Show that in ascending Loewy series, $S^r(R)=R$

Question

Let $R$ be an Artinian ring, $N$ its radical, and $r$ the smallest natural number such that $N^r=0$. Define an ideal $S^n(R)$ of $R$ recursively as follows:

$S^1(R)=soc(R)$

Assuming $S^i(R)$ is defined, $S^{i+1}(R)$ is an ideal containing $S^i(R)$ such that $S^{i+1}(R)/S^i(R)=soc(R/S^i(R)).$ Show that $S^r(R)=R$.

The socle of a ring is the sum of all minimal right ideals. The radical $N$ is the least ideal of $R$ such that $R/N$ is semisimple (i.e. when regarded as a module over itself, it can be expressed as a direct sum of simple submodules).

I need to show that $soc(R/S^{r-1}(R))=R$, and hence that $R/S^{r-1}(R)$ has no proper ideals. Hence, $R/S^{r-1}(R)$ is simple. If I could show that $N\subseteq S^{r-1}(R)$, then I'd be done. Any help on how I might be able to do that?

Answer 1

Here is an outline of the proof that I would write, with details for you to fill in:

1. Show that $\operatorname{soc}(R)$ is equal to $\operatorname{ann}_l(N) = \{x \in R \mid xN = 0\}$, the left annihlator of $N$.
2. Show that $\operatorname{ann}_l(N) = N^{r-1}$.
3. Now that you know $R/S^1(R) = R/N^{r-1}$, finish the argument using induction on $r$.