HW1

1. Let $R$ be a ring, and let $M$ be an R-module. Suppose that $S$ is a ring and $\theta :S\to R$ is a ring homomorphism satisfying $\theta \left(1_S\right)=1_R$. Let $S$ act on $M$ by $sx=\theta (s)x$, where $x\in M,s\in S$. Prove that $M$ is an $S$-module.

\begin{proof} By definition of the action, we have $s\cdot (x+y)=\theta (s)\cdot (x+y)$ and $s\cdot (x+y)=\theta (s)\cdot (x+y)$. Since $\theta$ is a ring homomorphism, there is $\theta (s+t)=\theta (s)+\theta (t)$ and so

$$\theta (s+t)\cdot x=(\theta (s)+\theta (t))\cdot x=\theta (s)\cdot x+\theta (t)\cdot x=s\cdot x+t\cdot x.$$

By definition of the action, we have $(st)\cdot x=\theta (st)\cdot x$. Since $\theta$ is a ring homomorphism, we know $\theta (st)=\theta (s)\theta (t)$. Thus

$$(st)\cdot x=\theta (st)\cdot x=\theta (s)\theta (t)\cdot x.$$

Since $M$ is an $R$-module, we have

$$s\cdot (t\cdot x)=\theta (s)\cdot (\theta (t)\cdot x)=(\theta (s)\theta (t))\cdot x=(st)\cdot x.$$

Since $\theta \left(1_S\right)=1_R$, we have $1_S\cdot x=\theta \left(1_S\right)\cdot x=1_R\cdot x=x$. Therefore, $M$ is a $S$-module. \end{proof}

2. Let $R$ be a ring, and let $M$ and $N$ be two right $R$-modules. Let ${\mathrm{Hom}}_{\mathbb{Z}}(M,N)$ be the set of group homomorphisms from $M$ to $N$ (as abelian groups). The action of $R$ on ${\mathrm{Hom}}_{\mathbb{Z}}(M,N)$ is given as follows: for any $a\in R,\phi \in {\mathrm{Hom}}_{\mathbb{Z}}(M,N)$, define

$$(\phi a)(x)=\phi (xa),\forall x\in M$$

Prove that ${\mathrm{Hom}}_{\mathbb{Z}}(M,N)$ is a left $R$-module.

\begin{proof} For any $x\in M$, we have:

$$(\phi +\psi )a(x)=(\phi +\psi )(xa)=\phi (xa)+\psi (xa)=(\phi a)(x)+(\psi a)(x).$$

Thus $(\phi +\psi )a=\phi a+\psi a$. We also have

$$(\phi (a+b))(x)=\phi (x(a+b))=\phi (xa+xb)=\phi (xa)+\phi (xb)=(\phi a)(x)+(\phi b)(x),$$

that is, $\phi (a+b)=\phi a+\phi b$. Furthermore, there is

$$((\phi (ab))(x)=\phi (x(ab))=\phi ((xa)b)=(\phi b)(xa)=(\phi b)a(x)$$

and $\left(\phi 1_R\right)(x)=\phi \left(x{1}_R\right)=\phi (x)=\phi (x)$. Therefore, $(\phi )(ab)=(\phi b)(a)$ and $\phi 1_R=\phi$. So ${\mathrm{Hom}}_{\mathbb{Z}}(M,N)$ is a left $R$-module. \end{proof}

3. Let $R$ be a ring, and let $M$ be an $R$-module. Denote ${\mathrm{Ann}}_{R}M=\{r\in R\mid rm=$ $0,\forall m\in M\}$. We call $M$ a faithful $R$-module if ${\mathrm{Ann}}_{R}M=\{0\}$. Define the action of the quotient ring $R/{\mathrm{Ann}}_{R}M$ on $M$ by $\bar{r}x=rx$, where $x\in M,r\in R,\bar{r}=r+{\mathrm{Ann}}_{R}M$. Prove that $M$ is a faithful $R/{\mathrm{Ann}}_{R}M$-module.

\begin{proof} Otherwise, $R/{\mathrm{Ann}}_{R}M$ is not faithful on $M$. Then there exists nonzero $\overline{r}\in R/{\mathrm{Ann}}_{R}M$ such that $\overline{r}m=0$ for all $m\in M$. Since $\overline{r}m=(r+{\mathrm{Ann}}_{R}M)m=rm=0$, we have $r\in {\mathrm{Ann}}_{R}M$ and so $\overline{r}=0$, leading to a contradiction. \end{proof}

4. Let $\phi :M\to M$ be a homomorphism of $R$-modules such that $\phi \phi =\phi$. Prove that

$$M=\ker \phi \oplus \mathrm{im}\phi .$$

\begin{proof} For any $m\in M$, $m$ can be written as $m=m-\phi (m)+\phi (m)$. Note that $\phi (m-\phi (m))=0$ and $\phi (\phi (m))=\phi (m)$, then we have $m-\phi (m)\in \ker \phi$ and $\phi (m)\in \mathrm{im}\phi$. By the arbitrary of $m$, $M\subseteq \ker \phi \cup \mathrm{im}\phi$. Since $\ker \phi \cap \mathrm{im}\phi =\{0\}$, $M=\ker \phi \oplus \mathrm{im}\phi$. \end{proof}

5. Show that any $R$-module is a homomorphism image of some free $R$-module.

\begin{proof} For any $R$-module $M$, assume that the generator set of $M$ is $\{m_i:i\in I\}$. Define $M_0$ as the free abelian group generated by the set $\{m_i:i\in I\}$, then there is a $R$-module $M_0$ induced naturally from the $R$-module $M$. By the universal property of the free abelian group, there exists a unique group homomorphism $\phi :M_0\to M$ with $\phi (m)=m$ for all $m\in M$. Note that $\phi (r{m}_i)=r{m}_i=r\phi (m_i)$, so $\phi :M_0\to M$ is a homomorphism of modules. Therefore, $M$ is a homomorphism image of free $R$-module $M_0$. \end{proof}

6. Let $R$ be a commutative ring, and suppose that $\phi$ is an endomorphism of the free $R$-module $R^m$. Prove that if $\phi$ is surjective, then $\phi$ is injective. Consider that: if $\phi$ is injective, is $\phi$ surjective?

\begin{proof} Recall the following proposition mentioned in class:

• In particular, for any $R$-module $M$, any epimorphism $\pi :M\twoheadrightarrow L$ has a section, that is, there is a morphism $\sigma :L\to M$ such that $\pi \sigma ={\mathrm{id}}_L$. In that case, $\sigma$ is injective, $\mathrm{im}(\sigma )=L$ and $M=\ker \pi \oplus \mathrm{im}(\sigma )$.

Link to original

By this proposition, if $\phi :R^m\to R^m$ is surjective, then there exists a morphism $\sigma :R^m\to R^m$ such that $\phi \sigma ={\mathrm{id}}_{R^m}$, $\sigma$ is injective and $R^m=\ker \phi \oplus \mathrm{im}(\sigma )$. Since $\mathrm{im}(\sigma )=R^m$, we have $\ker \phi =\{0\}$ and so $\phi$ is injective.

Conversely, if $\phi$ is injective, $\phi$ is not necessarily surjective. For example, define $\phi :R^m\to R^m,(r_1,\cdots ,r_m)\mapsto (2r_1,\cdots ,2r_m)$. Then $\phi$ is injective but not surjective. \end{proof}