Posted on January 2, 2019

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. \text{Spec}(Z)=\{(0)\} \cup \{(p) \ : \ p \ \text{prime}\};
  2. \text{Spec}(\mathbb{Z}/3\mathbb{Z})=\{(0)\} (note \mathbb{Z}/3\mathbb{Z} is a field);
  3. \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);
  4. \text{Spec}(\mathbb{Z}_{(3)})=\{(0),(3)\};
  5. \text{Spec}(\mathbb{C}[x])=\{(0)\}\cup\{(x-a) \ : \ a\in\mathbb{C}\};
  6. 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.