HW3

Text-Ex. 6.1.

• i) Let $M$ be a Noetherian A-module and $u:M\to M$ a module homomorphism. If $u$ is surjective, then $u$ is an isomorphism.
• ii) If $M$ is Artinian and $u$ is injective, then again $u$ is an isomorphism. (For (i), consider the submodules $\ker (u^n)$; for (ii), the quotient modules $\mathrm{coker}(u^n)$.)

\begin{proof} i) Since $u$ is surjective, $u^n:M\to M$ is also surjective for any $n\in {\mathbb{N}}_{+}$. Define $M_i=\ker u_i$, then we have a chain $M_1\subseteq M_2\subseteq \cdots \subseteq M_n\subseteq \cdots$. There exists $n$ such that $M_n=M_{n+m}$ for any $m\in \mathbb{N}$ as $M$ is Noetherian. It deduces that $u(m^{\prime })\ne 0$ for any nonzero $m^{\prime }=u^n(m)\in M$. Note that $u^n(M)=M$, so $u$ is injective. Therefore, $u$ is an isomorphism.

ii) Similarly, as $u^n$ is also injective and $M$ is Artinian, there exists $n$ such that $\mathrm{im}(u^n)=\mathrm{im}(u^{n+m})$ for any $m\in \mathbb{N}$. It deduces that $u^n{|}_{\mathrm{im}(u^n)}$ is an isomorphism. For any $m\notin \mathrm{im}{u}^n$, there is $u(m)\notin \mathrm{im}{u}^n$, otherwise there exists $m^{\prime }\in \mathrm{im}{u}^n$ such that $u(m^{\prime })=u(m)$, which is impossible by $u$ injective. Thus, $m\notin \mathrm{im}{u}^n$ yields $u(m)\notin \mathrm{im}{u}^n$ and so $u^n\notin \mathrm{im}{u}^n$, which is a contradiction. Therefore, $M=\mathrm{im}{u}^n$, $u^n$ is an isomorphism and so $u$ is surjective. \end{proof}

Text-Ex. 6.6. Prove that the following are equivalent:

• i) $X$ is Noetherian.
• ii) Every open subspace of $X$ is quasi-compact.
• iii) Every subspace of $X$ is quasi-compact.

\begin{proof} i)→ii) Assume that $X$ is Noetherian. For open subspace $O\subseteq X$ and its open cover $\mathcal{U}$, the chain $U_1\subseteq U_1\cup U_2\subseteq \cdots \subseteq {\cup }_{i\in I}{U}_i$ has to stop by a.c.c. Hence $\mathcal{U}$ has a finite subcover and so $O$ is quasi-compact.

ii)→iii) For any subspace $S\subseteq X$ and its open cover $\mathcal{V}$, notice that each $V\in \mathcal{V}$ is induced from an open set $U_V$ with $V=S\cap U_V$. Define $U={\cup }_{V\in \mathcal{V}}{U}_V$, then $U$ is open and $\{U_V:V\in \mathcal{V}\}$ is an open cover. There exists a finite subcover ${\cup }_{i=1}^n{U}_i=U$ with $U_i=U_{V_i}$ for some $V_i$. Then ${\cup }_{i=1}^n{V}_i={\cup }_{i=1}^n(S\cap U_i)=S\cap U=S$ is a finite subcover of $\mathcal{V}$. Hence every subspace of $X$ is quasi-compact.

iii)→i) If there exists a family of open sets $\{U_i:i\in I\}$ does not satisfy a.c.c., that is, $U_1\subseteq U_1\cup U_2\subseteq \cdots \subseteq {\cup }_{i\in I}{U}_i(\ast )$ is not stationary, then ${\cup }_{i\in I}{U}_i$ has an open cover $\mathcal{U}:=\{U_i:i\in I\}$ and $\mathcal{U}$ does not have a finite subcover (Otherwise the chain $(\ast )$ stops at some $n$). It is impossible. \end{proof}

Text-Ex. 6.7. A Noetherian space is a finite union of irreducible closed subspaces. (Consider the set $\Sigma$ of closed subsets of $X$ which are not finite unions of irreducible closed subspaces.) Hence the set of irreducible components of a Noetherian space is finite.

\begin{proof} Let $\Sigma$ be the set of closed subsets of $X$ which are not finite unions of irreducible closed subspaces. Since $X$ is Noetherian, there exists a minimal element $\sigma$ in $\Sigma$. Note that $\sigma$ is reducible and it can be written as a union of two closed subspace, that is, $\sigma ={\sigma }_1\cup {\sigma }_2$. Since $\sigma$ is the minimal element in $\Sigma$, both ${\sigma }_1$ and ${\sigma }_2$ can be written as finite union of irreducible closed subspaces. It deduces that $\sigma$ can be written as finite union of irreducible closed subspaces, which is a contradiction. \end{proof}

Proposition 2.9, (ii). Let $0\to N^{\prime }\stackrel{u}{\to }N\stackrel{v}{\to }{N}^{\prime \prime }(☺\stackrel{◯}{R})$ be a sequence. It is exact iff for any $A$-module $M$, the induced sequence

$$0\to \mathrm{Hom}(M,N^{\prime })\stackrel{\overline{u}}{\to }\mathrm{Hom}(M,N)\stackrel{\overline{v}}{\to }\mathrm{Hom}(M,N^{\prime \prime })\text{\hspace{1em}(☺R◯☺R◯)}$$

is exact.

\begin{proof} ii) "→" Assume that $(☺\stackrel{◯}{R})$ is exact. It is enough to show $\overline{u}$ is injective and $\ker \overline{v}=\mathrm{im}\overline{u}$. If $\overline{u}(f)=0$, then $u\circ f=0$. Since $u$ is injective, we have $f=0$ and so $\overline{u}$ is injective. Note that $\overline{v}\circ \overline{u}=0$ as $v\circ u=0$, so $\mathrm{im}\overline{u}\subseteq \ker \overline{v}$. For any $g\in \ker \overline{v}$, we aim to find $f$ such that $g=f\circ u$. For any $m\in M$, define $f(m)=u^{-1}(g(m))$. Since $g(m)\in \ker v=\mathrm{im}u$ and $u$ is injective, $f$ is well-defined and $g=f\circ u$. Hence $\mathrm{im}\overline{u}=\ker \overline{v}$.

"←" Assume that $(☺\stackrel{◯}{R}☺\stackrel{◯}{R})$ is exact. It suffices to show $u$ is injective and $\ker v=\mathrm{im}u$. If $u$ is not injective, then $\ker u\ne 0$. Let $M=\ker u$ and apply it to $(☺\stackrel{◯}{R}☺\stackrel{◯}{R})$. The inclusion map $\iota :\ker u\hookrightarrow N^{\prime }$ is included in $\mathrm{Hom}(M,N^{\prime })$ and $\overline{u}(\iota )=0$, which contradicts with $\overline{u}$ injective. Take $M=N^{\prime }$ and ${\mathrm{id}}_{N^{\prime }}\in \mathrm{Hom}(N^{\prime },N^{\prime })$ satisfies $\overline{v}\circ \overline{u}\circ f=v\circ u=0$, that is, $\mathrm{im}u\subseteq \ker v$. Conversely, take $M=\ker v$ and

$$0\to \mathrm{Hom}(\ker v,N^{\prime })\stackrel{\overline{u}}{\to }\mathrm{Hom}(\ker v,N)\stackrel{\overline{v}}{\to }\mathrm{Hom}(\ker v,N^{\prime \prime })$$

is exact. Since ${\iota }^{\prime }:\ker v\to N$ is an element of $\mathrm{Hom}(\ker v,N)$ and $\overline{v}({\iota }^{\prime })=0$, there is ${\iota }^{\prime }\in \ker \overline{v}=\mathrm{im}\overline{u}$ and so there exists $\varphi$ such that $u\circ \varphi ={\iota }^{\prime }$. Then for any $m\in \ker v$, we have $m={\iota }^{\prime }(m)=u(\varphi (m))$ and $m\in \mathrm{im}u$. Now we finish the proof.
\end{proof}

Proposition 2.10. Consider a commutative diagram where both rows are short exact.

$$\begin{array}{ccccccccc}0& \to & M^{\prime }& \stackrel{G}{\to }& M& \stackrel{H}{\to }& M^{\prime \prime }& \to & 0\\ & & f^{\prime }\downarrow & & f\downarrow & & f^{\prime \prime }\downarrow & & \\ 0& \to & N^{\prime }& \stackrel{g}{\to }& N& \stackrel{h}{\to }& N^{\prime \prime }& \to & 0\end{array}$$

Then we have an exact sequence

$$0\to \ker f^{\prime }\to \ker f\to \ker f^{\prime \prime }\to \mathrm{coker}{f}^{\prime }\to \mathrm{coker}f\to \mathrm{coker}{f}^{\prime \prime }\to 0.$$

Prove the sequence is exact at the two places $\ker f^{\prime \prime }$ and $\mathrm{coker}{f}^{\prime }$.

\begin{proof} Notice that $\ker f^{\prime }\to \ker f\to \ker f^{\prime \prime }$ are restriction of $M^{\prime }\to M\to M^{\prime \prime }$. We can check that $g$ and $h$ induce maps $\mathrm{coker}{f}^{\prime }\to \mathrm{coker}f$ and $\mathrm{coker}f\to \mathrm{coker}{f}^{\prime \prime }$. So it suffices to construct $d:\ker f^{\prime \prime }\to \mathrm{coker}{f}^{\prime }$. For any $x\in M^{\prime \prime }$, take $\hat{x}\in M$ and so $f(\hat{x})\in \ker h$. Then define $\check{x}\in N^{\prime }$ such that $g(\check{x})=\hat{x}$ and $\overline{\check{x}}\in \mathrm{coker}{f}^{\prime }$ is what we desired. It is easy to check this map is well-defined.

It remains to check the sequence is exact. For convenience, define

$$0\to \ker f^{\prime }\stackrel{G^{\prime }}{\to }\ker f\stackrel{H^{\prime }}{\to }\ker f^{\prime \prime }\stackrel{d}{\to }\mathrm{coker}{f}^{\prime }\stackrel{g^{\prime }}{\to }\mathrm{coker}f\stackrel{h^{\prime }}{\to }\mathrm{coker}{f}^{\prime \prime }\to 0.$$

For any $x\in \ker d⩽M^{\prime \prime }$, there exists $y\in M$ such that $H(y)=x$. By the definition of $d$, we know $f(y)=g(d(x))=0$ and so $y\in \ker f$. It deduces that $x\in \mathrm{im}(H^{\prime })$. Conversely, for any $x\in \mathrm{im}(H^{\prime })$, there exists $y\in \ker f$ such that $x=H^{\prime }(y)$. By the definition of $d$, there is $g(d(x))=f(y)=0$ and $d(x)=0$. Thus $x\in \ker d$. Now we have proved $\ker d=\mathrm{im}{H}^{\prime }$.

For any $x\in \ker g^{\prime }$, there is $g^{\prime }(x)=0$ and so $g(\hat{x})\in \mathrm{im}f$ for any given preimage $\hat{x}\in N^{\prime }$. Then there is $y\in M$ such that $f(y)=g(\hat{x})$. By the definition of $d$, we know $d(H(y))=x$. In addition, since $f^{\prime \prime }(H(y))=h(f(y))=h(g(\hat{x}))=0$, $H(y)\in \ker f^{\prime \prime }$. It yields that $x\in \mathrm{im}d$. Conversely, for any $x\in \mathrm{im}d$, there exists $y$ such that $d(y)=x$ and $z\in M$ such that $H(z)=y$ and $f(z)=g(\hat{x})$ for any given $x\in N^{\prime }$. By the definition of $d$, there is $g^{\prime }(x)=\overline{f(z)}\in \mathrm{coker}f$ and so $g^{\prime }(x)=0$, $x\in \ker g^{\prime }$. Now we have proved $\mathrm{im}d=\ker g^{\prime }$. \end{proof}

Text-Ex. 2.15. In the situation of Exercise 14, show that every element of $M$ can be written in the form ${\mu }_i\left(x_i\right)$ for some $i\in I$ and some $x_i\in M_i$. Show also that if ${\mu }_i\left(x_i\right)=0$ then there exists $j⩾i$ such that ${\mu }_{ij}\left(x_i\right)=0$ in $M_j$.

\begin{proof} For any $m\in M$, let $c=(x_i{)}_{i\in I}$ be the preimage of $m$, i.e. $\mu (c)=m$. Since $c\in {\oplus }_{i\in I}{M}_i$, only finitely many $x_i\ne 0$ and we define $J=\{i\in I:x_i\ne 0\}$. Notice that for any $i⩽j$, there is ${\mu }_i(x_i)={\mu }_j\circ {\mu }_{ij}(x_i)={\mu }_j({\mu }_{ij}(x_i))$. Let $k$ be the maximal element of $J$. Since $I$ is a direct set, $k$ is well-defined. Then for each $i\in J$, ${\mu }_i(x_i)={\mu }_k({\mu }_{ij}(x_i))$ and ${\mu }_{ik}(x_i)\in M_k$. It deduces that $m={\mu }_k({\sum }_{i\in J}{\mu }_{ik}(x_i))$ with ${\sum }_{i\in J}{\mu }_{ik}(x_i)\in M_k$.

If ${\mu }_i(x_i)=0$, then $\mu (x_i)=0$ and so $x_i\in D=⟨x_j-{\mu }_{jk}(x_j):j⩽k,x_j\in M_j⟩$. Since $D⩽C={\oplus }_{i\in I}{M}_i$, either $x_i=y_i-{\mu }_{ij}(y_i)$ with $i⩽j$ or $x_i=z_h-{\mu }_{hi}(z_h)$ with $h⩽i$. For the former case, ${\mu }_{ij}(y_i)=y_i-x_i\in M_i$ yields that $x_i=y_i$ and ${\mu }_{ij}(x_i)={\mu }_{ij}(y_i)=0$. For the latter one, $z_h=x_i-{\mu }_{hi}(z_h)\in M_h\cap M_i=0$ yields that $z_h=0$ and $x_i={\mu }_{hi}(z_h)=0$, and then ${\mu }_{ij}(x_i)=0$. \end{proof}

Text-Ex. 2.16. Show that the direct limit is characterized (up to isomorphism) by the following property. Let $N$ be an $A$-module and for each $i\in I$ let ${\alpha }_i:M_i\to N$ be an $A$ module homomorphism such that ${\alpha }_i={\alpha }_j\circ {\mu }_{ij}$ whenever $i⩽j$. Then there exists a unique homomorphism $\alpha :M\to N$ such that ${\alpha }_i=\alpha \circ {\mu }_i$ for all $i\in I$.

\begin{proof} Define $\beta :C\to N,(x_i{)}_{i\in I}\mapsto {\sum }_{i\in I}{\alpha }_i(x_i)$. Since $(x_i{)}_{i\in I}\in C$, only finitely many $x_i\ne 0$ and ${\sum }_{i\in I}{\alpha }_i(x_i)$ makes sense. Define $\alpha :M\to N,m\mapsto \beta ({\mu }^{-1}(m))$. Note that ${\mu }^{-1}(m)=c+D$ for some $c\in {\mu }^{-1}(m)$, and $\beta (D)=0$ by ${\alpha }_i={\alpha }_j\circ {\mu }_{ij}$, $i⩽j$. It deduces that $\alpha$ is well-defined. For any $i\in I$ and $x_i\in M_i$, we have $\alpha ({\mu }_i(x_i))=\beta (x_i)={\alpha }_i(x_i)$ and so ${\alpha }_i=\alpha \circ {\mu }_i$ for all $i\in I$. Now we finish the proof. \end{proof}