HW4

Let $R$ be a ring with $1$.

1. Prove: the following three conditions on an $R$-module $V$ are equivalent.

• (1) $V$ is a Noetherian module.
• (2) (Ascending chain condition.) For every infinite chain of $R$ submodules of $V$,

$$V_1\subseteq V_2\subseteq \cdots \subseteq V_n\subseteq \cdots ,$$

there is a number $m$ such that $V_m=V_{m+1}=\cdots$.

• (3) Every $R$-submodule of $V$ is finitely generated.

\begin{proof} It suffices to show the following proposition.

The followings are equivalent:

• Every non-empty family of submodule of $M$ has a maximal element. Namely, given $P=\{M_j{\}}_{j\in J}$, there exists $N\in P$ such that if $N\subseteq M_j$, then $M_j=N$.
• Every increasing sequence of submodule $\{M_i{\}}_{i=1}^{\infty }$ is stationary. Namely, there exists $N$ such that $M_i=M_N$ for all $i⩾N$.
• Every submodule of $M$ is finitely generated.

i)→iii) Let $E\subseteq M$ be a submodule. Define $\Phi =\{G\subseteq E:G\text{mbox}isfinitelygenerated\}$. Note that $(0)\in \Phi$. By i), there exists a maximal element $F\in \Phi$. Assume that $F\ne E$, then there exists $x\in F\setminus E$ such that $⟨F,x⟩\in \Phi$, which is a contradiction. Therefore, $F=E$ and so $E$ is finitely generated. By the arbitrary of $E$, we finish the proof of iii).

iii)→ii) Consider the increasing sequence $\{M_i{\}}_{i=1}^{\infty }$ and define $E={\cup }_{i=1}^{\infty }{M}_i$, then $E$ is still a submodule of $M$ (for any $x,y\in M$, verify $ax+by\in M$). By iii), $E$ is finitely generated. Suppose that $\{e_1,\cdots ,e_d\}$ is a set of generators of $E$. Assume that $e_i\in M_{n_i}$, then $\{e_1,\cdots ,e_d\}\in M_N$, where $N=\max \{n_1,\cdots ,n_d\}$. It follows that $E=M_N$ and $M_n=M_N$ for all $n⩾N$.

ii)→i) Consider $P=\{M_i{\}}_{i\in I}$. Assume that there does not exist maximal element, then there is a strictly increasing sequence

$$N_1⊊N_2⊊N_3\cdots$$

which contradicts with ii). \end{proof}

2. Let $V$ be a Noetherian $R$-module and $f:V\to V$ an epimorphism. Show that $f$ is an $R$-isomorphism.

\begin{proof} Note that $f^n:V\to V$ is also surjective, and $\ker (f^{n-1})\subseteq \ker (f^n)$ for any $n\in {\mathbb{N}}_{+}$. Then we have a chain $\ker (f)\subseteq \ker (f^2)\subseteq \cdots \subseteq \ker (f^n)\subseteq \cdots$. There exists $n$ such that $\ker (f^n)=\ker (f^{n+m})$ for any $m\in \mathbb{N}$ as $V$ is Noetherian.

It deduces that $f(v^{\prime })\ne 0$ for any nonzero $v^{\prime }=f^n(v)\in V$. Note that $f^n(V)=V$, so $f$ is injective. Therefore, $f$ is an isomorphism. \end{proof}

3. Are the following statements correct?

• (1) Any submodule of a semisimple module is also semisimple.
• (2) Any semisimple Noetherian module is Artinian.

\begin{proof} (1) Correct. Assume that $U$ is a semisimple module and $V\subseteq U$ is a submodule. Since $U$ is semisimple, each submodule of $U$ is a direct summand of $U$. Thus $U=V\oplus V^{\prime }$ for some $V^{\prime }$. For any submodule $V_1\subseteq V$, $V_1$ is also a submodule of $U$ so $U=V_1\oplus V_1^{\prime }$. Define $V_2:=V_1^{\prime }\cap V$, then $V_1\cap V_2=\{0\}$ and $V=V_1\oplus V_2$. Therefore, any submodule of $V$ is a direct summand of $V$ and so $V$ is semisimple.

(2) Correct. Assume that $W$ is a semisimple Noetherian module, and assume that $W={\oplus }_{i\in I}{S}_i$ with simple modules $S_i$. As

$$S_1\subseteq S_1\oplus S_2\subseteq \cdots \subseteq {\oplus }_{i=1}^n{S}_i\subseteq \cdots$$

is stationary, $|I|<\infty$ and so $W$ has a finite length composition series. Recall the following proposition

$M$ has a finite length composition series iff $M$ is Noetherian and Artinian.

Link to original

$W$ is Artinian. \end{proof}

4. Prove that: the following conditions on an $R$-module $V$ with composition series are equivalent.

• (1) $V$ is semisimple.
• (2) Every irreducible submodule of $V$ is a direct summand of $V$.
• (3) Every maximal submodule of $V$ is a direct summand of $V$.

\begin{proof} (1)→(2). Since $V$ is semisimple, $V={\oplus }_{i\in I}{S}_i$ for some simple modules $S_i$. For any irreducible submodule $W\subseteq V$, define $\mathcal{J}=\{J\subseteq I:{\oplus }_{j\in J}{S}_j\cap W=0\}$. By Zorn’s lemma, there exists maximal element $J_0\in \mathcal{J}$. If ${\oplus }_{j\in J_0}{S}_j+W\ne V$, then there exists $S_i⊈({\oplus }_{j\in J_0}{S}_j+W)$ and so $J_0\cup i\in \mathcal{J}$, contradiction. Hence ${\oplus }_{j\in J_0}{S}_j+W=V$ with ${\oplus }_{j\in J_0}{S}_j\cap W=0$, which deduces that $V={\oplus }_{j\in J_0}{S}_j\oplus W$ and so $W$ is a direct summand of $V$.

(2)→(3) Let $M\subseteq V$ be a maximal submodule. Let $\mathcal{S}$ be the set of irreducible submodules of $V$. For any $S\in \mathcal{S}$, either $S\cap M=0$ or $S\cap M=S$ by $S$ irreducible. Take $v\in V\setminus M$, then $v^R\in \mathcal{S}$ and $v^R\cap M=0$. Note that $v^R+M=V$, otherwise $V\supset v^R+M\supset M$ contradicts with $M$ maximal. Thus $M\oplus v^R=V$ and so $M$ is a direct summand of $V$.

(3)→(1) For any maximal submodule $M_1\subseteq V$, we have $V=M_1\oplus N_1$ for some irreducible $N_1$. (Otherwise there is $N_1^{\prime }\subseteq N_1$ and $M_1<⟨N_1^{\prime },M_1⟩<V$, which is impossible by $M_1$ maximal.) Take a maximal module $M_2$ containing $N_1$, and there exists irreducible $N_2$ such that $V=M_2\oplus N_2$. Then $N_1\cap N_2=0$ and $N_1\oplus N_2$ is a submodule of $V$. Take a maximal module $M_3$ containing $N_1\oplus N_2$, then $V=M_3\oplus N_3$ and we get $N_1\oplus N_2\oplus N_3$. Repeat this procedure and there is a chain

$$N_1\subseteq N_1\oplus N_2\subseteq \cdots \subseteq {\oplus }_{i=1}^m{N}_i\subseteq \cdots$$

where $N_i$ are irreducible modules. Since $V$ has a composition series, $V$ is Noetherian and so the chain is stationary. Therefore, there exists $n$ such that ${\oplus }_{i=1^n}{N}_i=V$ and so $V$ is semisimple. \end{proof}

5. Prove: $\mathrm{Soc}(U\oplus V)=\mathrm{Soc}(U)\oplus \mathrm{Soc}(V)$.

\begin{proof} Assume that $S=\mathrm{soc}(U\oplus V)$, then $S$ is the largest semisimple submodule of $U\oplus V$. Define ${\pi }_U:U\oplus V\to U$ and ${\pi }_V:U\oplus V\to V$ as two projections, then ${\pi }_U$ and ${\pi }_V$ are homomorphisms between modules. So ${\pi }_U(S)$ and ${\pi }_V(S)$ are semisimple. It deduces that ${\pi }_U(S)\subseteq \mathrm{soc}(U)$ and ${\pi }_V(S)\subseteq \mathrm{soc}V$. Now we have

$$\mathrm{soc}(U\oplus V)=S\subseteq {\pi }_U(S)\oplus {\pi }_V(S)\subseteq \mathrm{Soc}(U)\oplus \mathrm{Soc}(V).$$

Assume that $S_1=\mathrm{soc}(U)$ and $S_2=\mathrm{soc}(V)$, then $S_1\oplus S_2\subseteq U\oplus V$ is semisimple and so $\mathrm{soc}(U)\oplus \mathrm{soc}(V)=S_1\oplus S_2\subseteq \mathrm{soc}(U\oplus V)$. Now we finish the proof. \end{proof}