[ag/hw] Homework

Affine and projective spaces

Let $Y$ be the plane curve $y = x^2$ (i.e. $Y$ is the zero-set of the polynomial $f = y - x^2$). Show that $A(Y)$ is isomorphic to the polynomial ring in one variable over $k$.

Proof: We need to show that $A(Y) = k[x, y]/(y - x^2)$ is isomorphic to $k[x]$. To do this, consider $k[x][y]$ as a polynomial in $y$ adjointed with the solutions of $y = x^2$. The solutions lie in $k[x]$. Alternatively, the kernel of the isomorphism $k[x, y]/(y - x^2) \xrightarrow{\varphi} k[x]$ is $\text{Ker}(\varphi) = y - x^2$, by the first isomorphism theorem $\Box$

Let $Z$ be the plane curve $xy = 1$. Show that $A(Z)$ is not isomorphic to the polynomial ring in one variable over $k$.

Proof: $A(Z) = k[x, 1/x]$ contains an invertible element not in $k$, and is therefore not a polynomial ring over $k$ $\Box$

Let $Y \subseteq A^3$ be the set $Y = \{t, t^2, t^3 \mid t \in k\}$. Show that $Y$ is an affine variety of dimension $1$. Find the generators for the ideal $I(Y)$. Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$. We say that $Y$ is given by the parametric representation $x = t, y = t^2, z = t^3$, and this curve is called the twisted cubic.

Solution: $A(Y) = k[x, y, z]/\langle z - y^2, z - x^3 \rangle \cong k[x]$, since the solutions of the two polynomials lie in $A^1$. Generators of $I(Y)$ are therefore $\langle z - y^2, z - x^3 \rangle$ Futher, $Y \cong A^1$, and is hence an affine variety of dimension $1$ $\Box$

Show that the Zariski topology on $A^2$ is not the [product topology](/gn#product-topology) of the Zariski topologies on two copies of $A^1$.

For homogenous ideal $a \subseteq S$, show that the following conditions are equivalent:

1. $Z(a) = \phi$.
2. $\sqrt{a} =$ either $S$ or the ideal $S_+ = \oplus_{d \gt 0} S_d$.
3. $a \supseteq S_d$ for some $d \gt 0$.

Sheaves

Suppose $\mathscr{F}(U)$ are maps to $S = \underline{S}_{\text{pre}}(U)$ that are locally constant; i.e. for any $p \in U$, there is an open neighborhood of $p$ where the function is constant. Show that this is a sheaf.

Proof: In the axioms of a presheaf, let the restriction map be the identity map. The gluability axiom of a sheaf is satisfied trivially $\Box$

Show that the [pushforward](/ag/1#definition-of-presheaf-and-sheaf) $\pi_* \mathscr{F}$ is a sheaf, if $\mathscr{F}$ is.

Proof: Suppose $\pi : X \rightarrow Y$ is a continuous function, and $\mathscr{F}$ is a sheaf on $X$. Then, $\pi_* \mathscr{F}(V) = \mathscr{F}(\pi^{-1}(V))$, where $V$ is an open subset of $Y$. To show that this is a presheaf, let $U \subset X$ be the preimage of $V$ under $\pi^{-1}$. Then, we have a restriction map for the inclusion map $V \hookrightarrow U$. Further, let $\phi_* \mathscr{F}(U) = \mathscr{F}(\phi^{-1}(U))$, where $\phi : Z \rightarrow X$. We end up with the inclusion map $V \hookrightarrow U \hookrightarrow W$, where $W \subset Z$. This satisfies the commutative diagram for the restriction map, and the gluability axiom for a sheaf, since $\mathscr{F}$ does $\Box$

Pushforward induces maps of stalks. Suppose $\pi : X \rightarrow Y$ is a continuous map and $\mathscr{F}$ is a sheaf of sets, rings or $A$-modules. If $\pi(p) = q$, describe the natural morphism of stalks $(\pi_* \mathscr{F})_q \rightarrow \mathscr{F}_p$.

Solution: The task is to describe the map $\underset{\longrightarrow}{\text{Lim }} \mathscr{F}(\pi^{-1}(V)) \rightarrow \underset{\longrightarrow}{\text{Lim }} \mathscr{F}(U)$, where $p \in U \subset X, q \in V \subset Y$. This maps the colimit of the sheaf of the preimage of $V$ to that of the subset $U$; in other words, what we have is the identity map $\Box$

If $(X, \mathscr{O}_X)$ is a ringed space, and $\mathscr{F}$ is an $\mathscr{O}_X$-module, describe how for each $p \in X$, $\mathscr{F}_p$ is an $\mathscr{O}_{X, p}$-module.

Solution: For $\mathscr{F}$ to be an $\mathscr{O}_X$-module, the following diagram must commute, for $U \subset V$:

$\begin{xy} \xymatrix{ \mathscr{F}(V) \times \mathscr{O}_X(V)\ar[r]\ar[d]_{\text{res}_{V, U} \times \text{res}_{V, U}} & \mathscr{F}(V)\ar[d]^{\text{res}_{V, U}} \\ \mathscr{F}(U) \times \mathscr{O}_X(U)\ar[r] & \mathscr{F}(U) } \end{xy}$

Taking colimits for all open neighborhoods $U, V$ such that $U \subset V$, we get the required result $\Box$

Show that the sequence of presheaves $0 \rightarrow \mathscr{F}_1 \rightarrow \ldots \rightarrow \mathscr{F}_n \rightarrow 0$ is exact only if $0 \rightarrow \mathscr{F}_1(U) \rightarrow \ldots \rightarrow \mathscr{F}_n(U) \rightarrow 0$ is, for all $U$.

Suppose $\phi : \mathscr{F} \rightarrow \mathscr{G}$ is a morphism of sheaves. Show that the presheaf kernel $\text{ker}_\text{pre} \phi$ is in fact a sheaf. Further show that it satisfies the universal property of kernels.

Let $X$ be $\mathbb{C}$ with classical topology. Let $\underline{\mathbb{Z}}$ be the constant sheaf on $X$ associated to $\mathbb{Z}$, $\mathscr{O}_X$ the sheaf of holomorphic functions, and $\mathscr{F}$ the presheaf of functions admitting a holomorphic logarithm. Describe the exact sequence of presheaves on $X$:

$0 \rightarrow \underline{\mathbb{Z}} \rightarrow \mathscr{O}_X \rightarrow \mathscr{F} \rightarrow 0$

where $\underline{\mathbb{Z}} \rightarrow \mathscr{O}_X$ is a natural inclusion, and $\mathscr{O}_X \rightarrow \mathscr{F}$ is given by $f \mapsto \text{exp}(2 \pi i f)$. Show that $\mathscr{F}$ is not a sheaf.

Prove that a section of a sheaf of sets is determined by its germs; i.e. the natural map is injective: $\mathscr{F}(U) \rightarrow \prod_{p \in U} \mathscr{F}_p$

Show that $\text{Supp}(s)$ is a closed subset of $X$.

Prove that any choice of compatible germs for a sheaf of sets $\mathscr{F}$ over $U$ is the image of a section of $\mathscr{F}$ over $U$.

Sheafification is a functor. Use the universal property to show that for any morphism $\phi : \mathscr{F} \rightarrow \mathscr{G}$ of presheaves induces the natural map $\phi^\text{sh} : \mathscr{F}^\text{sh} \rightarrow \mathscr{G}^\text{sh}$ of sheaves. Hence, show that sheafification is a functor from presheaves on $X$ to sheaves on $X$.

Proof: Using the universal property of sheafification, we get:

$\begin{xy} \xymatrix{ \mathscr{F}\ar[r]^{\text{sh}}\ar[dr]_g & \mathscr{F}^{\text{sh}}\ar[d]^f\ar@{.>}[dr]|{\exists!} & \\ & \mathscr{G}\ar[r]^{\text{sh}} & \mathscr{G}^{\text{sh}} } \end{xy}$

completing our proof $\Box$

Show that $\mathscr{F}^\text{sh}$ forms a sheaf using tautological restriction maps.

Describe the natural map of presheaves $\text{sh} : \mathscr{F} \rightarrow \mathscr{F}^\text{sh}$.

Show that the sheafification functor is left-adjoint to the forgetful functor from sheaves on $X$ to the presheaves on $X$.

Show that $\mathscr{O}_X \xrightarrow{exp} \mathscr{O}^*_X$ describes $\mathscr{O}^*_X$ as a quotient sheaf of $\mathscr{O}_X$. Find the open set in which this map is not surjective.

Morphisms of sheaves correspond to morphisms of sheaves on a base. Suppose $\{B_i\}$ is a base for topology of $X$. A morphism $F \rightarrow G$ of sheaves on the base is the collection of maps $F(B_k) \rightarrow G(B_k)$ such that the following diagram commutes:

$\begin{xy} \xymatrix{ F(B_i)\ar[r]\ar[d]_{\text{res}_{B_i, B_j}} & G(B_i)\ar[d]^{\text{res}_{B_i, B_j}} \\ F(B_j)\ar[r] & G(B_j) } \end{xy}$

for all $B_j \hookrightarrow B_i$. Prove the following:

1. A morphism of sheaves is determined by an induced morphims of sheaves on a base.
2. A morphism of sheaves on a base gives a morphism of induced sheaves.

Suppose $X = \cup U_i$ is an open cover, and we have sheaves $\mathscr{F}_i$ on $U_i$ along with the isomorphisms $\phi_{ij} : \mathscr{F}_i|_{U_i \cap U_j} \rightarrow \mathscr{F}_j|_{U_i \cap U_j}$, that agree on triple overlaps; i.e. $\phi_{ik} = \phi_{jk} \circ \phi_{ij}$ on $U_i \cap U_j \cap U_k$. Show that these sheaves can be glued together into a sheaf $\mathscr{F}$ on $X$, such that $\mathscr{F}_i \cong \mathscr{F}|_{U_i}$, and the isomorphisms over $U_i \cap U_j$ are the obvious ones. Thus, we can glue sheaves together using limited patching information.

Stalk of a kernel is the kernel of stalks. For all $p \in X$, show that:

$(\text{Ker}(\mathscr{F} \rightarrow \mathscr{G}))_p \cong \text{Ker}(\mathscr{F}_p \rightarrow \mathscr{G}_p)$

Show that the cokernel of stalks is naturally isomorphic to the stalk of cokernels.

Suppose $\phi : \mathscr{F} \rightarrow \mathscr{G}$ is a morphism of sheaves of abelian groups. Then, show that the image sheaf, $\text{Im } \phi$ is the sheafification of the image presheaf. Moreover, show that the stalk of an image is the image of a stalk.

Show that taking stalks of a sheaf of abelian groups is an exact functor. More precisely, if $X$ is a topological space and $p \in X$ is a point, show that taking the stalk at $p$ defines the exact functor $\text{Ab}_X \rightarrow \text{Ab}$.

"Sections of U" is a left-exact functor. For topological space $X$, and $U \subset X$, if the sequence of sheaves $0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H}$ is exact, then so is: $0 \rightarrow \mathscr{F}(U) \rightarrow \mathscr{G}(U) \rightarrow \mathscr{H}(U)$

Moreover, show that if the sequence of sheaves $\mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{H} \rightarrow 0$ is exact, then this is not: $\mathscr{F}(U) \rightarrow \mathscr{G}(U) \rightarrow \mathscr{H}(U) \rightarrow 0$

Show that the pushforward is a left-exact functor.

Show that, if $(X, \mathscr{O}_X)$ is a ringed space, then $\mathscr{O}_X$-modules form an abelian category.

Tensor product of $\mathscr{O}_X$-modules. Then:

1. Suppose $\mathscr{O}_X$ is a sheaf of rings on $X$. Then, what is the tensor product of two $\mathscr{O}_X$-modules?
2. Show that the tensor product of a stalk is the stalk of a tensor product.

Schemes I

Show that a map of differentiable manifolds $\pi : X \rightarrow Y$ with $\pi(p) = q$ induces a morphism of stalks $\pi^\# : \mathscr{O}_{Y, q} \rightarrow \mathscr{O}_{X, p}$. Show that $\pi^\# m_{Y, q} \subset m_{X, p}$. In other words, if you pull back a function that vanishes at $q$, you get a function that vanishes at $p$.

Describe $\text{Spec } k[\epsilon]/(\epsilon^2)$, where $\epsilon \neq 0$ is very small. $k[\epsilon]/(\epsilon^2)$ is called the ring of dual numbers. This is the first example of a non-zero function which is not determined by its values at points; its value at all points is $0$.

Describe $\mathbb{A}_\mathbb{Q}^1$.

Deduce that for $k$ an algebraically closed field, the only maximal ideals are $(x_0 - a_0, \ldots, x_n - a_n)$, from the Nullstellansatz.

If $\phi : B \rightarrow A$ is a map of rings, and $\mathfrak{p}$ is a prime ideal in $A$, prove that $\phi^{-1}(\mathfrak{p})$ is a prime ideal in $B$.

Consider the map $\mathbb{C} \rightarrow \mathbb{C}$ sending $x \mapsto y = x^2$. The "source" $\mathbb{C}$ can be interpreted as the x-line, and the "target" $\mathbb{C}$ can be interpreted as the y-line. This can be pictured as the parabola $y = x^2$ in the xy-plane mapping to the y-line. Interpret the corresponding maps of rings as $\mathbb{C}[y] \rightarrow \mathbb{C}[x]$ given by $y \mapsto x^2$. Verify that the preimage, or fiber above $a \in \mathbb{C}$ are the points $\pm \sqrt{a} \in \mathbb{C}$.

Suppose $k$ is a field, and $f_1, \ldots, f_n \in k[x_1, \ldots, x_m]$ are given. Let $\phi : k[y_1, \ldots, y_n] \rightarrow k[x_1, \ldots, x_m]$ be the ring morphism defined by $y_i \mapsto f_i$.

1. Show that $\phi$ induces a map $\text{Spec } k[y_1, \ldots, y_n]/I \rightarrow \text{Spec } k[x_1, \ldots, x_m]/J$ for ideals $I \subset k[y_1, \ldots, y_n], J \subset k[x_1, \ldots, x_m]$ such that $\phi(J) \subset I$.
2. Show that the map of (a) sends the point $[(x_1 - a_1, \ldots, x_m - a_m)] \in \text{Spec } k[x_1, \ldots, x_m]$ to: $(f_1(a_1, \ldots, a_m), \ldots, f_n(a_1, \ldots, a_m)) \in k^n$

Check that the $x$-axis is contained in $V(xy, yz)$.

Solution: Since $y = z = 0$ is one solution of $xy = yz = 0$, the $x$-axis is indeed contained in $V(xy, yz)$.

Prove that $V(S) = V((S))$.

Proof: Given that $S \subset \mathfrak{p}$, we need to show that $(S) \subset \mathfrak{p}$. The ideal generated by vanishing points is a set of functions vanishing at those points, and this is contained in the same prime ideal $\Box$

$V(S)$ is closed for all $S$. Check this is a topology:

1. Show that both $\phi$ and $\text{Spec } A$ are open sets of $\text{Spec } A$.
2. If $I_i$ is a collection of ideals, show that $\bigcap_i V(I_i) = V(\sum_i I_i)$. Hence, union of any collection of open sets is open.
3. Show that $V(I_1) \cup V(I_2) = V(I_1 I_2)$. Hence, intersection of any finite number of open sets is open.

Solution: Let $\text{Spec } A = \bigcup{\lambda \in \Lambda} U_\lambda$ be the open cover of $\text{Spec } A$: since arbitary unions of open sets is open, $\text{Spec } A$ is open in $\text{Spec } A$. It is also closed, because it is the whole space. Similarly, $\phi$ is both open and closed in $\text{Spec } A$, finishing the proof for (a).

By showing that closed sets pull back to closed sets, show that $\pi$ is a continous map. Interpret $\text{Spec}$ as a contravariant functor $\textbf{Rng} \rightarrow \textbf{Top}$.

Radicals commute with finite intersections. Prove that, for ideals $I_1, \ldots, I_n$, $\sqrt{\cap_{i = 1}^n I_i} = \cap_{i = 1}^n \sqrt{I_i}$.

Suppose $I, S \subset B$ are an ideal and multiplicative subset respectively.

1. Show that $\text{Spec } B/I$ is a closed subset of $\text{Spec } B$, and that $\text{Spec } S^{-1} B$ is an open subset of $\text{Spec } B$. Further, show that, for arbitary $S$, $\text{Spec } S^{-1} B$ need not be open or closed.
2. Show that $\text{Spec } B/I$ and $\text{Spec } S^{-1} B$ are the subspace topology induced by inclusion in $\text{Spec } B$.

Suppose $I \subset B$ is an ideal. Show that $f$ vanishes on $V(I)$ iff $f \in \sqrt{I}$; i.e. $f^n \in I$ for some $n \gt 1$.

Proof: $V(I)$ is the set of points which are contained in $\mathfrak{p}$, a prime ideal in $B$. It remains to be shown that $f \in \mathfrak{p}$, given that $f^n \in \mathfrak{p}$. This is obvious from the fact that $\mathfrak{p}$ is prime $\Box$

Show that distinguished open sets form a base for the (Zariski) topology.

Solution: Since $V(S)$ forms a topology, it remains to be shown that, given a subset $S \subset A$, the complement of $V(S)$ is $\bigcup_{f \in S} D(f)$. From the definitions of $V(S) = \{[\mathfrak{p}] \in \text{Spec } A : S \subset \mathfrak{p}\}$ and $D(f) = \{[\mathfrak{p}] \in \text{Spec } A : f \notin \mathfrak{p}\}$, this is clear $\Box$

Suppose $f_i \in A$ as $i$ runs over some index set $J$. Show that $\bigcup_{i \in J} D(f_i) = \text{Spec A}$ iff $\{(f_i)_{i \in J}\} = A$, or equivalently, there are $(a_i)_{i \in J}$ all but finitely many zero, such that $\sum_{i \in J} a_i f_i = 1$.

Show that if $\text{Spec } A$ is an infinite union of distinguished open sets, $\bigcup_{i \in J} D(f_i)$, then in fact, it is a union of finitely many number of these; i.e. for subset $J' \subset J$, $\text{Spec } A = \bigcup_{i \in J'} D(f_i)$.

Show that $D(f) \cap D(g) = D(fg)$.

Show that $D(f) \subset D(g)$ iff $f^n \in (g)$ for some $n \geq 1$, and $g$ is an invertible element of $A_f$.

Show that $D(f) = \phi$ iff $f \in \mathfrak{N}$.

For $A = A_1 \times \ldots \times A_n$, describe homeomorhpism $\text{Spec } A_1 \sqcup \ldots \sqcup \text{Spec } A_n \rightarrow \text{Spec } A$ for which each $\text{Spec } A_i$ is mapped onto distinguished open set $D(f_i)$ of $\text{Spec } A$. Thus, $\text{Spec } \prod_{i = 1}^n A_i = \bigsqcup_{i = 1}^n \text{Spec } A_i$ as topological spaces.

1. Show that, in an irreducible topological space $X$, any nonempty open set is dense.
2. If $X$ is a topological space, and $Z \subset X$ is an irreducible subset, show that $\bar{Z}$ is also irreducible in $X$.

Solution: To prove (a), let $U$ be this nonempty open set, and $\bar{U} \neq X$. Then, $X = U' \cup \bar{U}$, leading to a contradiction since $X$ is irreducible. To prove (b), take closure of $Z \subset X$, leading to $\bar{Z} \subset \bar{X}$. Since $X \subset \bar{X}$, and $X \not\subset \bar{Z}$, we get $\bar{Z} \subset X \subset \bar{X}$ $\Box$

If $A$ is an integral domain, show that $\text{Spec } A$ is irreducible.

Solution: $A$ has no nilpotents other than $0$. Hence, $\mathfrak{N} = \{(0)\}$.

Show that an irreducible topological space $X$ is connected.

Solution: Since $X$ is irreducible, $X = Y \cup Z \implies Y = X \vee Z = X$, where $X, Y$ are proper closed subsets. Since all open subsets of irreducible spaces are dense, $Y = \bar{Y}, Z = \bar{Z}$. Taking closure and complement, we get $\phi = Y' \cap Z' \implies Y' \neq \phi \wedge Z' \neq \phi$; hence, for nonempty open subsets $Y', Z'$, we get $\phi = Y' \cap Z'$; this is exactly the condition for separability, as required by the definition of connectedness $\Box$

Prove that $\text{Spec } A$ is quasicompact.

Proof: Let $\text{Spec } A = \bigcup_{\lambda \in \Lambda} U_\lambda$ be an open cover, with $U_\lambda = \text{Spec } A \backslash V(I_\lambda)$. Now, \begin{align*} \text{Spec } A &= \bigcup_{\lambda \in \Lambda} (\text{Spec } A \backslash V(I_\lambda)) \\ &= \text{Spec } A \backslash \bigcap_{\lambda \in \Lambda} V(I_\lambda) \\ &= \text{Spec } A \backslash V(\sum_{\lambda \in \Lambda} I_\lambda) \end{align*}

Hence, $V(\sum_{\lambda \in \Lambda} I_\lambda) = \phi$ and $\sum_{\lambda \in \Lambda} I_\lambda = A$.

Since $1_A \in A$, there exists some finite subset $\mu \subset \Lambda$, $i_m \in I_m$ such that $1_A = \sum_{m \in \mu} i_m$. Then, for $a \in A$, $a . 1 = \sum_{m \in \mu} a . i_m \in \sum_{m \in \mu} I_m$. Hence, $V(\sum_{m \in \mu} I_m) = \phi$, and $\text{Spec } A = \bigcup_{m \in \mu} U_m$ $\Box$

1. If $X$ is any topological space which is a finite union of quasicompact spaces, show tht $X$ is quasicompact.
2. Show that every closed subset of a quasicompact space is quasicompact.

1. If $k$ is a field, and $A$ a finitely-generated $k$-algebra, show that the closed points of $\text{Spec } A$ are dense, by showing that if $f \in A$, and $D(f)$ is a non-empty distinguished open subset of $\text{Spec } A$, then $D(f)$ contains a closed point of $\text{Spec } A$.
2. Show that if $A$ is a $k$-algebra that is not finitely generated, then the closed poitns of $\text{Spec } A$ need not be dense.

If $X = \text{Spec } A$ show that $[\mathfrak{q}]$ is a specialization of $[\mathfrak{p}]$ iff $\mathfrak{p} \subset \mathfrak{q}$. Hence, show that $V(\mathfrak{p}) = \overline{\{[\mathfrak{p}]\}}$.

Verify that $[(y - x^2)] \in \mathbb{A}^2$ is a generic point for $V(y - x^2)$.

Suppose $p$ is a generic point for the closed subset $K$. Show that it is "near every point $q$ of $K$" (every neighborhood of $q$ that contains $p$), and not "near every point $r$ not contained in $K$".

Show that every point $x \in X$ is contained in an irreducible component of $X$.

Show that $\mathbb{A}_\mathbb{C}^2$ is Noetherian, and $\mathbb{C}^2$ in the classical topology is not.

Show that a ring is Noetherian iff every ideal is finitely-generated.

If $A$ is Noetherian, show that $\text{Spec } A$ is a Noetherian topological space. Describe $A$ when $\text{Spec } A$ is a non-Noetherian topological space.

Show that $V(I(S)) = \bar{S}$.

Solution: We know that $I(S) = I(\bar{S})$, since $I(S)$ is the set of functions vanishing on $S$, and we have defined this to be closed on a Zariski topology. Now, it remains to be shown that $V(I(\bar{S})) = \bar{S}$.

Prove that, if $J \subset A$ is an ideal, $I(V(J))) = \sqrt{J}$. To gain some intuition about this statement, consider the ideal $(x^2) \subset \mathbb{R}[x, y]$; operating on it with V, we get y-axis as the variety; operating on the variety with I, we get the ideal $(x)$, which is the radical of $(x^2)$.

Proof: Let $f \in J \subset k[x_1 ... x_n]$ $(a_1 ... a_n) \in V(J)$

Introduce a new variable $y$, so that $J' = (J, fy - 1) \subset k[x_1 ... x_n, y]$ $(a_1 ... a_n, b) \in V(J')$

such that $bf(a_1 ... a_n) = 1$. Then,

$V(J') = \phi \implies 1 = \sum g_i h_i + g_0 (fy - 1) \mid g_i \in k[x_1 ... x_n, y], h_i \in J$

Multiply both sides by $f^m$ to obtain:

$f^m = \sum g_i(x_1 ... x_n, fy) h_i + g_0(x_1 ... x_n, fy) (fy - 1)$

This identity holds if $fy = 1$, and hence $f^m \in J$, as required $\Box$

Show that $V(\bullet)$ and $I(\bullet)$ give a bijection between irreducible closed sets of $\text{Spec } A$ and prime ideals of $A$. From this conclude that, in $\text{Spec } A$, there is a bijection between points of $\text{Spec } A$ and irreducible closed subsets of $\text{Spec } A$, where a point determines an irreducible closed set by taking closure. Hence, any irreducible closed set has exactly one generic point: irreducible closed subset $Z$ can be written as $\overline{\{z\}}$.

If $A$ is any ring, show that the minimal prime ideals are in bijective correspondence with irreducible components of $\text{Spec } A$. In particular, $\text{Spec } A$ is irreducible iff $A$ has only one minimal prime ideal.

What are the minimal prime ideals of $k[x, y]/(xy)$ where $k$ is a field?

Schemes II

Show that $A_f \rightarrow \mathscr{O}_{\text{Spec } A}(D(f))$ is an isomorphism.