6 250401

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} See ^045478. \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 ${\mathrm{soc}}^n(U)/{\mathrm{soc}}^{n-1}(U)=\mathrm{soc}(U/{\mathrm{soc}}^{n-1}(U))$.

So ${\mathrm{Rad}}^n(U)=J(A{)}^{n}U$, ${\mathrm{soc}}^n(U)=\{u\in U:J(A{)}^{n}u=0\}$.

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.

We call ${\mathrm{Rad}}^{n-1}U/{\mathrm{Rad}}^{n}U$ radical layer/Loewy layer; ${\mathrm{soc}}^n(U)/{\mathrm{soc}}^{n-1}(U)$ socle layer.

Lemma

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

Theorem

Let $A$ be an Artinian ring. Let $U$ be an 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 and 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} See ^5ypk5s. \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} See ^u3of0p. \end{proof}

Definition

Let $I$ be a $2$-sided ideal of $R$. $f$ is an idempotent of $R$. $e=f+I$ is an idempotent in $R/I$. We say $f$ lifts $e$.

If for every idempotent $e$ of $R/I$ there is an idempotent $f\in R$ lifting $e$, then we say we can lift idempotents from $R/I$ to $R$.

Theorem

Let $I$ be a nilpotent ideal of $R$, then we can lift idempotent from $R/I$ to $R$. That is, for every idempotent $e$ of $R/I$, there is an idempotent $f\in R$ with $e=f+I$. If $e$ is primitive, so is any lift $f$.

\begin{proof} We define inductively idempotent $e_i\in R/I_i$ such that $e_i+I^{i-1}/I^i=e_{i-1}$ for each $i$ and $e_1=e$. Suppose $e_{i-1}$ is an idempotent of $R/I^{i-1}$. Pick any element $a\in R/I^i$ with $a+I^{i-1}=e_{i-1}$, then $a^2-a\in I^{i-1}/I^i$. Note that $a^2-a=0\in R/I^i$.

Put $e_i=3a^2-2a^3$, then $e_i+I_{i-1}=e_{i-1}$, and

$$e_i^2-e_i=-(3-2a)(1+2a)(a^2-a{)}^2=0$$

in $R/I^i$. If $I^r=0$, then $f=e_r$ is what we desired.

Suppose $e$ is primitive and $f$ has an idempotent decomposition $f=f_1+f_2$ Then $e=e_1+e_2$ with $f_i+I=e_i$. Since $e$ is primitive, one of $e_1,e_2$ is $0$ and so WLOG $f_1+I=0$ and so $f_1\in I$. It contradicts with $f_1$ idempotent.

Alternating proof. See ^84f4de. \end{proof}

Corollary

Let $I$ be a nilpotent ($2$-sided) ideal of a ring $R$ and let $1=e_1+\cdots +e_n$ be an idempotent decomposition in $R/I$. Then we have an idempotent decomposition $1=f_1+\cdots +f_n$ in $R$ with $e_i=f_i+I$. If $e_i$ are primitive, so are $f_i$.

\begin{proof} See ^3rmbjt. \end{proof}

Corollary

Let $f$ be an idempotent in a ring $R$ that has a nilpotent ideal $I$. Then $f$ is primitive in $R$ iff $f+I$ is primitive in $R/I$.

\begin{proof} One direction is easy.

Assume that $f$ is imprimitive, then consider the ring $fRf$, whose identity is $f$. \end{proof}

Lemma

Let $A$ be a semisimple Artinian ring.

• There is a primitive idempotent decomposition $1=e_1+\cdots +e_n$. This corresponds $A=A{e}_1\oplus \cdots \oplus A{e}_n$ where $A{e}_i$ is simple.
• For every simple $A$-module $S$, there is a primitive idempotent $e$ with $S\simeq Ae$.
• for any simple $A$-module $S$, there exists $1⩽i⩽n$, $e_{i}S\ne 0$ and if $T$ is simple $A$-module with $T\not\simeq S$, then $e_{i}T=0$.

\begin{proof} See ^drcz17. \end{proof}

Theorem

Let $A$ be an Artinian ring, and let $S$ be a simple $A$-module.

• There is an indecomposable projective $A$-module $P_S$ with $P_S/\mathrm{Rad}(P_S)\simeq S$ of the form $P_S\simeq A_f$ where $f$ is a primitive idempotent in $A$.
• The idempotent $f$ has the property that $fS\ne 0$, and if $T$ is any simple module with $T\not\simeq S$ then $fT=0$.
• $P_S$ is the projective cover of $S$, it is uniquely determined up to isomorphism by $S$.

\begin{proof} See ^koarlo. \end{proof}

Theorem

Let $A$ be an Artinian ring. Up to isomorphism, the indecomposable projective $A$-module are exactly the module $P_S$ that are the projective covers of simple modules, and $P_S\simeq P_T$ iff $S\simeq T$. Each $P_S$ appears as direct summand of ${}_{A}A$ (left regular module) with multiplicity equal to the multiplicity of $S$ as a summand of ${}_{A/J(A)}A/J(A)$. Precisely, ${}_{A}A\simeq {\oplus }_S(P_S{)}^{n_S}$, where $S$ runs through simple $A$-modules up to isomorphism, and $n_S={\dim }_{D}S$ with $D={\mathrm{End}}_A(S)$.

\begin{proof} See ^u11ppk. \end{proof}

Theorem

Let $A$ be an Artinian ring, and let $U$ be $A$-module. Then $U$ has a projective cover.

\begin{proof} See ^mjip80. \end{proof}

Let $A$ be an Artinian ring. Let $V$ be a finitely generated $A$-module. Then $V$ has a composition series, which is both Artinian and Noetherian. For any simple $A$-module $S$, the number of the composition factors isomorphic to $S$ is called the multiplicity of $S$ in $V$.

Theorem

Let $A$ be an Artinian ring, and let $f\in A$ be a primitive idempotent. Assume that $e=f+J(A)\in A/J(A)=\overline{A}$ is a primitive idempotent. Then the multiplicity of $\overline{A}e$ in any finitely generated $A$-module $V$ is equal to the composition length of $fV$ as an $fAf$-module.

\begin{proof} See ^3hxo3d. \end{proof}