Solutions to The Rising Sea (Vakil)

Proof:

Consider the multiplicative subset \( S = \{ g \in A \mid D(f) \subset D(g) \} \subset A\). By Exercise 3.5.E, \( D(f) \subset D(g) \Leftrightarrow f^n \in (g) \) for some \( n \geq 1 \Leftrightarrow g\) invertible in \(A_f\). By the last equivalent definition and the universal property of the localization, there is a natural map \( A_f \to S^{-1} A = \OO_{\spec A}(D(f)) \) which we will denote as \(\phi\). To see that this is an isomorphism, suppose \( \phi( \frac{g_1}{f^{r_1}} ) = \phi( \frac{g_2}{f^{r_2}} ) \) so that for each \( \pp \in D(f) \) ( i.e. \( f \notin \pp \) ) there is some \( h \notin \pp \) with \( h(g_1 f^{r_1} - g_2f^{r_2}) = 0 \). As this is true for each \( \pp \in D(f) \) so that \( V(h) \subset V(f) \), we have again by Exercise 3.5.E that \( h = f^k \) for some \( k \). Thus, the two are necessarily equal in \( A_f \). The map is clearly surjective since \( \{ 1, f, f^2, \dots \} \subset S \) so that \( S^{-1}A \subset A_f \).

Proof:

Using quasicompactness from Exercise 3.5.C, there exists some \(n \) with

By the previous exercise, we know that \( \OO_{\spec A}(D(f)) = A_f \) — thus, suppose \( \frac{s}{f^m} \in A_f \) is such that \( \res{D(f)}{D(h_i)} \frac{s}{f^m} = 0 \) — in other words, \( h_i^{k_i} \frac{s}{f^m} = 0 \) in \( A_{h_i} \). Now since \( \langle h_1, \dots, h_n \rangle = A_f \), there are some \( \frac{b_1}{f^{a_1}}, \dots, \frac{b_n}{f^{a_n}} \in A_f \) such that

so that

for large enough \(N\) — as \( f \) is invertible, this implies \( s = 0 \).

Proof:

As was the case in the previous exercise, this simply requires copying the above argument and multiplying by a large enough power of \( f \).

Proof:

We clearly have that every ring isomorphism \( A^\prime \to A \) induces an isomorphism \( \spec A \to \spec A^\prime \) as \( \spec (\cdot) \) is a contravariant functor. To see that an isomorphism \( \pi : \spec A \to \spec A' \) induces an isomorphism of rings \( A^\prime \to A \), notice that \( A = \OO_{\spec A} (\spec A) \) and \( A^\prime = \OO_{\spec A^\prime}(\spec A^\prime) \). Since \( \pi \) is an isomorphism of sheaves, we indeed get that the map \( \pi_* : \OO_{\spec A^\prime} \to \pi_* \OO_{\spec A} \) is an isomorphism of sheaves — let \( \phi = \pi_*(D(1)) :A^\prime \to A \) be the isomorphism of global sections. Letting \( \pp \in \spec A \) and \( \mathfrak{q} = \pi( \pp) \), by Exercise 2.2.I we get an isomorphism of stalks \( \phi_{\mathfrak{q}} : A^{\prime}_{\mathfrak{q}} \to A_\pp \) — pulling back prime ideals under \( \phi \), we get \( \pp = \phi^{-1}(\mathfrak{q}) \). But then \( \pi = \spec \phi \).

Proof:

Let \( \psi : A \to A_f \) be the localization map at \(f \in A \). By Exercise 4.1.A, \( \Gamma (D(f), \spec A) = A_f \) so that \( D(f) \) is homeomorphic to \( \spec A_f \). Following the hint, we wish to show that every distinguished open set in \( \spec A_f \) is a distinguished open set in \( \spec A\). If we suppose \( \pp \in A_f \) then \( D(\pp) = \{ \mathfrak{q} \in \spec A_f \mid \pp \not\subset \mathfrak{q} \} \) — however, since \( \spec A_f \cong D(f) \). Since \( \pp \) corresponds to a prime ideal \( \overline{\pp} = \psi^{-1} (\pp) \) in \( A \) not containing \(f\) and \( \psi \) preserves inclusion, we get \( D(\pp) = D(f) \cap D( \overline{\pp} ) = D( f \overline{\pp} ) \).

Proof:

We have by Vakil §2.2.8 that \( \OO_X \vert_U \) is indeed a sheaf of rings. Now if \( p \in U \), then as \( U \subset X \) there exists some open set \( V_p \subset X \) containing \( p \) such that \( (V_p, \OO_{V_p}) \cong \spec A^p, \OO_{\spec A^p} \) for some ring \( A^p \) ( I wasn't sure what notation to use since \( A_p \) looks like a local ring ). By quasicompactness Exercise 3.5.C, \( \spec A^p = \bigcup_{i=1}^n D(f_{i,p}) \) — in particular, \( p \in D(f_{j, p}) \) for some \( j \), so that \( D(f_{j, p}) \cap U \) is affine. In other words, \( U \) may be covered by distinguished open sets as well.

Proof:

Since every scheme \(X\) is covered by affine open schemes isomorphic to \( \spec A \), and the Zariski topology on \( \spec A \) has a base of distinguished open sets \( D(f) \), \(X\) may be covered by affine open schemes by Exercise 4.3.B.

Proof:

1. By Exercise 3.6.A we have that \( \coprod_{i=1}^n \spec A_i = \spec \prod_{i=1}^n A_i \), so that our sheaf is given on distinguished open subsets by localizations \(f \in \prod_{i=1}^n A_i \).

2. Consider \( X = \coprod_{n \geq 1} \spec A_n \) — then each \( \spec A_i \) is open in \( X \), however the collection \( \{ \spec A_i \} \) cannot have a finite subcover. Since every affine open set is quasicompact by Exercise 3.6.G, it must be the case that \( X \) quasicompact.

Proof:

Let us denote \( X = \spec A \), and suppose \( \frac{f}{g} \) with \( g \notin \pp \). Then \( [\pp] \in D(g) \) so that \( \frac{f}{g} \in A_g = \OO_X(D(g)) \). Thus, we consider the map \( \phi : A_\pp \to \OO_{X,\pp} \) given by \( \frac{f}{g} \mapsto (D(f), \frac{f}{g}) \).

To see that the map is injective, suppose that \( \phi( \frac{f}{g} ) = 0 \) in \( \OO_{X,\pp} \). Then there exists some \( g^\prime \in A \) with \( [\pp] \in D(g^\prime) \subset D(g) \) so that the restriction \( \res{D(g)}{D(g^\prime)} \frac{f}{g} = 0 \). Then there exist some \( n \) with \( (g^\prime)^n f = 0 \). Since \( D(g^\prime) \subset D(g) \), there exists some \( m \) with \( (g^\prime)^m = ag \) for some \( a \in A \). Multiplying by large enough powers, we get that \( a^kg^{mk} f = 0 \) so that \( \frac{f}{g} = 0 \)

To see surjectivity, fix \( \psi \in \OO_{X,\pp} \). Then we may take some local representative with a distinguished open subset \( D(g) \ni \pp \). Then \( \psi = \frac{f^\prime}{g} \) for some \( f^{\prime} \in A_g \) — since \( g \in A - \pp \) we have \( \frac{f}{g} \in A_\pp \) with \( \phi(\frac{f}{g}) = \psi \).

Proof:

1. This should simply follow from Exercise 2.4.B.

2. Though this was meant to be shown in part (a), we wish to show that the subset of \(X\) where the germ of \(f\) is invertible is open. Suppose \( f_\pp \) is invertible in \( \OO_{X,\pp} \) such that there exists \( g_\pp \in \OO_{X,\pp} \) with \( f_\pp g_\pp = 1_\pp \). Taking representatives, there exists some open set \( V \ni \pp \) and \( g \in \OO_{X}(V) \) with \( fg \vert_V = f\vert_V g\vert_V = 1_V \) — in fact, by the sheaf axiom we necessarily have that all stalks are invertible on \( V \), so that invertiblity is an open condition. Now if \( f \) doesn't vansish at \( \pp \), \( f_\pp \notin \mm_\pp \) so that \( f_\pp \) invertible ( if it weren't, \((f_\pp)\) would necessarily be contained in some maximal ideal which must be \( \mm_\pp \) ). We can conclude that if \( \textrm{Supp}\,(f) = \emptyset \), then \( f_\pp \) at all \( \pp \in X \) — using gluability on \(X\) we get that \( f \) is invertible.

Proof:

We construct our topological space in the same way to the prior discussion, via \( X = (\coprod_i X_i) / \sim \) with \( X_{ij} \ni x \sim f_{ij}(x) \) for all \( i, j \in I \). Then the topology on \(X\) is precisely the quotient topology — in particular, if \( \overline{\sigma}_i : X_i \hookrightarrow \coprod_{i} X_i \) are the canonical inclusions and \( \sigma_i : X_i \hookrightarrow X \) are the induced maps on the quotient, then \( U \subset X \) is open in \( X\) if and only if \( \sigma^{-1}_i (U) \) is open in \( X_i \) for all \(i\).

To construct the sheaf \( \OO_X \) over \( X \), we want \( \OO_{X} \) to agree with \( \OO_{X_i} \) trivially when there are no intersections, and \( \OO_X \) to be independent of choice of \( i, j \) when there is some overlap on \( X_{i} \) and \( X_{j} \). However, the cocycle condition guarantees that if there were some other set \( X_k \) intersecting \( X_i \cap X_j \) (really there should be an isomorphism of open subsets here), our choice of isomorphisms is independent of indices \( i, j \). In other words, we may construct a sheaf over a base of open subsets on \(X\) by \( (\sigma_i)_*( \OO_{X_i} ) \) and this extends to a sheaf on \(X\) by §2.5.

Proof:

The two schemes that we are gluing are \( X_1 = X_{11} = \spec k[t] \) and \( X_2 = X_{22} = \spec k[u] \), with open sets \( X_{12} = D(t) \cong \spec k[t, 1/t] \) and \( X_{21} = D(u) \cong \spec k[u, 1/u] \) and isomorphisms \( f_{12} : X_{12} \to X_{21} \), \( f_{21} : X_{21} \to X_{12} \) given by \( t \mapsto u \) and \( u \mapsto t \) respectively. By the previous exercise, this indeed gives us a scheme. Let \( s \in \Gamma(X, \OO_X) \) be a global section — by the construction of our structure sheaf from the previous exercise, this pulls back to global sections on \( X_1, X_2 \) i.e. polynomials \( g_1 \in k[t] \) and \( g_2 \in k[u] \) with the same coefficients. In other words, \( \Gamma(X, \OO_X) \cong k[t] \cong k[u]\). If we suppose to the contrary that \( X \cong \spec A \) for some ring \( A \), then by Exercise 4.3.A we would get some isomorphisms \( A \cong k[t] \) and \( A \cong k[u] \), which is impossible since \( D(u) \cong D(t) \) so that whatever point in \( A \) maps to \( u\) must also map to \(t\), and thus is 2-to-1.

Proof:

As before, we take \( X_1 = X_{11} = \spec k[x, y] \), \( X_2 = X_{22} = \spec k [u, v] \), \( X_{12} = D(x) \cup D(y) \cong \spec k[x, y, 1/x, 1/y] \) and \( X_{21} = D(u) \cup D(v) \cong k[u, v, 1/u, 1/v] \). Then we get an isomorphism \( f_{12} : X_{12} \to X_{21} \) by \( x \mapsto u, y \mapsto v \) (and vice-versa for inverse). By Exercise 4.4.A we obtain a scheme \(X\) which corresponds to two copies of the affine plane glued together.

By the same reasoning as the exercise above, \(X\) cannot be isomorphic to an affine scheme \(\spec A\) as that would imply by Exercise 4.3.A that we have an injective morphism but \( (x,y) \mapsto \pp \) and \( (u, v) \mapsto \pp \) for some prime \( \pp \).

Proof:

For the sake of notation and sanity, we will restrict our attention to \( \P^2_k \), and consider the three affine open subsets

Our open sets \( X_{ij} \) in this case are just the intersections of these open sets (i.e. \( X_{02} = X_0 \cap X_2 \)). In this case, our isomorphism \( f_{ij} : X_{ij} \to X_{ji} \) is given by multiplication by \( \frac{x_i}{x_j} \). Then the cocycle condition is fairly immediate since \( f_{ij} \circ f_{jk} \) is multiplication by \( \frac{x_i}{x_j} \) composed with multiplication by \( \frac{x_j}{x_k} \), i.e. multiplication by

which is precisely our isomorphisms \( f_{ik} \).

Proof:

Suppose we have \( f(x_{1/0}, \dots, x_{n/0}) \) on \( D(x_0) \) and \( g( x_{0/1}, \dots, x_{n/1} ) \) on \( D(x_1) \). In order for these to define a global section, they must agree on overlaps; by applying our transition function \( f_{01} : D(x_0) \to D(x_1) \) given by \( x_{i/0}/x_{1/0} \mapsto x_{i/1} \) means we must have

Assuming our functions are homogeneous (necessarily of the same degree since they must agree on overlaps) this implies

But then it must be the case that \( \operatorname{deg}(f) = 0 \) so that \( f \) and \( g \) are constants.

Proof:

Let \( P^n_k \) denote \( (k^{n+1} - \{ (0, \dots, 0) \}) / \sim \) (where \( \sim \) is the above relation). For any \( (a_0, \dots, a_n) \neq 0 \) in \( k^{n+1} \), there must exist some \( a_i \neq 0 \) so we define \( \phi( a_0, \dots, a_n ) = [( \frac{x_0}{x_i} - \frac{a_0}{a_i}, \dots, \frac{x_n}{x_i} - \frac{a_n}{a_i} )] \); since \( \frac{a_j}{a_i} = \frac{\lambda a_j}{\lambda a_i} \) for any \( \lambda \neq 0 \), this map preserves equivalence classes under \( \sim \) so it uniquely factors through to \( \overline{\phi} : P^n_k \to \P^n_k \).

We additionally wish to show that \( \overline{\phi} \) does not depend on choice of \( a_i \neq 0 \). If we suppose that \( a_j \neq 0 \) as well, then

Since \( \lambda = \frac{a_i}{a_j} \neq 0 \), our map \( \overline{\phi} \) is well-defined on intersections.

To construct the inverse, since \( k \) is algebraically closed our points are all of the form \( [ \frac{x_0}{x_i} - a_0, \dots, 1, \dots \frac{x_n}{x_i} - a_n ] \) on some affine open \( D(x_i) \). Then we define \( \overline{\psi} : \P^n_k \to P^n_k \) by

It is easy to check that these maps are inverse.

Proof:

From the hint, there is very little to do here. We consider the three affine schemes

Then our transition functions are the same as in the case of projective space (as we are simply considering a closed subset of projective space), so that these glue together in a natural way by Exercise 4.4.A.

Proof:

Given any homogeneous polynomial \( f \in A[x_{0/i}, \dots, x_n] \) of degree \(d\), we have that for any unit \( u \in A\), \(f[ux_0, \dots, u x_d] = u^{d} f[x_0, \dots, x_n] \) — thus, \( f \) factors through the quotient \( [x_0, \dots, x_n] \sim [ux_0, \dots, ux_n] \). Notice that in the ring \( A[x_0, \dots, x_n] \) our indeterminites \( x_i \) are non-nilpotent, so that on the affine open sets \( D(x_i) \) we have

is a polynomial (no longer homogeneous) of degree \(d\) in \( A[x_{0/i}, \dots, x_{n/i}] \) (which we will denote \( f_i \). In fact, when dividing through by \( x_i^{d} \) above, any term in \( x_{i/i} \) should become constant, so this is the same as a polynomial in \( A[x_{0/i}, \dots, x_{n/i}] / (x_{i/i} - 1) \). In particular, if \( f \) vanishes at some point \( P = [x_0, \dots, x_n] \), then it must also vanish at each \( [u x_0, \dots, u x_n] \) (for any unit \(u \in A\)). Thus, if \( P \in D(x_i) \) for some \(i\), we must necessarily have that \( f_i \) vanishes at \( P \). In fact, if \( D(x_j) \) is another affine open set which contains \( P \), then

so that our \( f_i \) and \( f_j \) necessarily glue under the transition functions.

As mentioned in Exercise 3.4.I, the vanishing set \(V (I) \) for \( I \subset B \) may be interpreted as \( \spec B / I \). Thus, we consider the locus of our homogeneous polynomial \( f \) in \( \P^n_A \) as the gluing together of affine sets \( \spec A[x_0, \dots, x_n] / (f_i) \) — by an identical argument to Exercise 4.4.D, these glue together using our standard transition functions.

Proof:

1. For the forward direction, suppose \( I \) a homogeneous ideal so that it is by definition generated by some homogeneous elements \( f_i \in S_\bullet \). We may suppose that each \( f_i \in S_{d_i} \) for some \( d_i \in \Z \). Then for any element \( f \in I \), \( f \) is a finite sum over our generators, so we may write

   $$ f = \sum_{i=1}^N a_i f_i $$

   for sufficiently large \(N\) and some \( a_i \in S_\bullet \). Now each \( a_i \) may be written \( a_i = \sum_{d \in \Z} (a_i)_{(d))} \). Thus, the degree \( n \) component of \( f \) may be expressed as

   $$ f_{(n)} = \sum_{i = 1}^N (a_i)_{(n - d_i))}\cdot f_i $$

   which is clearly generated by our \(f_i\) so that it is an element of \( I \).

   Conversely, if \( I \) is generated by some \( \{ f_i \}_{i \in J} \) (not necessarily finite), then each \( f_i \) may be decomposed into its degree \( n \) components \( f_i = \bigoplus_{d \in \Z} (f_i)_{(d)} \). By assumption, \( (f_i)_{(d)} \in I \) for each \( d \in \Z \) and \(i \in J\), so that \( I \) may be generated by \( \{ (f_i)_{(d)} \mid i \in J, d \in \Z \} \) — thus \( I \) is homogeneous.

2. Suppose \( I \) and \( J \) are homogeneous ideals in \( S_\bullet \) so that \( I \) is generated by some homogeneous elements \( I = \{ f_i \} \) and \( J = \{ g_j \} \). Then \( I + J \) is generated by \( \{ f_i\} \cup \{g_j\} \) and \( IJ \) is generated by \( \{ f_ig_j \} \) — the former is clearly homogeneous, and by the definition of a graded ring multiplication must respect grading, so \( f_i g_j \in S_{\deg f_i \cdot \deg g_j} \) and thus \( IJ \) homogeneous.

   To see that \( I \cap J \) is homogeneous, we utitlize part (a) above. If \( f = \sum_{d \in \Z} f_{(d)} \in I \cap J \), then as \( I \) and \( J \) are homogeneous ideals every degree \( d \) component of \( f \) is contained in \( I, J \) and thus \( I \cap J \). Then by part (a) we have that \( I \cap J \) is homogeneous.

   Lastly, we wish to show that \( \sqrt{I} \) is homogeneous by again utilizing part (a) above. Suppose \( f \sqrt{I} \) so that there exists some \( n \geq 1 \) with \( f^n \in I \). As \( f \) is an element in the direct sum, only finitely many of its homogeneous terms / components are non-zero — taking \( d_0\) to be the smallest such degree and \( N \) the largest, we have \( f = \sum_{d = d_0 }^N f_{(d)}\). Then \( f^n \) has lowest degree homogeneous term \( f^n_{(d_0)} \) of degree \( n \cdot d_0 \). As \( I \) is homogeneous, by part (a) \( f^n_{(d_0)} \in I \) so that \( f_{(d_0)} \in \sqrt{I} \) by definition. As \( \sqrt{I} \) is indeed an ideal, this implies that \( f - f_{(d_0)} \in \sqrt{I} \). Proceeding inductively, we can see that \( f_{(d)} \in \sqrt{I} \) for each degree \( d \) term of \( f \). Since \( f \in I\) was arbitrary, by part (a) we have \( \sqrt{I} \) is homogeneous.

3. Assume to the contrary we have \( fg \in I \) with \( f \notin I \) and \( g \notin I \), and write \( f = \sum_{ d = a }^n f_{(d)} \), \( g = \sum_{d = b}^m g_{(d)} \). Suppose we have chosen \( f, g \) with \( \deg f + \deg g \) minimal. Now the leading term of \( fg \) is \( f_{(n)}g_{(m)} \) — by part (a) we know that \( f_{(n)}g_{(m)} \in I \), and by assumption it must be the case that \( f_{(n)} \in I \) or \( g_{(m)} \in I \). If we assume without loss of generality it is the former, notice that it must be the case that \( f - f_{(n)} \notin I \) (as otherwise if \(f = (f - f_{(n)}) + f_{(n)} \) would be the sum of elements in \( I \) and thus in \(I\)). Then \( (f - f_{(d)}) \cdot g = fg - f_{(d)}g \in I \), though neither of the components are — however, this contradicts our minimality assumption of the degrees of \(f \) and \(g\).

Proof:

1. For the forward direction, suppose that \( S_\bullet \) is a finitely-generated graded ring over \(A\) so that the irrelevant ideal \( S_+ \) is finitely generated by some \( f_1, \dots, f_n \), and let \( f \in S \) be arbitrary. Using the grading on \(S_\bullet\), \(f\) may be written as \( f = a_0 + s_1 f_1 + \dots s_n f_n \) where \( a_0 \in S_0 = A \) and \( s_1, \dots, s_n \in S_{\bullet} \). Now for each \( s_i \), if \( d_i = \deg(s_i) > 0 \) then \( s_i \in S_+ \) so that it can be expressed as a linear combination \(f_1, \dots, f_n \) with coefficients having degree strictly less than \(d_i\). Proceeding by induction, it follows that \(f\) may be written as a polynomial in the \( f_1, \dots, f_n \) so that \( S_\bullet \) is a finitely-generated \(A\)-algebra.

   Conversely, if \( S_\bullet \) is a finitely generated \(A\)-algebra, with generators \( f_1, \dots, f_n \), then \( S_+ \) is generated by those \( f_i \) with positive degree.

2. For the forward direction, suppose that \( S_\bullet \) is Noetherian. Then every ideal is finitely generated, so in particular \( S_+ \) is finitely generated (i.e. \( S_+ \) is a finitely generated graded ring over \( A \)). Additionally, \( 0 \to S_+ \to S_\bullet \to S_0 \to 0 \) exact, so by Exercise 3.6.V \( S_0 \) is Noetherian.

   Conversely, suppose that \( A = S_0\) is Noetherian and \( S_\bullet \) is a finitely generated ring over \( A \). By part (a) above, this implies that \( S_\bullet \) is a finitely generated \(A\)-algebra. Then there is some surjection from \( S_\bullet \to A[x_0, \dots, x_m] \) for some \( m \geq 1 \). Since \( A \) is Noetherian, \( A[x_0, \dots, x_m] \) is necessarily Noetherian (Hilbert Basis Theorem), and thus \( S_\bullet \) is as well.

Proof:

1. For the forward direction, let \( \pp \subset A \) be a prime ideal and let \( i : A_0 \hookrightarrow A \) denote the inclusion morphism; by Exercise 3.2.M we know that \( i^{-1}(P) \) is necessarily a prime ideal in \( A_0 \)

   In the reverse direction, we utilize the hint given — that is, given a prime ideal \( \pp_0 \subset A_0 \), we define \( \pp = \bigoplus Q_i \) where \( a \in Q_i \Leftrightarrow a^{\deg f} / f^i \in \pp_0 \). Obviously if \( a \in Q_i \), then \( a^{\deg f} / f^i \in \pp_0 \) and thus \( a^{ 2\deg f} / f^{2i} \in \pp_0 \) so that \( a^2 \in Q_{2i} \); the converse is proved the same way. By homogeneity, if \( a \in Q_i \) and \( b \in Q_j \), then

   $$ (ab)^{\deg f} / f^{i + j} = \frac{a^\deg f}{f^i} \frac{b^\deg f}{f^j} \in \pp_0 $$

   so that \( ab \in Q_{i+j} \). It follows that if \( a_1, a_2 \in Q_i \), then \( 2a_ia_j \in Q_{2i} \) and thus \( (a_1 + a_2)^2 = a_1^2 + 2a_1 a_2 + a_2^2 \in Q_{2i} \) which implies \( a_1 + a_2 \in Q_i \). It follows that \( \pp \) is necessarily an ideal; it remains to show that it is prime. By the preceding exercise, it suffices to check only on homogeneous elements. Thus, if we have some \( a \in A_i, b \in A_j\) with \( ab \in P \) (in particular \( ab \in Q_{i + j} \) by looking at degrees), then

   $$ (ab)^{\deg f} / f^{i+j} = \frac{a^{\deg f}}{f^i}\frac{b^{\deg f}}{f^j} \in \pp_0 $$

   However, since \( \pp_0\) is assumed to be prime, this implies that either \( \frac{a^{\deg f}}{f^i} \in \pp_0 \) or \( \frac{b^{\deg f}}{f^j} \in \pp_0 \) from which the result follows.

   To see the two are inverse, notice that if \( \pp \subset A \) is a homogeneous prime ideal, then \( a \in \pp_i \Leftrightarrow \pp_i^{\deg f} \in \pp_{(\deg f) i} \Leftrightarrow a^{\deg f}/f^i \in \pp_0 \)

2. By part (a) above, the prime ideals of \( ((S_\bullet)_f)_0 \) correspond to the homogeneous prime ideals of \( (S_\bullet)_f \), which we identify with \( D(f) \subset \proj S_\bullet \)

Proof:

Using the definition from the preceeding paragraph, \( \pp \) is an element of \( D(f) \) iff \( \pp \) is not an element of \( V(f) \) iff \( f \notin \pp \) and \( \pp \) does not contain \( S_+ \). By an identical argument to the affine case, this corresponds precisely to the (now homogeneous) prime ideals in the localization \( (S_\bullet)_f \), which by the preceding exercise are in correspondence with the prime ideals in \( ((S_\bullet)_f)_0 \). Therefore the two may be identified (up to a bijective correspondence).

Proof:

Similar to the affine case, if \( U \subset \proj S_\bullet \) is an open subset in the Zariski topology then \( \proj S_\bullet \backslash U = V(T)\) for some homogeneous ideal \(T\). Then \( V(T) = \bigcap_{f \in T} V(f) \) so that \( U = \bigcup_{f \in T} D(f) \).

Proof:

1. Similar to the affine case (as indicated in the hint), we know that for each \( I \subset \pp \in \proj S_\bullet \) contained in \( S_+ \), \( V(I) \subset V(f) \Leftrightarrow f \in \pp \). Since \( I \subset \pp \) was arbitrary, this tells us \( V(I) \subset V(f) \Leftrightarrow f \in \bigcap_{I \subset \pp} \pp \). Thus, we wish to show that \( \sqrt{I} = \bigcap_{I \subset \pp} \pp \).

   For the forward direction, suppose that \( g \in \sqrt{I} \) — by Exercise 4.5.C, we may suppose without loss of generality that \(g\) is in fact homogeneous, as we may otherwise consider each degree \( n\) homogeneous component. Then there exists some \( n \geq 1 \) with \( g^n \in I \). Thus, for every homogeneous prime \( \pp \) (not containing \( S_+ \)) we have \( g^n \in \pp \). Using Exercise 4.5.C(c), this implies that \( g \in \pp \) for each prime \( \pp \) and thus \( g \in \bigcap_{I \subset \pp} \pp \).

   Conversely, suppose \( g \in \bigcap_{I \subset \pp} \pp \) — again, we may assume without loss of generality that \( g \) is homogeneous. Suppose to the contrary that \( g^n \notin I \) for any \( n \) so that the homogeneous prime \( (g) \) is disjoint from \( I \). By Zorn's lemma, this implies the existence of some ideal \( \qq \) maximal among ideals disjoint from \( (g) \). We must have \( \qq \) is prime, since otherwise we could find homogeneous \( ab \in \qq \) with \( a, b \notin \qq \) — by maximality of \( \qq \) we would have \( \qq + (a) \) and \( \qq + (b) \) intersect \( (g) \) so that \( g^m = q_1 + ra \), \( g^n = q_2 + sb\) for some \( m, n \geq 1 \) — since \( ab \in \qq \) we must also have \( g^{m + n} \in \qq \)

2. As in the affine case, we define \( I(Z) = \bigcap_{\pp \in Z} \pp \) — however, we now require that our prime (homogeneous) ideals now do not contain \( S_+ \) (notice this implies \( I(Z) \subset S_+ \)). By Exercise 4.5.C(b), this \( I(Z) \) is also necessarily a homogeneous ideal. It follows from definition that \( I(Z_1 \cup Z_2) = I(Z_1) \cap I(Z_2) \) since we may possibly repeat any prime ideals \( \pp, \pp^\prime \) in the intersection of \( Z_1 \cap Z_2 \) without affecting \( \bigcap_{\pp \in Z_1} \pp \cap \bigcap_{\pp^\prime \in Z_2} \pp^\prime \).

3. For reverse inclusion, by definition we have that for all \( \pp \in Z \), \( I(Z) \subset \pp \). Since \( V(I(Z)) \) is the union of all homogeneous prime ideals containing \( I(Z) \), this implies that \( Z \subset V(I(Z)) \). Since \( V(I(Z)) \) is closed, this implies \( \overline{Z} \subset V(I(Z)) \). Conversely, since \( \overline{Z} \) is closed we have that \( \overline{Z} = V(J) \supset Z \) for some homogeneous ideal \( J \) (by definition of the Zariski topology). Then \( J \subset I(V(J)) \subset I(Z) \) so that \( V(I(J)) \subset V(J) = \overline{Z} \) as desired.

Proof:

Notice that (a) and (b) are clearly equivalent from definitions, since if \( I = (f_i) \) so that \( V(I) = \bigcap V(f_i) \), then

To see that (b) implies (c) suppose that \( f \in S_+ \) so that \( D(f) \subset \bigcup D(f_i) \) and thus \( V(I) = V( (f_i)) \subset V(f) \). By the previous exercise, this is equivalent to \( f \in \sqrt{I} \). Since \( f \in S_+\) was arbitrary, this tells us \( S_+ \subset \sqrt{I} \).

Lastly, to see that (c) implies (a), suppose \( \pp \in V(I) \) so that \( \pp \) is a homogeneous prime ideal containing \( I \) that doesn't contain \( S_+ \). Since prime ideals are radical and radicalization preserves containment, we have \( \sqrt{I} \subset \pp \). By assumption, however, this implies \( S_+ \subset \sqrt{I} \subset \pp \) — a contradiction.

Proof:

Let \( R \) denote the graded ring \( (S_\bullet)_f \), and suppose \( V(I_0) \subset \spec R_0 \) is a closed subset. Since \( V(I) = V(\sqrt{I}) \) we may assume without loss of generality \( I \) is radical so that \( I_0 = \bigcap_{ I_0 \subset \pp_0 } \pp_0 \subset R_0 \) (where the prime ideals are not necessarily homogeneous here). From our corresponence in Exercise 4.5.E there exists some homogeneous \( \pp \in R \) for each \( \pp_0 \in R_0 \) such that \( \pp \cap R_0 = \pp_0 \) — if we let \( I \) denote the intersection of all such \( \pp \), then \( I \) is again homogeneous by Exercise 4.5.C(b) and more importantly \( I \cap R_0 = I_0 \). By part (b) of Exercise 4.5.E the (not necessarily homogeneous) prime ideals of \( R_0 \) corresponds to \(D(f)\) so that \( V(I_0) = V(I) \cap D(f) \).

Proof:

Since \( f \) is invertible in \( (S_\bullet)_{fg} \), by the universal property of the localization \( (S_\bullet)_f \) there exists a unique morphism \( \phi : (S_\bullet)_f \to (S_\bullet)_{fg} \) so that \( \phi(\frac{h}{f^n}) = \frac{hg^n}{(fg)^n} \) and thus \( \phi \) preserves grading. Thus, we get a morphism \( \phi_0 : ((S_\bullet)_f)_0 \to ((S_\bullet)_{fg})_0 \). Since

has inverse \( \frac{ f^{\deg f + \deg g} }{ (fg)^{\deg g} } \), its degree 0 parts become invertible and we may apply the universal property of localization again to get a unique morphism \( \phi_0^\prime : \frac{ g^{\deg f} }{ f^{\deg g} } \subset \spec ((S_\bullet)_f)_0 \to \spec ((S_\bullet)_{fg})_0 \). We will not show that this is necessarily an isomorphism, as it can be easily checked.

Proof:

If we now suppose \( f, g, h \in S_+\) nonzero homogeneous polynomials, we have by the previous exercise that \( \spec ((S_\bullet)_{fgh})_0 \) is isomorphic to \( D( \frac{ h^{\deg f + \deg g} }{(fg)^{\deg h}}) \) in \( \spec ((S_\bullet)_{fg}) \), \( D( \frac{ g^{ \deg f + \deg h } }{ (fh)^{\deg g} } ) \) in \( \spec ((S_\bullet)_{fh}) \), and \( D( \frac{ f^{\deg h + \deg g} }{ (gh)^{\deg f} } ) \) in \( \spec ((S_\bullet)_{gh}) \). Each of \( f, g \) and \(h\) are invertible in \( \spec ((S_\bullet)_{fgh}) \) (since, for example, \( \frac{1}{f} = \frac{gh}{fgh} \)), and thus the obvious isomorphism between the first and the second distinguished open sets is multiplication by \( \frac{ g^{\deg f + 2\deg h} }{ h^{\deg f + 2 \deg g}} \) — the isomorphisms are similar. Each of these isomorphisms induces an isomorphism on the degree 0 part of our graded schemes, so that these gluings agree on triple overlaps

Since we have well-defined structure sheaves for each distinguished open set \( \spec ((S_\bullet)_{ab})_0 \), by Exercise 2.5.D there exists a sheaf \( \OO_{\proj S_\bullet} \) which restricts to \( \OO_{\spec (S_bullet)_{x_i}} \) for each of our homogenous projective coordinate \( f, g, h = x_i \).

Proof:

I assume this is interpreted as \( \OO_{\proj S_\bullet} \) consisting of collections of germs \( (f_i / s_i)_{i \in I} \) where each \( f_i / s_i \in \OO_{\spec ((S_\bullet)_{s_i})_0} \) and the \( (s_i) \) generate \( S_\bullet \) (i.e. if \( S_\bullet = A[x_0, \dots, x_n] \) we take the \( s_i = x_i \) ) which are compatible in the sense that they agree on overlap.

Proof:

I assume this is interpreted as \( \OO_{\proj S_\bullet} \) consisting of collections of germs \( (f_i / s_i)_{i \in I} \) where each \( f_i / s_i \in \OO_{\spec ((S_\bullet)_{s_i})_0} \) and the \( (s_i) \) generate \( S_\bullet \) (i.e. if \( S_\bullet = A[x_0, \dots, x_n] \) we take the \( s_i = x_i \) ) which are compatible in the sense that they agree on overlap.

Proof:

This effectively follows from the discussion preceeding Exercise 4.5.E (in the text). We originally defined \( \P^n_A \) as \( n + 1 \) copies of \( \A_A^n \), each with local coordinates \( x_{0/i}, \dots, x_{n/i} \) and \( x_{i/i} = 1 \). From the discussion mentioned above, in the \( \proj \) construction we take \( S_\bullet = A[x_0, \dots, x_n] \) graded by degree and \( f = x_i \). Then \( \spec (S_\bullet)_f = \spec A[x_0/ x_i, \dots, x_n/ x_i] \) and taking the zero-degree components, we obtain scalar multiples of \( x_i \) on the \(i^{th} \) affine open set. On intersections, this agrees with \( D(\frac{x_i}{x_j}) \) by Exercise 4.5.K.

Our \( D(x_i) \) must necessarily cover \( \proj A[x_0, \dots, x_n] \) since if there were some point \( P \in \P^n_A \) not contained in any \( D(x_i) \), then \( P \) would vanish on \( S_+ \) implying that the corresponding prime ideal \( \pp \) contains \( S_+ \) — a contradiction from our definition of \( \proj\).

Proof:

If \( [a_0 : \dots : a_n] \) is a closed ("classical") point of \( \P_k^n \), then the corresponding ideal on \( \spec (k[x_0, \dots, x_n]_{x_i})_0 \) is \( \left( \frac{ x_0 }{x_i} - \frac{a_0}{a_i}, \dots, \frac{x_n}{x_i} - \frac{a_n}{a_i} \right) \). In \( S_\bullet = k[x_0, \dots, x_n]\), this corresponds to the homogeneous prime ideal \( (a_ix_0 - a_0x_i, \dots, a_ix_n - a_nx_i) \).

Proof:

Since \( f \) is homogeneous, \( S_\bullet / f \) is a graded ring so we define \( V_+(f) := \proj S_\bullet / (f) \). On every affine open set \( U = \spec R\), \( V_+(f) \cap U \) is clearly a closed subset by Exercise 4.5.J so that \( V_+(f) \cap U = \spec R / J\) for some ideal \( J \).

Proof:

If \( x \in \P V \) is a closed point corresponding to a homogeneous maximal ideal \( \mm \subset \textrm{Sym}^\bullet\,V^\vee \), which must correspond to the span of linear forms by maximality. By linearity, this corresponds to the vanishing set of some one-dimensional subspace. Conversely, any one dimensional subspace corresponds to the vanishing set of some linear forms. It is a standard linear algebra fact that this is independent of choice of dual basis.