\right\}.$$ Then $JA$ is a non-zero ideal in $A$ and since $A$ is simple, $JA=A$. (Understood)
3. Thus there exists $j_1,\ldots,j_k$ in $J$ and $a_1,\ldots,a_k$ in $A$ such that $1_A=j_1a_1+\cdots+j_ka_k$. We may clearly suppose that each $j_ia_i\ne0$, so that $Ja_i$ is a non-zero left ideal. (Understood)
4. Now the mapping $j\to ja_i$ is a left $A$-module homomorphism of $J$ onto $Ja_i$. (Understood)
5. By Schur's lemma, it is an isomorphism, and so each $Ja_i$ is also a simple left $A$-submodule. (This is where I first become stuck. How can I use Schur's lemma here if I don't know beforehand that $Ja_i$ is simple?)
6. Since $Ja_1,\ldots,Ja_k$ span $A$, $A$ is semi-simple. (Don't we also need to show that the sum is direct? How do we know that $J\cap Ja_i=\{0\}$?)

I would appreciate any insight from this community.

For reference, the version of Schur's lemma used here is stated as follows: Suppose that $T\in\text{Hom}_A(M_1,M_2)$, where $M_1$ and $M_2$ are simple left $A$-modules. If $T\ne0$, then $T$ is an $A$-isomorphism of $M_1$ onto $M_2$.

Answer 1

Step 5 is not really applying Schur's lemma but rather the same idea as the proof of Schur's lemma. The kernel of $j\mapsto ja_i$ is a submodule of $J$, so by simplicity it is either $0$ or all of $J$. The kernel can't be all of $J$ since $Ja_i$ is nonzero, so the kernel is $0$, so $j\mapsto ja_i$ is an isomorphism $J\to Ja_i$.

Step 6 is using the general fact that any module which is a sum (not necessarily direct) of simple submodules is semisimple. If you have only seen this fact for direct sums, you can deduce it for arbitrary sums because an arbitrary sum is a quotient of the direct sum, and a quotient of a semisimple module is semisimple.