Socle and Radical Series

Proposition

Let $A$ be an Artinian ring, and let $U$ be a finitely generated $A$-module.

• TFAE for an $A$-submodule of $U$:
  • $V=\mathrm{Rad}U$
  • $V$ is the smallest submodule of $U$ with semisimple quotient.
  • $J(A)U=V$
• TFAE for an $A$-submodule of $U$:
  • $V=\mathrm{soc}U$
  • $V$ is the largest semisimple submodule of $U$
  • $V=\{u\in U:J(A)u=0\}$.

\begin{proof} (1) iii)→ii) Let $V=J(A)U$, then $U/V$ is an $A/J(A)$-module by ^xy300k. Since $A$ is Artinian, we know $A/J(A)$ is semisimple by ^r8oe32. Since $U/V$ is a finitely generated $A/J(A)$-module, $U/V$ is also semisimple because $(A/J(A){)}^n\twoheadrightarrow U$ for some $n$ and $U$ is a quotient module of semisimple $(A/J(A){)}^n$. To show $V$ is the smallest submodule, assume that there exists $V^{\prime }$ with $V^{\prime }\subseteq V$ and $U/V^{\prime }$ semisimple. By ^p95eh4 i), we know $J(A)(U/V^{\prime })=0$ and $V=J(A)U\subseteq V^{\prime }$. It deduces that $V=V^{\prime }$ and we finish the proof.

ii)→i) By ^r8oe32.

i)→iii) It suffices to show $J(A)U=\mathrm{Rad}U$. Since $U/\mathrm{Rad}U$ is semisimple, $J(A)(U/\mathrm{Rad}U)=0$ and so $J(A)U\subseteq \mathrm{Rad}U$. On the other hand, since $U/J(A)U$ is a finitely generated $A/J(A)$-module and so is semisimple, we know $J(A)U\supseteq \mathrm{Rad}U$. Therefore, $J(A)U=\mathrm{Rad}U$.

(2) By definition of socle, i) and ii) are equivalent. By ^p95eh4 i), $\mathrm{soc}U\subseteq \{u\in U:J(A)u=0\}$ is trivial. Conversely, notice that $W:=\{u\in U:J(A)u=0\}$ is the largest submodule of $U$ annihilated by $J(A)$. Thus $W$ is a $A/J(A)$-module and so it is semisimple $A$-module. Thus $W\subseteq \mathrm{soc}U$ by definition. Now we finish the proof. \end{proof}

Definition

Let $U$ be a finitely generated $A$-module with $A$ Artinian. Define inductively

• ${\mathrm{Rad}}^n(U)=\mathrm{Rad}({\mathrm{Rad}}^{n-1}(U))$, and
• ${\mathrm{soc}}^n(U)/{\mathrm{soc}}^{n-1}(U)=\mathrm{soc}(U/{\mathrm{soc}}^{n-1}(U))$.

By ^045478, we have ${\mathrm{Rad}}^n(U)=J(A{)}^{n}U$ and ${\mathrm{soc}}^n(U)=\{u\in U:J(A{)}^{n}u=0\}$. Then define chains of submodules:

• ${\mathrm{Rad}}^0(U)\supseteq \mathrm{Rad}(U)\supseteq \cdots$ is called radical series/Loewy series.
• $0\subseteq \mathrm{soc}(U)\subseteq {\mathrm{soc}}^2(U)\subseteq \cdots$ is called socle series.

Furthermore, we call ${\mathrm{Rad}}^{n-1}U/{\mathrm{Rad}}^{n}U$ radical layer/Loewy layer, and call ${\mathrm{soc}}^n(U)/{\mathrm{soc}}^{n-1}(U)$ socle layer.

Lemma

${\mathrm{Rad}}^n(U\oplus V)={\mathrm{Rad}}^n(U)\oplus {\mathrm{Rad}}^n(V)$. ${\mathrm{soc}}^n(U\oplus V)={\mathrm{soc}}^n(U)\oplus {\mathrm{soc}}^n(V)$.

\begin{proof} See homework. \end{proof}

Theorem

Let $A$ be an Artinian ring, and let $U$ be a finitely generated $A$-module. The radical series of $U$ is the fastest descending series of submodules of $U$ with semisimple quotient, and the socle series of $U$ is the fastest ascending series with semisimple quotient.

The two series terminate. If $m$ and $n$ are the minimal integer with ${\mathrm{Rad}}^{m}U=0$ and ${\mathrm{soc}}^{n}U=U$, then $m=n$.

\begin{proof} Suppose that $\cdots \subseteq U_2\subseteq U_1\subseteq U_0=U$ is a series of submodules of $U$ with semisimple quotient. We show by induction on $r$ such that ${\mathrm{Rad}}^r(U)\subseteq U_r$. We prove it by induction. When $r=0$, it is trivial. Now assume that ${\mathrm{Rad}}^{r-1}(U)\subseteq U_{r-1}$. Then we have

$${\mathrm{Rad}}^{r-1}(U)/({\mathrm{Rad}}^{r-1}(U)\cap U_r)\simeq {\mathrm{Rad}}^{r-1}(U)+U_r/U_r\subseteq U_{r-1}/U_r$$

is semisimple and so ${\mathrm{Rad}}^{r-1}(U)\cap U_r\supseteq \mathrm{Rad}({\mathrm{Rad}}^{r-1}(U))={\mathrm{Rad}}^r(U)$. Similarly, we can prove the case of socle.

Since $A$ is an Artinian ring, its Jacobson radical $J(A)$ is nilpotent by ^a43f1d. Let $k⩾1$ be an integer such that $J(A{)}^k=0$. Then ${\mathrm{Rad}}^k(U)=J(A{)}^{k}U=0\cdot U=0$. And ${\mathrm{soc}}^k(U)=\{u\in U\mid J(A{)}^{k}u=0\}=\{u\in U\mid 0\cdot u=0\}=U$. Thus, both series terminate in at most $k$ steps.

Let $m$ be the minimal integer such that ${\mathrm{Rad}}^m(U)=0$, and let $n$ be the minimal integer such that ${\mathrm{soc}}^n(U)=U$. 

• Show $m⩽n$: Consider the socle series $0=V_0\subset V_1\subset \cdots \subset V_n=U$, where $V_i={\mathrm{soc}}^i(U)$. Define a descending series $U_i=V_{n-i}$ for $0⩽i⩽n$. Then $U=U_0\supset U_1\supset \cdots \supset U_n=0$, and the quotients $U_{i-1}/U_i=V_{n-i+1}/V_{n-i}$ are semisimple. We have proved that ${\mathrm{Rad}}^r(U)\subseteq U_r$ for all $r\le n$. Setting $r=n$, we get ${\mathrm{Rad}}^n(U)\subseteq U_n=V_0=0$. Since ${\mathrm{Rad}}^n(U)=0$ and $m$ is the minimal such integer, we must have $m⩽n$. 
• Show $n⩽m$: Consider the radical series $U=W_0\supset W_1\supset \cdots \supset W_m=0$, where $W_i={\mathrm{Rad}}^i(U)$. Define an ascending series $V_i=W_{m-i}$ for $0⩽i⩽m$. Then $0=V_0\subset V_1\subset \cdots \subset V_m=U$, and the quotients $V_i/V_{i-1}=W_{m-i}/W_{m-i+1}$ are semisimple. Recall that $V_r\subseteq {\mathrm{soc}}^r(U)$ for all $r⩽m$. Setting $r=m$, we get $V_m=W_0=U\subseteq {\mathrm{soc}}^m(U)$. Since ${\mathrm{soc}}^m(U)=U$ and $n$ is the minimal such integer, we must have $n\le m$.

Combining $m⩽n$ and $n⩽m$, we conclude $m=n$. \end{proof}

Lemma

• Let $A$ be a semisimple ring, then $A$ is Artinian iff it is Noetherian.
• If $A$ is Artinian ring, then it is Noetherian.

\begin{proof} i) is easy.

ii) For the left $A$-module ${}_{A}A$, we have $J^{n+1}=J(J^n)=\mathrm{Rad}(J^n)$ and so $J^n/J^{n+1}$ is semisimple. To show $A$ is Noetherian, as $A/J$ is Noetherian, it suffices to show $J$ is Noetherian. As $J/J^2$ is Noetherian, it suffices to show $J^2$ is Noetherian. Repeat this procedure, and finitely it suffices to show $J^r=0$ is Noetherian. Now we finish the proof. \end{proof}