HW2

We assume that all the rings below always contain an identity.

1. Let $\phi :M\to N$ is a homomorphism of $R$-modules. Prove that:

• (1) $\phi$ is a monomorphism if and only if for any $R$-module $T$ and any two $R$-module homomorphisms $\xi ,\eta :T\to M,\phi \xi =\phi \eta$ implies $\xi =\eta$;
• (2) $\phi$ is an epimorphism if and only if for any $R$-module $S$ and any two $R$-module homomorphisms $\rho ,\theta :N\to S,\rho \phi =\theta \phi$ implies $\rho =\theta$.

\begin{proof} i) Assume that $\phi$ is monomorphism and $\phi \xi =\phi \eta$. If there exists $t\in T$ such that $\xi (t)\ne \eta (t)$, then $\phi (\xi (t)-\eta (t))=0$, which is impossible as $\phi$ is injective. Conversely, assume that for any $R$-module $T$ and any two $R$-module homomorphisms $\xi ,\eta :T\to M,\phi \xi =\phi \eta$ implies $\xi =\eta$. If $\phi$ is not injective, then there exist distinct $m_1,m_2\in M$ such that $\phi (m_1)=\phi (m_2)$. Define $\xi :M\to M,m\mapsto m_1$ and $\eta :M\to M,m\mapsto m_2$. Then $\phi \xi =\phi \eta$ and $\xi \ne \eta$, contradiction.

ii) Assume that $\phi$ is epimorphism and $\rho \phi =\theta \phi$. If $\rho \ne \theta$, then there exists $n\in N$ such that $\rho (n)\ne \theta (n)$. Since $\phi$ is epimorphism, there exists $m\in M$ such that $\phi (m)=n$. Then $\rho \phi (m)\ne \theta \phi (m)$, which is a contradiction. Hence, $\rho =\theta$. Conversely, assume that any two $R$-module homomorphisms $\rho ,\theta :N\to S,\rho \phi =\theta \phi$ implies $\rho =\theta$. If $\phi$ is not an epimorphism, then $S:=N/\mathrm{im}(\phi )\ne 0$. Define $\rho :N\to S$ as the canonical map and $\theta :N\to S$ as the zero map, then $\rho \phi =\theta \phi$ and $\rho \ne \theta$, contradiction. \end{proof}

2. Let $M$ and $M_i(1\le i\le n)$ be $R$-modules. Show that $M\simeq {⨁}_{i=1}^n{M}_i$ if and only if for every $i\in \{1,\dots ,n\}$, there exists an ${\phi }_i\in {\mathrm{End}}_R(M)$ such that $\mathrm{im}{\phi }_i\simeq M_i,{\phi }_i{\phi }_j=0$, and ${\phi }_1+{\phi }_2+\cdots +{\phi }_n={\mathrm{id}}_M$.

\begin{proof} If $M\simeq {\oplus }_{i=1}^n{M}_i$, then there exists an isomorphism $\psi :M\to {\oplus }_{i=1}^n{M}_i$. Define ${\varphi }_i:{\oplus }_{i=1}^n{M}_i\to M_i$, then ${\phi }_i={\varphi }_i\circ \psi \in {\mathrm{End}}_R(M)$ satisfies $\mathrm{im}{\phi }_i\simeq M_i$ and ${\phi }_i{\phi }_j=0$ for all $i\ne j$. Furthermore, notice that

$${\phi }_1+\cdots +{\phi }_n=({\varphi }_1+\cdots +{\varphi }_n)\circ \psi ={\mathrm{id}}_M.$$

Conversely, assume that there exist ${\phi }_i\in {\mathrm{End}}_R(M)$ such that $\mathrm{im}{\phi }_i\simeq M_i,{\phi }_i{\phi }_j=0$, and ${\phi }_1+{\phi }_2+\cdots +{\phi }_n={\mathrm{id}}_M$. Note that any $m\in M$ can be written as $m={\phi }_1(m)+\cdots +{\phi }_n(m)$, as ${\phi }_1+{\phi }_2+\cdots +{\phi }_n={\mathrm{id}}_M$. Furthermore, ${\phi }_i(m)={\phi }_1({\phi }_i(m))+\cdots +{\phi }_n({\phi }_i(m))={\phi }_i({\phi }_i(m))$ yields that ${\phi }_i{|}_{\mathrm{im}{\phi }_i}=\mathrm{id}$.

Define $\pi :M\to {\oplus }_{i=1}^n{M}_i,m\mapsto {\pi }_1({\phi }_1(m))+\cdots +{\pi }_n({\phi }_n(m))$, where ${\pi }_i:\mathrm{im}{\phi }_i\to M_i$ are isomorphisms. We claim that $\pi$ is an isomorphism. If there exists $m_0$ such that ${\pi }_i({\phi }_i(m_0))=0$ for all $i$, then ${\phi }_i(m_0)=0$. Since ${\phi }_1+{\phi }_2+\cdots +{\phi }_n={\mathrm{id}}_M$, we have $m_0=0$ and so $\pi$ is injective. For any $k_1+\cdots +k_n\in {\oplus }_{i=1}^n{M}_i$, there exists ${\ell }_i\in \mathrm{im}{\phi }_i$ such that ${\pi }_i({\ell }_i)=k_i$ for all $i$ and define $m={\sum }_{i=1}^n{\ell }_i$. Then $\pi (m)={\pi }_1({\phi }_1({\ell }_1))+\cdots +{\pi }_n({\phi }_n({\ell }_n))=k_1+\cdots +k_n$ and so $\pi$ is surjective. Now we proved that $\pi$ is an isomorphism and so $M\simeq {\oplus }_{i=1}^n{M}_i$. \end{proof}

3.

• (1) Prove $\mathbb{Z}/m\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}/(m,n)\mathbb{Z}$, where $m,n\in \mathbb{Z}$.
• (2) Let $A$ be an abelian group, prove $A{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\simeq A/nA$.
• (3) Let $A,B$ be two finitely generated abelian groups. What is $A{\otimes }_{\mathbb{Z}}B$ ?

\begin{proof} i) Define $T:\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/(m,n)\mathbb{Z},(\overline{a},\overline{b})\mapsto \overline{a}b$, and we can verify it is a bilinear map. Thus there exists a $\mathbb{Z}$-morphism

$$\varphi :\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/(m,n)\mathbb{Z},\overline{a}\otimes \overline{b}\mapsto \overline{ab}.$$

On the other hand, define

$$\psi :\mathbb{Z}/(m,n)\mathbb{Z}\to \mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z},\overline{a}\mapsto \overline{1}\otimes \overline{a}.$$

Remark that $\psi$ is well-defined: by Bezout’s lemma, there exist $a,b\in \mathbb{Z}$ such that $am+bn=(m,n)$ and then

$$\psi (\overline{(m,n)})=\overline{1}\otimes \overline{(m,n)}=\overline{1}\otimes \overline{am+bn}=\overline{am}\otimes \overline{1}+\overline{1}\otimes \overline{bn}=0.$$

In addition, $\psi$ is a $\mathbb{Z}$-morphism and $\psi ={\varphi }^{-1}$. It deduces that $\mathbb{Z}/m\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}/(m,n)\mathbb{Z}$.

ii) Define $T:A{\times }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\to A/nA,a\times \overline{m}\mapsto \overline{ma}$. Since it is bilinear, there exists

$$\varphi :A{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\to A/nA,a\otimes \overline{m}\mapsto \overline{ma}.$$

Also define

$$\psi :A/nA\to A{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z},\overline{a}\mapsto a\otimes \overline{1}.$$

It is well-defined as $\psi (\overline{na})=na\otimes \overline{1}=a\otimes \overline{n}=0$ and it is a $\mathbb{Z}$-morphism with ${\psi }^{-1}=\varphi$. Therefore, $A{\otimes }_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\simeq A/nA$.

iii) By fundamental theorem of abelian groups, we may assume that

$$A\simeq A_1\oplus \cdots \oplus A_m\oplus {\mathbb{Z}}^{s_0},B\simeq B_1\oplus \cdots \oplus B_m\oplus {\mathbb{Z}}^{k_0}$$

where $A_i$, $B_i$ are $p_i$-groups and $A_i\simeq \mathbb{Z}/p_i^{{\alpha }_{i1}}\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}/p_i^{{\alpha }_{i{s}_i}}\mathbb{Z}$, $B_i\simeq \mathbb{Z}/p_i^{{\beta }_{i1}}\mathbb{Z}\oplus \cdots \oplus \mathbb{Z}/p_i^{{\beta }_{i{k}_i}}\mathbb{Z}$. Then by i) and ii), we have

$$A{\otimes }_{\mathbb{Z}}B\simeq \left(\underset{1⩽i,j⩽m}{⨁}{A}_i{\otimes }_{\mathbb{Z}}{B}_j\right)\oplus (A{\otimes }_{\mathbb{Z}}\mathbb{Z}{)}^{k_0}\oplus (\mathbb{Z}{\otimes }_{\mathbb{Z}}B{)}^{s_0},$$

where $A_i\otimes B_i\simeq {\oplus }_{1⩽j_1⩽s_i,1⩽j_2⩽k_i}\mathbb{Z}/p_i^{{\alpha }_{i{j}_1}}\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/p_i^{{\beta }_{i{j}_2}}\mathbb{Z}={\oplus }_{1⩽j_1⩽s_i,1⩽j_2⩽k_i}\mathbb{Z}/p_i^{\min ({\alpha }_{i{j}_1},{\beta }_{i{j}_2})}\mathbb{Z}$, $A_i\otimes B_j\simeq {\mathbb{Z}}^{s_i{k}_j}$, $A\otimes \mathbb{Z}\simeq A$ and $\mathbb{Z}\otimes B\simeq B$. \end{proof}

4. Let $\phi :\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/4\mathbb{Z}$ be the unique monomorphism. Prove that

$$\mathrm{id}\otimes \phi :\mathbb{Z}/2\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}{\otimes }_{\mathbb{Z}}\mathbb{Z}/4\mathbb{Z}$$

is a zero homomorphism.

\begin{proof} Since $2\phi (\overline{1})=\overline{0}$, there is $\phi (\overline{1})\equiv 2(\mathrm{mod}4)$ and $\phi (\overline{1})=\overline{2}$. Then

$$\mathrm{id}\otimes \phi (\overline{a}\otimes \overline{b})=\overline{a}\otimes 2\overline{b}=2\overline{a}\otimes \overline{b}=0,$$

that is, $\mathrm{id}\otimes \phi$ is a zero homomorphism. \end{proof}

5. Let $V_1$ and $V_2$ be two vector spaces over a field $F$ with dimension $n$ and $m$ respectively. Let $\left\{{ϵ}_1,\dots ,{ϵ}_n\right\}$ and $\left\{{\eta }_1,\dots ,{\eta }_m\right\}$ be bases of $V_1$ and $V_2$ respectively. Suppose that $\mathcal{A}$ (rep. $\mathcal{B}$ ) is an $F$-linear map of $V_1$ (rep. $V_2$ ), and it is represented by the matrix $A={\left(a_{ij}\right)}_{n\times n}$ (rep. $B={\left(b_{ij}\right)}_{m\times m})$ with respect to the above bases. Prove that

$$\left\{{ϵ}_i\otimes {\eta }_j\mid 1\le i\le n,1\le j\le m\right\}$$

is an $F$-basis of $V_1{\otimes }_F{V}_2$. Determine the matrix representing the $F$-linear map $\mathcal{A}\otimes \mathcal{B}$ under this basis.

\begin{proof} For any $v\in V_1$ and $w\in V_2$, we have $v={\sum }_{i=1}^n{a}_i{ϵ}_i$ and $w={\sum }_{j=1}^m{b}_j{\eta }_j$ for some $a_i,b_j\in F$. Then $v\otimes w={\sum }_{1⩽i⩽n,1⩽j⩽m}{a}_i{b}_j{ϵ}_i\otimes {\eta }_j$ and so $\left\{{ϵ}_i\otimes {\eta }_j:1⩽i⩽n,1⩽j⩽m\right\}$ is an $F$-basis of $V_1{\otimes }_F{V}_2$.

Furthermore, we can compute that

$$\mathcal{A}\otimes \mathcal{B}({ϵ}_s\otimes {\eta }_t)=(\sum _{i=1}^n{a}_{is}{ϵ}_i)\otimes (\sum _{j=1}^m{b}_{jt}{\eta }_j)$$

and so the the matrix representing the $F$-linear map $\mathcal{A}\otimes \mathcal{B}$ under this basis is

$$\left[\begin{array}{ccc}{b}_{11}A& \cdots & b_{1n}A\\ ⋮& & ⋮\\ b_{m1}A& \cdots & b_{mm}A\end{array}\right]$$

where the basis is in the following order

$${ϵ}_1\otimes {\eta }_1,{ϵ}_2\otimes {\eta }_1,\cdots ,{ϵ}_n\otimes {\eta }_1,{ϵ}_1\otimes {\eta }_2,\cdots ,{ϵ}_n\otimes {\eta }_m.$$