HW2

Text-Ex. 1.17. Let $A$ be a ring and let $X=\mathrm{Spec}(A)$. For $f\in A$, define $X_f=\mathrm{Spec}(A)\setminus V(f)$. Then $\{X_f:f\in A\}$ forms a basis of Zariski topology. Show that

• each $X_f$ is quasi-compact;
• an open subset of $X$ is quasi-compact iff it is a finite union of some $X_f$.

\begin{proof} i) Suppose $\{X_{g_i}:i\in I\}$ is an open cover of $X_f$, then

$${\cup }_{i\in I}(X_f\cap X_{g_i})={\cup }_{i\in I}{X}_{f{g}_i}=X_{\{f{g}_i:i\in I\}}=X_f$$

and so $V(\{f{g}_i:i\in I\})=V((f))$. It deduces that $r(⟨\{f{g}_i:i\in I\}⟩)=r((f))$. Since $f\in r((f))$, there exists $n\in {\mathbb{N}}_{+}$ and a finite set $J\subseteq I$ such that ${\sum }_{j\in J}f{g}_j=f^n$. Thus, $f^n$ can be generated by $\{f{g}_i:i\in J\}$ and $X_{f^n}={\cup }_{j\in J}{X}_{f{g}_j}$. For any $x\in r((f))$, $x^n\in r((f^n))$ and so $x\in r((f^n))$. Also, for any $x\in r((f^n))$, $x^m=a{f}^n=(a{f}^{n-1})f\in (f)$ yields $x\in r((f))$. Hence $r((f))=r((f^n))$ and then

$$X_f=X_{f^n}={\cup }_{j\in J}{X}_{f{g}_j}\subseteq {\cup }_{j\in J}{X}_{g_j}.$$

Therefore, $\{X_{g_j}:j\in J\}\subseteq \{X_{g_i}:i\in I\}$ is an open subcover of $X_f$ and so $X_f$ is quasi-compact.

ii) If an open set is a finite union of some $X_f$, we assume $O={\cup }_{k=1}^n{X}_{f_k}$. For any open cover $\mathcal{U}$, $\mathcal{U}$ is an open cover of $X_{f_k}$ for each $k$ and there is a finite subcover ${\mathcal{U}}_k$ of $X_{f_k}$. Then ${\cup }_{k=1}^n{\mathcal{U}}_k$ is a finite subcover of $O$.

Otherwise, if an open set $O$ is quasi-compact, then there exists an open cover $\{X_{f_i}:i\in I\}$. Since $\{X_f:f\in A\}$ forms a basis of Zariski topology, we can take $I$ such that ${\cup }_{i\in I}{X}_{f_i}=O$. As it has a finite subcover, there is finite $J\subseteq I$ with $O={\cup }_{j\in J}{X}_{f_j}$ and so $O$ is a finite union of some $X_f$. \end{proof}

Text-Ex. 1.18. For psychological reasons it is sometimes convenient to denote a prime ideal of $A$ by a letter such as $x$ or $y$ when thinking of it as a point of $X=\mathrm{Spec}(A)$. When thinking of $x$ as a prime ideal of $A$, we denote it by $p_x$ (logically, of course, it is the same thing). Show that

• i) the set $\{x\}$ is closed (we say that $x$ is a “closed point”) in $\mathrm{Spec}(A)\iff p_x$ is maximal;
• ii) $\overline{\{x\}}=V\left(p_x\right)$;
• iii) $y\in \overline{\{x\}}\iff p_x\subseteq p_y$;
• iv) $X$ is a $T_0$-space (this means that if $x,y$ are distinct points of $X$, then either there is a neighborhood of $x$ which does not contain $y$, or else there is a neighborhood of $y$ which does not contain $x$ ).

\begin{proof} i) If the set $\{x\}$ is closed, then there exists a subset $E$ such that $V(E)=\{x\}$. It deduces that $p_x$ is the unique prime ideal containing $E$. Assume that $p_x$ is not maximal, then there is a maximal ideal $p_y\supseteq p_x$ and so $V(E)=\{x,y\}$, which is a contradiction. Hence, $p_x$ is maximal.

Conversely, assume that $p_x$ is maximal, then $V(p_x)=\{y\in X:p_y\supseteq p_x\}=\{x\}$ and so $\{x\}$ is closed.

ii) Note that $V(p_x)\supseteq \{x\}$ and $V(p_x)$ is closed, then $V(p_x)\supseteq \overline{\{x\}}$. Since $\overline{\{x\}}$ is closed, there exists subset $E\subseteq A$ such that $V(E)=\overline{\{x\}}=\{y\in X:p_y\supseteq E\}$. It deduces that $p_x\supseteq E$ and so $V(p_x)\subseteq V(E)=\overline{\{x\}}$. Therefore, we have $\overline{\{x\}}=V\left(p_x\right)$.

iii) By ii) we have $y\in \overline{\{x\}}=V(p_x)$ is equivalent to $p_y\supseteq p_x$.

iv) For any distinct $x,y\in X$, at least one of $p_x⊈p_y$ and $p_y⊈p_x$ holds. WLOG $p_x⊈p_y$, then by iii) $y\notin \overline{\{x\}}$ and so $y\in (\overline{\{x\}}{)}^c$. As $(\overline{\{x\}}{)}^c$ is open and $x\notin (\overline{\{x\}}{)}^c$, $(\overline{\{x\}}{)}^c$ is a neighborhood of $y$ which does not contain $x$. \end{proof}

Text-Ex. 1.21. Let $\varphi :A\to B$ be a ring homomorphism. Let $X=\mathrm{Spec}(A)$ and $Y=\mathrm{Spec}(B)$. If $q\in Y$, then ${\varphi }^{-1}(q)$ is a prime ideal of $A$, i.e., a point of $X$. Hence $\varphi$ induces a mapping ${\varphi }^{\ast }:Y\to X$. Show that

• i) If $f\in A$ then ${\varphi }^{\ast -1}\left(X_f\right)=Y_{\varphi (f)}$, and hence that ${\varphi }^{\ast }$ is continuous.
• ii) If $a$ is an ideal of $A$, then ${\varphi }^{\ast -1}(V(a))=V\left(a^e\right)$.

\begin{proof} i) For any $q\in Y_{\varphi (f)}=Y\setminus V(\varphi (f))$, the prime ideal $q⊉\varphi (f)$ and so $\varphi (f)\notin q$. Then $f\notin {\varphi }^{-1}(q)$ and ${\varphi }^{-1}(q)\in X\setminus V(f)=X_f$. Therefore, ${\varphi }^{\ast -1}(X_f)\supseteq Y_{\varphi (f)}$.

Conversely, for any $q\in {\varphi }^{\ast -1}(X_f)$, there is ${\varphi }^{\ast }(q)\in X_f=X\setminus V(f)$ and ${\varphi }^{\ast }(q)⊉f$. Then $f\notin {\varphi }^{\ast }(q)={\varphi }^{-1}(q)$ and $\varphi (f)\notin q$. Therefore, $q⊉\varphi (f)$, $q\in Y_{\varphi (f)}$ and so ${\varphi }^{\ast -1}(X_f)\subseteq Y_{\varphi (f)}$.

Now we have proved that ${\varphi }^{\ast -1}(X_f)=Y_{\varphi (f)}$. Furthermore, since $X_f$ is a basis of Zariski topology and ${\varphi }^{\ast -1}(X_f)$ are open sets for all $f\in A$, ${\varphi }^{\ast }$ is continuous.

ii) For any $q\in {\varphi }^{\ast -1}(V(a))$, we have ${\varphi }^{-1}(q)={\varphi }^{\ast }(q)\supseteq a$ and so $q\supseteq \varphi (a)$. Thus $q\in V(\varphi (a))=V(a^e)$. Conversely, for any $q\in V(a^e)=V(⟨\varphi (a)⟩)=V(\varphi (a))$, there is $q\supseteq \varphi (a)$ and ${\varphi }^{\ast }(q)={\varphi }^{-1}(q)\supseteq a$. Hence ${\varphi }^{\ast }(q)\in V(a)$ and $q\in {\varphi }^{\ast -1}(V(a))$. Now we finish the proof. \end{proof}

Proposition 6.3. Let $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$ be a short exact sequence of $A$-module, then $M$ is Artinian iff $M^{\prime }$ and $M^{\prime \prime }$ are Artinian.

\begin{proof} Assume that $M$ is Artinian. For any descending chains $M^{\prime }=M_0^{\prime }⊋M_1^{\prime }⊋\cdots ⊋M_n^{\prime }⊋\cdots$ and $M^{\prime \prime }=M_0^{\prime \prime }⊋M_1^{\prime }⊋\cdots ⊋M_n^{\prime \prime }⊋\cdots$, they induce descending chains of $M$

$$\begin{array}{rl}& M={\beta }^{-1}(M_0^{\prime \prime })⊋{\beta }^{-1}(M_1^{\prime \prime })⊋\cdots ⊋{\beta }^{-1}(M_n^{\prime \prime })⊋\cdots ,\\ & M=\alpha (M_0^{\prime })⊋\alpha (M_1^{\prime })⊋\cdots ⊋\alpha (M_n^{\prime })⊋\cdots .\end{array}$$

Since $M$ is Artinian, there exist $s$ and $t$ such that ${\beta }^{-1}(M_k^{\prime \prime })={\beta }^{-1}(M_{k+1}^{\prime \prime })$ and $\alpha (M_t^{\prime })=\alpha (M_{t+1}^{\prime })$. Since $\beta$ is surjective and $\alpha$ is injective, it deduces that $M_k^{\prime \prime }=M_{k+1}^{\prime \prime }$ and $M_t^{\prime }=M_{t+1}^{\prime }$. Hence $M^{\prime }$ and $M^{\prime \prime }$ are Artinian.

Conversely, assume that $M^{\prime }$ and $M^{\prime \prime }$ are Artinian. If $M$ has a descending chain which is not stationary

$$M=M_0⊋M_1⊋\cdots ⊋M_n⊋\cdots$$

then we have infinitely many short exact sequences $0\to {\alpha }^{-1}(M_i)\to M_i\to \beta (M_i)\to 0$. Since $M_i⊋M_{i+1}$, at least one of $\{{\alpha }^{-1}(M_i),{\alpha }^{-1}(M_{i+1})\}$ and $\{{\beta }^{-1}(M_i),{\beta }^{-1}(M_{i+1})\}$ contain two distinct elements.

Consider the following chains of $M^{\prime }$ and $M^{\prime \prime }$

$$\begin{array}{rl}& M^{\prime }={\alpha }^{-1}(M_0)\supseteq {\alpha }^{-1}(M_1)\supseteq \cdots \supseteq {\alpha }^{-1}(M_n)\supseteq \cdots ,\\ & M^{\prime \prime }=\beta (M_0)\supseteq \beta (M_1)\supseteq \cdots \supseteq \beta (M_n)\supseteq \cdots .\end{array}$$

where some terms may repeat. Since $M^{\prime }$ and $M^{\prime \prime }$ are Artinian, these two chains must contain infinitely many same terms. That is, there exists $n_1$ and $n_2$ such that ${\alpha }^{-1}(M_n)={\alpha }^{-1}(M_{n+1})$ for all $n⩾n_1$ and $\beta (M_n)=\beta (M_{n+1})$ for all $n⩾n_2$. Take $n=n_1+n_2$, then ${\alpha }^{-1}(M_n)={\alpha }^{-1}(M_{n+1})$ and $\beta (M_n)=\beta (M_{n+1})$ hold, which is impossible. \end{proof}