Eisenbud-Harris

Here are my exercise solutions and notes for the beginning of Eisenbud and Harris’s The Geometry of Schemes.

Let $R$ be a ring and $\mathfrak{p}\subset R$ a prime ideal. Recall that we define $R_{\mathfrak{p}}=\{a/b \ : \ a\in R, b\in R-\mathfrak{p}\}$ , i.e., $R_{\mathfrak{p}}$ is the localization of $R$ with respect to the multiplicative set $R-\mathfrak{p}.$

1-1.

$\text{Spec}(Z)=\{(0)\} \cup \{(p) \ : \ p \ \text{prime}\}$ ;
$\text{Spec}(\mathbb{Z}/3\mathbb{Z})=\{(0)\}$ (note $\mathbb{Z}/3\mathbb{Z}$ is a field);
$\text{Spec}(\mathbb{Z}/6\mathbb{Z})=\{(2),(3)\}$ (note $(0)$ is not prime since $\mathbb{Z}/6\mathbb{Z}$ is not an integral domain);
$\text{Spec}(\mathbb{Z}_{(3)})=\{(0),(3)\}$ ;
$\text{Spec}(\mathbb{C}[x])=\{(0)\}\cup\{(x-a) \ : \ a\in\mathbb{C}\}$ ;
Recall the prime ideals of $R/I$ are precisely the quotients of prime ideals $J\supseteq I.$ Thus $\text{Spec}(\mathbb{C}[x]/(x^2))=\{(0),(x)\}.$

I-2. Note that $15$ sends an element $x$ of $\text{Spec}(\mathbb{Z})$ to $\overline{15}\in\kappa(x)=\mathbb{Z}/x.$ Thus $15$ takes the value $\overline{1}$ in $\kappa(7)$ and the value $\overline{0}$ in $\kappa(5).$

I-3.a. This follows since the map $p(x)\mapsto p(a)$ defines an isomorphism $\kappa((x-a))\to \mathbb{C}.$

I-3.b. Let $V\subset\mathbb{A}^n(K).$ The maximal ideals of $K[x_1,\ldots,x_n]$ are $(x_1-a_1,\ldots,x_n-a_n)$ for all points $(a_1,\ldots,a_n)\in\mathbb{A}^n(K).$ Thus the maximal ideals of $R=\Gamma(V)=K[x_1,\ldots,x_n]/I(V)$ are the images of these ideals under the natural map $K[x_1,\ldots,x_n]\to R$ and correspond the points of $V.$ Moreover, for a point $(a_1,\ldots,a_n)\in V$ the map $f(x_1,\ldots,x_n)\mapsto f(a_1,\ldots,a_n)$ is an isomorphism $\Gamma(V)\mapsto K.$ Therefore given a polynomial $f\in R$ we have $f(\mathfrak{p})=f(a_1,\ldots,a_n)$ where $\mathfrak{p}$ is the maximal ideal corresponding to the point $(a_1,\ldots,a_n).$

I-4.a. The points of $\text{Spec}(\mathbb{C}[x])$ are the ideals generated by irreducible polynomials, which in $\mathbb{C}[x]$ are the polynomials $x-a.$ Thus the closed points of $\text{Spec}(\mathbb{C}[x])$ are precisely the maximal ideals, i.e., the points corresponding to “actual points” in $\mathbb{C}.$ For a point $(x-a)$ to be open we would have some ideal $(f)\subset C[x]$ such that $(x-b)\supset (f)$ for all $b\ne a$ and $(x-a)\not\supset (f)$ ; this is clearly impossible, since $f$ would then have an infinite number of distinct factors.

I-4.b. The localization $K[x]_{(x)}$ is a DVR with uniformizing parameter $x.$ Thus it has prime ideals $(0)$ and $(x)$ , i.e., $\text{Spec}(K[x]_{(x)})=\{(0),(x)\}.$ The closed sets are precisely $\emptyset,\{(x)\},\{(0),(x)\}.$

I-5.a. Let $\mathscr{F}$ be a sheaf on $X.$ We may pick $\mathscr{F}(\{0\})$ and $\mathscr{F}(\{1\})$ to be arbitrary sets. Then $\mathscr{F}(\{0,1\})$ is the set consisting of a unique $f$ for each distinct $f_0\in\mathscr{F}(\{0\})$ and $f_1\in\mathscr{F}(\{1\})$ such that $f|_{\{0\}}=f_0$ and $f|_{\{1\}}=f_1.$ If we take $R=\mathbb{Z}/6\mathbb{Z}$ , then $\text{Spec}(R)=\{(2),(3)\}$ and the closed sets are $\emptyset,\{(2)\},\{(3)\},\{(2),(3)\}$ , i.e., every subset is closed. Thus the topology is discrete as desired.

I-5.b. If we take $R=K[x]_{(x)}$ then $\text{Spec}(R)$ has the desired topology (see exercise 1-4).

1-6. Let $\pi:E\to X$ be a vector bundle over a topological space $X.$ Let $\mathscr{F}$ be the sheaf of sections of $\pi$ on $X$ and let $\mathscr{G}$ be the sheaf of continuous functions on $X.$ Then $\mathscr{F}(U)$ is a $\mathscr{G}(U)$ -module for each open subset $U\subseteq X.$

I-8.a. We will first show that $\pi$ is continuous. Suppose that $U\subseteq X$ is open. Then $\pi^{-1}(U)=\{(x,s_x):x\in U, s\in\mathscr{F}(V)\text{ for some open } V\subseteq X\}.$ But if $s\in\mathscr{F}(V)$ then $t=s|_{V\cap U}\in\mathscr{F}(V\cap U)$ and $t_x=s_x$ for all $x\in U\cap V.$ So we in fact have $\begin{aligned}[t] \pi^{-1}(U) &=\{(x,s_x):x\in U, s\in\mathscr{F}(V)\text{ for some open } V\subseteq U\}\\ &= \bigcup_{V\subseteq U}\bigcup_{s\in\mathscr{F}(V)}\mathscr{V}(V,s). \end{aligned}$ So $\pi^{-1}(U)$ is open and $\pi$ is continuous.

Now we will show that $\sigma$ is a continuous section of $\pi$ over $U.$ By the definition of $\sigma$ we have that $\pi\circ\sigma$ is the identity on $U$ ; $x\mapsto (x,s_x)$ under $\sigma$ and $\pi$ projects $(x,s_x)\mapsto x.$ It suffices to show $\sigma^{-1}(\mathscr{V}(V,s))$ is open for each open $V$ and $s\in\mathscr{F}(V)$ since $\mathscr{V}(V,s)$ form a basis. Observe that $\sigma^{-1}(\mathscr{V}(V,s))=\{x\in U : (x,s_x)\in\mathscr{V}(V,s)\}=\{x\in U : x\in V\}=U\cap V.$

I-8.b. For each $x\in U$ pick an open set $W_x=\mathscr{V}(V_x,s^x)$ , $V_x\subset U$ , in $\overline{\mathscr{F}}$ containing $\sigma(x).$ Observe that since $\pi\circ\sigma$ is the identity on $U$ , we must have $\sigma^{-1}(W_x)=V_x\cap U.$ Note that $\{V_x\}_{x\in U}$ is an open cover of $U.$ Moreover, if $z\in V_x\cap V_y$ , then $\sigma(z)\in\mathscr{V}(V_x,s^x)\cap\mathscr{V}(V_y,s^y).$ So $s^x|_{V_x\cap V_y}=s^y|_{V_x\cap V_y}.$ Thus by the sheaf axiom, there exists $s\in\mathscr{F}(U)$ such that $s|_{V_x}=s^x.$ So $\sigma$ is in fact defined by $x\mapsto (x,s_x).$

Note. So $\mathscr{F}$ is in fact isomorphic to the sheaf of germs of continuous sections of $\pi.$

I-9. Suppose that $\varphi(U)$ is injective for each open $U\subset X.$ Let $x\in X$ and $s_x,t_x\in\mathscr{F}_x$ such that $\varphi_x(s_x)=\varphi_x(t_x).$ Then there exists an open $U$ containing $x$ and $s,t\in \mathscr{F}(U)$ and $\varphi(U)(s)=\varphi(U)(t).$ But then $s=t$ since $\varphi(U)$ is injective. So $s_x=t_x.$ If $\varphi(U)$ is also surjective for each $U\subset X$ , then $\phi_x$ is surjective for each $x.$

Conversely, suppose that $\varphi_x$ is injective for each $x.$ Suppose that $U\subset X$ is open and $s,t\in\mathscr{F}(U)$ such that $\varphi(U)(s)=\varphi(U)(t).$ Then $\varphi_x(s_x)=\varphi_x(t_x)$ for all $x\in U.$ So $s_x=t_x$ for all $x\in U$ and $s=t.$ So $\varphi(U)$ is injective.

It remains to show that if $\varphi_x$ is bijective for each $x$ , then $\varphi(U)$ is also surjective for each $U.$ Let $s\in\mathscr{G}(U).$ Since each $\varphi_x$ is surjective for each $x\in U$ there exists an open $V_x\subset U$ containing $x$ and $t^x\in\mathscr{F}(V_x)$ such that $\varphi(V_x)(t^x)=s|_{V_x}.$ Observe that for each $x,y\in U$ we have $\varphi(V_x\cap V_y)(t^x)=\varphi(V_x\cap V_y)(t^y).$ But since $\varphi(V_x\cap V_y)$ is injective, $t^x|_{V_x\cap V_y}=t^y|_{V_x\cap V_y}.$ Thus by the sheaf axiom there exists $t\in\mathscr{F}(U)$ such that $t|_{V_x}=t^x$ for each $x$ , and hence $\varphi(U)(t)=s.$

I-13. Suppose that $\{U_\alpha\}$ is an open cover of an open set $U\subseteq X$ , and for each $\alpha$ we have $s^\alpha\in\mathscr{F}(U_\alpha)$ with the collection of sections satisfying the property that $s^\alpha|_{U_\alpha\cap U_\beta}=s^\beta|_{U_\alpha\cap U_\beta}$ for every $\alpha,\beta.$ We must show that there is a unique element $s\in\mathscr{F}(U)$ such that $s|_{U_\alpha}=s^\alpha.$ Note that $s^\alpha=(s^\alpha_{V})_{V\subset U, V\in\mathcal{B}}\in\varprojlim_{V\subset U,V\in\mathcal{B}}\mathscr{F}(V).$

Let $V\subset U$ be such that $V\in\mathcal{B}.$ Define a basic open cover of $V$ $S=\{B\in\mathcal{B} \, : \, B\subset V, B\subset U_\alpha \ \text{for some} \ \alpha\}.$ For each $B\in S$ we define a section $t_B\in\mathscr{F}(B)$ by picking some $\alpha$ such that $B\subset U_\alpha$ and defining $t_B=s^\alpha_B.$ Observe that this is well defined as if $B$ is also contained in another set in the cover $U_\beta$ the requirement that $s^\beta|_{U_\alpha\cap U_\beta}=s^\alpha|_{U_\alpha\cap U_\beta}$ ensures that $s^\beta_B=s^\alpha_B.$ Moreover, if $B_1,B_2\in S$ then for any basic open set $B_3\subset B_1\cap B_2$ a similar argument shows that $s_{B_1}|_{B_3}=s_{B_2}|_{B_2}.$

Since $\mathscr{F}$ was a $\mathcal{B}$ -sheaf, and $S$ was a cover of $U$ , our argument above actually defines a section $s_B\in\mathscr{F}(B)$ for every basic open set $B\subseteq U$ (not just those in $S$ ) and these sections satisfy the condition that for any two basic open sets $B_1,B_2\subseteq U$ the sections $s_{B_1}$ and $s_{B_2}$ restrict to the same section on any basic open set in their intersection. Therefore $s=(s_B)_{B\subset U,B\in\mathcal{B}}$ is a well defined element of $\mathscr{F}(U)$ and by construction $s|_{U_\alpha}=s_\alpha.$ It is clear that this construction was unique and completely determined by the sections $s^\alpha.$