HW4

Text-Ex. 2.12. Let $M$ be a finitely generated $A$-module and $\varphi :M\to A^n$ a surjective homomorphism. Show that $\mathrm{Ker}(\varphi )$ is finitely generated. (Let $e_1,\dots ,e_n$ be a basis of $A^n$ and choose $u_1\in M$ such that $\varphi \left(u_i\right)=e_1$ $(1⩽i⩽n)$. Show that $M$ is the direct sum of $\mathrm{Ker}(\varphi )$ and the submodule generated by $u_1,\dots ,u_n$.)

\begin{proof} Let $e_1,\cdots ,e_n$ be a basis of $A^n$. Since $\varphi$ is surjective, there exist $u_1,\cdots ,u_n$ such that $\varphi (u_i)=e_i$ for $i=1,\cdots ,n$. Define $N=⟨u_1,\cdots ,u_n⟩$, which is a $A$-submodule of $M$. For any $m\in M$, $\varphi (m)\in A^n$ can be written as $\varphi (m)={\sum }_{i=1}^n{a}_i{e}_i$ with $a_i\in A$. It deduces that $m^{\prime }:={\sum }_{i=1}^n{a}_i{u}_i\in M$ satisfies $\varphi (m^{\prime })=\varphi (m)$ and $m-m^{\prime }\in \ker \varphi$. Hence, $M=N+\ker \varphi$. Also note that $N\cap \ker \varphi =0$ as ${\sum }_{i=1}^n{a}_i\varphi (u_i)=0$ iff $a_i=0$ for all $i\in \{1,\cdots ,n\}$. Therefore, $M=N\oplus \ker \varphi$.

Because $M$ is finitely generated, we suppose $M=⟨w_1,\cdots ,w_k⟩$, then each $w_i=x_i+y_i$ with $x_i\in N$ and $y_i\in \ker \varphi$. Notice that $N$ and $\ker \varphi$ are $A$-submodules of $M$, then we have $⟨x_i:1⩽i⩽k⟩=N$ and $⟨y_i:1⩽i⩽k⟩=\ker \varphi$. So $\ker \varphi$ is finitely generated. \end{proof}

Text-Ex. 2.13. Let $f:A\to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B=B{\otimes }_{A}N$. Show that the homomorphism $g:N\to N_B$ which maps $y$ to $1\otimes y$ is injective and that $g(N)$ is a direct summand of $N_B$. (Define $p:N_B\to N$ by $p(b\otimes y)=by$, and show that $N_B=\mathrm{Im}(g)\oplus \mathrm{Ker}(p)$.)

\begin{proof} Define $p:N_B\to N,b\otimes y\mapsto by$. Note that $\mathrm{im}(g)$ and $\ker (p)$ are $B$-submodules of $N_B$. For any $b\otimes y\in N_B$, we have $p(b\otimes y-1\otimes by)=0$ and $1\otimes by\in \mathrm{im}(g)$. Therefore,

$$b\otimes y=(b\otimes y-1\otimes by)+1\otimes by\in \ker (p)+\mathrm{im}(g)$$

yields that $N_B=\mathrm{im}(g)+\ker (p)$. Take $b_0\otimes y_0\in \mathrm{im}(g)\cap \ker (p)$, then $b_0{y}_0=0$ and $b_0\otimes y_0=1\otimes y_1$ for some $y_1\in N$. Remark that $b_0\otimes y_0=f(a_0)\cdot 1\otimes y_0=1\otimes b_0{y}_0=0$ with $f(a_0)=b_0$, then $b_0\otimes y_0=0$ and so $N_B=\mathrm{im}(g)\oplus \ker (p)$. Hence $g(N)=\mathrm{im}g$ is a direct summand of $N_B$.

For any $y\in Y$, we have $g(y)=1\otimes y$ and $p(g(y))=y$, so $p\circ g={\mathrm{id}}_N$. If $g(y)=0$, then $y=p(g(y))=p(0)=0$. Therefore, $g$ is injective. \end{proof}

Tensoring Back: Round-trip and Structure

Let $f:A\to B$ be a ring homomorphism, and let $N$ be a $B$-module. We view $N$ as an $A$-module via restriction of scalars, then form the $B$-module

$$N_B=B{\otimes }_{A}N.$$

Define $g:N\to N_B$ by $g(y)=1\otimes y$, and define $p:N_B\to N$ by $p(b\otimes y)=by$. Then

$$p\circ g={\mathrm{id}}_N,$$

so $g$ is injective and splits. Hence,

$$N_B=\mathrm{Im}(g)\oplus \ker (p),$$

i.e., $N$ embeds as a direct summand in $N_B$.

Summary: This “pullback-then-pushforward” operation reveals that $N$ survives inside $B{\otimes }_{A}N$, not just as a submodule, but as a direct summand. It’s a structural reflection of how scalar restriction and extension interact.

Text-Ex. 19. A sequence of direct systems and homomorphisms

$$\mathbf{M}\to \mathbf{N}\to \mathbf{P}$$

is exact if the corresponding sequence of modules and module homomorphisms is exact for each $i\in I$. Show that the sequence $\mathbf{M}\to \mathbf{N}\to \mathbf{P}$ of direct limits is then exact. (Use Exercise 15.)

\begin{proof} We aim to show, if $\left(M_i,{\mu }_i\right),\left(N_i,{\nu }_i\right)$ and $\left(P_i,{\rho }_i\right)$ are direct systems and there are families $\{\varphi {\}}_{i\in I}$ and $\{\psi {\}}_{i\in I}$ of $A$-module homomorphisms such that

$$M_i\stackrel{{\varphi }_i}{\to }{N}_i\stackrel{{\psi }_i}{\to }{P}_i$$

are exact for all $i\in I$, then the sequence of direct limits

$$\mathbf{M}\stackrel{\varphi }{\to }\mathbf{N}\stackrel{\psi }{\to }\mathbf{P}$$

is also exact. It suffices to show $\ker \psi =\mathrm{im}\varphi$.

If $\overline{n}\in \ker \psi$, then $\psi (\overline{n})\in ⟨p_i-{\rho }_{ij}(p_i):i,j\in I⟩$. By Exercise 15, each element of $\mathbf{P}=C_P/D_P$ can be written in the form ${\rho }_i(x_i)$. For any ${\rho }_i(x_i)\in \psi (\overline{n})$,

Assume $n=(n_i{)}_{i\in I}$ satisfies $\overline{n}\in \ker \psi$, then $\psi (\overline{n})=\overline{({\psi }_i(n_i){)}_{i\in I}}=0$ and so $({\psi }_i(n_i){)}_{i\in I}\in D_P$, where $D_P=⟨p_i-{\rho }_{ij}(p_i):i,j\in I⟩$. By Exercise 15, each element of $\mathbf{P}=C_P/D_P$ can be written in the form ${\rho }_i(x_i)$.

If $\overline{n}\in \mathrm{im}\varphi$, then there exists $\overline{m}\in \mathbf{M}$ such that $\varphi (\overline{m})=\overline{n}$. By the definition of $\varphi$ and $\psi$ (see exercise 18), the diagram

commutes. By Exercise 15, $\overline{m}={\mu }_i(x_i)$ for some $i$ and then $\varphi (\overline{m})=\varphi ({\mu }_i(x_i))={\nu }_i({\varphi }_i(x_i))$. It deduces that

$$\psi (\varphi (\overline{m}))=\psi ({\nu }_i({\varphi }_i(x_i)))={\rho }_i({\psi }_i({\varphi }_i(x_i)))=0.$$

Therefore, $\mathrm{im}\varphi \subseteq \ker \psi$.

If ${\overline{n}}^{\prime }\in \ker \psi$, by exercise 15, there exists $y_i^{\prime }$ such that ${\overline{n}}^{\prime }={\nu }_i(y_i)$ and then ${\rho }_i\circ {\psi }_i(y_i)=0$. Hence ${\psi }_i(y_i)=z_i-{\rho }_{ik}(z_i)$ or ${\psi }_i(y_i)=z_m-{\rho }_{mi}(z_m)$ for $m⩽i⩽k$. Then ${\psi }_i(y_i)=z_i$ with ${\rho }_{ik}(z_i)=0$ or ${\psi }_i(y_i)=0$. For the former case, ${\nu }_{ik}(y_i)\in \ker {\psi }_k=\mathrm{im}{\varphi }_k$ and so there exists $x_k^{\prime }\in M_k$ such that ${\varphi }_k(x_k^{\prime })={\nu }_{ik}(y_i)$. Hence $\varphi ({\mu }_k(x_k^{\prime }))={\overline{n}}^{\prime }$ and so ${\overline{n}}^{\prime }\in \mathrm{im}\varphi$.

For the latter case, $y_i\in \ker {\psi }_i=\mathrm{im}{\varphi }_i$ and so there exists $x_i^{\prime }\in M_i$ such that ${\varphi }_i(x_i)=y_i$. Hence $\varphi ({\mu }_i(x_i))={\overline{n}}^{\prime }$ and so ${\overline{n}}^{\prime }\in \mathrm{im}\varphi$. Therefore, $\mathrm{im}\varphi \supseteq \ker \psi$. \end{proof}