Let $I$ be maximal such that $V \cap \oplus_{i \in I} S_i = 0$. We claim that \[ U = V + \bigoplus_{i \in I} S_i . \] If not then some $S_j$ is not contained in $V + \oplus_{i \in I} S_i .$ We may adjoin $j$ to $I$ contradicting the maximality of $I$.

(1)$\Rightarrow$(2) follows from the Lemma, and (2)$\Rightarrow$(1) is straightforward.

First we show that a submodule $N$ of a semisimple module $M$ is semisimple. Let $U \subset N$ be a submodule. Find a submodule $V$ of $M$ such that $M = U \oplus V$. Then we have $N = U \oplus (V \cap N)$. Indeed, any element $x$ of $N$ may be written uniquely as $u + v$ with $u \in U$ and $v \in V$, and since $x, u \in N$ we have $v \in V \cap N$. This proves that every submodule of $N$ is a direct summand so $N$ is semisimple.

If $M$ is semisimple and $Q$ is a quotient, write $Q = M / N$ for some submodule $N$. Since $M$ is semisimple, $M = N \oplus N'$ for some submodule $N'$, and $Q \cong N'$. But by the first part $N'$ is semisimple.

First suppose that $U$ is semisimple. We may write $U = \bigoplus S_i$ as direct sum of simple modules. Let $M_j = \bigoplus_{i \neq j} S_i$. Then the $M_j$ are maximal since $U / M_j \cong S_j$ is simple. The intersection of the $M_j$ is zero, proving one direction.

Conversely, suppose that the intersection of all maximal submodules is zero. Then we may find a finite collection $M_i$ of maximal submodules with intersection zero. Then the canonical map $U \longrightarrow \bigoplus U / M_i$ is injective, so $U$ is a submodule of a direct sum of simple modules, hence semisimple.

Assume that $A$ is semisimple as an $A$-module. And so is any free module. Since an arbitrary module is a quotient of a free module we've proved (1)$\Rightarrow$(2) and the other direction is obvious.

If $I$ and $I'$ are nilpotent so is $I + I'$, so $A$ has a largest nilpotent ideal $J$.

Let $J'$ be the intersection of all maximal left ideals of $A$.

We note that if $I$ and $I'$ are left ideals of $A$ such that $A / I$ and $A / I'$ are semisimple, then $A / (I \cap I')$ is semisimple, since there exists an injective homomorphism $A / (I \cap I') \longrightarrow A / I \oplus A / I'$. Thus $A / (I \cap I')$ is semisimple. It follows that there exists a smallest left ideal $J''$ such that $A / J''$ is semisimple.

We need to show that $\operatorname{rad} (A) = J' = J''$.

First let us show $\operatorname{rad} (A) = J$, the largest nilpotent ideal. If $S$ is simple, then $J S = S$ or $J S = 0$. Since $J S = S$ would imply $J^n S = S$ contradicting the nilpotence of $J$, we have $J S = 0$ and so $J \subseteq \operatorname{rad} (A) .$ On the other hand, if $A = A_0 \supset A_1 \supset A_2 \supset \cdots \supset A_n = 0$ is a composition series, then $\operatorname{rad} (A)^k \subseteq A_k$ so $\operatorname{rad} (A)$ is nilpotent. This proves that $\operatorname{rad} (A) = J$.

Next let us show that $\operatorname{rad} (A) = J'$, the intersection of all maximal left ideals. Every simple module is of the form $A /\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. The annihilator of $S$ is then \[ \operatorname{Ann} (A /\mathfrak{m}) =\{x \in A| x A \subseteq \mathfrak{m}\} \subseteq \mathfrak{m}. \] (It is easy to see that this is the largest two-sided ideal contained in $\mathfrak{m}$.) So \[ J = \bigcap \operatorname{Ann} (A /\mathfrak{m}) \subseteq \bigcap \mathfrak{m}= J', \] where the intersection runs over maximal left ideals. Conversely, we show that $J' \subset J$. If not there is a simple module $S$ such that $J' S = S$; indeed $J' s = S$ for some $0 \neq s \in S$. So write $x s = - s$ for $x \in J'$. Thus $(x + 1) s = 0$. But $A (x + 1) = A$ since otherwise $A (x + 1) \subset \mathfrak{m}$ for some maximal $\mathfrak{m}$. We have $1 + x, x \in \mathfrak{m}$ because $J' \subseteq \mathfrak{m}$. Therefore $1 \in \mathfrak{m}$ which is a contradiction. \ This proves that $J' \subset J$ and so $J' = J = \operatorname{rad} (A)$.

Next let us show that $A / \operatorname{rad} (A)$ is semisimple. Since as a finite-dimensional algebra over a field $A$ is Artinian, there exist a finite number of maximal left ideals $\mathfrak{m}_1, \cdots, \mathfrak{m}_n$ such that $J' = \bigcap \mathfrak{m}_i$. We have an injective homomorphism \[ A / J' \rightarrow \bigoplus (A / M_i), \] namely the map that sends the coset of $a \in A$ to $(a + M_1, a + M_2, \cdots)$. Each $A / M_i$ is simple so $A / J'$ is a submodule of a semisimple module. Therefore $A / J'$ is semisimple.

We have shown that $A / J'$ is semisimple, but we need to show that $J'$ is the smallest submodule with this property. Let $I$ be any left ideal such that $A / I$ is a semisimple module. Let $\overline{J'}$ be the image of $J'$ in $A / I$. Then $\overline{J'}$ is complemented, so $A / I = \overline{J'} \oplus \overline{W}$ for some $A$-submodule $\overline{W}$ of $A / I$. The preimage $W$ of $\overline{W}$ is a left ideal of $A$. If $W$ is proper, then it is contained in a maximal left ideal $\mathfrak{m}$. But $J'$ is also contained in $\mathfrak{m}$, while $J' + W = A$, which is a contradiction. Therefore $W = A$ which means that $\overline{J'} = 0$, proving that $J' \subseteq I$. This proves that any submodule $I$ such that $A / I$ is semisimple contains $J'$.

By Theorem 1.1 (2), the intersection of maximal submodules of $A$ is zero so $A$ is semisimple by Proposition 1.3. To see that the radical of $A / \operatorname{rad} (A)$ is zero, observe that if $\overline{\mathfrak{m}}$ is a maximal left ideal of $A / \operatorname{rad} (A)$ then its preimage $\mathfrak{m}$ is a maximal left ideal of $A$, and this correspondence is a bijection between the maximal left ideals of $A / \operatorname{rad} (A)$ and $A$. Since $\bigcap \mathfrak{m}= \operatorname{rad} (A)$ we have $\bigcap \overline{\mathfrak{m}} = 0$. So the radical of $A / \operatorname{rad} (A)$ is zero, and the ring is semisimple.

If $\operatorname{rad} (A) M = 0$, then $M$ is a module of $A / \operatorname{rad} (A)$, which is a semisimple ring. Thus $M$ is semisimple. Conversely, assume that $M$ is semisimple. Then $M$ is a direct sum of simple modules, which are annihilated by $\operatorname{rad} (A)$, and so $\operatorname{rad} (A) M = 0$.

Let $J$ be the intersection of all maximal submodules of $U$. Then if $V$ is a submodule of $U$, by Proposition 1.3 the quotient $U / V$ is semisimple if and only if $V \supseteq J$. We have to show that $J = \operatorname{rad} (A) U$. If $V$ is a submodule of $U$, then $U / V$ is semisimple if and only if $\operatorname{rad} (A)$ annihilate it, that is, if and only if $\operatorname{rad} (A) U \subset V$. Thus by criterion (2), $\operatorname{rad} (A) U = J$.

Observe that if $V, W$ are semisimple submodules of $U$ then $V + W$ is q quotient of the external direct sum $V \oplus W$, hence is semisimple. Therefore $U$ has a maximal semisimple submodule $S$. Now $\operatorname{soc} (U)$ is an $A / \operatorname{rad} (A)$-module. Since $A / \operatorname{rad} (A)$ is semisimple, $\operatorname{soc} (U)$ is semisimple and $\operatorname{soc} (U) \subset S$. On the other hand, by Proposition 1.6 part (2) the radical of $S$ is zero, so by Proposition 1.6 part (1) $\operatorname{rad} (A)$ annihilates $S$ and so $S \subset \operatorname{soc} (U)$.