Hartshorne Solution

Chapter 2

1.1 Let $A$ be an abelian group, and define the constant presheaf associated to $A$ on the topological space $X$ to be the presheaf $U \mapsto A$ for all $U \neq \varphi$, with restriction maps the identity. Show that the constant sheaf $\mathcal{A}$ defined in the text is the sheaf associated to this presheaf.
Solution:
Let $\mathcal{F}$ denote the given presheaf. Then we construct the associated sheaf , $\mathcal{F^+}$ of the given presheaf $\mathcal{F}$ by defining $\mathcal{F^+}(U)$ to be set of continuous functions from $U\rightarrow \bigcup_{p\in U} \mathcal{F_p}$ satisfying two conditions in the definition. Now recall how constant sheaf was defined that by taking set of continuous functions $U\rightarrow A$. It is clear that constant sheaf $\mathcal{A}$ satisfies conditions of associated sheaf and by uniqueness constant sheaf is associated sheaf $\mathcal{F^+}$.

1.2
a) For any morphism of sheaves $\varphi:\mathcal{F} \mapsto \mathcal{G}$ show that for each point $P$, $(ker\varphi)_P=ker\varphi_P$ and $(im\varphi)_P=im\varphi_P$.

Solution:
Consider the following $\varphi:\mathcal{F}\rightarrow \mathcal{G}$ then we have the induced map at stalks $\varphi_p:\mathcal{F_p}\rightarrow \mathcal{G_p}$ coming back to the question,
let $s_p\in ker\varphi_p$ then $[\varphi(s),U]=0$.It implies that there exists open subset $V$ of $U$ such that $\varphi(s)|_V=0|_V$.
Hence $s|_V \in ker\varphi(V)$ and $[s|_V, V]\in (ker\varphi)_p$. For other direction we have this induced map $i_p:(ker\varphi(U))_p\rightarrow \mathcal{F}_p$. so $\varphi_p \circ i_p=0$ and $(ker\varphi)_p\subseteq ker\varphi_p$.
For second part, $im\varphi$ is associated sheaf of presheaf image of $\mathcal{F}$.
$im(\varphi_p)$— image of stalk
$(im\varphi)_p$— Stalk of image sheaf=Stalk of preimage sheaf
$y_p\in (im\varphi)_p \implies y_p$ belongs to the stalk of presheaf image and it is contained in $\mathcal{G_p}$.So $y_p=[V,s]$ where $p\in V$ and $s\in preim\varphi(V)$.
So there exists $t\in \mathcal{F}(V)$ such that $\varphi(t)=s$ and $y_p\in im\varphi_p$.
Now consider $y_p\in im\varphi_p$ implies there exists $t_P\in \mathcal{F_p}$ such that $\varphi_p(t_p)=y_p$. Suppose $p\in U$open and $t\in \mathcal{F}(U)$ .$[U,t]\in \mathcal{F_p}$ and $\varphi(U)(t)\in$ presheaf image $\varphi(U)$.Implies $y_p=(U,\varphi(U)(t))\in (im\varphi)_p$.

b) Show that $\varphi$ is injective (resp. Surjective) if and only if induced map on stalks $\varphi_P$ is injective (resp. surjective) for all $P$.
Solution:
It follows from the above exercise because $ker\varphi=0$ if and only if $(ker\varphi)_p=0$ for all $p$ if and only if $ker\varphi_p=0$ for all $p$.
Similarly $\varphi$ is surjective if and only if $im\varphi=\mathcal{G}$ if and only if $(im\varphi)_p=\mathcal{G_p}$ for all $p$ if and only if $im\varphi_p=\mathcal{G_p}$ for all p.

c) Show that a sequence $\cdots \rightarrow \mathcal{F^{i-1}} \xrightarrow {\varphi^{i-1}} \mathcal{F^{i}} \xrightarrow {\varphi^{i}} \mathcal{F^{i+1}} \rightarrow \cdots$ of sheaves and morphisms is exact if and only if for each $P\in X$ the corresponding sequence of stalks is exact as a sequence of abelian groups.
Solution:
$ker\varphi_p^{i}=im\varphi_p^{i-1}$ if and only if $(ker\varphi^i)_p=(im\varphi^{i-1})_p$ for all $p$.
[$F,G$ be subsheaves of $H$ then $F=G\Leftrightarrow F_p=G_p \forall p$.]

1.3 Let $\varphi: \mathcal{F} \rightarrow \mathcal{G}$ be a morphism of sheaves on X. Show that $\varphi$ is surjective if and only if the following condition holds: for every open set $U \subseteq X$, and for every $s\in\mathcal{G}(U)$, there is a covering ${ U_i}$ of U, and there are elements $t_i\in \mathcal{F}(U_i)$, such that $\varphi(t_i) = s|_{U_i}$, for all $i$.
Solution: Suppose $\varphi$ is surjective then by definition $im\varphi=\mathcal{G}$.
Hence $\mathcal{G}(U)=im\varphi(U)$(definition of associated sheaf of presheaf image) for $U$ open subset of $X$.
Let $s\in \mathcal{G}(U)$ with $V_i's$ forming open cover of $U$ and $s_i\in im_p\varphi(V_i)$(presheaf image) such that $s|_{V_i}=s_i\in im_p\varphi(V_i)$.
$s_i\in im_p\varphi(V_i)$ implies there exists $t_i\in \mathcal{F}(V_i)$ such that $\varphi(t_i)=s_i=s|_{V_i}.$
Conversely it is enough to show that $\varphi_p$ is surjective. Let the said condition hold .
Let $s_p \in \mathcal{G_p}$ implies $s_p=[U,s]\in \mathcal{G_p}$ where $p\in U$ and $s\in \mathcal{G}(U)$. By condition, there is covering ${U_i}$ of $U$ and there are elements $t_i\in \mathcal{F}(U_i)$ such that $\varphi(t_i)=s|_{U_i}\in \mathcal{G}(U_i)$ for $p\in U_i$$. So
$[U_i,\varphi(t_i)]=[U,s]\implies [U_i,s|_{U_i}]=[U,s]$. It will imply $\varphi_p[U_i,t_i]=[U_i,\varphi(t_i)]= [U_i,s|_{U_i}]=[U,s]$. Therefore $\varphi_p$ is surjective.

1.4
a) Let $\varphi:\mathcal{F} \rightarrow \mathcal{G}$ be a morphism of presheaves such that
$\varphi(U):\mathcal{F}(U) \rightarrow \mathcal{G}(U)$ is injective for each U. Show that the induced map $\varphi^+: \mathcal{F}^+ \rightarrow \mathcal{G}^+$ of associated sheaves is injective.
Solution:
We show that $\varphi_p^+: \mathcal{F}_p^+ \rightarrow \mathcal{G}_p^+$ is injective. Note that
$\mathcal{F}_p^+=\mathcal{F}_p$ and $\mathcal{G}_p^+=\mathcal{G}_p$. Let $\varphi_p^+(U,s)=0$ , then there exists open neighborhood of $p$ say $V$ contained in $U$ such that $\varphi_V^+(s|_V)=0$.By using the definition of associated sheaf and injectivity of $\varphi$ we get $[U,s]=0$.

b) Use part (a) $\varphi:\mathcal{F} \rightarrow \mathcal{G}$ to show that if is a morphism of sheaves, then $im\varphi$ can be naturally identified with a subsheaf of $\mathcal{G}$ as mentioned in the text.
Solution:
We have $im_p\varphi\rightarrow \mathcal{G}$ is injective. So by above result
$im\varphi \rightarrow \mathcal{G}$ is injective. So the $im\varphi$ can be naturally identified by subsheaf of $\mathcal{G}$.

1.5 Show that a morphism of sheaves is an isomorphism if and only if it is both injective and surjective.
Solution:
Isomorphisms means there must exist two sided inverse. Also $\varphi$ is isomorphism between sheaves $\mathcal{F}$ and $\mathcal{G}$ if and only if the induced morphism at stalks is isomorphisms and $\varphi_p$ is injective and surjective iff $\varphi$ is injective and surjective.

1.8 For any open subset $U\subseteq X$, show that the functor $\Gamma(U, .)$ from sheaves on $X$ to abelian groups is a left exact functor.
Solution:
We have $0 \rightarrow \mathcal{F'} \xrightarrow {\varphi'} \mathcal{F} \xrightarrow{\varphi} \mathcal{F''} \rightarrow 0$ is exact sequence. So induced sequence at stalks is exact. Two things to note are $\varphi'$ is injective so $\varphi'(U)$ is injective for all open subsets $U$ of $X$. So it is enough to prove $im\varphi'(U)=ker\varphi(U)$ and $im_p\varphi'=im\varphi'$ ( $im_p\varphi'$ is presheaf image).For $U\subseteq X$ open, let $s\in im\varphi'$ then there exists $t\in \mathcal{F'}(U)$ such that $\ \varphi'_U(t)=s$. So $t_p\in \mathcal{F_p}$ and
$[U,t]\mapsto [U,\varphi'_U(t)]=[U,s]=s_p$. It implies
$s_p=[U,s]\in$ $im\varphi'_p=ker\varphi_p$. So $\varphi_p(s_p)=0$.
$[U,\varphi_U(s)]=[U,0] \implies$ $\varphi_U(s)|_{U_i}=0$ and these ${U_i}$ form open cover for $U$. Hence by Sheaf axiom $\varphi_U(s)=0$ and $s\in ker\varphi_U$. For other direction proof follows similar to proposition 1.1.

1.13. Espace Etale of a Presheaf.
Solution
Detailed solution can be found here.

1.14. Support. Let $\mathcal{F}$ be a sheaf on X, and lets $s\in \mathcal{F}(U)$ be a section over an open set U. The support of $s$, denoted $Supp$ $s$, is defined to be $\{ P\in U|s_p \neq 0\}$, where $s_p$ denotes the germ of $s$ in the stalk $\mathcal{F}_p$. Show that $Supp$ $s$ is a closed subset of $U$. We define the support of $\mathcal{F}$ , $Supp$ $\mathcal{F}$, to be $\{P\in X| \mathcal{F}_p \neq 0\}$. It need not be a closed subset.
Solution:
We show $W=\{p\in U|s_p\neq 0\}$ is open in $U$. Let $p\in W$ then $[U,s]=0$ implies there exists $V\subseteq U$ open with $p\in V$ and
$s|_V=0$ and $s|_V \in \mathcal{F}(V)$. Let $q\in V$ then $V\subseteq U$ and $s|_V=0$ means $s_{q}=0\implies V\subset W$.
Therefore $W$ is open.
Consider the following example for last part.
Let X be a irreducible topological space.Let $p\in X$.Define for any open subset $U$:
$\mathcal{F}(U)=0$ if $U$ is empty or $p\in U$ otherwise $\mathcal{F}(U)=A$ with restriction maps as identity or zero. Then $F$ forms a Sheaf.
If $q \in \overline{\{p\}}\implies q\in U$ and $U\cap \{p\} \neq \emptyset\implies p\in U$ and $\mathcal{F}(U)=0$.Hence $\mathcal{F_q}=0$.
If $q\notin \overline{\{p\}}\implies \mathcal{F_q}=A$.
$Supp\mathcal{F}=\{q\in X|\mathcal{F_p}\neq 0\}=X$\ $\overline{\{p\}}$ which is open.

1.16. Flasque Sheaves. A sheaf $\mathcal{F}$ on a topological space $X$ is Falsque if for every inclusion $V\subseteq U$ of open sets, the restriction map
$\mathcal{F}(U)\rightarrow \mathcal{F}(V)$ is surjective.

a) Show that a constant sheaf on an irreducible topological space is flasque. See (I, §I) for irreducible topological spaces.
Solution:
$X$ is irreducible space so by definition of irreducibility $U$ is connected and so we have $\mathcal{F}(U) \cong A$ and $\mathcal{F}(V) \cong A$. Hence restriction mapping is just identity and we get surjectivity.

b) If $0 \rightarrow \mathcal{F'} \xrightarrow{\varphi''} \mathcal{F} \xrightarrow{\varphi} \mathcal{F''} \rightarrow 0$ is an exact sequence of sheaves, and if $\mathcal{F'}$ is Flasque , then for any open set $U$, the sequence
$0\rightarrow \mathcal{F'}(U)\rightarrow \mathcal{F}(U) \rightarrow \mathcal{F''}(U)\rightarrow 0$ of abelian groups is also exact.
Solution:
To show $0\rightarrow \mathcal{F'}(U)\xrightarrow{\varphi'_U}\mathcal{F}(U) \xrightarrow{\varphi_U} \mathcal{F''}(U)\rightarrow 0$
Taking sections is left exact. It is enough to show that $\varphi_U$ is surjective. Let $t\in \mathcal{F''}(U)$. Since the following sequence is exact
$0 \rightarrow \mathcal{F'} \xrightarrow{\varphi''} \mathcal{F} \xrightarrow{\varphi} \mathcal{F''} \rightarrow 0$ . So $\varphi$ is surjective. It implies that for $U\subseteq X$ open, there exist open covering $\{U_i\}$ of $U$ for $t_i\in \mathcal{F}(U_i)$ such that $\varphi(t_i)=t|_{U_i}$. Let $Z$ be the set of pair $(U_i,t_i)$ with the following partial order $(U_i,t_i)<(U_j,t_j)\Leftrightarrow U_i\subseteq U_j$ such that $t_j|_{U_i}=t_i$. Let $\mathcal{M}$ be chain in $Z$ and let $V=\bigcup_CU_i$ where
$C=\{i|(U_i,t_i)\in Z\}$.By the given partial order relation, for $i\in C$ each $t_i\in \mathcal{F}(U_i)$ agree on intersection, so by sheaf axiom there exist $s\in \mathcal{F}(V)$ such that $s|_{U_i}=t_i$ and $\varphi(s)=t|_V$.Therefore $(V,s)\in Z$ and is the upper bound of the chain $\mathcal{M}$. By Zorn’s lemma, there exist maximal element say $(V',s')$ of $Z$. Our Claim is that
$V'=U$. Now, suppose otherwise then there exists $x\notin V'$. So we have $V'\subset V'\bigcup U_k=U$ for some $U_k$ and $\varphi(s')=t|_{V'}$ and $\varphi(t_k)=t|_{U_k}$.Consider, $V'\bigcap U_k$ then $s'|_{V'\bigcap U_k}\in \mathcal{F}(V'\bigcap U_k)$ and $t|_{V'\bigcap U_k}\in \mathcal{F}(V'\bigcap U_k)$.
So $\varphi(s'|_{V'\bigcap U_k})=t|_{V'\bigcap U_k}=\varphi(t_k|_{V'\bigcap U_k})$.It implies
$s'|_{V'\bigcap U_k}-t_k|_{V'\bigcap U_k}\in ker\varphi(V'\bigcap U_k)$.So there exists $w|_{V'\bigcap U_k}\in \mathcal{F'}(V'\bigcap U_k)$ such that $\varphi'(w|_{V'\bigcap U_k})=s'|_{V'\bigcap U_k}-t_k|_{V'\bigcap U_k}.$
$\mathcal{F'}$ is given to be flasque and $w|_{V'\bigcap U_k}\in \mathcal{F'}(V'\bigcap U_k)\implies$ there exists $w\in \mathcal{F'}(V')$ such that $w\mapsto w|_{V'\bigcap U_k}$.Let $m=\varphi'(w)\in im\varphi'(V')=ker\varphi(V')$ and $m|_{V'\bigcap U_k}=s'|_{V'\bigcap U_k}-t_k|_{V'\bigcap U_k}$ (Draw diagram for better understanding).Hence, $(m-s')|_{V'\bigcap U_k}=t_k|_{V'\bigcap U_k}$ with $m-s'\in \mathcal{F}(V')$ and $t_k\in \mathcal{F}(U_k)$. Also, $U=V'\bigcup U_k$. So by sheaf axiom there exist $t''\in \mathcal{F}(U)$ that restricts to $m-s'$ and $t_k$ on $V'$ and $U_k$ respectively with $\varphi_U(t'')=t|_U$. So $(U,t'')\in Z$ contradicting maxamilty of $(V',s')$.So $U=V'$ and the result follows

c) $0 \rightarrow \mathcal{F'} \rightarrow \mathcal{F} \rightarrow \mathcal{F''} \rightarrow 0$ is an exact sequence, and if $\mathcal{F'}$ and $\mathcal{F}$ are Flasque then $\mathcal{F''}$ is Flasque.
Solution:
Since $0 \rightarrow \mathcal{F'} \rightarrow \mathcal{F} \rightarrow \mathcal{F''} \rightarrow 0$ is an exact sequence and $\mathcal{F'}$ is flasque we have upper sequence and lower sequence of the following diagram as exact, also $V\subseteq U$

$\varphi_V$ is surjective. So start with $w|_V\in \mathcal{F''}(V)$ and use the fact that diagram is commutative and $\mathcal{F}$ is flasque to complete the proof.

d) If $f:X \rightarrow Y$ is a continuous map, and if $\mathcal{F}$ is a flasque sheaf on $X$ , then $f_{*}\mathcal{F}$ is a flasque sheaf on $Y$.
Solution:
For $V\subseteq Y$ is open then $f^{-1}(V)$ is open in X.
$f_*\mathcal{F}(V)=\mathcal{F}(f^{-1}(V))$ and $\mathcal{F}$ is given to be flasque.

e) Let $\mathcal{F}$ be any sheaf on $X$. We define a new sheaf $\mathcal{G}$, called the sheaf of discontinuous sections of $\mathcal{F}$ as follows. For each open set $U \subseteq X$ , $\mathcal{G}(U)$ is the set of maps $s:U\rightarrow \bigcup_{p\in U} \mathcal{F_p}$ for all $p\in U$, $s(p)\in \mathcal{F_p}$. Show that $\mathcal{G}$ is flasque sheaf and that there is a natural injective morphism of $\mathcal{F}$ to $\mathcal{G}$.
Solution:
$\mathcal{G}$ is a sheaf just by taking restrictions. Let $V\subseteq U$ and
$\mathcal{G}(U)\longrightarrow \mathcal{G}(V)$. For $t\in \mathcal{G}(V)$ then define $w\in \mathcal{G}(U)$ such that $w|_V=t$ and $w=0$ elsewhere.

1.17 Skyscraper Sheaves.
Solution:
Stalks can be checked with same method as in the example given in solution of exercise 1.14. with necessary changes.

2.1. Let $A$ be a ring, let $X=Spec A$, let $f\in A$ and let $D(f)\subseteq X$ be the open complement of $V((f))$. Show that locally ringed space $(D(f),\mathcal{O}_X|_{D(f)})$is isomorphic to $Spec A_f.$
Solution:
Similar to the solution of proposition 2.5(b) and 2.18.

2.2. Let $(X,\mathcal{O}_X)$ be a scheme, and let $U\subseteq X$. Show that $(U,\mathcal{O}_{X}|_U)$ is a scheme.We call this induced scheme structure on the open set $U$, and we refer to $(U,\mathcal{O}_X|_U)$ as open subscheme of $X$.
Solution:
A scheme by definition is something which locally looks like an affine scheme. Let $x\in U\subseteq X$ and let $V\cong SpecA$ be the affine neighborhood of $x$, since $D(f)$ form basis for topology on $V$, where $f\in A$. So $x\in D(f)\subset U$ for some $f\in A$ and from previous exercise $(D(f),\mathcal{O}_{SpecA}|_{D(f)})$ is isomorphic to $(Spec A_f,\mathcal{O}_{Spec A_f})$.

2.3. Reduced schemes. A scheme $(X,\mathcal{O}_X)$ is called reduced if for every open set $U\subseteq X$, the ring $\mathcal{O}_X(U)$ has no nilpotent elements.
Show that $(X,\mathcal{O}_X)$ is reduced if and only if for every $P\in X$, the local ring $\mathcal{O}_{X,P}$ has no nilpotent elements.
Solution:
Suppose $(X,\mathcal{O}_X)$ is reduced then $\mathcal{O}_X(U)$ has no nilpotent elements for any open set $U\subseteq X$. For $P\in X$ let $[U,s]$ be germ of $\mathcal{O}_{X,P}$ such that it is nilpotent. Then
$s^n|_W=0$ for $P\in W\subseteq U$ and $0=s^n\in \mathcal{O}_X(W)$ but it has no nilpotent so $s$ is zero on $\mathcal{O}_X(W)$ hence on $U$.
Conversely, suppose $s\in \mathcal{O}_X(U)$ such that $s^n=0$ for some $n$. Then $s^n_x=0$ in $\mathcal{O}_{X,x}$ where $x\in U$. But there are no nilpotent elements in $\mathcal{O}_{X,x}$.

2.4. Let $A$ be a ring and let $(X, \mathcal{O}_X$ be a scheme. Given a morphism $f\rightarrow SpecA$, we have asscociated maps on sheaves $f^\# :\mathcal{O}_{SpecA}\rightarrow f_{*} \mathcal{O}_X.$
Taking global sections we obtain a homomorphism
$A\rightarrow \Gamma(X,\mathcal{O}_X)$.Thus there is natural map
$$\alpha:Hom_{\mathfrak{Sch}}(X,SpecA)\rightarrow Hom_{\mathfrak{Rings}}(A, \Gamma(X,\mathcal{O}_X))$$

Show that $\alpha$ is bijective.
Solution:
There are few subtle things to note before giving the solution: First of all, on right side we are dealing rings unity but not necessarily commutative. Then we take spectrum of commutative ring with unity. Also we make use of the following resulting when giving the solution:

If $X$ and $Y$ are two schemes, then giving a morphism from $X\rightarrow Y$ is equivalent to giving an open cover $\{Ui\}_{i\in I}$ of $X$, together with morphism $f_i:U_i\rightarrow Y$, where $U_i$ has the induced open subscheme structure, such that the restrictions of $f_i$ and $f_j$ to $U_i\cap U_j$ are the same, for each $i,j\in I$.

We try to give the inverse map to $\alpha$. So consider
$g\in Hom_{\mathfrak{Rings}}(A, \Gamma(X,\mathcal{O}_X))$
Since we have the propostion 2.3. we try to give glueing argument to get scheme $X$.
Since $X$ is scheme let $X$ be covered by affine opens $\{U_i\}_{i\in I}$. Since $U_i$ is affine we have $U_i\cong Spec$ $C_i$.
Therefore, $\mathcal{O}_X(U)=A$.
$A\xrightarrow{g}\mathcal{O}_X(X)\xrightarrow{\rho_{XU}}\mathcal{O}_X(U)=C$ and taking Spec we have a map $f_i:U_i\cong Spec$ $C\rightarrow Spec$ $A$.
By the above result we only have to show that $f_i|_{Ui\cap U_j}=f_j|_{U_i\cap U_j}$. But in order to use functoriality of Spec(proposition 2.3), we consider affine covering of
$U_i\cap U_j$ say $\{V_k\cong SpecZ_k\}$. So its enough to show $f_i$ and $f_j$ agree on $V_k.$ We have the following:
$V_k\subset U_i\cap U_j \subset U_i\subset X$ and $V_k\subset U_i\cap U_j \subset U_j\subset X$.
So we have the morphisms of ring:

Taking spec we get,

Hence the result follows

2.5. Describe $Spec Z$, and show that it is a final object for the category of schemes.
Solution:
Since $Z$ is initial object in category of rings with 1 ($f:Z\rightarrow R$ by $n\mapsto n\cdot 1$) . By using the above result we get the desired result.

2.9. If $X$ is a topological space, and $Z$ an irreducible closed subset of $X$, a generic point for $Z$ is point $\zeta$ such that $Z=\{\zeta\}$ $\bar{}$. If $X$ is a scheme show that every (nonempty) irreducible closed subset has a unique generic point.
Solution:
Remark: In $SpecA$ there is one to one correspondence between points of $SpecA$ and irreducible closed subsets of $Spec A.$ Now, let $Z$ be irreducible closed subset of $X$. Since $X$ is a scheme, $U$ be affine open set such that
$U\cap Z=\emptyset$ (there exist such a $U$). So $U\cap Z$ is irreducible( since open subset of $Z$) closed subset of $U$. So $W=U\cap Z=V(p) (=Spec(A/p))=\bar{\{p\}}$ for some prime ideal. $\bar{\{p\}}$ is generic point of $W$. $\bar{\{p\}}$ is generic point of $Z$. By definition of denseness, $p\in V\cap Z$ where $V$ is nonempty affine open subset. Taking closure of $\{p\}$ in $V\cap Z$ is $Z$. So $\bar{\{p\}}=V\cap Z$ and by uniqueness of generic point of affine scheme (see above remark) we have that it is the only generic point for any $V\cap Z$. Hence, for $Z$.

2.13. A topological space is quasi-compact if every open cover has a finite subcover.
a) Show that a topological space is noetherian (I, §1) if and only if every open subset is quasi-compact.
Solution:
A topological space $X$ is noetherian then every collection of open subset has maximal element. Let $Z$ be open subset such that $Z=\bigcup_i U_i.$Then $Z\subset U_1\cup U_2\cup \cdots \cup U_m$ (said maximal element). Conversely, let every open subset be quasi compact. Consider the following sequence of open subsets: $Z_1\subset Z_2\subset Z_3\subset \cdots$ and suppose $X$ does not satisfy ACC for open subsets. Consider $Z=\bigcup_i Z_i$ where $Z_i$ is open and so is$Z$. Hence $Z=\bigcup_iU_i$ and we get contradiction.

b) If $X$ is an affine scheme, show that $sp(X)$ is quasi-compact, but not in general noetherian. We say a scheme $X$ is quasi-compact if $sp(X)$ is.
Solution:
$X\cong SpecA$ and $D(f)$ form basis for topology on $SpecA$.So any cover of $Spec A$ is given by $D(f_i)$.
Let $SpecA=\bigcup_iD(f_i)$ then
$\emptyset=(\bigcup_iD(f_i))^c=\bigcap_iV(f_i)=V(\sum f_i).$It implies that $1\in \sum f_i$ so $1=\sum_{finite} a_if_i$. Hence,
$\emptyset=V(\sum_{finite}f_i)=\bigcap_{finite}V(f_i)$. Taking complement we get $SpecA=\bigcup_{finite}D(f_i)$.
For other part take $Spec$ $k[x_1,x_2,x_3,\cdots].$

c) If $A$ is a noetherian ring, show that $sp(Spec$ $A)$ is a noetherian topological space.
Solution:
Let $Z(T_1)\subset Z(T_2)\subset \cdots$ be the chain of closed subset of $Spec$ $A$ that does not satisfy DCC.
Take $I(\cdot)$ to get ascending chain of ideals in $A$.
Let $\mathcal{I}=\bigcup_iI(Z(T_i))$ then $\mathcal{I}$ is ideal in $A$ which is noetherian then $\mathcal{I}$ is finitely generated by say $f_i$'s and these $f_i$'s are contained in some $I(Z(T_i))$. Hence $\mathcal{I}$ is contained in $I(Z(T_i))$ and we get contradiction.

2.18. Let $A$ be a ring, $X=Spec A$, $f\in A$. Show that $f$ is nilpotent if and only if $D(f)$ is empty.
Solution:
Suppose $f$ is nilpotent. Then $f^n=0$ for some positive $n$. Then $f^n\in p$ ($p$ is prime ideal). So $f\in p$ and $D(f)$ is empty.
Suppose $f$ is not nilpotent. Then $f^k\neq0$ for any $k$. Consider the following multiplicative set,
$S=\{1,f,f^2,f^3,\cdots\}$ then we localize to at this set.
Define $l:A\rightarrow A_f$ by $x\mapsto \frac{x}{1}$.
Then $Spec(l):Spec(A_f)\rightarrow Spec(A)$
And we know prime ideals of localization correspnds to prime ideal $p$ of ring $A$ such that $p\cap S=\emptyset$, that is $p$ is a prime ideal such that $f\notin p$.Therefore, $Spec(A_f)\cong D(f)$.