Socle and Radical of Module

Socle

Definition

The unique largest semisimple submodule of $U$ is called the socle of $U$, denote $\mathrm{soc}(U)$.

Radical of Modules

Definition

A maximal submodule $N$ of $M$ is a proper submodule $N$ satisfying $N\subseteq U\subseteq M$ $⟹$ $U=N$ or $U=M$. Equivalently, $M/N$ is simple.

Define $\mathrm{Rad}(M)={\cap }_N\{\text{mbox}maximalsubmodulesN\subseteq M\}$. If $M$ does not have maximal submodules, then define $\mathrm{Rad}(M)=M$.

Lemma

Let $W$ be a proper $R$-submodule of ${}_{R}V$. If $V/W$ is finitely generated over $R$, then there exists a maximal $R$-submodule of $V$ containing $W$. In particular, if $V$ is finitely generated, then there exists a maximal submodule of $V$ containing $W$.

\begin{proof} Let $\overline{V}=V/W={\sum }_{i=1}^{n}R{\overline{v}}_i$ with $v_i\in V$. Let $\mathcal{O}$ be the set of proper $R$-submodule of $V$ containing $W$. It is a poset with inclusion.

Claim that an arbitrary totally ordered subset $\{W_{\lambda }:\lambda \in \Lambda \}$ in $\mathcal{O}$, it has a upper bound. Let $U={\cup }_{\lambda \in \Lambda }{W}_{\lambda }$. If $U=V$, then for each $i$, there exists $W_{{\lambda }_i}$ such that $v_i\in W_{{\lambda }_i}$. Let $W_{\alpha }$ be the largest among $\{W_{{\lambda }_i}:i=1,\cdots ,n\}$. Then $W_{\alpha }=V$, which is a contradiction.

So $U\ne V$, and $U\in \mathcal{O}$ is a upper bound for $W_{\lambda }$. By Zorn’s lemma, we finish the proof. \end{proof}

Corollary

For every proper left ideal $I$ of $R$, there exists a maximal left ideal of $R$ containing $I$.

Lemma

Let $M,N$ be finitely generated $R$-modules, and let $f:M\to N$ be $R$-morphism.

• $f(\mathrm{Rad}(M))\subseteq \mathrm{Rad}(N)$.
• If $f$ is an epimorphism with $\ker f\subseteq \mathrm{Rad}(M)$, then $\mathrm{Rad}(N)=f(\mathrm{Rad}(M))$.

\begin{proof} i) Let $x\in \mathrm{Rad}(M)$. We show $f(x)\in \mathrm{Rad}(N)$. Suppose $K\subseteq N$ is a maximal submodule. Consider the composite map:

$$g:M\stackrel{f}{\to }N\stackrel{\pi }{\to }N/K$$

where $\pi$ is the quotient map. Since $N/K$ is simple, $\ker (g)$ is either $M$ or a maximal submodule of $M$.

• If $\ker (g)=M$, then $g(x)=0⟹\pi (f(x))=0⟹f(x)\in K$.
• If $\ker (g)$ is maximal, then $x\in \mathrm{Rad}(M)\subseteq \ker (g)$ (as $\mathrm{Rad}(M)$ is the intersection of all maximal submodules). Hence $g(x)=0⟹f(x)\in K$.

In either case, $f(x)\in K$. Since $K$ was arbitrary, $f(x)\in \mathrm{Rad}(N)$. Thus, $f(\mathrm{Rad}(M))\subseteq \mathrm{Rad}(N)$.

ii) It suffices to show $\mathrm{Rad}(N)\subseteq f(\mathrm{Rad}(M))$.

Let $K=\ker f\subseteq \mathrm{Rad}(M)$. By the isomorphism theorem, $N\cong M/K$. For any quotient module $M/K$, we have

$$\mathrm{Rad}(M/K)=(\mathrm{Rad}(M)+K)/K$$

Since $K\subseteq \mathrm{Rad}(M)$, one have $\mathrm{Rad}(M/K)=\mathrm{Rad}(M)/K$. Under the isomorphism $M/K\cong N$, it deduces that $\mathrm{Rad}(N)=f(\mathrm{Rad}(M))$. \end{proof}

Lemma

Let $U$ be an $R$-module.

• Suppose that $M_1,\cdots ,M_n$ are maximal submodules of $U$. Then there is a subset $I\subseteq \{1,\cdots ,n\}$ such that

  $$U/(M_1\cap \cdots \cap M_n)\simeq {\oplus }_{i\in I}U/M_i$$

• Suppose that $U$ is Artinian, then $U/\mathrm{Rad}U$ is a semisimple and $\mathrm{Rad}U$ is the unique smallest submodule of $U$ with semisimple quotient.

\begin{proof} i) Let $I$ be the maximal subset of $\{1,\cdots ,n\}$ with the property that the quotient homomorphism ${\phi }_i:U/{\cap }_{i\in I}{M}_i\to U/M_i$ induce an isomorphism $U/{\cap }_{i\in I}{M}_i\to {\oplus }_{i\in I}U/M_i,u+{\cap }_{i\in I}{M}_i\mapsto ({\phi }_i(u+{\cap }_{i\in I}{M}_i){)}_{i\in I}$. Note that $I$ exists, because $\{1\}$ is a subset such that the property holds.

We will show ${\cap }_{i\in I}{M}_i=M_1\cap \cdots \cap M_n$. If not, there exists $j$ such that ${\cap }_{i\in I}{M}_i⊈M_j$. Consider the homomorphism

$$f:U\to ({\oplus }_{i\in I}U/M_i)\oplus (U/M_j),u\mapsto ((u+M_i{)}_{i\in I},u+M_j),$$

and $f$ has kernel $M_j\cap ({\cap }_{i\in I}{M}_i)$. It suffices to show $f$ is surjective, because this implies the larger set $I\cup \{j\}$ has the same property as $I$. Let

$$g:U\to (U/{\cap }_{i\in I}{M}_i)\oplus (U/M_j).$$

Observe that $M_j+({\cap }_{i\in I}{M}_i)=U$. If $x\in U$, then $x=y+z$ with $y\in {\cap }_{i\in I}{M}_i$ and $z\in M_j$ such that $g(y)=(0,x+M_j)$ and $g(z)=(x+{\cap }_{i\in I}{M}_j,0)$. So $g$ is surjective and $f$ is surjective.

ii) Since $U$ is Artinian, $\mathrm{Rad}U$ is the intersection of finitely many maximal submodules. So $U/\mathrm{Rad}U$ is semisimple by i).

If $V$ is a submodule of $U$ with semisimple $U/V$, then $U/V$ is Artinian and $U/V\simeq S_1\oplus \cdots \oplus S_n$ where $S_i$ is simple. Let $M_i$ be the kernel of the composition $U\to U/V\to S_i$, then $M_i$ is maximal and ${\cap }_{i=1}^n{M}_i\subseteq V$. Also $V\subseteq M_i$ as $V$ is the kernel of $U\to U/V\to S_i$ for all $i$. It deduces that $M_1\cap \cdots \cap M_n=V$ and so $\mathrm{Rad}U\subseteq V$. \end{proof}

Radical and Essential

Nakayama's lemma

Let $U$ be Noetherian(or finitely generated). Then $U\to U/\mathrm{Rad}U$ is essential. Equivalently, if $V$ is a submodule of $U$ with $V+\mathrm{Rad}U=U$, then $V=U$.

\begin{proof} If $V\ne U$, there exists maximal $W\supseteq U$ by ^fbacdf and $V+\mathrm{Rad}U\subseteq W\ne U$, contradiction. \end{proof}

Remark. The role of the Jacobson radical in Nakayama lemma is analogous to that of the Frattini subgroup in group theory. Both concepts capture the idea of “non-generators,” where elements from these substructures are superfluous and can be disregarded when considering generating sets.

Proposition

• Suppose that $f:U\to V$ and $g:V\to W$ are $R$-morphisms of $R$-modules. If two of $f,g,gf$ are essential, then so is the third.
• Let $f:U\to V$ be a homomorphism of Noetherian modules. Then $f$ is an essential epimorphism iff the radical quotient $U/\mathrm{Rad}U\to V/\mathrm{Rad}V$ is an essential epimorphism. In fact, if it holds, then the radical quotient is an isomorphism.
• Let $f_i:U_i\to V_i$ be homomorphisms of Noetherian modules for $i=1,\cdots ,n$, then $f_i$ are all essential epimorphisms iff $\oplus f_i:\oplus U_i\to \oplus V_i$ is an essential epimorphism.

\begin{proof} i) Suppose $f$ and $g$ are essential, then $gf$ is surjective. If $U_0\subseteq U$ is a proper submodule, then $f(U_0)⊊V$ and $gf(U_0)⊊W$. Hence $gf$ is surjective.

Suppose $f$ and $gf$ are essential, then $g$ is surjective. Let $V_0⊊V$. If $g(V_0)=W$, then $f^{-1}(V_0)\ne U$ satisfying $gf(f^{-1}(V_0))=W$ and $gf$ is not essential, which is impossible. So $g$ is essential.

Suppose $g$ and $gf$ are essential. If $f$ is not surjective, then $f(U)⊊V$ and $gf(U)=g(f(U))=W$. It contradicts with $g$ essential and so $f$ is surjective. If $f$ is not essential, then there exists $U_0⊊U$ such that $f(U_0)=V$. It deduces that $gf(U_0)=g(V)=W$, leading to a contradiction. Thus, $f$ is essential.

ii) Consider the commutative diagram

By ^3f1b99, $\downarrow$ ‘s are essential. So $f$ is essential iff $\overline{f}$ is essential.

Furthermore, we claim that if $\overline{f}:U/\mathrm{Rad}U\to V/\mathrm{Rad}V$ is an essential epimorphism, then $\overline{f}$ is an isomorphism. To show it, it suffices to show $\overline{f}$ is injective. Otherwise, if $\ker \overline{f}\ne \{0\}$, then semisimple $U/\mathrm{Rad}U$ can be written as $U/\mathrm{Rad}U=\ker \overline{f}\oplus K^{\prime }$. Then $\overline{f}(K^{\prime })=V/\mathrm{Rad}V$ contradicts with $\overline{f}$ essential. Therefore, $\overline{f}$ is an isomorphism.

iii) It suffices to consider $\oplus U_i/\mathrm{Rad}(\oplus U_i)\to \oplus V_i/\mathrm{Rad}(\oplus V_i)$. Note that $\oplus (U_i/\mathrm{Rad}{U}_i)\simeq \oplus U_i/\mathrm{Rad}(\oplus U_i)$ and $\oplus (V_i/\mathrm{Rad}{V}_i)\simeq \oplus V_i/\mathrm{Rad}(\oplus V_i)$, then we have done. \end{proof}

Corollary

If $P$ and $Q$ are projective Noetherian modules over a ring, then $P\simeq Q$ iff $P/\mathrm{Rad}P\simeq Q/\mathrm{Rad}Q$.

\begin{proof} By ^3f1b99, $P\to P/\mathrm{Rad}P$ and $Q\to Q/\mathrm{Rad}Q$ are essential. Thus $P$ (rep. $Q$) is a projective cover of $P/\mathrm{Rad}P$ (rep. $Q/\mathrm{Rad}Q$). By ^9so7ao, projective cover is unique up to isomorphism and so we finish the proof. \end{proof}