Proof of $R$ is semi simple $\iff $ every finitely generated modules over $R$ is semi-simple. Let $R$ a ring. Let denote ${}_RR$ the ring $R$ seen as a left $R-$module. I want to prove that $R$ is semi simple $\iff$ every finitely generated module over $R$ is semi simple. 
Proof
$\Leftarrow :$ In my course it's written that it's obvious but I don't understand why. I recall that by definition, $R$ is semi simple if ${}_RR$ is semi simple.
$\Rightarrow$: Let $M$ a finitely generated module. By definition, there is a surjection $$\varphi:\bigoplus_{i=1}^d{}_RR\longrightarrow M.$$
Q1) Where does this surjection come from ?
By definition of semi simplicity,
$$\bigoplus_{i=1}^dM_i=\sum_{i=1}^d M_i,$$
where $M_i$ are simple.
Q2) To me $$\bigoplus_{i=1}^d M_i=\prod_{i=1}^d M_i.$$
I don't understand why $$\prod_{i=1}^d M_i=\sum_{i=1}^d M_i.$$
Therefore, $$\varphi(M)=\sum_{i=1}^d \varphi(M_i).$$
Q3) Do we always have $$\varphi\left(\sum_{i=1}^d M_i\right)=\sum_{i=1}^d \varphi(M_i)$$
for a module homomorphism $\varphi$ or it's a consequence of the fact that $$\bigoplus_{i=1}^d M_i=\sum_{i=1}^dM_i\ \ ?$$
Q4) Why using $\oplus_{i=1}^d M_i$ instead of $\prod_{i=1}^d M_i$ since both are the same ? It looks really weird to me this notation !
The rest is fine. Sorry for all those question, but I'm a beginner in module.
 A: I assume you define $R$ to be semisimple if the regular left module $_RR$ is semisimple (that is, a sum of simple submodules).
It is obvious that assuming all finitely generated modules are semisimple then $R$ is semisimple, because $_RR$ is finitely generated.
The harder part is the converse. First let's prove that if $R$ is semisimple, then every cyclic module is semisimple. Suppose $M=Rx$, for $x\in M$. Then we have the surjective homomorphism of $R$-modules $\mu_x\colon R\to M$ defined by $\mu_x(r)=rx$. Since $R=\sum_{i\in I}S_i$, where $S_i$ is simple, we can say that
$$
M=\mu_x(R)=\sum_{i\in I}\mu_x(S_i)
$$
and each $\mu_x(S_i)$ is either $0$ or isomorphic to $S_i$, since $S_i$ is simple. Therefore $M$ is semisimple.
Next, let $M=x_1R+x_2R+\dots+x_nR$ be finitely generated. Since each $x_iR$ is semisimple, also $M$ is a sum of simple submodules.

Let's see your questions.
(Q1) Saying that $M$ is finitely generated is the same as saying there exists a surjective homomorphism $\bigoplus_{i=1}^n R\to M$. With notations as above, the homomorphism is $(r_1,r_2,\dots,r_n)\mapsto r_1x_1+r_2x_2+\dots+r_nx_n$.
(Q2) Finite direct sums are the same as finite direct products. In the infinite case they differ.
(Q3) Just argue elementwise. An element of $\sum_{i=1}^nM_i$ has the form $\sum_{i=1}^n m_i$, so $\varphi(\sum_{i=1}^n m_i)=\sum_{i=1}^n \varphi(m_i)$ is an element of $\sum_{i=1}^n\varphi(M_i)$. Can you do the converse?
(Q4) Because direct sums are the proper concept to use here.
