# [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:

- \(Z(a) = \phi\).
- \(\sqrt{a} =\) either \(S\) or the ideal \(S_+ = \oplus_{d \gt 0} S_d\).
- \(a \supseteq S_d\) 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:

- A morphism of sheaves is determined by an induced morphims of sheaves on a base.
- 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:

- Suppose \(\mathscr{O}_X\) is a sheaf of rings on \(X\). Then, what is the tensor product of two \(\mathscr{O}_X\)-modules?
- 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 \(y_i \mapsto f_i\).

- 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 \(\phi(J) \subset I\).
- 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:

- Show that both \(\phi\) and \(\text{Spec } A\) are open sets of \(\text{Spec } A\).
- 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.
- 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.

- 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.
- 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.

- Show that, in an irreducible topological space \(X\), any nonempty open set is dense.
- 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 \(\text{Spec } A = \bigcup_{m \in \mu} U_m\) \(\Box\)

- If \(X\) is any topological space which is a finite union of quasicompact spaces, show tht \(X\) is quasicompact.
- Show that every closed subset of a quasicompact space is quasicompact.

- 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 \(\text{Spec } A\).
- 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.