Solutions to The Rising Sea (Vakil)

Proof:

To see that the class of open embeddings is local on the target, first suppose that \( \pi : X \to Y \) is an open embedding and \( V \subset Y \) is open. If \( \pi^{-1}(V) = \emptyset \) then the claim is trivial. Otherwise, \( \pi^{-1}(V) \) is nonempty and open in \(X\). By definition \( \pi \) factors as \( (X, \OO_X) \iso (U, \OO_Y\vert_U) \hookrightarrow (Y, \OO_Y) \), so that the restriction map factors as \( ( \pi^{-1}(V), \OO_X\vert_{\pi^{-1}(V)} ) \iso (U \cap V, \OO_Y\vert_{U \cap V}) \hookrightarrow (Y, \OO_Y) \). Since \( U \cap V \) is open, the restriction map is therefore an open embedding.

Simialrly, assume \( \pi : X \to Y \) is a morphism, \( \{ V_i \} \) is an open cover of \(Y\), and for each \( i \) we have \( \pi_i : \pi^{-1}(V_i) \to V_i \) is an open immersion. Since the \(V_i\) cover \( Y \), \( \pi^{-1}(V_i) \) form an open cover of \(X\). Moreover, each \( \pi_i \) factor through some isomorphism \( \rho_i : ( \pi^{-1}(V_i), \OO_X \vert_{\pi^{-1}(V_i)} ) \iso ( \widetilde{V_i}, \OO_Y\vert_{\widetilde{V_i}} ) \) (where \( \widetilde{V_i} \subset V_i \) open subset since \( \pi \) need not be a surjection). By our usual sheaf axioms, the \( \OO_X\vert_{U_i} \) glue to \( \OO_X \) and the \( \OO_Y \vert_{ \widetilde{V_i} } \) glue to \( \OO_Y \vert_{\textrm{Im}(\pi)} \). However, since \( \textrm{Im}\,(\pi) = \bigcup_i \widetilde{V_i} \) these glue to an open subscheme of \( (Y, \OO_Y) \) so that \( \pi \) is an open immersion.

To see that property (ii) holds, let \( \pi : X \to Y \) and \( \rho : Y \to Z \) be open embeddings. Then \( \pi \) factors through an isomorphism \( (X, \OO_X) \iso (U, \OO_Y\vert_U) \) to an open subset \( U \subset Y \). By (i), the restriction \( \rho\vert_U \) is also an open embedding and thus factors through an isomorphism \( (U, \OO_Y\vert_U) \iso (V, \OO_Z\vert_V) \hookrightarrow (Z, \OO_Z) \) ( where \( V \subset Z \) open subset). Since the composition of isomorphisms is an isomorphism, our map \( \rho \circ \pi \) factors through the iso \( (X, \OO_X) \iso (V, \OO_X\vert_V) \) and is thus an open embedding.

Proof:

Without loss of generality, we may assume \( U \subset Z \). Following the hint, we wish to show that (\rho^{-1}(U), \OO_Y\vert_{\rho^{-1}(U)} ) indeed satisfies the universal property of the fiber product. Note that we must first come up with maps \( (\rho^{-1}(U), \OO_Y\vert_{\rho^{-1}(U)} ) \to (U, \OO_Z \vert_U) \) and \( (\rho^{-1}(U), \OO_Y\vert_{\rho^{-1}(U)} ) \to (Y, \OO_Y) \) that make the cartesian square commute in order for the diagram to make sense. Indeed the obvious map \( \rho^{-1}(U) \to U \) is simply \( \rho \vert_{\rho^{-1}(U)} \), so that the top arrow in

Must simply be the inclusion \( j : ( \rho^{-1}(U), \OO_Y\vert_{\rho^{-1}(U)} ) \hookrightarrow (Y, \OO_Y) \). To see that the universal property is satisfied, let \( (W, \OO_W)\) be another scheme along with morphisms \( \alpha : W \to U\) and \( \beta : W \to Y \) such that \( \iota \circ \alpha = \rho \circ \beta \). Since \( \textrm{Im}\alpha \subset U \) we must also clearly have \( \textrm{Im}(\rho \circ \beta) \subset U \) so that \( \textrm{Im}(\beta) \subset \rho^{-1}(U) \). But then the unique map \( (W, \OO_W) \to (\rho^{-1}(U), \OO_Y\vert_{\rho^{-1}(U)} ) \) must simply be \( \beta \) itself. Therefore the fiber product exists.

The inclusion \( j : ( \rho^{-1}(U), \OO_Y\vert_{\rho^{-1}(U)} ) \hookrightarrow (Y, \OO_Y) \) is clearly an open embedding, as \( \rho^{-1}(U) \subset Y \) is open by virtue of \( \pi \) being continuous.

Proof:

Since \( \pi : X \to Y \) an open embedding, it factors through some isomorphism \( \rho: (X, \OO_X) \iso (U, \OO_Y\vert_U) \). First suppose \( Y \) is locally Noetherian such that \( Y \) can be covered by \( V_i = \spec B_i \) where each \( B_i \) is Noetherian. Then clearly \( U \) can be covered by such sets as well, say \( U \cap V_i = \spec (B_i)_{\qq_i} \) — notice \( B_i \) Noetherian implies \( (B_i)_{\qq_i} \) Noetherian by §3.6.16. Since \( \rho \) is an isomorphis, we may pull such a cover back to \( (X, \OO_X) \) and be guaranteed that our rings are still Noetherian (since an isomorphism of schemes induces an isomorphism of sheaves). The proof showing \(Y\) Noetherian implies \(X\) Noetherian is identical.

To see that \( Y \) quasicompact does not imply \(X\) quasicompact, we may use the example of Exercise 3.6.G(b) almost verbatim. If we consider \( A = \spec k[x_1, x_2, \dots] \) along with the maximal ideal \( \mm = (x_1, x_2, \dots) \), then \( \spec A \) is clearly quasicompact as it is affine. However, we showed in the linked exercise that the compliment of \( V(\mm) \) was not quasicompact, so we simply let \( \pi \) denote the inclusion of \( V(\mm)^c \) into \( \spec A \).

Proof:

Consider the double cover \( \A^1_k \coprod \A^1_k \to \A^1_k \) given by the diagonal morphism \( k[x] \to k[u] \times k[v] \), \( f(x) \mapsto (f(u), f(v)) \). Take the obvious affine cover \( \{\spec k[u], \spec k[v]\} \) of \( \A^1_k \coprod \A^1_k \). Then the restriction maps \( k[x] \to k[u] \) and \( k[x] \to k[v] \) are clearly isomorphisms, so that \( \A^1_k \coprod \A^1_k \to \A^1_k \) restricted to either of its components is obviously an isomorphism (and thus an open embedding as \( \A^1_k \) is open in itself). However, these maps do not glue to an isomorphism as the diagonal embedding is surely not surjecive.

Proof:

Following the hint, we may suppose without loss of generality that \( B \subset A \) is a subring. Letting \(a \in A\) be arbitrary, we have for each \( 1 \leq i \leq n \) that there exists some \( m_i > 0 \) (the result would be trivial if there were some \( m_i = 0 \)), \( \beta_{i, j} \in B\) (for \( 1 \leq j \leq m_i \)) and \( t_{i, j} \geq 0 \) such that

Letting \( t_i \) be the maximum of all \( t_{i, j} \), we have that \( b^{t_i} a^{m_i + \beta_{i, m_i - 1} a^{m_i-1} + \dots + \beta_{i, 0} = 0 \) — in other words, \( b^{t_i} a^{m_i} \in B + Ba + \dots + Ba^{m_i - 1} \). By taking \( m \) to be the maximum of all \( m_i \) and \( t \) the maximum of all \( t_i \), we have \( b^t_i a^m \in B + Ba + \dots + Ba^{m - 1} \) for each \( i \). Similar to Theorem 4.1.2, we know that as \( \spec B = \bigcup D(f_i) = \bigcup D(f_i^t) \), so \( (b_1, \dots, b_n) = 1 \) implies \( (b_1^t, \dots, b_n^t) = 1 \). Therefore, we may find \( r_i \in B\) such that \( 1 = r_i b_1^t + \dots + r_n b_n^t \) and thus

Proof:

1. Let \( \phi : B \to A \) be integral and first consider the case that \( T \) is a multiplicative subset ( so that \(\phi(T)\) is as well ). Let \( \overline{a} = \frac{a}{\phi(t)} \in \phi(T)^{-1} A \) be arbitrary. Since \( a \in A \) and \( \phi : B \to A\) is integral, we can find some \( n > 0 \) and \( b_i \in B \) such that \( a^n + \phi(b_{n-1})a^{n-1} + \dots + \phi(b_0) = 0 \). Dividing the equation by \( \phi(t)^n \) gives us

   $$ \begin{align} 0 &= \frac{a^n}{\phi(t)^n} + \frac{\phi(b_{n-1})}{\phi(t)} \frac{ a^{n-1} }{ \phi(t)^{n-1} } + \dots + \frac{ \phi(b_0) }{ \phi(t)^n } \\&= \left(\frac{ a }{\phi(t)} \right)^n + \phi \left( \frac{b_{n-1}}{\phi(t)} \right) \left( \frac{a}{\phi(t)} \right)^{n-1} + \dots + \phi \left( \frac{b_0}{t^n} \right) \end{align} $$

   So that \( \frac{a}{\phi(t)} \) integral over \( T^{-1}B \)

   Next, let \( J \subset B \) be an ideal. Notice that for any \( a \in A \), since \( \phi\) is integral there exist \( n > 0 \) and \( b_i \in B \) such that \( a^n + \phi(b_{n-1})^{n-1}a^{n-1} + \dots + \phi(b_0) = 0 \). This implies all \( b_i \in J \) iff \( a^n \in \phi(J) \) so that integrality is well-defined and preserved on the quotient map.

   Lastly, let \( I \subset A \) be an ideal and \( \pi : A \to A /I \) denote the canonical projection map. Then for any \( \overline{a} \in A / I \) and representative \( a \in \pi^{-1}(\overline{a}) \), we know that there exists some \( n > 0 \) and \( b_i \in B \) such that \( a^n + \phi(b_{n-1})a^{n-1} + \dots + \phi(b_0) = 0\). Applying our map \( \pi : A \to A/ I \) gives the desired result.

   To see, however, that integrality is not preserved by localization of \(A\) we follow the example given: the identity \( k[t] \to k[t] \) is trivially integral, however the induced localization map \( k[t] \hookrightarrow k[t]_{(t)} \) given by \( f(t) \mapsto \frac{f(t)}{1} \) is certainly not integral. To see this, notice that by taking \( a = \frac{1}{1 + t} \), for any \( f_{n-1}(t), \dots, f_0(t) \) we have

   $$ \frac{1}{(1 + t)^n} + \frac{ (f_{n-1}(t))^{n-1} }{1} \frac{1}{(1 + t)^{n-1}} + \dots + \frac{ f_0(t) }{1 } \neq 0 $$

   since we cannot cancel out the first \( \frac{1}{(1 + t)^n} \) term.

2. This follows from part (a) since if \( \phi : B \to A \) is integral, then \( \phi(T) : T^{-1} B \to \phi(T)^{-1} A \) is integral (additionally by algebra if \( \phi \) is injective then \( \phi(T) \) is injective ). Additionally, we have already seen that integrality need not hold for localization of \(A\) — to see that integral extensions need not be preserved follows from the simple fact that \( \phi : B \to A \) injective does not imply \( B \to A \to A / I \) is injective.

3. We know from part (a) again that \( B / J \to A / JA \) is integral so it suffices to show it is injective. Since \( J = I \cap B \subset I\), \( B /J \to A/I \) factors through \( A / JA \). Since \( R^\prime \to R^{\prime \prime} \) and \( R \to R^\prime \to R^{\prime \prime} \) injections imply \( R \to R^\prime \) is an injection (easy proof by contradiction), the result follows.

Proof:

Let \(a \in A\) be arbitrary; since the map \( B \to A \) is integral, \( a \) is contained in a subalgebra \( R \subset A\) with is finitely generated as a \(B\)-module, say by \( b_1, \dots, b_n \). Since the map \( C \to B \) is integral, each \( b_i \) is contained in some subalgebra \( S_i \subset B\) that is finitely generated as a \(C\)-module. Then \(\prod S_i \) is also finitely generated so that \( R \) must also be a finitely generated \(C\)-module. Since \( a \in A \) was arbitrary the result follows.

Proof:

Suppose \( s, t \in A\) are integral over \(B\). By Lemma 7.2.1, we have that \( s \in M \) and \( t \in N \) for some finitely generated \( B \)-modules \( M, N \subset A \). But then \( MN \) is also a finitely generated \( B \)-module which contains \( s + t \) and \( st \) so the result follows.

Proof:

With the Lying over theorem, the exercise is immediate from the fact that a ring is a field if and only if \( (0) \) is the only prime ideal ( the proof in the reverse direction requires the fact that \(\mm\) maximal implies \( R / \mm \) field and \( R / (0) \cong R\) ). To see this, notice that if \( \pp \subset B \) is a prime ideal then there exists some \( \qq \subset A \) prime such that \( \pp = \qq \cap B \). But \(A\) is a field so we must have \( \qq = (0) \) and thus \( \pp = (0) \).

However, since the proof of the Lying Over theorem below relies on this exercise, we should probably avoid circular logic. Following the hint, let \( 0 \neq b \in B \) — we wish to show that \( b^{-1} \in B \). Since \( A \) is a field, \( b^{-1} \in A \) and thus we can find some \( n \in N \) and \( \beta_i \in B \) such that

Multiplying through by \( b^n \), we may rewrite the equation as

Thus, \( b^{-1} = -\beta_{n-1} - \dots - \beta_0 b^{n-1} \) so that \( B \) is also a field.

Proof:

1. By induction we only consider the case that \( m = 1 \) and \( n = 2 \); that is to say, suppose that \( \qq_1 \subset \qq_2 \) is a chain of primes in \(B\) and \( \pp_1 \subset A \) with \( \qq_1 = \pp_1 \cap B \) (we may assume without loss of generality that \(B\) is a subring of \(A\)). Then the induced map \( \overline{\phi} : B / \qq_1 \to A / \pp_1 \) is clearly integral, since for any \( \overline{a} \in A /\pp_1 \) and representative \( a \) we may choose \( b_i \) such that \( a^n + b_{n-1}a^{n-1} + \dots + b_0 = 0 \) and reduce mod \( \pp_1 \). Moreover, the map must also be injective since if \( \overline{b_1} = \overline{b_2} \) in \( A / \pp_1 \) for some \( b_1, b_2 \in B \) then \( b_1 - b_2 \in \pp_1 \cap B = \qq_1 \) so that they are indeed equivalent in \( B / \qq_1 \). Therefore, the map is an integral extension so by the Lying Over theorem there exists some \( \overline{\pp_2} \subset A / \pp_1 \) lying over \( \qq_2 / \qq_1 \subset B / \qq_1 \). If we let \( \pi : A \to A/ \pp_1 \) denote the canonical projection, then \( \pp_2 := \pi^{-1} (\overline{\pp_2}) \) is the desired prime lying over \( \qq_2 \).

2. Take \( B = \Q[x, y] \), \( A = \Q[x, y, z] / (z^2 + 1) \) and consider \( (0) \subset (x) \subset (x, y) \) in \( B \) and \( (z) \subset (x, z) \) in \( A \).

Proof:

Notice that the assumption implies that \( N + IM = M \). Thus, by considering the quotient module \( M / N \), we have that \( M / N = ( N + IM ) / N = I(M / N) \). By Version 2 of Nagata's lemma, we have \( M / N = 0 \) so that \( M = N \).

Proof:

Take \( N = \langle f_1, \dots, f_n\rangle \subset M\). By assumption, \( N / \mm N \to M / \mm M \) is surjective ( since \( N \to N / \mm M \) is surjective and the composition \( N \to M / \mm M \) is surjective by assumption ) we have by the previous exercise that \( M = N \).

Proof:

Let \( m_1, \dots, m_n \) be a generating set of \( M \) as an \(S\)-module and suppose \( rm_i = \sum a_{ij} m_j \) for some \( a_{ij} \in M \). Taking \( Z = (a_{ij}) \) this implies that

Multiplying through by \( \textrm{adj}(rI_{n\times n} - Z) \) implies that

Now since \(\det(r I_{n \times n} - Z) \) has entries that are polynomials in \( M \) and \( M \) is faithful \( S[r] \)-module, we must have that \( \det(r I_{n \times n} - Z) = 0\) so that \( r \) is integral over \( S \).

Proof:

This is immediate from \( K(\widetilde{A}) = K(A) \).

Proof:

Let \( \phi : X \to Y \) and \( \psi : Y \to Z \) be quasicompact morphisms and let \( U \subset Z \) be affine. Then \( \psi^{-1}(U) \) is quasicompact by definition, so by Exercise 5.1.D \( \psi^{-1}(U) = \bigcup_{i=1}^n \spec B_i \). Since \( \phi \) is also quasicompact, for each \( i\) we have that \( \phi^{-1}(\spec B_i) \) is quasicompact. Thus, \( (\psi \circ \phi)^{-1}(U) \) is the finite union of quasicompact spaces and thus quasicompact (see Exercise 3.6.H).

Proof:

1. Let \( \phi : X \to Y \) be a morphism of schemes with \( X \) Noetherian and suppose \( U \subset Y \) is affine. Since \( \phi \) is continuous we have that \( \phi^{-1}(U) \subset X \) is open. By Exercise 3.6.T since \(X\) is Noetherian as a topological space we have that \( \phi^{-1}(U) \) is open.

2. Let \( \phi : X \to Y \) be a morphism of schemes with \(X\) quasiseparated and suppose \( U \subset Y \) is affine. Then \( \phi^{-1}(U) \) is open; we wish to show that it is in addition quasiseparated. Let \( W_1, W_2 \) be affine open subsets of \( \phi^{-1}(U)\); then as open affines of \( X \), we know that \( W_1 \cap W_2\) is a finite union of open affine sets (by Exercise 5.1.F).

Proof:

1. Suppose that \( V \subset Y \) is an affine open subset; for each \( i \) we have \( V \cap U_i \) is a distinguished open of \( U_i \) and is therefore also affine. Since \( \pi\vert_{\pi^{-1}(U_i}) \) is quasicompact (as a morphism), we have that \( \pi^{-1}(U_i \cap V) \) is also quasicompact and thus a finite union of open affines. Taking the union over all \( i\) is still a finite union of open affines so that \( \pi : X \to Y\) is quasicompact.

2. In a similar fashion, let \( V \subset Y \) be an affine open subset so that \( V \cap U_i \) is a distinguished open (and thus affine) for each \(i\). Since \( \pi\vert_{\pi^{-1}(U_i)} \) is quasiseparated for each \(i\), we have that \( \pi^{-1}(U_i \cap V) \) is quasicompact. Thus, \( \pi^{-1}(V) \) is the finite union of quasicompact sets and thus quasicompact.

Proof:

Let \( \pi : X \to Y \) be an affine morphism and \( U \subset Y \) an affine open set. Then \( \pi^{-1}(U) \) is affine, so by Exercise 3.6.G and Exercise 5.1.G it is quasicompact and quasiseparated (respectively).

Proof:

Recall from Exercise 1.3.D that \( \Gamma(X,\OO_X)_s \) is initial among all \( \Gamma(X, \OO_X) \)-algebras such that \( s \) is invertible. Since \( X_s \) is the locus where \( s \in \Gamma(X, \OO_X) \) doesn't vanish, by Exercise 4.3.G(b) we know that \( s \) is invertible on \( \Gamma(X_s, \OO_X) \). Thus, by the universal property noted above we have that there is a map \( \Gamma(X, \OO_X)_s \to \Gamma(X_s, \OO_X) \) which we can give explicitly by

Proof:

By assumption we may cover \( \spec A \) with distinguished opens \( U_i = D(f_i) \) such that \( U_i \cap Z = V(g_i) \) for some \( g \in A_{f_i} \). Consider the inclusion \( \iota : Y \hookrightarrow \spec A \); for each \( i \) we notice that \( U_i \cap W = D(g_i) \) so \( \iota \) restricted to (the preimage of) \( U_i \) is affine. Then by proposition 7.3.4 we have that \( Y = \iota^{-1}(\spec A) \) is affine.

Proof:

Let \( \spec B \) be an arbitrary affine open subset — if \( \spec B \cap \spec A \) for some affine in our open cover then we may assume it is a union of distinguished open, say \( D(f_i) \). Suppose \( \spec B = \pi^{-1}(\spec A) \); by assumption \( B \) is a finite \( A \)-algebra, so we must also have \( B_{\pi^\sharp f_i} \) is a finite \( A_{f_i} \)-algebra for each \( i \). In other words, \( \pi\vert_{\spec B} \) is clearly a finite morphism. Thus it suffices to show that if \( (f_1, \dots, f_n) = A \) and \( B_{\pi^\sharp f_i} \) a finite \( A_{f_i} \)-algebra (for all \( i \)) implies \( B \) is a finite \(A\)-algebra.

For each \( i \), suppose \( g_{i,1}, \dots, g_{i,n_i} \) is a generating set of \( B_{\pi^\sharp f_i} \) (as an \( A_{f_i} \)-module) for some \( g_{i, j} \in B \) . Taking large enough \( N \), we may define a map \( \phi : B^{\oplus N} \to A \) such that at each \( f_i \) we have \( (\textrm{coker}\phi)_{f_i} = 0 \). Then for any \( x \in \textrm{coker} \phi \) there exist \( r_i \geq 0 \) such that \( f_i^{r_i} x = 0 \). Taking \( r = \prod r \), notice that since \( (f_1, \dots, f_n) = 1 \) we also have \( (f_1^r, \dots, f_n^r) = 1 \)

Therefore, \( \textrm{coker}\phi = 0 \) so that \( B \) is a finitely generated \(A\)-algebra.

Proof:

By definition we must have \( \pi^{-1}(\spec k) \) is affine such that \( X = \spec A \) where \( A \) is a finite \(k \)-algebra. Now for any \( \pp \in \spec A \) we have that \( A/\pp \) is an integral domain which is also necessarily a finite \( k \)-algebra and thus a field so that the only ideal is \( (0) \) --- by the correspondence theorem, this implies that \( \pp \) must be maximal (additionally, it also follows that \( K(A/\pp) \) must be a finite extension of \( k \), possibly by using something similar to Exercise 7.2.J). By the Chinese Remainder theorem, if \( \mm_1, \dots, \mm_n \) is a distinct set of \( n \) maximal ideals in \(A\) then

Since \(A\) is a finite \(k\)-algebra, we have that \( n \leq \dim_k A \) --- in particular, \( X = \spec A \) is a collection of no more than \( \dim_k A \) points. To see that \( \spec A \) has the discrete topology, it suffices to show that every point is both open and closed. Now we know that points corresponding to maximal ideals are closed, but since there are only finitely many points \( [\mm_j] \) we may take the finite intersection of the compliment of \( V(\mm_i) \) for each \( i \neq j \).

Proof:

Let \( \phi : X \to Y \) and \( \psi : Y \to \Z \) be finite morphisms. Indeed the composition of affine morphisms is affine so we know that \( \psi \circ \phi \) is indeed affine. Let \( U = \spec A \subset Z \), \( V = \spec B = \psi^{-1}(U) \) and \( W = \spec C = \phi^{-1}(V) \) be the affine preimages. Since \( A \) is a finite \( B \)-algebra, we get a surjective morphism \( \pi : B^{\oplus n} \to A \). Similarly, as \( B \) is a finite \( C \)-algebra, we get a surjective morphism \( \rho : C^{\oplus m} \to B \). Thus, we obtain a surjective map \( C^{\oplus mn} \to A \) so that \( A \) is a finite \(C\)-algebra as desired.

Proof:

The \(\Z\)-grading on \( S_\bullet \) occurs in a natural way; \( S_0 = A \) acts on each \( S_n = R \) as \( R \) is an \( A \)-algebra. Moreover, for \( a \in R = S_i \) and \( b \in R = S_j \) we declare \( ab \in S_{i + j} \) as this is required to make this a graded \( A \)-algebra (one should keep in mind that \( S_\bullet \) looks similar to \( R[x] \), so as \( \proj R[x] \cong \spec R \) we hope to find a bijection \( S_\bullet \leftrightarrow R[x] \) — see Exercise 6.4.A). Notice from our \(A\)-structure on \(S_\bullet\), it is ambiguous how to embed any element \(f \in R\) into \( S_\bullet \) as \( [f]_n = [f]_1 [1_R]_1^{n+1} \); however, as we only care about prime ideals, this allows us to construct a well-defined bijection \(\spec (S_{[f]_1})_0 \to \spec R \). For each \( f \in R \), we define \( R \to (S_{[f]_1})_0 \) by

In fact, this gives us an isomorphism \( R \cong (S_{[1]_1})_0 \) in quite an obvious way. Notice that if \( \pp \in \proj S_\bullet \) with \( \pp \notin D_+([1]_1) \) then \( \pp \) must contain \( S_+ \) since for any \( [f]_n = [f]_{n-1}[1]_{n-1} \). Thus, \( \proj S_\bullet = D_+ ([1]_1) \)

Proof:

Fix \( p \in Y \). As fibers are a local construction, it suffices to consider \( Y = \spec B\) so that \( p = [\pp] \) and \( spec A = X = \pi^{-1}(Y) \). Similar to Exercise 7.3.H, we know that \( A / \pp \) is a domain — thus, for any \( q = [\qq] \in \pi^{-1}(\pp) \), we know that \(\kappa(q) = K(B/\qq) \) is a finite \( \kappa(p) = K(A / \pp) \) field-extension. Thus, \( \pi^{-1}(\qq) \) can only have finitely many points.

Proof:

The open embedding \( \spec D(x) \hookrightarrow \( \spec \C[x] \) \) corresponds to the ring morphism \( \C[x] \to \C[x]_{(x)} \). Indeed this is affine and has finite fibre (as the preimage of any point is a point), but \( \C[x]_{(x))} \) is certainly not a finite \( \C[x] \)-algebra as \( \frac{1}{1 + x} \) cannot be expressed by \( 1, x, \dots, x^n \) for any \( n \).

Proof:

Let \( \pi : X \to Y \) be an integral morphism and let \( Z \subset X \) be closed. Since integral morphisms are affine, we may assume without loss of generality that \( X = \spec A \) and \( Y = \spec B \) so that \( Z = V(I) \) for some \( I \subset A \). If we take \( J = (\pi^{\sharp})^{-1}(I) \) then clearly \( B \to A \to A / I \) factors through \( B / J \). Since \( B / J \to A/ \pi^{\sharp} JA \) and \( A / \pi^\sharp(J) A \to A / I \) integral, \( B/J \hookrightarrow A/I \) integral. By the lying over theorem, for any \( \qq \in V(J) \) we have \( J \subset \qq\) so that \( I \subset \pi^{\sharp}(\qq) \). Since \( V(I) = Z \), \( \pi(Z) = V(J) \) is clearly closed.

Proof:

Let \( \phi : B \to A \) and \( \psi : B \to C \) be our ring morphisms giving \( A \) and \(C\) the structure of \(B\)-algebras. Recall that \( A \otimes_B C \) comes with the \( B \)-algebra structure induced by the commutative diagram

In particular, for every \( b \in B \) we have that \( \phi(b) \otimes 1 = 1 \otimes \psi(b) \). Now let \( a \otimes c \in A \otimes_B C\) be a pure tensor; as \( a \) is integral over \( B \), there exist \( n > 0 \) and \( b_i \in B \) such that \( a^n + \phi(b_{n-1})a^{n- 1} + \dots + \phi(b_0)= 0 \). Tensoring with \( c^n \), this tells us

Since each \( 1 \otimes \psi(b_i)c^{n-i} \) lies in the image of the natural map \( C \to A \otimes_B C \), we have that \( a \otimes c \) is integral over \( C \).

Proof:

Suppose \( \spec A \subset Y\) is an affine open neighborhood. Then \( \{ \spec A \cap \spec B_i \} \) is an affine open cover of \( \spec B_i \) by distinguished opens. As \( \spec A \) is quasicompact, we may choose some finite affine open cover \( D(f_i) \) of \( \spec A \) (so that \( (f_1, \dots, f_n ) = A\) as an algebra). If we let \( \spec C \subset \pi^{-1}( \spec A ) \), then we get maps \( A_{f_i} \to C_{\pi^\sharp f_i} \) for each \( i \) — by an argument similar to Exercise 7.3.G, this implies that \( C \) is a finitely generated \(A\)-algebra as desired.

Proof:

1. This is trivial by definition since if \( \pi : X \to Y \) is a finite morphism and \( \spec A \subset Y \), then \( \pi^{-1}(\spec A) = \spec B \) with \( B \) a finite \(A\)-algebra and thus finitely generated \(A\)-algebra.

2. The forward direction follows from the previous part (and finite algebras are integral). Conversely, suppose \( \pi : X \to Y \) is integral and of finite-type. Since integral maps are affine, we may restrict to the case \( Y = \spec B \) and \( X = \spec A \) so that \( A \) is integral and finite type over \(B\). To see that \( A \) is a finite \(B\)-algebra, let \( x_1, \dots, x_n \) be the generators of \( A \) (as a \(B\)-algebra). Now \( x_1\) is integral over \(B\) so \( B[x_1] \) is a \(B\)-subalgebra finite over \(B\) by Lemma 7.2.1. Similarly, \( x_2 \) is integral over \( B[x_1] \) so by induction the result follows.

Proof:

1. This follows since if we restrict to \( X = \spec A \) and \( U \to \spec A \) with \( U = \bigcup D(f_i) \) then each \( A_{f_i} \) is finitely generated by \( f_i^{-1} \) over \(A\). If \( X \) is locally Noetherian, then we may modify the above assumption \( X = \spec A \) to the case that \( A \) is a Noetherian ring. By Exercise 3.6.T, every open subset of a Noetherian topological space is quasicompact; thus, \( U \to X \) is a quasicompact morphism and therefore of finite type.

2. Suppose \( \phi : X \to Y \) and \( \psi : Y \to Z \) are locally of finite type and let \( U = \spec A \) be an affine open. Using the equivalent definition of locally of finite type, \( \psi^{-1}(\spec A) \) may be covered by affine opens \( \spec B_i \) such that each \( B_i \) is a finitely generated \(A\)-algebra. Similarly for each \( i \), \( \phi^{-1}(\spec B_i) \) may be covered by affine opens \( \spec C_{ij} \) such that \( C_{ij} \) is a finitely generated \( B_i \)-algebra. Since each \( B_i \) is a finitely generated \(A\)-algebra, we get that \( C_{ij} \) is a finitely generated \(A\)-algebra.

3. Since \( Y \) is locally Noetherian, it may be covered by open affines \( \spec A \) where each \( A \) is Noetherian. Since \( \pi : X \to Y \) is locally of finite type, \( \pi^{-1}(\spec A) \) may be covered by \( \spec B_i \) where \( B_i \) is a finitely generated \(A\)-algebra. By Exercise 3.6.X, since \( B_i \) is finitely-generated as a module and \(A\) is Noetherian, \( B_i \) must be Noetherian as well. By running over our affine cover \( \spec A \), we may cover \( X \) by \( \spec B_i \) with \( B_i \) Noetherian so that \( X \) is locally Noetherian.

   Now if \( \pi : X \to Y \) is of finite type, then it is locally of finite type and quasicompact. Assuming \(Y \) is Noetherian, we may cover \( Y \) by finitely many \( \spec A \) with \( A \) Noetherian. Since \( \pi\) is quasicompact, \( \pi^{-1}(\spec A) \) is quasicompact so we can cover by finitely many of the \( \spec B_i \) as above (each a finitely generated \( A \)-algebra and thus Noetherian). Since there were originally finitely many \( \spec A \), we can thus cover \( X \) by a finite number of \( \spec B_i \) so that \(X\) is also Noetherian.

Proof:

1. Recall that \( F^r \) as a map of sets is simply the map \( \spec k[x_1, \dots, x_n] \ni \pp \mapsto (\phi^{r})^{-1}(\pp) \). Now if \( \phi^r(f) = f^{p^r} \in \pp \) then \( f \in \pp \) (since \( \pp \) is prime) so that \( \pp = (\phi^{r})^{-1} (\pp) \) — therefore, the map of sets is indeed a bijection.

   Now by construction, \( \mathbb{F}_q \) is the splitting field of \( X^{p^r} - X \) over \( \mathbb{F}_p \) so that every element \( a \in k \) satisfies \( a^{p^r} = a \). However, this need not hold for elements \( f \in k[x_1, \dots, x_n] - k \) (for example, \( x_1 \) is free so \( \phi^r(x_i) = x_i^{p^r} \neq x_i \)).

2. On the level of sets, \( F \) is clearly a bijection since \( F^{- 1} = F^{r-1} \). However, \( F \) cannot be an isomorphism since \( F^\ast = \phi : k[x_1, \dots, x_n] \to k[x_1, \dots, x_n] \) is not surjective (pick \( x^s \) with \(s\) coprime to \( p \))

3. If we consider \( S \) to be the set of all polynomials of the form \( x^{p^n} - x \) for some \( n > 0 \) then \( K = \overline{ \mathbb{F}_p } \) is the splitting field of \(S\) (a common Galois theory exercise). To see that \( F \) is a bijection requires a little more work than before, as it need not hold that \( \phi(f) \) is a power of \( f \) for any \( f \in K[x_1, \dots, x_n] \). However, if we write \( f = \sum_{\nu} a_\nu x^\nu \) for some multi-index \( \nu \), then for each \( \nu \) there exists some \( r_\nu \) such that \( a^{p^{r_\nu}} = a \) and thus \( \phi^{r_\nu}(a_\nu x^\nu) = a_\nu (x^\nu)^{p^{r_\nu}} = (a_\nu x^\nu)^{p^{r_\nu}} \). Taking \( r = \prod_\nu r_\nu \), we have that \( \phi^r(f) = f^{p^r} \). Now any prime \( \pp \in K[x_1, \dots, x_n] \) is finitely-generated (as \( K[x_1, \dots, x_n] \) is obviously Noetherian) so that we may apply this argument to a finite set of generators of \( \pp \) giving us some \( s >> 0 \) with \( (\phi^{s})^{-1}(\pp) = \pp \). Thus, our map \( \A^n_K \to \A^n_K \) must be a bijection. However, our integer \( s \) depends on each point \( [\pp] \in \A^{n}_K \) so that no \(s\) works universally. That is to say if we choose some arbitrary \( r^\prime > 0 \) and consider the map \( F^{r^\prime}( x_1 - a ) = x_1^{p^{r^\prime}} - a \) then we could simply choose \( a \in K \) such that \( a\) is not a root of \( X^{p^{r^\prime}} - X \).

Proof:

If we let \( \{ \spec A_i \} \) be any affine cover, then by definition each \( A_i \) is a \( \mathbb{F}_p \)-algebra so we get a map \( \OO_X(\spec A_i) \to \OO_X(\spec A_i) \) given by \( f \mapsto f^p \), which by the previous exercise induces the identity on points of \( \spec A_i \). Then these maps clearly glue together since our open sets glue.

Proof:

This is somewhat immediate from the definition. Let \( \pi : X \to Y \) be a morphism locally of finite presentation and \( V= \spec A \subset Y \); we first wish to show that \( \pi^{-1}(V) \to V \) is locally of finite presentation. If \( \spec C \subset V \) is an affine open of \( V \), then it is also an affine open of \( Y \) so by virtue of \( \pi \) being locally of finite type, \( \pi^{-1}(\spec C) \) can be covered by open affines \( \spec B_i \) with each \( B_i \) a finitely presented \(C\)-algebra.

For the converse, suppose \( \{ \spec A_i \} \) is an affine cover for \(Y\) with each \( \pi_i : \pi_i^{-1}(\spec A_i) \to \spec A_i \) locally of finite presentation, and let \( \spec B \subset Y \) be an arbitrary open affine. Then whenever \( \spec A_i \cap \spec B \) is non-empty, we may assume it is of the form of some distingished open \( D(f_i) \) for some \( f_i \in B \). As this is also a distinguished open of \( \spec A_i \) (for some different \( \widetilde{f}_i \in A_i\) ), we have \( \pi_i^{-1}(D(\widetilde{f}_i)) = \bigcup_j \spec C_{ij} \) with each \( C_{ij}\) a finitely-presented \( (A_i)_{\widetilde{f}_i} \)-algebra (and thus \( B_{f_i} \)-algebra). Thus, it suffices to show that if \( (f_1, \dots, f_n) = B \) with each \( C_{ij} \) a finitely-presented \( B_{f_i} \) module, then \( C_{ij} \) is a finitely presented \( B \)-module. Similar to before, we know that since \( C_{ij} \) finitely generated over each \( B_{f_i} \) and \( B = (f_1, \dots, f_n) \) there exist a surjective map \( \phi : B^{\oplus N} \to C_{ij} \); by looking at the cokernel of this map, it must be the case that \( \coker \phi = 0 \) since it vanishes at each localization. Similarly, the kernel must be finitely generated, as it is finitely generated at each localization.

Proof:

This ammounts to showing that if \( (f_1, \dots, f_n) = A \) and \( A_{f_i} \) a finitely-presented \(B\)-algebra for each \( i \), then \( A \) is a finitely-presented \(B\)-algebra. I will outsource this proof to [Stacks 00EO].

Proof:

Let \( \iota : U \hookrightarrow X \) be our open embedding. Indeed we may reduce to the case \( X = \spec A \) and consider the identity morphism \( \textrm{id} : X \to X \). Taking \( U = D(f_i) = \spec A_{f_i} \subset X \), we simply apply the previous exercise.

Proof:

It suffices to reduce to the affine case, so that we have ring maps \( \phi : A \to B \) and \( \psi : B \to C \) where \( B \) is a finitely presented \( A \)-algebra and \( C \) is a finitely presented \( B \) algebra. Thus, there exist integers \( m, n > 0 \) and surjections \( \pi : A[x_1, \dots, x_n] \to B \) and \( \rho : B[y_1, \dots, y_m] \to C\) such that \( \ker(\pi) \) and \( \ker(\rho) \) are finitely-generated ideals of \( A[x_1, \dots, x_n] \) and \( B[y_1, \dots, y_m]\) respectively. Then \( \pi \) may be extended to a morphism \( \widetilde{\pi} : A[x_1, \dots, x_n, y_1, \dots, y_m] \to B[y_1, \dots, y_m] \) in the natural way, so that \( \ker (\widetilde{\pi}) = \ker(\pi)[y_1, \dots, y_m] \) is finitely generated in \( A[x_1, \dots, x_n, y_1, \dots, y_m] \). By taking compositions, we see that \( \rho \circ \widetilde{\pi} \) is clearly a surjection, so that \( C \) is a finitely-generated \(A\)-algebra. To see that it is finitely presented, we must show that \( \ker(\rho \circ \widetilde{\pi}) \) is finitely generated.

Suppose \( x \in \ker( \rho \circ \widetilde{\pi} ) \), or in other words \( \widetilde{\pi}(x) \in \ker \rho \). Since \( \ker \rho \) is assumed to be finitely generated in \( B[y_1, \dots, y_m] \) (say by some \( v_1, \dots, v_k \)) we have that \( \widetilde{\pi}(x) = \sum g_i v_i \) for \( g_i \in B[y_1, \dots, y_m] \). Since \( \widetilde{\pi} \) surjects onto \( B[y_1, \dots, y_m] \), we can find some \( f_{ij}, w_{ij} \in A[x_1, \dots, x_n, y_1, \dots, y_m] \) with \( \widetilde{\pi}( \sum_{i, j} f_{ij}w_{ij} ) = \sum_i g_iv_i \). By linearity, this implies \( \widetilde{\pi}( x - \sum_{i,j} f_{ij}w_{ij} ) = 0 \) or \( x - \sum_{i, j}f_{ij}w_{ij} \in \ker \widetilde{\pi}\). Since this is finitely generated by \( \ker \pi \) and \( y_1, \dots, y_m \), we must have that \( x \) is generated by \( \ker (\pi), y_1, \dots, y_m \) and \( \ker \rho \).

Proof:

For the forward direction, suppose \( Y \) is a finite disjoint union of locally closed subsets; that is to say \( Y = \coprod_i Z_i \cap U_i \) with \( Z_i \) closed and \( U_i \) open. In fact if we just assume the sets are disjoint we may write \( Y = \bigcup_i Z_i \cap U_i \). Now since every open set is in the family \( \mathfrak{F} \) of constructible sets and \( \mathfrak{F} \) is closed under taking compliments, all closed sets are in \( \mathfrak{F} \). Similarly, as finite intersections and complements are in \( \mathfrak{F} \) we must also have finite unions are in \( \mathfrak{F} \). Thus each \( Z_i \cap U_i \) and consequently \( Y \) is in \( \mathfrak{F} \).

Now suppose we let \( \mathfrak{F}^\prime \) denote the collection of subsets of \(X\) that can be written as a finite disjoint union of locally closed subsets; by the previous paragraph we have \( \mathfrak{F}^\prime \subset \mathfrak{F} \). To show that \( \mathfrak{F}^\prime = \mathfrak{F} \), it suffices to show that \( \mathfrak{F}^\prime \) also satisfies properties (i), (ii), and (iii). Clearly \( \mathfrak{F}^\prime \) contains all open sets \( U \subset X \) since \( U \cap X = U \) and \( X \) is closed. To see that \( \mathfrak{F}^\prime \) is closed under intersections, notice that if \( \coprod_i Z_i \cap U_i, \coprod_j Z_j^\prime \cap U_j^\prime \in \mathfrak{F}^\prime \), then the intersection is \( \coprod_{i, j} (Z_i \cap Z_j^\prime) \cap (U_i \cap U_j^\prime) \) which is also clearly an element of \( \mathfrak{F}^\prime \). To prove that \( \mathfrak{F}^\prime \) is closed under compliments, we let \( \mathfrak{F}^\prime_n \) denote the collection of subsets of \(X\) which can be expressed as a disjoint union of exactly \(n \) locally-closed sets. First notice that for any \( Y \in \mathfrak{F}^\prime_1 \) we have \( Y = Z \cap U \). Then \( Y^c = Z^c \cup U^c = (U^c \cap X) \coprod (Z^c \cap U) \) which is clearly in \( \mathfrak{F}^\prime \). Proceeding inductively, suppose that compliments in \( \mathfrak{F}^\prime_i \) are contained in \( \mathfrak{F}^\prime\) for \( i < n \). If we let \( Y \in \mathfrak{F}_n^\prime \) then we can write \( Y = Y_{n-1} \coprod Y_1 \) for some \( Y_{n-1} \in \mathfrak{F}^\prime_{n-1} \) and \( Y_1 \in \mathfrak{F}_1 \). Then \( Y^c = Y^c_{n-1} \cap Y^c_1 \); by inductive hypothesis both \( Y^c_{n-1} \in \mathfrak{F}^\prime \) and \( Y^c_1 \in \mathfrak{F}^\prime \) and we have shown \( \mathfrak{F}^\prime \) is closed under intersection. Since \( \mathfrak{F}^\prime = \bigcup_n \mathfrak{F}^\prime_n \), this completes the proof.

Proof:

Let \( \eta = [(0)] \) denote the generic point; as \( \{ \eta \} \) is dense, every open subset \( U \subset X \) intersects \( \{\eta\} \) so that \( \{ \eta \} = U \cap \{ \eta \} \) for all \( U \). Therefore, by the previous theorem we would have \( \eta \) is constructible if and only if \( \{ \eta \} \) is closed. This would hold if and only if \( \{ \eta \} = \overline{ \{\eta }\} = \A^1_k \) which is impossible by Exercise 3.2.D as \( \A^1_k \) is infinite.

Proof:

1. For the "only if" direction, suppose that \(Z\) is closed and let \( y_1 \in \overline{\{y_2\}} \) with \( y_2 \) in \(Z\). Since \( Z\) is closed, \( \overline{\{y_2\}} \subset Z \) so that \( y_1 \in Z \) as well. Conversely, suppose \( Z \) is constructible and stable under specialization; as \(Z\) is constructible, we may write \( Z = \coprod_{i=1}^n U_i \cap Z_i \) by Exercise 7.4.A with each \( U_i \) open and \(Z_i\) closed. As \(X\) is Noetherian, we know that each \( Z_i \) may be written as \( Z_i = Z_{i, 1} \cup \dots \cup Z_{i, n_1} \) for \( Z_{i, j} \) irreducible components of \(Z\) (by Proposition 3.6.15). Now by using an inductive argument, if \( Z_i = Z_{i, 1} \cup Z_{i, 2} \) then we may write \( Z_i \cap U_i = (Z_{i, 1} \cap (Z_{i,2}^c \cap U) ) \cup (Z_1 \cap U) \) — however, \( Z_{i, 1} \) now may not be reducible in the Noetherian subspace \( (Z_{i,2}^c \cap U) \), so we must again break it into finitely many irreducible components (as a helpful example, take \( U \) to be an open neighborhood of the origin and consider \( Z=V(xy) \) — by removing \(V(x)\) we have split \(V(y)\) into two irreducible components). After finitely many steps, this allows us to write \( Z = \coprod Z_i \cap U_i \) where each \( Z_i \) is irreducible. Let \( \eta_i \) denote the unique generic point of \( Z_i \) - since \( \{\eta_i\} \) nontrivially intersects any open set intersecting \( Z_i \), we know \( \eta_i \in U_i \) so that \( \eta_i \in Z \). Thus, for any \( x \in \bigcup_{i=1}^n Z_i \) we know that \( x \in \overline{\{\eta_j\}} \) for some \( \eta_j \); as we just showed \( \eta_j \in Z\) this implies that \( x \in Z \) since \( Z \) is stable under specialization. In other words, \( \bigcup_{i=1}^n Z_i \subset Z = \coprod_{i=1}^n Z_i \cap U_i \). However since \( U_i \cap Z_i \subset Z_i \) for all \( i \) we get the reverse containment so that \( Z \) is closed.

2. Much like the previous part, it is immediate that open sets are stable under generization: if \( Z \) is open and \( y_1 \in \overline{\{y_2\}} \) with \( y_1 \in Z \), then by definition every open neighborhood of \( y_1 \) nontrivially intersects \( \{y_2\} \). Since \( Z \) is such an open neighborhood, \( y_2 \in Z \). For the reverse direction, if \( Z \) is a constructible set stable under generization, then \( Z^c \) is a constructible set stable under specialization and therefore closed. Thus \(Z\) is open.

Proof:

By definition, \( X = \pi^{-1}(\spec k) \) is a finite set; moreover, we may write \( X = \bigcup_{i=1}^n \spec A_i \) by quasicompactness. As the morphism is locally of finite type, we know that each \( A_i\) is a finitely generated \( k \)-algebra (say by \( x_1, \dots, x_m \)). By the same argument to that in the proof of the Nullstellensatz, each \( A_i \) must in fact then be a finite \(k\)-algebra, so that \( \spec A_i \) has support which is a point. Thus \( X \) is discrete and thus the spectrum of the product of our \( A_i\) so that \( \pi^{-1}(\spec k) \) is affine.

Proof:

For the forward direction, if \( \pi \) is surjective it is certainly surjective on closed points. Conversely, suppose that \( \pi \) is surjective on closed points. By Chevalley's theorem, the image of \( \pi(X) \) is constructible so \( Y \backslash \pi(X) \) is constructible. We wish to show \( Y \backslash \pi(X) = \emptyset \); if we suppose to the contrary that there exists some \( p \in Y \backslash \pi(X) \) then it must be the case that \( p \) is not a closed point (as all closed points are in the image \( \pi(X) \) by assumption). Now as \( Y \backslash \pi(X) \) is constructible, by Exercise 7.4.A we may write \( Y \backslash \pi(X) = \coprod_i Z_i \cap U_i \) so that \( p \in Z_j \cap U_j \). We may think of \( Z_j \) as an irreducible affine subvariety, so that \( Z_j \cap U_j \) is a non-empty open subset. But then \( Z_j \cap U_j \) must contain a closed point by Exercise 3.6.J as the closed points are dense, which must then also be a closed point of \( Y \) — a contradiction.

Proof:

Let \( M \) be a finitely-generated \( B \)-module with generators \( x_1, \dots, x_n \). Then for any multiplicatively closed subset \( S \subset B \), \( S^{-1} M \) is generated by \( x_1/1, \dots, x_n/1 \) (as a \( S^{-1}B \)-module). By taking \( S = B - \{ 0 \} \), since \( B \) is an integral domain we have that \( S^{-1}B \) is a field and \( S^{-1}M \) is a \( S^{-1}B \)-vector space. Thus, we can extract a basis \( x_1/1, \dots, x_m / 1 \) for some \( m \leq n \); then for each \( m + 1 \leq j \leq n \) we can write

for \( a_{i,j}, b_{i, j} \in B \). Then by taking \( f = \prod_{i,j} b_{i,j} \), each relation above holds in \( M_f \) so \( M_f \) has a linearly independent basis and is thus free.

Proof:

If \( A \) is a finitely generated \(B\)-algebra, then it is isomorphic to \( B[T_1, \dots, T_n] / I \) for some \( n \geq 0 \) and ideal \( I \subset B[T_1, \dots, T_n] \). Thus, a finitely generated \( A \)-module \(M\) is a finitely-generated \( B[T_1, \dots, T_n] \)-module with \( IM = 0 \). By assuming the statement in the exercise, one may use induction to show that \( M \) as a \( B[T_1, \dots, T_n] \)-module must have some corresponding \( f \in B \) such that \( M_f \) is a free \( B_f \)-module.

Proof:

By inductive construction of our \( M_n \), we can simply express our \( A \)-submodules as

Then for any \( r > 0 \), we have that \( M_i / M_{i-1} \cong T^i \langle S \rangle S \), from which it is obvious that multiplication by \( T \) gives a surjection \( \psi_n : M_n / M_{n-1} \to M_{n+1}/ M_n \).

Proof:

The hint is quite literally the answer. As \( A \) is a finitely generated algebra over a Noetherian ring, it is Noetherian. Since the \( M_i\) form an ascending chain, they eventually terminate so \( M_{n}/M_{n-1} \) will eventually be trivial in which case \( \phi_n \) will also be trivial (and thus an isomorphism).

Proof:

Since we are assuming \( A \) satisfies \( (\dagger) \), each \( M_{i+1}/M_i \) is a finitely-generated \(A\)-module so that there exists some nonzero \( f_i \in B \) such that \( (M_{i+1}/M_i)_{f_i} \) is a free \( B_{f_i} \)-module. Since there are only finitely many isomorphism classes by the previous exercise, we may take \( f = f_1f_2\dots f_n \) for sufficiently large \( n \gg 0 \) so that the claim follows.

Proof:

We first wish to construct the compatible isomorphisms \( \phi_n : \bigoplus_{i=0}^{n-1} M_{i+1}/M_i \to M_n \). The base case \( n = 1 \) is obvious since \( M_0 = 0 \). Now suppose \( \phi_j \) is an isomorphism for \( j \leq n \), and consider the short exact sequence

Since \( M_{n+1} / M_{n} \) is free, the sequence splits so that we get a (non-natural) isomorphism \( \psi : M_{n} \oplus M_{n+1} / M_n \to M_{n+1} \). By inductive hypothesis, we have an isomorphism \( \phi_n : \bigoplus_{i=0}^{n-1} M_{i+1}/M_i \to M_n \) — composing \( \phi_n \) in the first component of \( \psi \) gives the desired isomorphism. Since \( \varprojlim \) is a functor, it takes isomorphisms ( over a filtered category, so in our case the entire directed system of isomorphisms \( \{ \phi_n \} \) ) to isomorphisms. Since the submodules \( M_i \hookrightarrow M \) all have directed arrows to \( M \), it is fairly easy to see that \( \varprojlim M_i = M \); thus we obtain the desired isomorphism

Proof:

Since \( \pi \) is finite type, it suffices to check on our finite affine cover; that is we restrict to \( X = \spec A \) and \( Y = \spec B \) as noted in the hint. Moreover, Generic Freeness Lemma implies there is some nonzero \( f \in B \) with \( A_f \) a free \( B_f \)-module. In the case that \( A_f \) is rank zero it is necessarily torsion — that is to say we can find some \( g \in B \) such that \( A_{fg} = 0 \). Since no prime ideal may contain a zero-divisor and \( f \) becomes a zero-divisor in \( A \), we have that the composition \( \spec A \twoheadrightarrow \spec A_f \to \spec B_f \) is zero.

Otherwise when \( A_f \) has positive rank we get a surjective morphism \( \spec A_f \to \spec B_f \), so that composing with the surjective morphism \( \spec A \to \spec A_f \) we get that \( \spec B_f \) is actually in the image of \( \spec A \). Since \( Y = \spec B \) is assumed to be irreducible, we get that \( \spec B_f = D(f)\) is open and dense ( Exercise 3.6.B ).

Proof:

We assume the statement above (image of finite type + noetherian is constructible) and wish to show that this implies that for any \( Z \) constructible, \( \pi(Z) \) is constructible. Let \( Z = \coprod Z_i \cap U_i \) for each \( U_i \) open; since the finite disjoint union of constructible sets is constructible, it suffices to show that each \( \pi(Z_i \cap U_i) \) is constructible. But the inclusion \( Z_i \cap U_i \hookrightarrow X \) is finite type by Exercise 7.3.Q so that composing with \( \pi \) gives us the restriction \( \pi\vert_{Z_i \cap U_i} \) which must then also be a finite type morphism of Noetherian schemes. By assumption, this tells us \( \pi\vert_{Z_i \cap U_i} = \pi(Z_i \cap U_i) \) is constructible so that \( \pi(Z)\) is as well.

Proof:

Since \( Y\) is a Noetherian scheme, it is quasicompact by Exercise 3.6.T; thus we may take some finite affine cover \( \spec B_i \) for \( Y\). But then

Since the finite union of constructible subsets is constructible, it suffices to consider \( \pi\vert_{\pi^{-1}(\spec B_i)} \) --- in other words, simply consider \( Y = \spec B \).

Since \( \pi \) is finite type, it is locally finite type and quasicompact. Since we are assuming \( Y = \spec B \), \( \pi^{-1}(Y) \) must be quasicompact and covered by finitely many open affines. Since the finite union of constructible sets is constructible, we may reduce to \( X = \spec A \).

Proof:

Following the hint, let \( A = k[W_1, \dots, W_m, X_1, \dots, X_n] \) so that \( V(f_1, \dots, f_p) = \spec A / (f_1, \dots, f_p) \hookrightarrow \spec A = \A^{n + m} \) is finite type, as \( A/(f_1, \dots, f_n) \) is finitely-generated over \(A\). Now for each \( D(g_i) \), since \( A_{f_i} \) is a finitely-generated \( A \)-algebra, \( D(g_i) \hookrightarrow V(f_1, \dots, f_p) \) is of finite type. Since there are finitely many \( g_i \), we get that the morphism \( D(g_1, \dots, g_q) \hookrightarrow V(f_1, \dots, f_m) \) is of finite type and thus the composition. Thus, the map \( D(g_1, \dots, g_q) \hookrightarrow \A^{m+n} \) is of finite type so by Chevalley's theorem, it has a constructible image. Since \( \overline{k}[W_1, \dots, W_m, X_1, \dots, X_n] \) is finitely generated over \( \overline{k}[W_1, \dots, W_m] \), the morphism \( \A^{m+n} \to \A^m \) is of finite-type. Applying the Nullstellensatz cleans things up a bit.