Socle length via semisimple filtration I've been working through Auslander, Reiten, and Smalø's Representation Theory of Artin Algebras, and I've gotten quite stuck on exercise II.6, which involves not-necessarily-Artin rings. The exercise states:

Let $\Lambda$ be any ring and $M$ a $\Lambda$-module of finite length. Let $0 = M_0\subset M_1 \subset M_2 \subset \cdots \subset M_n = M$ be a filtration such that $M_{i+1} / M_i$ is semisimple for $i = 0,\ldots, n-1$ and such that $M_{i+1} / M_i'$ is not semisimple if $M_i'$ is a proper submodule of $M_i$.

and part (c) asks

Prove that $M_i\not\subset \operatorname{soc}^{i-1} M$ for any $i \geq 1$.

where $\operatorname{soc}^{i}(M)$ is defined inductively by $\operatorname{soc}^1(M) = \operatorname{soc}(M)$, and $\operatorname{soc}^i(M)$ is the preimage of $\operatorname{soc}(M / \operatorname{soc}^{i-1}(M))$ in $M$.
I am stumped on how to approach this exercise. Directly charging into a proof by contradiction, if we suppose that $M_i \subset \operatorname{soc}^{i-1} M$, this is the same as saying that $M_i / (M_i\cap \operatorname{soc}^{i-2}(M)) \subset \operatorname{soc} (M / \operatorname{soc}^{i-2}(M))$, which I think is equivalent to saying that $M_i / (M_i\cap \operatorname{soc}^{i-2}(M))$ is semisimple. This would raise a contradiction if I could show that $M_i\cap \operatorname{soc}^{i-2}(M)\subsetneq M_{i-1}$, but I don't see why this should be true; a priori, it seems like the filtration of $M$ by the $M_i$'s and the filtration by the socle series might behave differently from each other, so I'm not sure how to make this comparison.
The essential issue I'm getting hung up on is that the hypothesis given in the setup of the problem naturally lends itself to reasoning about the filtration from the top down, since it says that each term is minimal with respect to giving a semisimple quotient of the next higher one. On the other hand, the inductive construction of the socle series assembles it from the bottom up, by making each term maximal with respect to having a semisimple quotient by the next lower one. Because the ring is arbitrary, I don't have a duality I can use to match these two perspectives.
In attempting to deal with this, I've tried to use induction on the length of the filtration of $M_i$'s. I can show that $\operatorname{soc}^{i}(M_{n-1}) = \operatorname{soc}^{i}(M)\cap M_{n-1}$. However, comparing the socle series of $M$ to those of the lower $M_i$ doesn't seem to be helpful either, because they all cut off at different points. If $M\subset \operatorname{soc}^{n-1}(M)$, then they are equal, and equivalently $M / \operatorname{soc}^{n-2}(M)$ is semisimple. This implies that $M_{n-1} / \operatorname{soc}^{n-2}(M_{n-1})$, which I can identify with a submodule, is also semisimple; but this is true regardless, and not strong enough to obtain a contradiction anyway.
I feel like I may be overthinking this problem. Can anyone provide assistance?
 A: Recall first that the socle of a module $M$, denoted $\mathrm{soc}(M)$, is the sum of all simple submodules of $M$, and is the largest semisimple submodule of $M$. Iterating, we write $\mathrm{soc}^i(M)$ for the preimage of the socle of $M/\mathrm{soc}^{i-1}(M)$.
If $\Lambda$ is a ring with Jacobson radical $J$ such that $\Lambda/J$ is a semisimple ring, then $\mathrm{soc}^i(M)=\{m\in M:J^im=0\}$.
Dually, the radical $\mathrm{rad}(M)$ is the intersection of all maximal submodules of $M$, and is the smallest submodule of $M$ such that the quotient $M/\mathrm{rad}(M)$ embeds in a direct product of simple modules. Iterating, we write $\mathrm{rad}^i(M)$ for the radical of $\mathrm{rad}^{i-1}(M)$.
Again, if $\Lambda/J$ is semisimple, then $\mathrm{rad}(M)=JM$ and $M/JM$ is actually semisimple.

Now, the setup of the question has a finite filtration $0=M_0\leq M_1\leq\cdots\leq M_n=M$ such that $M_{i+1}/M_i$ is semisimple, whereas $M_{i+1}/U$ is not semisimple for any proper submodule $U<M_i$.
It follows that $M_i=\mathrm{rad}(M_{i+1})$. For, by definition we have $\mathrm{rad}(M_{i+1})\leq M_i$. If this were not equality, then we would have a maximal submodule $N<M_{i+1}$ not containing $M_i$, but then $U:=M_i\cap N$ would be a proper submodule of $M_i$ and $M_{i+1}/U$ would embed in the semisimple module $(M_{i+1}/M_i)\times (M_{i+1}/N)$, so would itself be semisimple, a contradiction.
The claim is that $M_i$ is not contained in $\mathrm{soc}^{i-1}(M)$ for all $i$.
Suppose for contradiction that $M_{i+1}$ is contained in $\mathrm{soc}^i(M)$. Then $M_{i+1}/(\mathrm{soc}^{i-1}(M)\cap M_{i+1})$ is semisimple, so $\mathrm{soc}^{i-1}(M)\cap M_{i+1}$ contains $\mathrm{rad}(M_{i+1})=M_i$, so $M_i\leq\mathrm{soc}^{i-1}(M)$. Repeating, we obtain $M_2\subset\mathrm{soc}(M)$, so is semisimple, a contradiction. This proves the claim.

Of course, if we are working over a ring $\Lambda$ such that $\Lambda/J$ is semisimple, then the argument simplifies. We have $J(M_{i+1}/M_i)=0$ but $J(M_{i+1}/U)\neq0$ for all proper $U<M_i$, so $\mathrm{rad}(M_{i+1})=JM_{i+1}=M_i$. Then $M_i=J^{n-i}(M)$ for all $i$, so $J^{i-1}(M_i)=M_1$ for all $i$. Since $\mathrm{soc}^{i-1}(M)$ consists of those elements killed by $J^{i-1}$, we see that $M_i$ is not contained in $\mathrm{soc}^{i-1}(M)$.
