Solutions to The Rising Sea (Vakil)

Proof:

Let \( Z = \pi(X) \); notice that we may check whether a set is closed with respect to a given base; in our case, each \( \spec B_i \backslash (Z \cap \spec B_i) = \spec B_i \backslash V(I) \) (for some ideal \(I \subset B\)) is clearly open in \( \spec B_i \) (by construction of the Zariski topology) and thus open in \( Y \). Since the \( \spec B_i \) cover \( Y \), it follows that \( Y \backslash Z \) is open in \( Y \) so that \( Z \) is closed.

Proof:

Closed embeddings are by definition affine, so it suffices to consider \( X = \spec A \) and \( Y = \spec B \). However by definition of a closed embedding, we have that \( A = B/I \) for some ideal \( I \subset B \). Since \( B / I \) is finitely generated as a \(B\)-module, the ring structure and natural projection make it a finite \( B \)-algebra so that \( \pi : X \to Y \) is a finite morphism. The second claim follows from Exercise 7.3.P(a)

Proof:

Let \( \pi : X \to Y \) and \( \rho : Y \to Z \) be closed embeddings. Since closed embeddings are affine, it suffices to consider the case that \( Z = \spec C \) and \( Y = \spec B \). Since \( \rho \) is assumed to be a closed embedding, we must have that \( B = C/I \) for some ideal \( I \subset C \); in a similar fashion, since \( \pi \) is a closed embedding we have \( X = \spec (C/I) / J^\prime \) for some \( J^\prime \subset C / I \). Letting \( J \) denote the preimage of \( J^\prime \) under the natural quotient map, the correspondence theorem tells us \( J \subset C\) is an ideal and \( (C/I)/J^\prime \cong C/(I + J) \). Then the composition of \( \rho \circ \pi \) is simply the map \( \spec C/(I + J) \to \spec C\) which is by definition a closed embedding.

Proof:

By definition we know that if \( \{\spec B_i\} \) is an affine cover of \( Y \) and \( \pi : X \to Y \) a closed embedding, then we can cover \(X\) by \( \pi^{-1}(\spec B_i) = \spec A_i \) such that \( \pi\vert_{\spec A_i} : \spec A_i \hookrightarrow \spec B_i \) is a closed embedding with \( \spec A_i \cong V(J_i) \) for some \( J_i \subset B_i \).

Let \( \spec B \subset Y \) be some arbitrary open affine. Since the property of being an affine morphism is affine-local on the target by Proposition 7.3.4, we may reduce to the case \( Y = \spec B \) and \( X = \spec A \). It suffices to show that if \( (f_1, \dots, f_k) = A \) and \( \pi_{f_i} : \spec A_{f_i} \to \spec B_{g_i} \) closed embedding for each \(i \) (with \( \pi^\sharp (g_i) = f_i \)) then our orignal map is a closed embedding. However, notice that by considering \( \coker \pi^\sharp \), we have that at each \( f_i \) the cokernel of the localization map \( (\coker \pi)_{f_i} = \coker (\phi)_{g_i} = 0\). Therefore, we must have \( \coker \pi^{\sharp} = 0 \) so that by the first isomorphism theorem we may simply take \( J = \ker \phi \) to get \( A / J \cong B \).

Proof:

This is quite similar to the method used in the previous exercise; letting \( \pi : \spec A \to \spec B \) denote our affine morphism and \( \pi^\sharp : B \to A \) the surjective map of rings, the first isomorphism theorem tells us \( A \cong B/\ker \pi^\sharp \). By Exercise 4.3.A, we get \( \spec A \cong \spec B/\ker \pi^\sharp \) so that our morphism factors as \( V(\ker \pi^\sharp) \hookrightarrow \spec B \).

Proof:

From the discussion in the preceding paragraph, let us suppose to the contrary that there is some closed subscheme \( Z \subset \spec k[x]_{(x)} \) such that \( \pi : Z \hookrightarrow \spec k[x]_{(x)} \) is a closed embedding and \( \OO_Z = \pi_\ast \OO_X \) is the cokernel of the map \( \mathscr{I} \to \OO_X \). Now as \( \spec k[x]_{(x)} \) is irreducible and \( Z \subset \spec k[x]_{(x)} \) is closed, it is also irreducible so we may consider \( \eta \) as the unique generic point for \( Z \) as well. Now on the open subset \( X \), \( \pi_\ast \OO_X(X) = \OO_Z(Z) \) is the cokernel of \( \{ 0 \} \hookrightarrow k[x]_{(x)} \) — thus, our map of global sections \( \OO_X \to \pi_\ast \OO_X \) is surjective which is as expected. However, on the open set \( \{\eta\} \) (compliment of the unique closed point \( [(x)] \)) we have that \( \mathscr{J}(\eta) = k(x) = \OO_X(\eta) \) so that our map is an isomorphism and thus the cokernel must be trivial. But then \( \OO_Z(\eta) \) must be the zero ring, which is a contradiction since this must be \( K(Z) \).

Alternatively, we could notice that the only proper closed subset is $$ Z = [(x)] = V((x)) = \spec k[x]_{(x)} / xk[x]_{(x)} \cong \spec ( k[x] / (x) )_{(x)} \cong \spec k $$ Indeed at the generic point we have \( \OO_Z([(0)]) = k \) which is non-trivial.

Proof:

Since closed embeddings are affine, we may write \( \spec A = \pi^{-1}(\spec B) \) (where \(\pi\) denotes our closed embedding). Let \( \pi^\sharp : B \to A \) denote the morphism of rings. Then we get a short exact sequence

Since localization is exact (by Exercise 1.6.F(a)) we get a sequence

However, as \( I(B_f) \) is defined to be the kernel of the morphism \( B_f \to A_{\pi^\sharp(f)} \) and kernels are unique up to unique isomorphism ( universal property ), we have that \( I(B)_f \to I(B_f) \) is an isomorphism.

Proof:

Suppose \( \spec B_i \) and \( \spec B_j \) are two affine open subsets of \( Y \), and consider their intersection \( \spec B_i \cap \spec B_j \). Since the distinguished opens form a base for the topology on \( \spec B_i \) and \( \spec B_j \), consider a distinguished open \( D_{B_i}(f) = D_{B_j}(g) \) in the intersection — notice that this implies \( (B_i)_f \cong (B_j)_g \). By considering the restriction of the closed subschemes \( \spec (B_i) / I(B_i) \) and \( \spec B_j / I(B_j) \) on this open subset, notice that since \( (B_i)_f \cong (B_j)_g \) we must have \( I(B_i)_f = I((B_i)_f) \cong I((B_j)_g) = I(B_j)_g \) and thus

Thus, if we denote the affine closed subscheme \( X_i = \spec B_i / I(B_i) \) for each \( i \) and \( X_{ij} = X_i \cap \spec B_j \), then we have just shown that we obtain isomorphisms \( \phi : X_{ij} \to X_{ji} \) with \( \phi_{ii} \) the obvious identity map. Showing that our closed subschemes \( X_i \) agree on triple overlaps is no different than the argument above so that by Exercise 4.4.A, the \( X_i \) glue to a well-defined subscheme \( X \hookrightarrow Y\).

Proof:

1. As noted in the hint, for any affine open \( \spec B \subset Y \) we define \( s_B = \res{Y}{\spec B}(s) \) and set \( V(s)_i = \spec B / (s_B) \). Now for any \( f \in B \) we have by our scheme axioms that \( s_{B_f} = \res{Y}{\spec B_f}(s) = \res{Y}{\spec B} \circ \res{\spec B}{D(f)} (s) \). Now if we consider the ideal \( (s_B)_f = \{ \frac{a s_B^i}{f^j} \vert a \in B \} \) it is easy to see that this is also the restriction of the ideal \( (s_B) \) to \( D(f) \) via \( \res{\spec B}{D(f)} \) since any coefficient \( \frac{a}{f^j} \) may be considered as an element of \( B_f \). Therefore, we get \( (s_B)_f \cong (s_{B_f}) \) so that by the previous exercise the \( V(s)_i \) agree on overlaps giving us a well-defined closed subscheme \( V(s) \hookrightarrow Y \)

2. If \( u \in \Gamma(Y, \OO_Y) \) is an invertible function, then for every \( \spec B \subset Y\) we must have \( u_B = \res{Y}{\spec B}(u) \) is an invertible element of \(B\) since the restriction maps are morphisms of rings — by the same reasoning we also have \( u_B s_B = (us)_B \). But then \( ((us)_B) = (u_B s_B ) = (s_B) \), so we have \( V(us)\vert_{\spec B} = V(s)\vert_{\spec B} \) on every open affine, which must glue by part (a) above.

3. As was the case for affine schemes, we want \( V(S) = V((S)) \) for any set and \( \cap_i V(I_i) = V(\sum I_i) \) — thus, one defines

   $$ V(S) = \bigcap_{s \in S} V(s) $$

   Since each \( V(s) \) is closed, \( V(S) \) must be closed as well. The next exercise will make this precise.

Proof:

1. Let \( Z_1, \dots, Z_n \) denote our closed subschemes and for each affine open \( \spec B \subset X \) let \( I_i(B) \subset B \) be the ideal sheaf of \( Z_i \). By Exercise 8.1.G we know that for any \( f \in B \) there is a natural isomorphism \( I_i(B_f) \cong I_i(B)_f \). To see that localization commutes with finite intersections (not proved in Vakil), suppose \( M_1, M_2 \) are \( B \)-modules and \( S \subset B \) is a multiplicatively closed subset. Then an element \( \frac{m}{s} \) is in \( S^{-1}M_1 \cap S^{-1} M_2 \) if and only if \( \frac{m}{s} \in S^{-1}M_1 \) and \( \frac{m}{s} \in S^{-1} M_2 \) if and only if we can find \( s_1, s_2 \in S \) such that \( s_1 m \in M_1 \) and \( s_2 m \in M_2 \) iff \( s_1s_2 m \in M_1 \cap M_2 \) iff \( \frac{m}{s} \in S^{-1}(M_1 \cap M_2) \). Thus, we obtain an isomorphism \( I_1(B_f) \cap I_2(B_f) \cong (I_1(B) \cap I_2(B))_f \) so that by Exercise 8.1.H, the \( \bigcap_{i=1}^n I_i(B) \) glue to a well-defined subscheme \( Z_1 \cup \dots \cup Z_n \).

   To define the arbitrary intersection, recall that we want \( \cap_i V(I_i) = V(\sum_i I_i) \) as in the affine case. Thus, if \( Z_1, \dots \) denote our closed subschemes and on \( \spec B \) we let \( I_i(B) \) denote the ideal sheaf of \( Z_i \) on \(B\), we define \( I(B) = \sum_i I_i(B) \). Since this may be considered as a quotient of \( \bigoplus_i I_i(B) \) and localization is exact and commutes with arbitrary direct sums by Exercise 1.3.F(b), we have that \( I(B_f) = I(B)_f \) so by Exercise 8.1.H, the \( I(B) \) glue to a well-defined subscheme \( \bigcap_i Z_i \)

2. \( V(y - x^2) \cap V(y) = V( (y - x^2, y) ) = \spec \Z[x, y] / (x^2, y) = \spec \Z[x] / (x^2) \). As described in Figure 4.5, the non-reducedness can be interpreted as "fuzz" in the \(x\)-direction which we may think of similar to our derivative at the double point. Since \( (y - x^2) \) and \( (y) \) are coprime ideals we have \( (y - x^2) \cap (y) = (y^2 - yx^2) \) so that \( Z_1 \cup Z_2 = V(y^2 - yx^2)\).

3. It clearly suffices to show this on affine open sets (since all poitns lie in an affine neighborhood by the definition of schemes). But then this follows from §3.4.

4. Notice that we can write \( Z = V(y^2 - x^2) \) as the scheme-theoretic union \( V(y- x) \cup V(y + x) = Z_1 \cup Z_2 \) since \( y - x \) and \( y + x \) are coprime in \( \Z[x, y] \). If we let \( W = V(y) \), then \( W \cap Z = \spec \Z[x, y] / (y^2 - x^2, y) = \spec \Z[x]/(x^2) \) which has multiplicity 2. However, \( W \cap Z_1 = V((y - x, y)) = V((x, y)) = V((y + x, y)) = W \cap Z_2 \) so \( (W \cap Z_1) \cup (W \cap Z_2) \) is simply \( V((x, y)) = \spec \Z \) which clearly has multiplicity 1. Thus, \( W \cap (Z_1 \cup Z_2) \neq (W \cap Z_1) \cup (W \cap Z_2) \).

Proof:

As mentioned in the hint, surjectivity of \( \pi^\sharp \) may be checked on an affine open base by Exercise 2.5.E so we may assume without loss of generality that \( Y = \spec B \). Now Vakil's definition assumes that \( \pi^{-1}(\spec B) \cong \spec B / I \) for some ideal \( I \) so that \( \pi \) is trivially homeomorphism with the closed subscheme \( \spec B / I \). Since localization is exact, it is easy to see that this means the maps of stalks are surjective as well.

For the reverse direction, we have a closed immersion \( \pi : X \to \spec B \) which by Exercise 6.3.F uniquely determines a morphism of rings \( \pi^\sharp(\spec B) : B \to \Gamma(X, \OO_X) \). Since \( \pi^\sharp \) is assumed to be surjective, we may let \( I = \ker (\pi^\sharp (Y)) \) denote the kernel so that there is a ring isomorphism \( \Gamma(X, \OO_X) \cong B / I \). Again using Exercise 6.3.F, this allows us to factor the map of schemes \( X \to \spec B \) through \( \spec B / I \). If we let \( \phi: X \to \spec B / I \) denote the first map, and \( \iota : \spec B / I \to \spec B \) the canonical closed immersion, it suffices to show that \( \phi \) is an isomorphism. Now since \( \pi^\sharp = \phi^\sharp \circ \iota^\sharp \) is a surjection, we must have \( \phi^\sharp \) is a surjection. To see that it is injective, let \( \mathcal{I} = \ker \iota^\sharp \) and \( \mathcal{J} = \ker \pi^\sharp \) denote the ideal sheaves on \( Y \) of our closed embeddings — by definition we have \( I = \mathcal{I}(Y) = \mathcal{J}(Y)\). Moreover, since \( \pi^\sharp = \phi^\sharp \circ \iota^\sharp \), we have \( \mathcal{I}(U) \subset \mathcal{J}(U) \) for all open sets \( U \). By checking on distinguished opens, we have that for any \( f \in B \) that \( \mathcal{J}(D(f)) \) is the kernel of the map \( \pi^\sharp(D(f)) : A_f \to \Gamma( \pi^{-1}(D(f)), \OO_X ) \) and \( \pi^{-1}(D(f)) = X_{\pi^\sharp(f)} \). Since \(X\) is homeomorphic to a closed subset of an affine scheme, it is quasicompact and quasiseperated. By the qcqs lemma, this implies \( \Gamma(X, \OO_X)_{\pi^\sharp(f)} \cong \Gamma(X_{\pi^\sharp(f)}, \OO_X) \) so that \( \pi^\sharp(D(f)) \) is simply the localization of \( \pi^\sharp(Y) \) so that it has kernel \( \mathcal{J}(D(f)) = I_f = \mathcal{I}(D(f)) \). Since \( \mathcal{J} \) and \( \mathcal{I} \) agree on our base of distinguished opens, they must agree everywhere by the identity axiom, and thus \( \phi \) is an isomorphism.

Proof:

As the property of being locally finite type is affine local on the target, suppose \( \pi : X \to \spec B \) is a locally closed embedding. By definition, we can factor \( \pi \) as \( \tau \circ \rho \) where \( \tau \) is an open embedding and \( \rho \) is a closed embedding. By Exercise 7.3.Q(a) we know that open embeddings are locally of finite type. By Exercise 8.1.B, closed embeddings are finite and thus of finite type. By Exercise 7.3.Q(b) we know that the composition of locally of finite type is locally of finite type, so that \( \pi \) is locally of finite type.

Proof:

Following the diagram approach mentioned in the hint, (i) clearly implies (ii) since if such a diagram exists we simply use the map \( V \to K \to X \). In the reverse direction, suppose we only have a diagram

We may assume without loss of generality that \( V \subset K \subset X \) since we are working "up to isomorphism". Since the topology on \( K \) is the subspace topology and \( V \subset K \) is open in the subspace topology, there exists some open subset \( U \subset X \) such that \( U \cap K = V \). Now \( U \hookrightarrow X \) is obviously an open embedding; to see that \( U \cap K \hookrightarrow U \) is a closed embedding we may assume \( U = \spec B \) is an affine open. Thus, \( V = K \cap U \) is of the canonical form \( \spec B/I \hookrightarrow \spec B \). Thus (i) and (ii) are equivalent. Since (i) clearly implies (iii) by following \( V \to U \to X \).

Proof:

Let \( \pi_1 : U \to X \) and \( \pi_2 : X \to Y \) be locally closed embeddings. By definition, we can factor \( \pi_1 \) into \( \pi_1 = \tau_1 \circ \rho_1 \) where \( \rho_1 : U \hookrightarrow V \) is a closed embedding and \( \tau_1 : V \hookrightarrow X \) is an open embedding. Similarly, we can factor \( \pi_2 = \tau_2 \circ \rho_2 \) where \( \rho_2 : X \hookrightarrow W \) is a closed embedding and \( \tau_2 : W \hookrightarrow Y \) is an open embedding. Then \( \rho_2 \circ \tau_1 \) is an open embedding followed by a closed embedding, so by the previous exercise (ii) implies (iii), we can find a closed embedding \( \sigma : V \to K \) and open embedding \( \eta : K \to W\) such that \( \rho_2 \circ \tau_1 = \eta \circ \sigma \). But then

Since the composition of open embeddings is open by Exercise 7.1.A we know that \( \tau_2 \circ \eta \) is an open embedding. Similarly by Exercise 8.1.C we know that \( \sigma \circ \rho_1 \) is a closed embedding, so that \( \pi_2 \circ \pi_1\) is a locally closed embedding by definition.

Proof:

1. We have already seen in Exercise 3.6.F that \( I = (wz - xy, x^2 - wy, y^2 - xz) \) is prime in \( k[x, y, z, w] \) and thus gives an irreducible affine scheme. By Exercise 4.5.E, on each distinguished open this defines an irreducible affine scheme. In order to show that the \( \proj \) construction makes sense, we wish to show that the affine opens glue.

   Let \( D(x), D(y), D(z), \) and \( D(w) \) denote the usual affine cover. On \( D(x) \), we may define local variables \( a = y/x \), \( b = z/x \), \( c = w/x \); on \( D(y) \) we have \( \widetilde{a} = x / y, \widetilde{b} = z / y, \widetilde{c} = w/y \) and so forth. To see that these agree on overlap requires explicitly checking the overlap of each chart. This is a bit tedious, but on \( D(x) \cap D(y) \) we have relations \( \widetilde{c} = b, \widetilde{b} = c \) and \( \widetilde{a} = a \). Similarly if we were to define the coordinates on \( D(z) \) by \( a' = x/z, b' = y/z, c' = w/z \) we would get the transition functions from \( D(x) \) to \(D(z)\) given by \( a' = 1/b \), \( b' = a /b \) and \( c' = c/b \). The transition from \( D(z) \) to \( D(w) \) would then have to be given by \( a'' = b', b'' = a', c'' = c' \) where one can easily guess what \( a'', b'', \) and \( c'' \) need to be. By the discussion at the beginning of the section, this gives us a closed subscheme.

2. To see that our curve is isomorphic to \( \P^1_k \), we consider the morphism of graded rings \( \phi k[x, y, z, w] \to k[s, t] \) given by \( x = s^2t, y = st^2, z = t^3, w = s^3 \). Since \( \{1, x, y, z, w \} \) are the \(k\)-algebra generators of \( k[x, y, z, w] \) this uniquely determines a \(k \)-algebra morphism. In fact this determines a morphism of graded rings since each monomial is sent to a degree 3 monomial (in the language of §6.4, this is a map of graded rings with \( d = 3 \). Since the Veronese subring of degree 3 \( \proj (k[x,y])_{3n} \hookrightarrow \proj (k[x,y])_n \) is an isomorphism by Exercise 6.4.D, we simply need to show that \( \ker \phi = ( wz - xy, x^2 - wy, y^2 - xz ) \). For ( \( \supseteq \) ), it is easy to see that \( \phi(wz - xy) = \phi( x^2 - wy ) = \phi( y^2 - xz ) = 0 \). For forward inclusion, we have shown that \( \phi \) factors through \( R_\bullet = k[x,y,z,w]/(wz - xy, x^2 - wy, y^2 - xz) \) so that there exists some \( \widetilde{\phi} : R_\bullet \to k[z, w]\) with \( \widetilde{\phi} = \phi \circ \pi \) where \( \pi \) is the natural projection. To show that \( \ker \widetilde{\phi} = (0) \), suppose \( f \in \ker \widetilde{\phi} \) and let us write \( f = p_1 + p_2 x + p_3 y \) for some \( p_1, p_2, p_3 \in k[z, w] \). Then

   $$ \phi(f) = p_1(t^3, s^3) + p_2(t^3, s^3) s^2 t + p_3(t^3, s^3) st^2 $$

   Then all the terms in \( p_1(t^3, s^3) \) have degree 0 mod 3 in \(t\), all the terms in \( p_2(t^3, s^3)s^2t \) have degree 1 mod 3 in \(t\), and all the terms in \( p_3(t^3, s^3)st^3 \) have degree 2 mod 3 in \(t\). Thus, in order for \( f \) to be in the kernel we must have \( p_1 = p_2 = p_3 = 0 \) so that \( \ker \phi = (wz - xy, x^2 - wy, y^2 - xz) \). By the isomorphism theorem, this tells us \( R_\bullet \) is isomorphic to a subring of \( (k[z, w]_\bullet \) which we know to be \( (k[z,w])_{3\bullet} \).

Proof:

Let \( \phi^\sharp : S_\bullet \twoheadrightarrow R_\bullet \) denote our surjection. Since \( \phi^\sharp \) preserves grading we can find some \( d > 0 \) such that \( \phi^\sharp(S_n) \subset R_{dn} \) — however, since \( \phi^\sharp \) is surjective it must be the case that \( d = 1 \), so in particular \( \phi^\sharp(S_+) = R_+ \). Then \( V(\phi^\sharp(S_+)) = V(R_+) = \emptyset \) so by Exercise 6.4.A this defines a morphism \( \phi : \proj R_\bullet \to \proj S_\bullet \). To see that it defines a closed embedding, by Exercise 8.1.D it suffices to check on our affine cover. However, for each \( f \in S_+ \) we have \( \phi^{-1}(D_+(f)) = D_+(\phi^\sharp(f) ) \) which corresponds to the ring map \( ((\phi^\sharp)_f)_0 : ((S_\bullet)_f)_0 \to ((R_\bullet)_{\phi^\sharp(f)})_0 \). Since \( \phi^\sharp \) is a surjection, \( (\phi^\sharp)_f \) is a surjection (as localization is exact), so \( ((\phi^\sharp)_f)_0 \) is as well proving the claim.

Proof:

We may in fact assume without loss of generality that \( S_\bullet \) is generated in degree 1 by Exercise 6.4.G. If we let \( A = S_1 \) and \( x_0, \dots, x_n \) denote our generators as a graded \(A\)-algebra, then we know that \( S_\bullet \cong A[x_0, \dots, x_n] / I \) for some ideal \( I \). Since \( \phi : X \hookrightarrow \proj S_\bullet \) is a closed embedding, it determines an ideal sheaf \( \mathscr{I}_{X/\proj S_\bullet} \) as the kernel of the map of \( S_\bullet \)-modules \( \phi^\sharp : \OO_{\proj S_\bullet} \to \phi_\ast \OO_X \). If we denote \( D_+(x_i) = \spec ((S_\bullet)_{x_i})_0 \) in the usual way, then it suffices to show that the \( \mathscr{I}_{X/\proj S_\bullet} (D_+(x_i))\) is locally determined by some homogeneous ideal.

If we let \( I_d \) denote the set

Then \( I = \bigoplus_{d > 0} I_d \) is a homogeneous ideal, and by construction we have \( ((I)_{x_i})_0 \subset \mathscr{I}_{X/\proj S_\bullet} (D_+(x_i)) \). Conversely if \( g \in \mathscr{I}_{X/\proj S_\bullet} (D_+(x_i)) \) is of the form \( g(x_{0/i}, \dots, x_{n/i}) = \sum_{m=0}^d g_m(x_{0/i}, \dots, x_{n/i}) \), then \( f = \sum_{m=0}^d x_i^{d - m} g_m(x_{0/i}, \dots, x_{n/i} \) is homogeneous of degree \(d\) and \( f/x_i^d = g \) so that \( g \in ((I)_{x_i})_0 \). Thus, \( X \cong \proj (S_\bullet / I) \).

Proof:

Recall from §4.5 that we defined \( \P V := \proj ( \operatorname{Sym}^\bullet V^\vee ) \) for any finite-dimensional vector space \( V\). If we let \( i : V \hookrightarrow W \) denote our \(k \)-linear inclusion, then the dual functor \( (\cdot)^\vee \) is simply the contravariant \( \textrm{Hom}(-, k) \) functor on the category of \( k \)-vector spaces. Thus, the induced map \( \iota_\ast \) is simply \( \iota_\ast(g) = g \circ \iota \) for \( g : W \to k \). In the finite-dimensional case, it is easy to show that \( \iota \) being injective implies \( \iota_\ast \) is surjective: any basis \( e_1, \dots, e_k \) of \(V\) can be extended to a basis \( e_1, \dots, e_n \) for \(W\) (here we are implicitly assuming \(V \subset W\)), so we may define a map \( g : W \to V \) by \( e_i \mapsto e_i \) for \( i \leq k \) and \( e_j \mapsto 0 \) for \( j > k \). Thus, \( g \circ \iota = id_V \) so that \( \iota_\ast \circ g_\ast = id_{V^\vee} \). But then for any \( f : V \to k \), we have define \( g_\ast(f) = f \circ g \in W^\vee\) and \( \iota_\ast(g_\ast(f)) = f \) so that \( \iota_\ast : W^\vee \to V^\vee \) is surjective. Then this clearly induces a surjective map \( \iota_{Sym} : \sym^\bullet W^\vee \to \operatorname{Sym}^\bullet V^\vee \) in the obvious way. By Exercise 8.2.B this gives us a closed embedding \( \P V = \proj(\sym^\bullet V^\vee) \to \proj ( \sym^\bullet W^\vee) = \P W \).

Proof:

Up to an algebraic change of coorindates, we may assume without loss of generality that \( \ell \) is the first copy of \( \P^1 \) in coordinates \( x_0, x_1 \) so that \( I(\ell) = (x_2, \dots, x_n) \). Then \( V(f) \cap V(I(\ell)) = V(f, x_2, \dots, x_n) \). Simplifying \( (f, x_2, \dots, x_n) \) as \( (\sum_{i=0}^d a_i x_0^{d-i}x_1^{i}, x_2, \dots, x_n \), we may simply restrict our attention to \( V( \sum_{i=0}^d a_i x_0^{d-i}x_1^{i} ) \) as a closed subscheme of \( \P^1_k \). Then on \( D_+(x_0) \), if we assume \( d^\prime \) is the largest index with \( a_i \neq 0 \) then

In other words, \( \sum_{i=0}^{d^\prime} a_i x_{1/0}^i \) splits into \( r \) irreducible factors \( f_j(x_{1/0}) \) each of degree \( d_j \) and multiplicity \( m_j \) such that \( \sum_{j=0}^r m_j d_j = d^\prime \).

Now on \( D_+(x_1) \), we may write \( \sum_{i=0}^d a_i x_0^{d-i}x_1^{i} \) as \( \sum_{i=0}^{d^\prime} a_i x_{0/1}^{d-i} = x_{0/1}^{d - d^\prime} \sum_{i=0}^{d^\prime} x_{0/1}^{d^\prime - i} \). By maximality of \( d^\prime \) such that \( a_{d^\prime} \) doesnt vanish, \( \sum_{i=0}^{d^\prime} x_{0/1}^{d^\prime - i} \) has a non-zero constant term and thus doesnt vanish at the closed point \( V(x_{0/1}) \). But then all of the zeros of \( \sum_{i=0}^{d^\prime} x_{0/1}^{d^\prime - i} \) lie in \( D_+(x_0) \) which were already accounted for, so the only unaccounted zero of \( x_{0/1}^{d - d^\prime} \sum_{i=0}^{d^\prime} x_{0/1}^{d^\prime - i} \) is the zero of degree \( d - d^\prime \) at \( V(x_{0/1}) \). Thus, there are \( d \) points counted with multiplicity.

Proof:

This was in fact the exact embedding used to solve Exercise 8.2.A(b), just with the order of the coordinates switched to match the previous equations.

Proof:

The problem statement essentially solves everything, since by Exercise 8.2.B a surjection of graded rings gives us a closed embedding.

Proof:

Let \( p = [a_0 , \dots, a_n] \in \proj S_\bullet \) be a closed point; there must exist some \( a_i \neq 0 \) so that \( p \in D_+(x_i) \), and thus since \( p \) is closed, \( k = \overline{k} \), and \( D_+(x_i) = \spec ((S_\bullet)_{x_i})_0 \) we may locally write \( p = [(x_{0/i} - a_{0/i}, \dots, x_{n/i} - a_{n/i} )] \). But then this corresponds to the line \( V((a_i x_0 - a_0x_i, \dots, a_i x_n - a_n x_i)) \) in \( \spec S_\bullet \) which clearly contains the origin \( 0 = [(x_0, \dots, x_n)] \).

Conversely, let \( \ell \subset \spec S_\bullet \) be a line going through the origin, and \( p \) a point in \( \ell \) other than the origin. Since \( k = \overline{k} \), we may write \( p = [ (x_0 - a_0, \dots, x_n - a_n) ] \) with some \( a_i \neq 0 \). Since \( \ell \) is determined by both \( p \) and the origin, we may write it as \( I(\ell) = (a_ix_0 - a_0x_i, \dots, a_ix_n - a_n x_i) \), which as before corresponds to the closed point \( [(x_{0/i} - a_{0/i}, \dots, x_{n/i} - a_{0/i}) ] \in \spec ((S_\bullet)_f)_0 \subset \proj S_\bullet\).

Proof:

Let \( \phi : k[u, v, w] \to S_{2\bullet} \) denote our map \( u \mapsto x^2 \), \( v \mapsto xy \), \( w \mapsto y^2\). This is clearly surjective since \( x^2, xy, y^2 \) generate \( S_{2\bullet } \). Moreover, \( \phi(uw - v^2) = x^2 y^2 - (xy)^2 = 0 \) so \( (uw - v^2) \subset \ker \phi \). For the reverse inclusion, suppose \( f \in \ker \phi \) and let \( p : k[u, v, w] \to k[u, v, w] / (uw - v^2) \) denote the natural projection. Note that since \( (uw - v^2) \subset \ker \phi \), \( \phi \) factors through \( p \) as some \( \widetilde{\phi} : k[u, v, w] / (uw - v^2) \to S_{2\bullet} \) with \( \phi = \widetilde{\phi} \circ p \). Notice that since \( \phi \) is surjective and \( p \) is surjective, \( \widetilde{\phi} \) must be as well. Moreover, if we let \( \widetilde{f} = p(f) \), then by definition we have \( \widetilde{\phi}(\widetilde{f}) = \widetilde{\phi}(p(f)) = \phi(f) = 0 \). However, since \( \overline{v}^2 = \overline{u}\overline{w} \) in \( k[u, v, w] / (uw - v^2) \), we may write

Then

However, since the first term is degree 0 mod 2 in \(x, y\) and the second term is degree 1 mod 2 in \(x, y\) it must be the case that \( \widetilde{f} = \widetilde{f}_1 = \widetilde{f}_2 = 0 \). In other words, \( f \in \ker p = (uv - w^2) \) so that \( \widetilde{\phi} \) is the desired isomorphism.

Proof:

We simply generalize the previous argument. Since \( S_{d\bullet} = k[x^d, x^{d-1}y, \dots, xy^{d-1}, y^d] \), consider the ring (changing notation a little since \(y\) is already used) \( k[z_0, \dots, z_n] \) along with the \( k\)-linear map \( \phi : k[z_0, \dots, z_d] \to S_{d\bullet} \) given by \( z_i \mapsto x^{d-i} y_i \) for each \(i\). This is clearly surjective since the \( x^{d - i} y_i \) generate \( S_{d\bullet} \). Let \( I \) denote the ideal generated by minors of the \( 2\times 2 \) matrix

Then any minor is of the form \( z_i z_j - z_{i+1} z_{j-1} \). Applying \( \phi \), we see that

so that \( I \subset \ker \phi \). Now let \( f \in \ker \phi \) and let \( p : k[z_0, \dots, z_d] \to k[z_0, \dots, z_d] / I \) denote the natural projection. By the equation above we know that \( \phi \) factors uniquely through \( p \) as some \( \widetilde{\phi} \) such that \( \phi = \widetilde{\phi} \circ p \). By writing \( \widetilde{f} = p(f) \), we have by construction that \( \widetilde{\phi} (\widetilde{f}) = \widetilde{\phi}(p(f)) = \phi(f) = 0 \). Letting \( \overline{z}_i \) denote the image of our generators in the quotient, we know that \( \overline{z}_j^2 = \overline{z}_{j+1}\overline{z}_{j-1} \) for any \( 1 \leq j \leq d - 1 \); thus, we may decompose \( \widetilde{f} \) as

Applying \( \widetilde{\phi} \), we get

Looking at the degree of \( y \) modulo \(d\), we have that the first term must be 0 mod \(d\), the second term \( 1 \) mod \(d\) and so forth. Therefore, we must have \( \widetilde{f}_i = 0 \) for all \( i \) and thus \( \widetilde{f} = 0 \) so that \( f \in \ker p = I \).

Proof:

We proceed by induction on the number of variables \( n \) in \( k[x_0, \dots, x_n] \). For the base case \( n = 1 \), we know from Exercise 8.2.J above that there are

homogeneous monomials of degree \( d \).

For the inductive step, assume the claim holds for \( k[x_0, \dots, x_{n-1}] \), and consider the number of possible monomials \( x_0^{\alpha_0}\dots, x_n^{\alpha_n} \) of degree \(d\) in \( k[x_0, \dots, x_n] \). We make judicious use of Pascal's identity

Notice that there are two posibilities for our monomial \(x_0^{\alpha_0}\dots, x_n^{\alpha_n}\) of degree \(d\): either it doesn't contain \( x_n \) (i.e. \( \alpha_n = 0 \)), in which case by inductive assumption we know that there are \( {{n+d}\choose{d}} \) choices — this accounts for the first term in Pascal's identity above. Now if \( x_0^{\alpha_0}\dots, x_n^{\alpha_n} \) does in fact contain \( x_n \), then we may divide by \( x_n \) to obtain a polynomial of degree \( d - 1 \) in \( n + 1 \) variables. Once we show that the claim holds for \( n + 1 \) variables, this will account for the \( {{n + 1 + d - 1}\choose{d-1}} = {{n + d}\choose{d-1}} \) term in Pascal's identity above. Since we have just divided by \( x_n \), we again reach two possibilities that either \( x_0^{\alpha_0}\dots, x_n^{\alpha_n - 1} \) contains \( x_n \) or doesn't contain \( x_n \) — in the first case we apply the inductive assumption, and in the second we repeat the argument. Repeating exhaustively, we arrive at the case that \( d - k = 1 \): there are obviously \( \#\{ x_0, \dots, x_n \} = n + 1 \) monomials of degree \(1\) which proves the claim.

Proof:

From the previous exercise, we know that there is a natural map \( \proj S_{2\bullet} \to \P^5 \) given by the graded ring morphism \( k[y_1, \dots, y_5] \to k[x_0, x_1, x_2]_{2\bullet} \)

Therefore, the equation above becomes

which implies all \( 2 \times 2 \) minors vanish. Thus, we get the following quadratic equations in \( \P^5\):

Proof:

1. The statement above may be formalized by stating that there is a well-defined map \( \phi : \P^1 \to X \) given by

   $$ [c, d] \mapsto [a_0c, b_0c, a_0d, b_0d] $$

   This is well-defined as \( [c, d] \) are not both 0 so that at least one of the coordinates in the image must be non-vanishing. In addition, it is easy to see that \( (a_0 c)(b_0 d) - (b_0c)(a_0d) = 0 \) so that the line lies on \(X\).

   There additional ruling is obtained by fixing \( c_0 \), \( d_0 \), not both \( 0 \) and allowing our original point to vary

   $$ [a, b] \mapsto [ac_0, bc_0, ad_0, bd_0] $$

2. Suppose \( p = [w_0, x_0, y_0, z_0] X \) is a \(k\)-point on \( X \subset \P_k^3 \). Without loss of generality, assume \( w_0 \neq 0 \) so that \( p \) is of the form \( [1 , x_0, y_0, x_0y_0] \). By taking the point \( [1 : x_0] \in \P^1 \), our first family of lines given by

   $$ [c, d] \mapsto [c , x_0 c, d, x_0 d] $$

   Passes through \( p \) at \( [c, d] = [1 , y_0] \). Similarly, by fixing the point \( [1 , y_0] \in \P^1\), our second family of lines given by

   $$ [a, b] \mapsto [a, b, a y_0, b y_0] $$

   Passes through \( p \) at \( [1, x_0] \).

   This shows that at least one line from each family passes through \( p \) — to see that it is exactly one line, notice that \( [a_0 , b_0] \neq [a_0^\prime, b_0^\prime] \) implies \( [a_0c, b_0c, a_0d, b_0d] \neq [a_0^\prime c, b_0^\prime c, a_0^\prime d, b_0^\prime d] \) for all \( [c, d] \in \P^1 \).

3. Following the hint, suppose \( L \) is a line on \( X \). By restricting to the affine patch \( \{ x_0 = 1 \} \), notice from part (b) that our two rulings may be described by the families of lines \( (x_0, t, x_0t) \) and \( t, y_0, y_0 t \) for any \( t \in \A^1 \). Now if we pick distinct points \( [1, x, y, z] \) and \( [1, x^\prime, y^\prime, z^\prime] \in X \cap D(x_0) \) we have that \( z = xy \) and \( z^\prime = x^\prime y^\prime \). Since the entire line \( L \) lies on \( X \), this also implies \( (1 + t) (z + tz^\prime) - (x + tx^\prime)(y + ty^\prime) = 0 \) for all \( t \). Taking \( t = -1 \) tells us \( (x - x^\prime)(y - y^\prime) = 0 \), so as \( k[x, y] \) is a UFD we either have \( x = x^\prime \) or \( y = y^\prime \). The first case tells us that \( y^\prime = ty \) --- replacting \( yt \) with \(t\) gives the first family in \(\A^1\). The second case tells us \( x^\prime = tx \) giving us the second family.

Proof:

Suppose \( x \) is given weight \( m \) and \( y \) is given weight \(n\). Let \( k[s, t] \) denote the usual weighting on the polynomial ring in 2 variables. Take \( d = \gcd(m, n) \) so that \( m = dm_0 \) and \( n = d n_0 \) where \( m_0, n_0 \) are relatively prime, and let \( k = m_0n_0 d \). Then there is a ring isomorphism \( k[s, t] \to k[x, y]_{k} \) given by \( s \mapsto x^{n_0} \), \( t \mapsto y^{m_0} \) which induces an isomorphism \( \proj k[x, y]_{k \bullet} \cong \P^1 \). By Exercise 6.4.D, \( \P(n, m) \cong \proj k[x, y]_k \) as desired.

For the second part, following the hint we notice that the degree 2 parts of \( k[x_0, x_1, x_2] \) (where \(x_2\) is given weight 2) are generated by \( x_0^2, x_0x_1, x_1^2, x_2\). Extending the isomorphism \( k[u, v, w] / (uw - v^2) \to k[x_0, x_1]_{2\bullet} \) given in Exercise 8.2.I to an isomorphism \( k[u, v, w, z] / (uw - v^2) \to k[x_0, x_1, x_2]_{2\bullet} \) in the obvious way \( z \mapsto x_2 \) gives the desired isomorphism \( \P(1, 1, 2) \cong \proj k[u, v, w, z] / (uw - v^2) \)

Proof:

By definition, \( \proj S_\bullet \) may be covered by affine sets \( \spec ((S_\bullet)_f)_0 \) for each \( f \in S_+ \). There is an obvious inclusion of rings \( (( S_\bullet )_f)_0 \hookrightarrow (S_\bullet)_f \) giving a map \( \spec (S_\bullet)_f \to \proj S_\bullet \). By an argument almost identical to Exercise 4.5.K, these maps must clearly glue to give a map \( \spec (S_\bullet) \backslash V(S_+) \to \proj S_\bullet \). It is easy to see why \( V(S_+) \) should be a point in the case that \( S_\bullet \) is a finitely-generated graded \(k\)-algebra, as the discussion following Exercise 6.4.G shows that we may assume \( S_\bullet \) is generated in degree 1 and \( S_\bullet \cong k[x_0, \dots, x_n] / I \) — thus \( V(S_+)\) corresponds to the classical origin \(V(x_0, \dots, x_n) \).

Proof:

This follows immediately from the previous problem, with the additional simplification added at the end that we may assume \( S_\bullet \) is generated in degree 1 with \( S_\bullet \cong A[x_0, \dots, x_n] / I \).

Proof:

By considering the natural projection \( S_\bullet[T] \twoheadrightarrow S_\bullet \) (obtained by setting \(T = 0\)), we have by Exercise 8.2.B that \( \proj S_\bullet \hookrightarrow \proj S_\bullet[T] \) is a closed subscheme.

Proof:

Since \(X\) is reduced, by Theorem 8.3.4 the scheme-theoretic image \( (\im \pi)_{sch} \) may be computed affine locally, so we may assume without loss of generality that \( Y = \spec B \) so that \( (\im \pi)_{sch} := \spec B / I(B) \) where \( I(B) = \ker( B \to \Gamma(\spec B, \pi_\ast(\OO_X)) ) \). Now if \( \overline{f}^k = 0 \) for some \( k \) on \( B / I(B) \), then taking some representative we have \( f^k \in I(B) \) so that \( \pi^\sharp(f^k) = \pi^\sharp(f)^k = 0 \). Since \( X \) is reduced, we must have \( \pi^\sharp (f) = 0 \) so that \( f \in I(B) \) and thus \( \overline{f} = 0 \). Thus, \( (\im \pi)_{sch} \) is reduced.

Proof:

First consider the affine case that \( \pi : \spec A \to \spec B \) is induced by some ring morphism \( \pi^\sharp : B \to A \). By the isomorphism theorem, we know that \( \pi^\sharp \) induces an isomorphism on \( \cong B / \ker \pi^\sharp \) with its image in \(A\). Since \( A \) and \(B\) are both Noetherian, the associated primes are simply the minimal prime ideals — since we have an induced isomorphism above, we know that minimal primes must be mapped to minimal primes, so we obtain an associated prime in \(X\).

In the more general case, if \( z \) is an associated point of the scheme theoretic image of \( \pi \), then we may take some affine neighborhood \( U = \spec B \) of \(z\) so that \( z \) is indeed an associated point of \( U \). Since \( \pi \) is quasicompact, we may consder finitely many \( \spec A_i \) covering \( \pi^{-1}(\spec B) \). Since the scheme theoretic image of \( \coprod_i \spec A_i \to \spec B \) is the same as the scheme theoretic image of \( f \) and the finite disjoint union of affine schemes is affine, we may again reduce to the affine case.

Proof:

By definition of a locally closed embedding (i.e., assuming (iii) in Exercise 8.1.M), we have that \( \pi : V \to X \) factors as

For some closed embedding \( \rho \) and open embedding \( \tau \). Then by taking \( K = (\im \pi \)_{sch} \) we obtain a closed embedding \( \iota : K \hookrightarrow X \) such that \( \pi \) factors through \( K \). Now by Corollary 8.3.5 we know that \( K \) is the closure of the set theoretic image of \(V\). Thus, \( V = U \cap K = U \cap \overline{V} \) so that \( U \cap K \hookrightarrow K \) is certainly an open embedding.

Proof:

First, note that a locally closed embedding onto a locally Noetherian target is necessarily quasicompact — as it suffices to check this condition affine locally, suppose \( Y = \spec B \) for some Noetherian ring \(B\) and \( \pi : X \hookrightarrow \spec B \) is a locally closed embedding. Then by Exercise 3.6.T we know that every open subset of \( \spec B \) is quasicompact. Without loss of generality, we may consider \( X \subset \spec B \) as a closed subscheme of an open subscheme of \(\spec B\) — indeed the open subscheme must be quasicompact, and any closed subset of a quasicompact set is quasicompact (c.f. Exercise 3.6.H(b)).

From the previous paragraph, we know that by Exercise 8.3.B that every associated point of \( (\im \pi)_{sch} \) is the image under \( \pi \) of some associated point \( x \in X \), which handles one direction of the correspondence. Conversley, suppose \( x \in X \) is an associated point. By Corollary 8.3.5, \( (\im \pi)_{sch} \) is the closure of the set-theoretic image. By the previous exercise, we may consider \( X \) as an open subset of \( (\im \pi )_{sch} \). Since associated points may be computed affine-locally (c.f Exercise 5.5.F) and open immersions preserve stalks, \( x \) must also be an associated point of \( (\im \pi)_{sch} \) as the stalk remains the fraction field at that point.

Proof:

Since we are assuming the scheme structure \(X\) over \(X^{set}\) is reduced, it suffices to check affine locally on the target \( Y = \spec B\). Now (1) and (3) can clearly be shown to be equivalent from Corollary 8.3.5 since a closed subscheme \(Z \hookrightarrow Y\) must be of the form \( \spec B / J \) for some ideal \(J\). If \( X^\prime = \spec B / I^\prime \subset \spec B / I(X^{set}) = X\) is a smaller closed subscheme containing \(X^{set}\) then we can find some \( f \in I^\prime \) with \( f \notin I(X^{set}) \). Then \( \overline{f} = 0 \) in \(X^\prime\) though as a function it does not vanish everywhere on \(X^{set}\) implying that \( X^\prime \) has some non-reduced structure, a contradiction.

To see that the second (2) definition is equivalent to (3), notice that since the property of being reduced is stalk-local (c.f. Exercise 5.2.A) \( W = \coprod_{p \in X^{set}} \spec \kappa(p) \) is a reduced scheme. By Corollary 8.3.5, this implies that \(X\) is just the set-theoretic closure of the image of \( \rho \) sending \(p \) to \(p\). It is a standard topological fact that the closure of \( \im \rho \) may also be interpreted as the smallest closed set containing \( \im \rho \). By construction, the points in the image of \( \im \rho \) are precisely the points of \( X^{set}\) as desired.

Proof:

For the latter claim, recall that another characterization of \(X\) reduced is that if \( \OO_X(U) \) is reduced for all \( U \) affine. By the first characterisation of the reduced subscheme structure on \( X^{set} \subset Y \), for each \( U = \spec B \subset Y \) we have that \( \OO_X( U) = B / I(X^{set}) \). Since \( I(X^{set}) \) is by construction a radical ideal, the quotient \( B / I(X^{set}) \) is reduced and thus \( \OO_X(U) \) is reduced proving the latter claim.

To see that the underlying set of \( X \to Y \) contains \( X^{set} \), we know from the third characterisation that the underlying set of \(X\) necessarily contains \( X^{set} \). To see that the reverse inclusion holds, suppose to the contrary that there is some point \( q \in \textrm{Set}(X) \) with \( q \notin X^{set} \) (the notation here is terrible, but \(X\) without the superscript should be understood as the scheme). Then by taking \( W^\prime = \coprod_{p \neq q} \kappa(p) \) and considering the usual map \( \rho : W \to Y \) sending \( p \) to \( p\), \( (\im \rho)_{sch} \) is now a closed subscheme whose underlying set contains \( X^{set} \) and is strictly smaller than \(X\) — a contradiction.

Proof:

There doesn't seem to be anything to show here; the "ideal of nilpotents" in the second definition is simply the nilradical, and as reducedness can be checked stalk-locally (c.f. Exercise 5.2.A) and the stalks must necessarily agree, our sheaf axioms tell us that the two construction must be the same.

Proof:

Suppose to the contrary the image \( t/1 \) is a zerodivisor in \( A_\pp \). Then there exists some \( s = a/g \in A_\pp \) with \( st = at / g = 0 = (0/1) \). This implies that there exist some \( g^\prime \notin \pp \) such that \( g^\prime at = 0 \) in \( A \), which further implies that \( t \) is a zerodivisor.

Proof:

Notice that our assumption shows that \( V(t) \) is globally principle (which indeed implies locally principal: by taking any \( \spec A \subset X \), we have \( V(t) \cap \spec A = V( \res{X}{\spec A}(t) ) \)). Thus, \( V(t) \) being an effective Cartier divisor is equivalent to \( t \) not being a zero-divisor. However, by property (C) of section §5.5, on any affine open \( \spec A \subset X \) we have that \( \res{X}{\spec A}(t) \) is a zero-divisor if and oly if it vanishes at some associated point, thus proving the claim.

Proof:

By the Nullstellensatz we have that \( \sqrt{(t)} = \sqrt{ (t^\prime) } \). However, this need not imply in the set-theoretic setting that \( (t) = (t^\prime) \) — simply take \( A = k[x] \) and \( (x), (x^2) \). Instead, since \( V(t) = V(t^\prime) \) in the scheme-theoretic sense we must also have that \( \Gamma(D(f), \spec A/(t)) = \Gamma(D(f), \spec A /(t^\prime)) \) for all \( f \in A \) — that is \( (A / (t))_f = (A/(t^\prime))_f \). In fact, by just looking at the localization at prime ideals we can immediately conclude \( (t) = (t^\prime) \) as desired.

Proof:

This follows by a slight generalization of Exercise 8.4.A above to \(A\)-modules: suppose \( S \subset A \) is a multiplicatively closed subset and \( x \in A \) is a zerodivisor of \( S^{-1} M \). Then we can find some non-zero \( \frac{m}{s} \) such that \( x \cdot \frac{m}{s} = 0 \) in \( S^{-1}M \); but then there exists some \( s^\prime \in S \) such that \( (s^\prime x) \cdot m = 0 \) in \( M \). Assuming \( A\) is commutative this impies \( x \) is a zero-divisor of \(M\). Thus, if \(x\) is a non-zero divisor of \(M\) then it is a non-zero divisor under any localization proving the claim.

Proof:

Part (ii) of definition 8.4.4 follows immediately: since \( (x^N, y) \subset (x, y) \) we know that \( (x^N, y)M \subseteq (x,y)M \subsetneq M \). For part (i), notice that \( x^N \) cannot be a zero-divisor of \(M\) since \( x \) is assumed to be a non-zerodivisor. Following the hint, we proceed by induction on \( N\); the base case \( N = 1 \) is trivial. Now assume we have shown that \( y \) is not a zerodivisor of \( M / (x^k M) \) for \( 1 \leq k \leq N - 1\). If \( y \) is a zerodivisor in \( M / (x^N M) \) then there exists some \( m \neq 0 \) such that \( ym = 0 (\mod x^N) \). Then \( ym = x^N k \) for some \( k \in M \) — since \( (x, y) \) is an \( M \)-regular sequence, \( ym = 0 (\mod x) \) and \( y \) is a nonzero divisor so that \( m = x^j \). Since we assumed \( m \neq 0 \mod x^n \), we must have \( j \leq N - 1 \). But then this contradicts the inductive assumption since we may take \( r = \max{j, N - j} \leq N - 1 \) and consider \( y x^j = x^N k \) modulo \( x^r \). Then \( y = x^{N - r} k \) in \( M / (x^r M) \) which is certainly a zero divisor.

Proof:

Since any permutation is generated by transpositions, it actually suffices to prove the case that if \( x_1, \dots, x_i, x_{i+1}, \dots, x_n \) is \( M \)-regular then \( x_1, \dots, x_{i+1}, x_i, \dots, x_n \) is \( M \)-regular. Since the regularity of \( x_1, \dots, x_{i-1}\) and \( x_{i+2}, \dots, x_n \) is immediate, we simply reduce to the case that \( a, b \) is \( M \)-regular implies \( b, a \) is \( M \)-regular (under the same hypotheses of \( M \) finitely generated over a Noetherian local ring). Part (ii) of definition 8.4.4 is satisfied automatically.

First we must show that \( b \) is a non-zerodivisor on \( M \). Suppose to the contrary that \( b m = 0 \) for some \( m \neq 0 \). By assumption, \( b \) is a non-zerodivisor on \( M / aM \) so that \( m \in aM \), say \( m = am_1 \). Rewriting our equation above, this gives us \( bm = b(a m_1) = a (bm_1) = 0 \); since \( a \) is a non-zerodivisor of \( M \) by hypothesis, we get that \( bm_1 = 0 \). Continuing in this fashion, we get that \( m = am_1 = a^2 m_2 = a^3m_3 = \dots \) so that \( m \in \mm^j M \) for each \( j \geq 0 \). By a corollary of Nakayama's lemma known as Krull's intersection theorem, one obtains \( \bigcap_j \mm^j M = 0 \) so that \( m = 0\) — a contradiction.

Showing that \( a \) is a non-zerodivisor on \( M / (b) \) is much easier: suppose to the contrary that \( a m = 0 \mod b \) for some \( m \neq 0 \mod b \) . Then \( am = bk \) for some \( k \in M \); that is to say that \( bk = 0 \mod a \). Since \( b \) is not a zero-divisor of \( M / aM \) by assumption, we must have that \( k = 0 \mod a \), say \( k = a k^\prime \). Then \( am = ab k^\prime \) so that \( a ( m - b k^\prime) = 0 \) in \( M \). By assumption, \( a \) is not a zerodivisor on \( M \) so we must have \( m = bk^\prime \) which contradicts the fact that \( m \neq 0 \mod b \).

Proof:

We prove the last hint first that we may recover an open neighborhood of \( [\pp] \) such that if the images of \( f_1, \dots f_r \) generate \( I_\pp \) in \( \OO_{Y, p} \) then the restriction of \( f_1, \dots, f_r \) generate \( I \) restricted to \( D(a) \) for some \( a \in \pp \). From the hint, we may clearly restrict to the case that if \( I \subset J \) and \( I_\pp = J_\pp \) then \( I_a = J_a \) for some \( a \notin \pp \). By taking \( K = J/I \) and applying the Noetherian hypothesis, we know that \( K \) is finitely generated by some \( k_1, \dots, k_n \). Then the image of each \( k_i \) in \( K_\pp \) is \( \frac{k_i}{1} = 0 \) so by definition there must be some \( a_i \notin \pp \) such that \( a_i k_i = 0 \) in \( K \). By taking \( a = \prod a_i \) the locality result follows.

Now to show that a regular sequence remains regular in a neighborhood, notice that \( x_1, \dots, x_r \) is regular at \( \pp \) (we have reduced to the affine case here as in the hint since locally closed embeddings are affine local on the target) if and only if \( (\phi_i)_\pp : B_\pp / (x_1, \dots, x_{i-1})_\pp \to B_\pp / (x_1, \dots, x_{i-1})_\pp \) is injective, where \( (\phi_i) \) is multiplication by \( x_i \). But this holds if and only if \( \ker ((\phi_i)_\pp) = (\ker \phi_i)_\pp = 0 \). By the previous paragraph, this holds on some open neighborhood \( D(a) \) of \( [\pp] \).

Proof:

For the forward direction, suppose \(\pi : X \hookrightarrow Y\) is an effective Cartier divisor. Then by definition, there is a cover \( \spec B_i \) of \(Y\) such that \( X \vert_{\spec A_i} = V(t_i) \) for some \( t_i \in B_i \). Thus, for every point \( p \in X \) we have that \( p = [\pp_i] \in \spec B_i\) for some \(i\), and the local ring \( \OO_{Y,p} \cong (B_i)_{\pp_i} \) is generated by the image of \( t_i \) so that \( \pi \) is a regular embedding of codimension 1.

Conversely, if \( \pi : X \hookrightarrow Y \) is a regular embedding of codimension 1, then we may take some affine cover \( \spec B_i \) of \( Y \) and apply the previous exercise. That is, for each \( p = [\pp] \in \spec B_i \), we may take some neighborhood \( D(f_{i,j}) \) on which the ideal of \( X \) is a regular sequence of length 1. By letting \( i \) vary over our open cover, this produces an affine subcover \( \spec A_k \) such that \( X\vert_{\spec A_k} \) is locally of the form \( V(t_k) \) where \( t_k \) is the regular sequence in \( D(f_{i. j}) \).