Solutions to The Rising Sea (Vakil)

Proof:

First we wish to show that the coproduct \( X \coprod Y \) exists; for every \( p \in X \coprod Y \), we have that \( p \in X \) or \( p \in Y \) in which case there is some affine neighborhood \( \spec A_j \subset X \) or \( \spec B_j \subset Y \) of \( p \). By Exercise 4.3.E, \( X \coprod Y \) may be covered by affine opens of the form \( \spec A_i \times B_j \). Since \( A_i \times B_j \) is the product in the category of rings and \( \spec\) is a contravariant functor to affine schemes, we have that each \( \spec A_i \times B_j \) is universal as the coproduct locally on \( X \coprod Y \). Gluing gives the desired coproduct.

Proof:

Let \(W\) be an arbitrary scheme along with morphisms \( \alpha : W \to \spec A \), \( \beta : W \to \spec B \) such that

commutes. By Exercise 6.3.F these correspond to morphisms \( A \to \Gamma(W, \OO_W) \) and \( B \to \Gamma(W, \OO_W) \). By the universal property of the tensor product as the fibred coproduct in the category of rings, there exists a unique morphism \( A \otimes_C B \to \Gamma(W, \OO_W) \); using Exercise 6.3.F again in the reverse direction, this corresponds to a unique morphism \( Z \to \spec (A \otimes_C B) \) so that \( \spec (A \otimes_C B) \) satisfies the universal property of the fibred product. Since universal objects are unique up to unique isomorphism, this implies \( \spec (A \otimes_C B) \cong \spec A \times_{\spec C} \spec B \).

Proof:

Since \( A \) is considered as a \(B\)-algebra, let \( \phi : B \to A \) denote the canonical map and let \( j : B \hookrightarrow B[t] \) and \( j^\prime : A \hookrightarrow A[t] \) denote the natural inclusions. Define a \(B\)-linear map \( \widetilde{\phi} : B[t] \to A[t] \) by \( \widetilde{\phi}(bt) = \phi(b)t \). Since this obviously preserves constant functions, we have that

commutes. Now suppose \(C\) is another \(B\)-algebra together with \(B\)-linear maps \(\alpha : A \to C \) and \( \beta : B[t] \to C \) such that \( \rho(\phi(b)) = \sigma(b) \) for any \( b \in B \). We define a map \( \psi : A[t] \to C \) by \( at \mapsto \alpha(a)t \); note that this clearly commutes with \( \alpha \) on constants and is uniquely determined by \( \alpha \). By construction of \( \widetilde{\phi} \), \(\psi\) also commutes with \(\sigma\) since the maps only depend on the image of constants in \( B[t] \) and \( \rho(\phi(b)) = \sigma(b) \). Therefore, \( A[t] \) satisfies the universal property of \( A \otimes_C B[t] \) — since universal objects are unique up to unique iso, this gives us a unique isomorphism \( A \otimes_C B[t] \cong A[t] \).

Proof:

Following the hint, we consider the short exact sequence

By right-exactness of \( \otimes_B A \), we obtain an exact sequence

Now the middle term is clearly \( A \); additionally, \( I^e \subset A \) may be described as the smallest ideal in \(A\) containing \( \phi (I) \). Since \( B \otimes_B A \cong A \) via \( b \otimes a \mapsto \phi(b)a \), \( I \otimes_B A \) may be considered as \( \phi(I)A = I^e\). Thefore, our sequence is of the form

By exactness, the result follows.

Proof:

1. By Exercise 8.1.D, the property of being a closed embedding is affine-local on the target, and from the proof of Theorem 9.1.1 the fibred product may also be defined on open affine sets — therefore we consider \( X = \spec B \). Let \( \sigma_i : Z_i \hookrightarrow \spec B \) denote two closed embeddings with respective ideal sheafs \( I_1, I_2 \subset B\) so that \( Z_i = V(I_i) \). From Exercise 8.1.J we defined the intersection \( Z_1 \cap Z_2 := V( I_1 + I_2 ) \cong \spec B / (I_1 + I_2) \). Then we get natural surjections \( B / I_i \twoheadrightarrow B / (I_1 + I_2) \) so that there is a unique morphism \( (B/I_1) \otimes_B (B/I_2) \to B/(I_1 + I_2) \) thus giving a natural map \( Z_1 \cap Z_2 \to Z_1 \times_{\spec B} Z_2 \) by Exercise 9.1.A.

   Conversely, if \( \pi_i : Z_1 \times_{\spec B} Z_2 \to Z_i \) denote the canonical projections with \( \sigma = \sigma_1 \circ \pi_1 = \sigma_2 \circ \pi_2 : Z_1 \times_{\spec B} Z_2 \to \spec B \), then by Exercise 8.1.C we know that \( \sigma \) is a closed embedding. By the previous exercise, we know that \( (B/I_1) \otimes_B (B/I_2) \cong (B/I_2) / I_1 (B/I_2) \) which by the isomorphism theorems is \( (B/I_2) / (I_1 + I_2)/I_2 \cong B / (I_1 + I_2) \) (by commutivity we may have also exchanged the order of \( I_1 \) and \(I_2\)). This gives us the desired isomorphism \( Z_1 \cap Z_2 \cong Z_1 \times_{\spec B} Z_2 \).

2. Let \( \begin{CD} X @>{\rho}>> Z @>{\tau}>> Y \end{CD}\) be a locally closed embedding, such that \( \rho \) is a closed embedding and \( \tau \) is an open embedding. Suppose \( \phi : W \to Y \) is a morphism of schemes. By Exercise 7.1.B, we know that the natural map \( Z \times_Y W \to W \) is an open embedding. By Exercise 1.3.Q we know that the tower of Cartesian diagrams

   $$ \begin{CD} X \times_{Z} (Z \times_Y W) @>>> Z \times_Y W @>>> W \\ @VVV @VVV @VV{\phi}V \\ X @>>{\rho}> Z @>>{\tau}> Y \end{CD} $$

   Is itself a Cartesian diagram so that \( X \times_{Z} (Z \times_Y W) \cong X \times_Y W \). By remark 9.2.2 we know that the map \( X \times_Z (Z \times_Y W) \to Z \times_Y W \) is a closed embedding, so that \( X \times_Y W \to W\) may be considered as a locally closed embedding.

3. It suffices to show that if \( \sigma_i : X_i \hookrightarrow Y \) are locally closed embeddings for \( i = 1, 2 \) then \( X_1 \times_Y X_2 \to Y \) is a locally closed embedding. Notice that \( X_1 \times_Y X_2 \) is indeed well-defined, and gives canonical projections \( \pi_i : X_1 \times_Y X_2 \to X_i \). By part (b) above, we know that \( \pi_1 \) and \( \pi_2 \) are also locally closed embeddings, so by Exercise 8.1.N, \( \sigma_1 \circ \pi_1 = \sigma_2 \circ \pi_2 \) is a locally closed embedding.

Proof:

Let \( \phi: X \to \spec A \) and \( \psi : Y \to \spec A \) denote the structure morphisms, and consider affine open covers \( \{\spec B_i\} \), \( \{ \spec C_j \} \) of \( X \) and \(Y\) (respectively) such that each \( B_i, C_j \) is a finitely-generated \( A \)-algebra. From the proof of Theorem 9.1.1, we know that the fibre product may be computed affine-locally; that is \( \spec( B_i \otimes_A C_j ) \) cover \( X \times_A Y \) and may be glued on overlaps / triple overlaps. It is a standard algebra fact that if \( B, C \) are finitely generated \(A\)-algebras, then \( B \otimes_A C \) is a finitely-generated \(A\)-algebra, completing the first part of the proof. By adding the additional hypothesis of quasicompactness to our morphisms \( \phi, \psi \) (i.e. dropping the "locally" part of the hypothesis), we get that our covers \( \{ \spec B_i \} \) and \( \{ \spec C_j \} \) are finite, so that \( \{ \spec(B_i \otimes_A C_j) \} \) is also clearly finite.

Proof:

We closely follow the approach of this StackExchange answer; first, it is worth mentioning that the profinite topology on \( G = \gal( \overline{\Q} / \Q )\) is by definition the topology induced by the limit \( \varprojlim G_i\) for some finite groups \( G_i \) with the discrete topology. In our case, these \( G_i \) must be \( \gal( K / \Q ) \) where \( K \) is some finite extension (thus making \( \gal( K / \Q) \) finite). For example, if \( K \) is a finite extension of \( \Q \) with minimal polynomial \( f \in \Q[x] \) of degree \( n \), then \( K \cong \Q[x] / f \) so that by Exercise 9.2.A we have that

Thus, we get \( \spec( \overline{\Q} \otimes_\Q K ) \cong \coprod_{i=1}^n \spec \overline{\Q} \) — as a finite set of \( n \) points, this clearly has the discrete topology which agrees with the topology on the finite group \( \gal(K/\Q) \).

Since we may describe \( \overline{\Q} \) as the colimit over \( \mathscr{K} \) where the elements \( K \in \mathscr{K} \) are finite extensions of \( \Q \) and the morphisms are inclusions of fields, we have

As we may consider \( \spec \) as a contravariant functor from the category of \( \overline{Q} \)-algebras to topological spaces, it takes colimits to limits and thus

Proof:

Recall that the isomorphism \( A \otimes_B B \cong A \) is given by \( a \otimes b \mapsto \phi(b) a \). Thus, the \( B \)-linearity in our tensor product gives \( a \otimes b \cong \phi(b)a \otimes 1_B \). But then \( A \otimes_B (S^{-1}B) \cong \phi(S)^{-1}A \otimes_B B \cong \phi(S)^{-1} A \) in the obvious way: \( a \otimes \frac{b}{s} \mapsto \frac{a \phi(b)}{\phi(s)} \). This is clearly injective since if \( a \otimes \frac{1}{s} \mapsto 0 \), then there exists some \( \phi(t) \in \phi(S) \) such that \( \phi(t)a = 0 \) in \(A\) (here, we may assume without loss of generality that \( b \) is \(1\) by possibly replacing \( a \) with \( a\phi(b) \)). Thus, \( a \otimes \frac{1}{s} = a \otimes \frac{t}{ts} = \phi(t)a \otimes \frac{1}{ts} = 0 \). It is easy to see that the map is surjective, giving the desired isomorphism.

Proof:

Suppose \( \iota: X \hookrightarrow Y \) is an open embedding. Since the notion of being an open embedding is affine-local on the target, we may assume \( Y = \spec B \) so that \( X \) is (isomorphic to) an open subscheme of \( \spec B \). Now suppose \( Z \) is another scheme and \( \mu_1, \mu_2 : Z \to X \) are morphisms of schemes such that \( \iota \circ \mu_1 = \iota \circ \mu_2 \). As \( X \) may be considered an open subset of \(Y\), these maps clearly agree on points. To see that the maps of sheaves agree, notice that at any \( p \in Z \) the stalk maps \( (\mu_i^\sharp \circ \iota^\sharp)_p : \OO_{Z,p} \to \OO_{X, \mu_i(p)} \to \OO_{\spec B, \mu_i(p)} \) agree. However, since \( \iota^\sharp : \OO_{X} \to \OO_{\spec B} \) is defined to be an isomorphism, we must have that the stalk maps \( \OO_{Z, p} \to \OO_{X, \mu_i(p)} \) all agree — thus \( \mu_1 = \mu_2 \).

As closed embeddings are also affine local on the target, suppose \( Y = \spec B \) and \( X = V(I) \) for some ideal \( I \subset B \). If \( Z \) is another scheme and there are maps \( \mu_1, \mu_2 : Z \to X \) such that \( \iota \circ \mu_1 = \iota \circ \mu_2 \). As \( X \) may be considered as a (closed) subset of \( Y \), these maps must clearly agree on points. To see that they agree on stalks, notice that since \( \iota^\sharp : \OO_Y \to \OO_X \) is defined to be a surjection, it is necessarily an epimorphism so that as \( \mu_1^\sharp \circ \iota^\sharp = \mu_2^\sharp \circ \iota^\sharp \) we have the maps of sheaves agree.

Lastly, for a multiplicative subset \( S \subset A \), the map \( A \to S^{-1}A \) induces an injection on the level of sets by Exercise 3.2.K — again, we will denote \( \iota : \spec S^{-1}A \hookrightarrow \spec A \). Thus, if \( Z \) is an arbitrary subscheme and \( \mu_1, \mu_2 : Z \to \spec S^{-1} A \) are morphisms of schemes such that \( \iota \circ \mu_1 = \iota \circ \mu_2 \), then they must necessarily agree at all stalks. Though it need not be true that \( A \to S^{-1}A \) is an injection (especially when \(S\) contains a zero divisor), we may always consider the localization at the nilradical. Thus, if \( \mu_1^\sharp \circ \iota^\sharp = \mu_2^\sharp \circ \iota^\sharp \) have stalks that agree then \( \mu_1 = \mu_2 \).

Proof:

As \( \Z \) is initial in the category of rings, every ring \( A \) may be considered as a \(Z\)-algebra in the natural way, giving a unique isomorphism \( \Z \otimes_\Z A \cong A \). By Exercise 9.2.A, this tells us that there is a unique isomorphism \( \Z[x_1, \dots, x_n] \otimes_\Z A \cong A[x_1, \dots, x_n] \) for every ring \( A \). Thus, we get

In a similar fashion, we may consider \( A \) as a graded \( \Z \)-algebra where all elements are graded in degree \(0 \). Thus, the ring isomorphism \( \Z[x_0, \dots, x_n] \otimes_\Z A \cong A[x_0, \dots, x_n] \) is in fact an isomorphism of graded \( \Z \)-algebras of degree \(1\) giving us the desired isomorphism of projective schemes.

Proof:

To be perfectly clear what we mean by the induced maps \( X_\ell \to Y_\ell \), notice that we get a diagram

where \( \sigma_X, \sigma_Y \) denote the structure morphisms, \( j : \spec \ell \to \spec k \) the induced map from our field extension, \( p_1, p_2, q_1, q_2 \) the canonical projections from our fibre product. Notice that since \( \rho, \pi \) are morphisms of \( k \)-schemes, we necessarily have that \( \sigma_X = \sigma_Y \circ \rho = \sigma_Y \circ \pi \). Thus,

So by the universal property of \( X_\ell = X \times_{\spec k} \spec \ell \), we get a unique morphism \( \pi_\ell : X_ \ell \to Y_\ell \) such that

commutes. A similar argument may be made for \( \rho\).

To see that \( \pi \) and \( \rho \) must necessarily agree on the level of sets we must make use of Exercise 9.4.D as mentioned in the hint: suppose \( x \in X \) is an arbitrary point — as \( p_1 \) is surjective, there exists some \( \overline{x} \in X_\ell \) with \( p_1(\overline{x}) = x \). Then \( \pi(x) = q_1(\pi_\ell(\overline{x})) = q_1(\rho_\ell(\overline{x})) = \rho(x) \) as desired.

To prove the result, we must also show that \( \pi^\sharp = \rho^\sharp \) as morphisms of sheaves. As it suffices to show this on the level of open affines, we may restrict to the case that \( X = \spec B \) and \( Y = \spec A \). The punchline here is that field extensions are faithfully flat, as a field extension may be considered as a vector space over our field which is necessarily free. Thus, the map \( p_1^\sharp : B \to B \otimes_k \ell \) is necessarily injective (as well as \( q_1^\sharp : A \to A \otimes_k \ell \)). Thus, for any \( a \in A \) we have that

which implies \( \pi^\sharp(a) = \rho^\sharp(a) \).

Proof:

Similar to the previous problem, since \( \ell \) is faithfully flat over \( k \) we know that \( \pi_\ell^\sharp : A \otimes_k \ell \to B \otimes_k \ell \) is surjective if and only if \( \pi^\sharp \) is surjective.

Proof:

Recall that \( \Q(t) = S^{-1}\Q[t] \) for \( S = \Q[t] - \{ 0 \} \). By Exercise 9.2.F we get that

Then by Exercise 3.2.K \( \spec \Q(t) \otimes_\Q \C \cong \spec S^{-1}\C[t] \) is in bijection with those prime ideals \( \pp \) that do not intersect \( \Q[t] - \{ 0 \} \). Since \( \C \) is algebraically closed we know that our prime / maximal ideals are of the form \( \pp = (x - a) \) for some \( a \in \C \). However, since these cannot be roots of \( \Q[t] \), they are precisely the transcendental numbers.

Proof:

Showing that the diagram

commutes is straightforward: the top map is the natural inclusion map, so that the topology on \( f^{-1}(p) \) is simply the subspace topology making \( j \) continuous. To see that this is indeed the fibred product, suppose \( W \) is another topological space with maps \( \alpha : W \to \{ p \} \) and \( \beta : W \to Y \) such that \( \alpha = f \circ \beta \) (we omit the obvious inclusion of our point). Notice that since \( \alpha(W) \subseteq \{ p\} \) we must have \( f \circ \beta (W) \subseteq \{ p \} \) so that \( \beta(W) \subset f^{-1}(p) \). In order to define a unique continuous map \( \psi : W \to f^{-1}(p) \), notice that it must satisfy \( f \circ \psi (W) = \alpha(W) = f \circ \beta(W) \). Since \( \beta(W) \subset f^{-1}(p) \), the obvious choice is simply \( \psi = \beta \) — this uniquely depends on our map \( \beta \) so that \( f^{-1}(p) \) satisfies the universal property of the fibre product.

Proof:

As before, let \( f : X \to Y \) denote our morphism and consider a point \( p \in Y \). By definition of \(Y\) as a scheme, we may take some affine neighborhood \( \spec B \subset Y \) of \( p = [\pp] \) and compute locally. Let us denote \( X_p = \spec \kappa(p) \times_{\spec B} X \) as the scheme theoretic fibre and \( f^{-1}(p) \) the topological fibre from the previous exercise. If we let \( q_1 : X_p \to X \) denote the first canonical projection and \( q_2 : X_p \to \spec \kappa(p) \) the second, we wish to show that \( q_1 \) is a homeomorphism onto \( f^{1}(p) \).

Similar to the natural morphisms obtained in Exercise 6.3.J, suppose \( U = \spec A \subset X \) is an affine open neighborhood in \( f^{-1}(\spec B) \) so that \( U_p = \spec \kappa(p) \times_{\spec B} \spec A \). Since \( q_1 : X_p \to X \) locally is induced by the ring morphism \(A \to A \otimes_B B_\pp \to A \otimes_B \kappa(p) \) and we have \( A \otimes_B B_\pp \cong f^{\sharp}(S)^{-1}(A) \) for \( S = B - \pp \), \( q_1 : X_p \to X \) locally induces a bijection with prime ideals \( \qq \subset A \) with \( \qq \cap f^{\sharp}(S) = \emptyset \) and \( \pp A \subseteq \qq \) (the latter from the fact that \(\kappa(p)\) may be defined by the quotient by extension of \( \pp \) ). By construction these are precisely the ideals \( \qq \) with \( f([\qq]) = [\pp] = p \) so that affine locally we get a homeomorphism \( X_p \cong f^{-1}(p) \); however since the topology is determined on open sets and the decomposition of \( q_1 : X_p \to X \) glues by Exercise 6.3.J, we obtain the desired homeomorphism.

Proof:

We again make use of Exercise 1.3.Q to obtain a tower of Cartesian diagrams:

The small right Cartesian square exists by the problem statement, and the leftmost square should be interperted as the construction of the fibre of \( X \times_Z Y \to X \) over \( p \) ( that is, \( \{ p \} \times_X (X \times_Z Y) \)). By the above mentioned exercise, the overall diagram is a Cartesian square and so we get an isomorphism \( \{ p \} \times_X (X \times_Z Y) \cong \{ \tau(p) \} \times_Z Y \) since the composition of the bottom two arrows may be seen as the inclusion \( \{ \tau(p) \} \hookrightarrow Z \). But this is precisely the fibre of \( \pi : Y \to Z \) over \( \tau(p) \) as desired.

Proof:

From the discussion above Exercise 9.3.B, the inclusion of our point \([(p)] \) into \( \spec \Z \) should be considered as the map \( \spec \kappa(p) \to \spec \Z \). Now the residue field of \( \pp = (p) \) is precisely \( \Z_\pp / \pp\cdot \Z_\pp = ( \Z / \pp )_\pp \cong \mathbb{F}_p \) since \( \mathbb{F}_p := \Z / (p) \) is already a field. Thus, the fibre

Corresponds to the tensor \( \spec \big( \Z / (p) \otimes_\Z \Z[x]/(x^2 + 1) \big) \cong \spec \mathbb{F}_p[x] / (x^2 + 1) \). Now consider the group \( \mathbb{F}_p^\times \) of units in \( \mathbb{F}_p \) — it is a standard fact in Galois theory that this is a cyclic group of order \( p - 1 \), say generated by some \( \alpha \in \mathbb{F}_p \). Assuming \( p \neq 2 \), we know that \( -1 \in \mathbb{F}_p[x] / (x^2 + 1) \) has order 2, so it must be of the form \( -1 = \alpha^{(p - 1) / 2} \). Notice that this element is a square precisely when \( p - 1 = 4n \) for some \( n \), in which case \( x^2 + 1 \) has two distinct roots if \( p \cong 1 \mod 4 \). Thus, in the case that \( p \cong 1 \mod 4 \), we have that the fibre above \( (p) \) is two reduced points.

In the case that \( p \cong 3 \mod 4 \), \( x^2 + 1 \) does not split so we get a single reduced point of degree 2 above \( (p) \). We cannot have \( p \equiv 0 \mod 4 \) as it is prime, and if \( p \equiv 2 \mod 4 \) then it must be the case that \( p = 2 \). In this instance, \( \mathbf{F}_p [x] / (x^2 + 1) = \mathbb{F}_p[x] / (x + 1)^2 \) which is a non-reduced point of degree 2.

Lastly, when our prime is in fact the generic point \( [(0)] \), we have that \( \kappa(p) = \Z_{(0)} = \Q \) so that the fibre is \( \spec \Q[x]/(x^2 + 1) \), which is again a single reduced point of degree 2.

Proof:

The map \( u \mapsto t^2 \) allows us to think of \( k[t] \) as a \( k[u] \)-algebra where we add a square root of \( u \): \( k[t] \cong k[u, x] / (x^2 - u) \). Thus,

Since \( (x^2 - y^2) = (x + y)(x-y) \) where \( (x + y) \), \( (x - y) \) are minimal primes, \( X \times_Y X \) has two irreducible components.

Proof:

Let \( p \in \P^1_k \) be a closed point, and suppose without loss of generality that \( p \in D(u) \) so that we may write it of the form \( p = [1 : \alpha] \) for some \( \alpha \in k \). Then locally \( \textrm{Bl}_{(0,0)} \A^2_k \) looks like the affine scheme \( \spec k[x, y, v] / (xv - y) \) and the fibre of the map \( \textrm{Bl}_{(0, 0)} \A^2_k \to \P^1_k \) may be identified with the fibre of \( \spec k[x, y, v] / (xv - y) \to \spec k[v] \) over \( \spec k[v] / (v - \alpha) \). Thus, we obtain the fibre \( \{p\} \times_{\P^1_k} \textrm{Bl}_{(0,0)} \A^2_k \) is precisely

That is to say, the fibre is precisely the collection of points on the plane lying on the linear subspace \( y = \alpha x \).

Now consider the map \( \textrm{Bl}_{(0,0)} \A^2_k \to \A^2_k \) — again we may look affine-locally on the affine open set \( D(u) \) where our map looks like \( \spec k[x, y, v] / (xv - y) \to \spec k[x, y] \) (induced by the obvious ring map \( k[x, y] \to k[x, y, v] / (xv - y) \)). Away from the origin (i.e. on the open set \( D(x, y) \)), we have that this map is of the form

We must ignore what happens on the \( y \)-axis since that would correspond to \( u = 0 \), i.e. outside our current affine neighborhood \(D(u)\). Therefore, it is safe to assume \( x \) does not vanish so we have \( v = y/x \), giving the desired isomorphism. Notice that this may be interpreted by saying that given a point \( (x_0, y_0) \in \A^2_k\), there is a single unique line connecting \( (0, 0) \) to \( (x_0, y_0) \) with slope precisely \( y_0/x_0 \) — this is a standard geometric fact.

It is not difficult to see that the fibre over \( (0, 0) \) is precisely \( \P^1_k \). Indeed on our open affine \( D(u) \) the fibre corresponds to \( k[x, y]/(x, y) \otimes_{k[x, y]} k[x, y, v] / (xv - y) \) which locally looks like \( k[v] \). These glue in the same manner that the affine patches of \( \P^1_k \) glue, so that we obtain a copy of \( \P^1_k \). Hartshorne pp. 28 gives a good interpretation of this as all lines through the origin must, of course, pass through te origin.

Proof:

Following the hint, we first consider an affine open of an irreducible component; that is we may assume \( Y = \spec B \) for some integral Noetherian \(B\). As \( \pi : X \to \spec B\) is affine and finite type, we have \( X = \spec A \) for some Noetherian \( A\) that is a finitely generated \(B \)-algebra: write \( A = B[x_1, \dots, x_n] / I \). By assumption, the generic fibre \( \spec (A \otimes_B K(B)) \) is finite which implies that \( A \otimes_B K(B) \cong K(B)[x_1, \dots, x_n] / I\) is a finite \(K(B) \)-module by Exercise 7.4.D. Thus, we must necessarily have that our generators \( x_i \) are integral / algebraic over \( K(B) \), so for each \( i \) there exists some monic \( f_i(t) \in K(B)[t] \) such that \( f_i(x_i) = 0 \). As noted in the hint, if we let \(b\) denote the product of denominators occuring across all the \(f_i(t)\), then by replacing \( B \) with \(B_b\) and \( A\) with \(A_b\) we may assume \( f_i(t) \in B[t] \). Then \( f_i(x_i) = 0 \) in \( A \otimes_B K(B) = S^{-1}A \) where \( S = B - \{ 0 \} \) so that for each \( i \), there exists some \( s_i \) such that \( s_i f_i(x_i) = 0 \) in \(A\). By letting \( s \) denote the product of all the \(s_i\), we may again localize at \( s \) such that \( f_i(x_i) = 0 \) in \(A\). Since the \( f_i \in B[t] \), this implies each generator \( x_i \) is integral over \(A\) and thus \(A\) is an integral extension. Since it is an integral, finite-type extension, it is finite by Exercise 7.3.P.

In the more general case, if we let \(Y_i\) denote the irreducible components of \( Y \) and \( \eta_i \) their respective generic points, then as the definition of generically finiteness is determined on the \( \eta_i \) we may take some affine neighborhoods \( U_i = \spec B_i \subset Y_i \) containing the generic point on which \( \pi \vert_{\pi^{-1}(U_i)} \) is finite. By taking the (necessarily finite) union of all such neighborhoods, we obtain an open set containing all generic points where \( \pi \) is finite.

Proof:

Let \( \psi : X \to Z \) be an arbitrary morphism and let \( \pi : Y \to Z \) be a locally principally closed subscheme, so that we may take some affine cover \( \{ U_i \} \) on which \( \pi^{-1}(U_i) \subset Y \) is of the form \( V(s_i) \). As we may consider smaller affine neighborhoods in the preimage of each \( \psi^{-1}(U_i) \), we may simply assume without loss of generality that \( Z = \spec B \) and \( Y = \spec B / (s) \) for some \( s \in B \). Let \( \spec A \subset \psi^{-1}(\spec B) \) be an affine neighborhood in \(X\). From Exercise 9.2.B, we know that \( \pi^\prime : X \times_Z Y \to X \) is a closed embedding, so that \( (\pi^\prime)^{-1}(\spec A) = \spec A / I \) for some ideal \( I \subset I \). Then the corresponding diagram on the level of commutative rings is of the form

However, from Exercise 9.2.B we know that \( I \) is precisely the ideal generated by \( t = \psi^\sharp(s) \), so that our closed embedding \( X \times_Z Y \to X \) is again locally principal.

Proof:

1. Let \( \pi : X \to Y \) be quasicompact and suppose \( \psi : Z \to Y \) is arbitrary. Since quasicompactness is affine local on the target by Exercise 7.3.C, we may assume \( Y = \spec B \) and take some \( \spec C \subset \psi^{-1}(\spec B) \). Since \( \pi \) is assumed to be q.c, we may write \( Y = \bigcup_{i=1}^n \spec A_i \) (taken as the scheme-theoretic union). If we let \( p : Z \times_Y X \to Z \) denote the natural projection, then we know by Theorem 9.1.1 that \( Z \times_Y X \) may be constructed affine locally and is thus of the form \( \bigcup_{i=1}^n \spec (C \otimes_B A_i) \), which is of course quasicompact.

2. Using the notation above, assume \( \pi : X \to Y \) is instead quasiseperated; again by Exercise 7.3.C this is affine-local on the target, so we may continue to assume \( Y = \spec B \) and take \( \spec C \subset \psi^{-1}(\spec B) \). Let \( \spec D_1, \spec D_2 \subset p^{-1}(\spec C) \) be affine opens in \( Z \times_Y X \), where \( p : Z \times_Y X \to Z \) again denotes the natural projection. Since \( Z \times_Y X \) may be constructed by gluing affines by Theorem 9.1.1, we may assume without loss of generality that \( \spec D_i = \spec (C \otimes_B A_i) \) for some \( \spec A_i \subset \pi^{-1}(\spec B) \). If \( \bigcup_{\alpha} \spec \widetilde{D_\alpha} \) is an affine cover of \( \spec D_1 \cap \spec D_2 \), then by the same logic these are of the form \( \spec (C \otimes_B \widetilde{A}_i) \). Since each \( \spec \widetilde{D}_\alpha \subset \spec D_1, \spec D_2 \), we necessarily have that \( \spec \widetilde{A}_\alpha \subset \spec A_1, \spec A_2 \) so that they cover the intersection. As \( \spec A_1 \cap \spec A_2\) is quasicompact, all but finitely many may be thrown away so that by considering the corresponding \( \spec D_\alpha \), we have that \( \spec D_1 \cap \spec D_2 \) is quasicompact.

3. This again will follow from Theorem 9.1.1, as the fibre product may be constructed affine locally and glued. Using the same notation, we now have\( X = \pi^{-1}(\spec B) = \spec A \) since \( \pi \) is necessarily affine. As a morphism being affine is affine-local on the target by Proposition 7.3.4, we may take \( \spec C \subset \psi^{-1}(\spec B) \). Then by the construction of the fibred product we have \( p^{-1}(\spec C) = \spec ( C \otimes_B A ) \) which is obviously affine.

4. This will in fact follow from (e) and (g) since finite = finite type + integral by Exercise 7.3.P.

5. This follows from Exercise 7.3.N.

6. Using the same notation as above, if \( \spec C \subset \psi^{-1}(\spec B) \) and \( \spec A \subset \pi^{-1}(B) \), then by assumption \( A \) is a finitely-generated \(B\)-algebra so that we may assume \( A = B[x_1, \dots, x_n] / I \). By the local construction of \( Z \times_Y X \), we have that the pullback of \( \spec C \) is covered by affine sets of the form \( \spec C \otimes_B A = \spec C[x_1, \dots, x_n] / I^e \) which is clearly finitely-generated over \(C\).

7. Finite type = locally finite type + quasicompact, so parts (f) and (a) respectively.

8. By a similar line of logic to part (f) above, if we instead suppose \( A = B[x_1, \dots, x_n] / I \) where \( I \) is now finitely generated. Then for any \( B\)-algebra \(C\) (given by some ring map \( \psi^\sharp : B \to C \)), we have that \( A \otimes_B C = C[x_1, \dots, x_n] / I^e \) as above, where \( I^e \) again denotes the extension of \( I \) to \( C[x_1, \dots, x_n] \) (i.e. the smallest ideal in \( C[x_1, \dots, x_n] \) containing \( \phi(I) \)). By right-exactness of the tensor product, since \( I^e \) is generated by \(I\) and \( I \) is finitely generated, we get that \( A \otimes_B C \) is finitely-presented over \(C\).

9. We simply apply part (a) to the previous part.

Proof:

Let \( \pi : X \to Z \) be a quasi-finite morphism and \( \psi : Z \to Y \) be arbitrary. Since finite-type is preserved by base change from the previous exercise, we know that the map \( \widetilde{\pi} : Z \times_Y X \to Z \) is finite-type. To show that this map has finite fibers, pick some \( z \in Z \) and consider the composition

where from our definitions we have \( \widetilde{\pi}^{-1}(z) := \spec \kappa(z) \times_Z (Z \times_Y X) \cong \spec \kappa(z) \times_Y X \). Now if we let \( y = \psi(z) \), we may also write

Which corresponds to an extension of fields \( \kappa(y) \hookrightarrow \kappa(z) \). By Exercise 7.4.D we know that \( \pi^{-1}(y) \to \spec \kappa(y) \) is finite; as this is stable under base-change, \( \widetilde{\pi}^{-1}(z) \to \spec \kappa(z) \) must also be finite. By Exercise 7.3.H we know that \( \widetilde{\pi}^{-1}(z) \) must be a finite union of points so that \( \widetilde{\pi} \) has finite fibres.

Proof:

Let \( \pi : X \twoheadrightarrow Y \) be a surjection and \( \psi : Z \to Y \) be arbitrary; in addition, let \( \widetilde{\pi} : Z \times_Y X \to Z \) denote the base-change map. Showing that \( \widetilde{\pi} \) is surjective is equivalent to showing that each \( z \in Z \) has non-zero fibre. If we pick some \( z \in Z \) and let \( y = \psi(z) \) denote its image, then as \( \pi : X \to Y \) is surjective there exists some \( x \in X \) with \( \pi(x) = y \). Omitting all the compositions, we may construct a diagram

Then our fibred product is of the form \( \spec \kappa(z) \otimes_{\kappa(y)} \kappa(x) \); indeed as any \(\kappa(y)\)-morphism is an embedding (as the kernel is necessarily the only ideal \( (0) \)) we have that \( \kappa(z) \) and \( \kappa(x) \) are both field extensions of \( \kappa(y) \), and that \( A = \kappa(z) \otimes_{\kappa(y)} \kappa(x) \) is a \( \kappa(y) \)-algebra. Now we know that it must be non-empty since field extensions are faithfully flat. By Zorn's lemma there exists a unique maximal ideal \( \mm \) so that \( A / \mm \) is a field extension of \( \kappa(y) \). As \( \kappa(z), \kappa(x) \hookrightarrow A / \mm \) (as otherwise \( A = \mm \) which would contradict \( \mm \) being a maximal ideal), we get canonical maps \( \spec A / \mm \to \spec \kappa(z) \) and \( \spec A / \mm \to \spec \kappa(x)\) making the overall diagram commute. By the universal property, there is a unique morphism \( \spec A / \mm \to Z \times_Y X \), so that by letting \( w \) denote the image of the unique point under this map, we get that \( \widetilde{\pi}(w) = z \).

Proof:

Since \( X \to \spec \overline{k}\) is finite type and finite type is preserved by base change, we have that \( X \times_{\overline{k}} Y \to Y\) is finite type. Since \( Y \to \spec k \) is finite type and finite type is preserved by composition by Exercise 7.3.Q, we have that \( X \times_{\overline{k}} Y\) is a finite type \( \overline{k} \) scheme. To see that \( X \times_{\overline{k}} Y \) is integral, we follow the hint and refer to Easy Exercise 9.5.L (proved later) to claim we may reduce to the case that \( X = \spec A \) and \( Y = \spec B \), with \( A, B \) integral domains. Writing \( X \times_{\overline{k}} Y = \spec (A \otimes_{\overline{k}} B ) \), suppose to the contrary that \( A \otimes_{\overline{k}} B \) is not a domain, so that there exist some \( \sum a_i \otimes b_i \) and \( \sum a_{j^\prime} \otimes b_{j^\prime} \) whose product is \( 0 \). By restricting to the span of the \( \{ b_i \}, \{ b^\prime_j \} \) over \( \overline{k} \) we may assume the two sets are linearly independent, and moreso that \( a_1, a_1^\prime \neq 0 \Rightarrow a_1a_1^\prime \neq 0 \) in \(A \). In particular, \( D(a_1a_1^\prime) \neq \emptyset \) as \( a_1a_1^\prime \) cannot be nilpotent, so as closed points are dense we can find some \( [\mm] \in D(a_1a_1^\prime) \). As \( A/\mm \) must be a finite \( \overline{k} \) extension by the Weak Nullstellensatz and \( \overline{k} \) is closed, we simply have \( A / \mm = \overline{k} \). By Exercise 9.2.B we have that

Since \((\sum a_i \otimes b_i) (\sum a_{j}^\prime \otimes b_{j}^\prime) = 0 \in \mm^e\) we have that (WLOG) \( \sum a_i \otimes b_i \in \mm^e \). Reducing modulo \( \mm^e \), we have that \(0 = \sum \overline{a}_i \otimes b_i \in (A \otimes_{\overline{k}} B)/ \mm^e \cong B \). Since \( (A \otimes_\overline{k}B) / \mm^e \cong (A / \mm) \otimes_A (A \otimes_{\overline{k}} B)/\mm \), we may take \( \alpha_i \) to be the image of the \( \overline{a_i} \) in \( B \) so that \( \sum \alpha_i b_i = 0 \) in \( B \). But since the \( \{b_i\} \) are assumed to be linearly independent, we must have that \( \alpha_1 = 0 \) which would imply that \( a_1 = 0\) — a contradiction.

Proof:

Consider the overall diagram

Where \( P \) denotes a morphism having property \( P\), where four of the six follow from assumption of \( P \) preserved under pullback. Notice by definition of morphism of \(S\)-schemes that \( X \to Y \to S \) is the same as the structure morphism \( X \to S \). By the universal property of the fibred product, since we have maps \( X \times_S X^\prime Y \) and \( X \times_S X^\prime \to Y^\prime \) which commute in the necessary sense, we get a unique map \( X \times_S X^\prime \to Y \times_S Y^\prime \) which factors through the two maps. However, as this is made from the top left square where all morphisms have property \(P\), then \( P \) being closed under composition implies that \( X \times_S X^\prime \to Y \times_S Y^\prime \) has property \( P \).

Proof:

By mimicing the argument of Exercise 9.3.E, we may consider the map \( t \mapsto u^p \) to construct a \( k(t) \)-algebra homomorphism \( k(u) \cong k(t, x) / (x - t^p) \). Thus,

As \( k \) is characteristic \(p\), the Frobenius homomorphism tells us that \( (x- y)^p = x^p - y^p \) so that \( x - y\) is a nilpotent element.

Proof:

This becomes fairly obvious after drawing a fibred diagram: first let \( \pi : X \to Y \) be the morphism in question, and \( \psi : Z \to Y \) the change of base morphism, such that there is an induced map \( \widetilde{\pi} : Z \times_Y X \to Z \). If we let \( \spec \overline{k} \to Z \) be a geometric point, then by composing with \( \psi \) we also get that \( \spec \overline{k} \to Y \) is a geometric point. From the diagram

we get that \( \pi^{-1}(\spec \overline{k}) \cong \widetilde{\pi}^{-1}(\spec \overline{k}) \) — by hypothesis this tells us that \( \widetilde{\pi}^{-1}(\spec \overline{k}) \) shares the same properties as \( \pi^{-1}(\overline{\spec k}) \).

Proof:

The idea is precisely the same as Example 9.2.3. If \( \spec \overline{k} \to \spec \R \) is a geometric fibre, then our fibred diagram corresponds to the tensor product

which again splits into two distinct, irreducible factors giving us \( \spec \overline{k} \coprod \spec \overline{k} \). Clearly \( \spec \overline{k} \) is reduced by Exercise 5.2.B.

Proof:

1. By taking \( k = \mathbb{F}_2(t) \) and considering the scheme \( X = \spec k[x] / (x^2 + t) \), by passing to \( X_\overline{k}\) we get \( X_\overline{k} = \spec \overline{k}[x] / ((x + \sqrt{t})^2) \) which is clearly non-reduced.

2. Consider \(X = \spec \R [x] / (x^2 + 1) \). Then \( X_\mathbb{C} = \spec \C[x] / (x - i) \coprod \spec \C[x] / (x + i) \)

3. Suppose \( a \in k \) is not a square and consider \( \spec k[x, y] / (x^2 + ay^2) \). Then

   $$ \begin{CD} V(x - \sqrt{-a} y) \cup V(x + \sqrt{-a} y) @>>> \spec k[x, y] / (x^2 + ay^2) \\ @VVV @VVV \\ \spec \overline{k} @>>> \spec k \end{CD} $$

   Since irreducibility is lost, we have that integrality is lost as well.

Proof:

Let \( \overline{k} \) be an algebraically closed field of characteristic other than 2, and let \( a \in \overline{k} \) be the image of \( x \in \Q[x] \) under the map \( \Q[x] \to \overline{k} \) (corresponding to \( \spec \overline{k} \hookrightarrow \Q[x] \)). Similar to Exercise 9.3.E we may consider \( \Q[y] \) as a \( \Q[x] \) algebra, and see that it is isomorphic to \( \Q[x, w] / (x - w^2) \). Taking the twnsor product, we get

Now as \( \overline{k} \) is algebraically closed, this splits as two distinct roots so that by the Chinese remainder theorem we get two isomorphic copies \( \spec \overline{k} \coprod \spec \overline{k} \).

Proof:

1. This follows from the fact that field extensions are faithfully flat (though only flatness is needed).

2. To see that \( (R_b) \) implies \( (R_a) \), we skip the ambiguity (the wording of this exercise is far more clear in Hartshorne) and jump to \( (R_c) \Rightarrow (R_a)\). We may assume without loss of generality that our arbitrary field \( K \) is an extension of \( \overline{k} \). Since being a nilpotent element implies being a zero-divisor, we may refer to part (b) below. The fact that \( (R_c) \Rightarrow (R_d) \) is immediate from \( (R_c) \Rightarrow (R_a) \).

3. Similar to above, the ambiguity is somewhat annoying here since \( k \) (prior to the problem statement was already arbitrary) — again, it suffices to show \( (R_c) \Rightarrow R_a \). We may again suppose that \( K \) is an extension of \( \overline{k} \). Notice that \( A \otimes_{\overline{k}} K \) having a zero-divisor is equivalent to there being a system of equations with coefficients in \( \overline{k} \) having a solution over \( K \). But as \( \overline{k} \) is algebraically closed, such a system must have its solution in \( \overline{k} \) implying that \(A\) is not a domain — a contradiction.

4. Connectedness only relies on the property of being a separable extension by Corollary 9.5.11, so a similar technique to above with the choice of separable extension suffices.

Proof:

To show that geometrically irreducible quartics are open, we show that quartics which are either reducible or become reducible after base change to the algebraic closure form a closed set. If we consider \( X \) as a reducible quartic over \( \P^2_{\overline{k}} \), then on each distinguished open our quartic \( V_+(f) \) may be decomposed as \( f = gh \) where either both \( g, h \) are quadratic equations or a line and a cubic. In the first case we get a map \( \P^5_{\overline{k}} \times \P^5_{\overline{k}} \to \P^{14} \) given by multiplication, and in the second case we get a map \( \P^2_{\overline{k}} \times \P^9_{\overline{k}} \to \P^{14} \). Since being an open map is a local condition (i.e. locally open implies open) and surjective open maps are also closed (easy set theory exercise), Proposition 9.5.4 implies that the set of quartics in \( \P^{14} \) which are the product of two quadratics is closed. We get a similar result for the product of a cubic and a line. By Lemma 9.5.6, we also get that the set of geometrically reducible quartics is closed. Taking the complement of the two aforementioned sets gives the desired result.

Proof:

Notice that Let \( p \in X \) be a point, and consider the class \( \mathcal{S} \) consisting of connected sets containing \( p \). Indeed the set \( \mathcal{S} \) is non-empty since \( \{ p \} \in \mathcal{S} \) and is bounded above by \( X \). Thus by Zorn's lemma, there is a maximal connected set containing \( p \) which is by definition a connected component.

To show that connected components are closed, notice that \( \overline{C} \) is connected: if \( \overline{C} \) has a disconnection, then there exist disjoint open sets \(U, V \) with \( \overline{C} = U \cup V \) — but then \( C = \C \cap U \cup C \cap V \), a contradiction. By maximality of \( C \), we must have \( C = \overline{C} \).

Proof:

First note that having non-empty fibres is equivalent to being surjective. Suppose \( Y^o \subset Y \) is a connected component. If we suppose that \( \phi^{-1}(Y^o) = U \cap V \) for some disjoint opens in \(X\), then for each \( y \in Y^o \) we have \( \phi^{-1}(\{y\}) = U \cap \phi^{-1}(y) \cup V \cap \phi^{-1}(y) \) — since \( \phi^{-1}(y) \) is connected, either \( \phi^{-1}(y) \subset U \) or \( \phi^{-1}(y) \subset V \). Therefore, \( Y^0 = Y_1 \cup Y_2\) for some closed disjoint \(Y_1, Y_2 \) with \( \phi^{-1}(Y_1) = U \) and \( \phi^{-1}(Y_2) = V \). However, as \( \phi \) is open and surjective and \( Y^o \) connected we must have \( V = \emptyset\) or \( U = \emptyset \).

Proof:

The hint effectively proves the claim, but to make the forward direction slightly more rigorous note that we may define our section \( e \in \Gamma(X, \OO_X) \) affine-locally (since they are determined at the level of stalks) — thus, we may assume \( X = \spec A \coprod \spec B = \spec (A \times B) \). Then the obvious choice of section here is \( e = (0, 1)\) which is of course not the identity nor zero section.

Proof:

Following the hint, we may reduce to the case that \( A, B \) are finite type over \( k \) by what Vakil calls the tensor-finiteness trick. Then for each prime \( \pp \in B \) we have that \( \kappa(B/\pp) \) is a separable finite extension of \( k \). Then by the previous proposition each \( \spec (A \otimes_k \kappa(B/\pp)) \) is irreducible and the result follows. An alternative approach may be obtained by reading through V.17 in Bourbaki.

Proof:

The forward direction is trivial as we may take the singleton cover \( \{X\} \).

Conversely suppose that \( \{U_i\} \) is a mutually non-disjoint irreducible open cover, and recall that irreducibility of a topological space can equivalently be defined as any two nonempty opens have nonempty intersection (this is an easy one-line set theory exercise). Now let \( V, W \) be non-empty open subsets of \(X\); since \( \{ U_i \} \) covers \(X\) we may find some \( j, k \) with \( U_j \cap V \neq \emptyset \) and \( U_k \cap W \neq \emptyset \). Now by assumption we have that \( U_j \cap U_k \neq \emptyset \), and we may consider both \( U_j \cap U_k \) and \( U_j \cap V \) as open subsets of \( U_j \). Since \( U_j \) is assumed to be irreducible, this implies that \( U_j \cap U_k \cap V \neq \emptyset\) i.e. that \( V \) also intersects \( U_k \). Thus \( V \cap U_k \) and \( W \cap U_k \) are non-empty open subsets of \( U_k \), so as \( U_k \) is irreducible we must have \( ( U_k \cap V) \cap (U_k \cap W) = U_k \cap V \cap W \neq \emptyset \Rightarrow \) \( V \cap W \neq \emptyset \).

Proof:

The proof should follow fairly closedly to Exercise 9.4.E: suppose \( x = \sum_i a_i \otimes b_i \) and \( y = \sum_j a_j^\prime \otimes b_j^\prime \) with \( xy = 0 \) in \( A \otimes_k B \). We may assume without loss of generality that \( \{ b_i \}, \{ b_j^\prime \} \) are both linearly independent sets and that \( a_1, a_1^\prime \neq 0 \) so that \( D(a_1a_1^\prime) \neq \emptyset\) — as before, this implies that we can find some closed point \( [\mm] \in D(a_1a_1^\prime) \). By the weak Nullstellensatz, \( K = A/\mm \) is a finite extension of \( k \) — since \( B \) is geometrically integral,

is an integral domain. The rest of the proof now follows the same way.

Proof:

Similar to our previous proofs, suppose to the contrary that there exists some nonzero \( x = \sum_{i} a_i \otimes b_i \) and \( y = \sum_j a^\prime_j \otimes b^\prime_j\) with \( xy = 0 \). Again, we may suppose that \( \{b_i\}, \{b^\prime_j\} \) are linearly independent. Since showing that \( A \otimes_{\overline{k}} B \) is a domain is equivalent to showing that \( A^\prime \otimes_{\overline{k}} B^\prime \) is a domain for every finitely generated subalgebra. Thus, we may assume without loss of generality that \( A \) is finitely generated by the \( a_i, a_j^\prime \). Now we can find some maximal ideal \( \mm \) not containing \( a_1, a_1^\prime \) (if \( a_1, a_1^\prime \) are contained in every maximal ideal then they are contained in the nilradical and thus 0 since \(A\) is a domain). By the Nullstellensatz, we have that \( A / \mm = \overline{k} \) so that

Similar to Exercise 9.4.E this will imply that after reducing modulo \( \mm \) we have that \( a_1 \) or \( a_1^\prime \) is \( 0 \) — a contradiction.

Proof:

Letting \( P \) denote the class of universally injective objects, we first wish to show that \( P\) is stable under composition. Suppose \( f : X \to Y \) and \( g : Y \to Z \) are universally injective and \( \pi : W \to Z \) is arbitrary. Since \( g \) is universally injective, the canonical map \( \widetilde{g} : W \times_Z Y \to Y \) is injective. Letting \( \rho : W \times_Z Y \to Y \) denote the other projection, we know that sine \( f \) is universally injective, the canonical map \( \widetilde{f} : W \times_Z X \to W \times_Z Y \) is injective. Since the composition of injective functions is injective, we have that \( g \circ f \) is universally injective.

It is immediate from the definition that universally injective objects are stable under base change, as the base change of a base change is just another base change. By products I believe Vakil means our usual fibre products, which should follow from base change.

To show the notion is local on the target, we simply notice that the property of a map of sets being injective can be checked on any open cover. Indeed, if we let \( f : X \to Y \) be universally injective, \( \{ U_i \} \) an affine open cover of \( Y \), and \( \pi : Z \to Y \) arbitrary, we may cover \( \pi^{-1}(U_i) \) with affine open \( V_{ij} \). Letting \( \widetilde{f} : Z \times_Y X \to Z \) denote the pullback of \( f \), its an easy set-theoretic exercise to see that \( \widetilde{g} \) is injective iff it is injective on all \( V_{ij} \).

Proof:

Consider the transition maps \( D(x_i) \to D(x_{i^\prime}) \) and \( D(y_j) \to D(y_{j^\prime}) \) given by multiplication by \( x_{i/i^\prime} = x^{i^\prime / i}\) and \( y_{j/j^\prime} = y_{j^\prime/j} \), respectively. Then applying the transition functions followed by the Segre morphism, we obtain

However, since the transition map \( D(z_{ij}) \to D(z_{i^\prime j^\prime}) \) is given by multiplication by \( z_{ij/i^\prime j^\prime} = z_{i^\prime j^\prime / ij}^{-1} \), we get a commutative diagram

which shows that the morphisms indeed glue.

Proof:

Let \( I \) denote the ideal generated by \( 2 \times 2 \) minors of

If we let \( \phi : A[a_{00}, \dots, a_{mn}] \to A[x_0, \dots, x_m] \otimes_A A[y_0 \dots, y_n] \) denote the ring map \( a_{ij} \mapsto x_i y_j \) as mentioned in the problem statement (corresponding to the Segre morphism on some affine local component), then showing \( \ker \phi = I \) will prove the claim.

For forward inclusion, if \( a_{ij}a_{k\ell} - a_{i\ell}a_{kj} \) denotes any minor of our matrix, then clearly we have \( \phi(a_{ij}a_{k\ell} - a_{i\ell}a_{kj}) = x_iy_j x_ky_{\ell} - a_iy_{\ell}x_ky_j = 0 \) so \( I \subseteq \ker \phi \).

For the reverse direction, let \( f \in \ker \phi \). Notice that mod \(I\), \( a_{i_0 j_0}\dots a_{i_r j_r} = a_{i_0j_{\sigma(0)}} \dots a_{i_r j_{\sigma(r)}} \) for any permutation \( \sigma \), since by definition this holds on the transpositions. Thus, every monomial \( a_{i_0 j_0} \dots a_{i_r j_r} \) in \( A[a_{00}, \dots, a_{mn}] / I \) may be written as a monomial with non-decreasing indices in both the \( i\) and \(j\) simultaneously. If we let \( B \) denote the submodule of \( A[a_{00}, \dots, a_{mn}] \) generated by monomials with increasing indices, then we have just showed that \( A[a_{00}, \dots, a_{mn}] = B + I\) (since reducing modulo \( I \) gives equality). Now it is easy to see that \( \phi \) is injective on \( B \), since each monomial is uniquely determined by the non-decreasing indices. By writing \( f = b + g \) for \( b \in B \) and \( g \in I \), we have that

Since \( I \subset \ker \phi \) from the forward inclusion, \( \phi(g) = 0\) and thus \( \phi(b) = 0 \). However, since \( \phi\vert_B \) is injective this implies \(b = 0\), or in other words \( f = g \in I \). Since \( f \in \ker \phi \) was arbitrary, this gives us \( \ker \phi \subset I \) as desired.

Proof:

We first explain what happens in a non-canonical manner (with fixed basis). Since \(\P V\) is only defined in 4.5.12 for finite-dimensional \(k\)-vector spaces, we will assume here that both \(V\) and \(W\) are finite dimensional (which will give us \( (V \otimes W)^\vee \cong V^\vee \otimes W^\vee \) in theis case). If we assume \( V \) has basis \( v_1, \dots, v_m \) and corresponding dual basis \( v^1, \dots, v^m \) (use superscripts for ease of notation), and \( W^\vee \) has basis \( w^1, \dots, w^n \) then clearly \( V^\vee \otimes W^\vee \) has basis \( v^i \otimes w^j \). Since \( \sym^\bullet V^\vee \cong k[v^1, \dots, v^m] \), \( \sym^\bullet W^\vee \cong k[w^1, \dots, w^n]\) and \( \sym^\bullet V^\vee \otimes W^\vee \cong k[v^1 \otimes w^1, \dots, v^m \otimes w^m] \), it becomes obvious (when the basis is fixed) that the map

is really the same as the \( k \)-algebra morphism \( z_{ij} \mapsto x_iy_j \) by associating \( v^i \otimes w^j \) with \( z_{ij} \). Since the former map is precisely the definition of the Segre morphism, this shows that when a basis is fixed the two constructions are one in the same.

In a more canonical manner, we may still use the fact that \( \sym^\bullet V^\vee \) is always generated in degree one, regardless of bases chosen. By looking affine locally at \( \spec ((\sym^\bullet V^\vee \otimes W^\vee)_{v^i \otimes w_j})_0 \), any element is of the form \( (f\otimes g) / (v^i \otimes w^j)^r \) where \( \deg f + \deg g = r \). If we assume without loss of generality that \( \deg g \geq \deg f \), then we get a map

given by \( (f \otimes g) / (v^i \otimes w^j)^r \mapsto \frac{ f (v^i)^{\deg g - \deg f} }{(v^i)^{\deg g}} \otimes \frac{ g }{ (w^j)^\deg } \). Since this is well-defined and clearly surjective on the pure tensors, we may extend it linearly; by the universal property of the coproduct, this map factors through \( \bigoplus_{i=0}^\infty ((\sym^{\deg g} V^\vee)_{v^i})_0 \otimes ((\sym^{\deg g} W^\vee)_{w^j})_0 \) to give us the desired surjection and thus embedding.

Proof:

By Exercise 6.4.G we know that since \( S_\bullet \), \( T_\bullet \) are finitely generated we may assume withoult loss of generality that they are generated in degree one, say by \( s_1, \dots, s_m \) and \( t_1, \dots, t_n \). Then the argument is the exact same as the previous exercise.

Proof:

Since \(A\) is a domain, \( \spec A \) only has one irreducible component and thus \( \nu \) is a bijection of irreducible components. To see that \( \nu \) is a dominant morphism, recall from Exercise 6.5.A \( \nu \) is dominant if and only if it sends the generic point of \( \spec \widetilde{A} \) to the generic point of \( \spec A \). As \( \nu \) is induced from the ring map \( \iota : A \hookrightarrow \widetilde{A} \) we thave that this holds precisely when \( \ker \iota \subset \mathfrak{N}(A) \). Since \( \iota \) is clearly an injective ring morphism, \( \nu \) is dominant.

To see that it satisfies a universal property, suppose \( \mu : X \to \spec A \) is a dominant morphism with \( X \) normal that induces a bijection on irreducible components (i.e. \(X\) is irreducible). In particular on every affine open subset \( \spec B \subset X \) we get an injective ring morphism \(f : A \to B \) with \( B \) integrally closed over \(A\). We claim that there is an obvious injective ring morphism \( \widetilde{f} : \widetilde{A} \to B \): indeed any element \( a \in A \subset \widetilde{A} \) is necessarily sent to \( f(a) \) so that \( \widetilde{f}\vert_A = f \). More generally, if \( \alpha \) is any element of \( \widetilde{A} \) we know that it is necessarily the root of some \( a_n x^n + \dots + a_0 = 0 \) with coefficients in \( A \). Since \( B \) is integrally closed in \(A\), \( f(a_n) x^n + \dots + f(0) = 0 \) has a solution, say \( \beta \in B\). Then we define \( \widetilde{f}(\alpha) = \beta \). Since \( f \) and \( \iota \) are injective, it is easy to see that \( \widetilde{f} \) is as well — moreover, it uniquely depends on \(f\) alone. In particular this gives us the desired unique dominant morphism \( X \to \spec \widetilde{A} \).

Proof:

The idea, much like the fibred product, is to build affine locally and show that the constructions glue using Exercise 4.4.A (i.e. the cocycle condition) — the universal property proved in the exercise above will be crucial. Suppose \( \spec A_i \) is an open cover of our integral scheme \( X \). If we suppose \( \spec A_i \cap \spec A_j \) is covered by open sets of the form \( A_{ij} = D(f_j) \cong D(g_i) = A_{ji} \) for \( f_j \in A_i \) and \( g_i \in A_j \). Letting \( \widetilde{A_i} \) denote the integral closure as before, it is a standard fact in commutative algebra (see Atiyah-MacDonald 5.12 for proof) that \( (\widetilde{A_i})_{f_j} \) is the integral closure of \( (A_i)_{f_j} \) — in other words, it makes sense to think of \( \spec \widetilde{A_{ij}} \) as an open subset of \( \spec \widetilde{A_i} \). Then restricting to intersections, we get maps

where the vertical arrows are our natural injections. By the universal property from the previous exercise, the \( \widetilde{\phi} \)'s are the unique maps making each square commute. As the structure sheaf on \( X \) necessarily satisfies the cocycle condition, the \( A_{ij} \) must as well so we get that \( \phi_{jk}\vert_{ A_{ji} \cap A_{jk} } \circ \phi_{ij}\vert_{A_{ij} \cap A_{ik}} = \phi_{ik}\vert_{A_{ij} \cap A_{ik}} \). Then this uniquely lifts so that the \( \widetilde{\phi_{ij}} \) satisfies the same cocycle condition, and thus uniquely defines a scheme \( \widetilde{X} \).

Proof:

Since the property of a morphism being integral is affine local (as described in section 7.2), the first claim is immediate. To see that such a morphism is surjective, we simply follow the hint and work affine-locally. For any point \( p \in X \) and affine neighborhood \( \spec B \) with \( p = [\pp] \), let \( \iota : B \hookrightarrow \widetilde{B} \) denote the integral extension. By the Lying over theorem, there exists some prime ideal \( \qq \subset \widetilde{B} \) such that \( \qq \cap B = \iota^{-1}(\qq) = \pp \). If we let \(q\) denote the point \( [\qq] \), this tells us that the induced morphism \( \widetilde{X} \to X \) sends \( q \mapsto p \).

Proof:

Similar to the approach of Exercise 9.7.A and Exercise 9.7.B, we may define such a property affine locally first and show that the properties glue together nicely. By assumption, we know affine locally that \( \spec A \subset X \) has finitely many irreducible components, call them \( Y_i = \spec A / J_i \) for some ideals \( J_i \subset A \). If we let \( \widetilde{Y_i} = \spec \widetilde{A / J_i} \) denote the normalization of each, then by a similar approach to Exercise 5.4.B we cannot have any two of the \( \widetilde{Y_i}, \widetilde{Y_j} \) overlap (for example in the case that we have two irreducible components \( V(f), V(g) \) with nonempty overlap, each stalk \( \OO_{X,p} \) would have non-zerodivisors \(f, g\)). Therefore, the natural choice is to consider the normal scheme \( \widetilde{Y} = \coprod_{i=1}^n \widetilde{Y_i} \); indeed this has one connected, irreducible component for each irreducible component of \( \spec A \), so the map \( \nu\vert_{\spec A} : \widetilde{Y} \to \spec A\) clearly gives a bijection of irreducible components. It is also easily seen to be dominant since each \( \iota_i : A/J_i \hookrightarrow \widetilde{A/J_i} \) implies \( \iota_1 \times \dots \times \iota_n : A/J_1 \times \dots \times A/J_n \to \dots \) is injective.

The fact that this satisfies a universal property simply follows from the fact that any other normal scheme \( Z \) with dominant morphism \( Z \to \spec A \) producing a bijection of irreducible componentes must indeed be a disjoint union, and thus we may apply the previous universal property on each irreducible component individually. Similar to the previous exercise, we may use this universal property to show that the cocycle condition lifts, giving us a unique global normalization \( \widetilde{X} \).

Proof:

For the sake of notation let \( A = k[x, y]/(y^2 − x^2(x + 1)) \). First notice that any element \( f(x, y) \in A\) may be factored as \( f(x, y) = (y^2 - x^2(x + 1))a(x, y) + b(x) y + c(x) \). Then it must be the case that the map \(A \to k[t]\) is injective since

and \( b(t^2 - 1) \cdot (t(t^2 - 1))\) has odd degree in \(t\) while \( c(t^2 - 1) \) has even degree in \(t\), implying both must be 0. Since localization is exact, this tells us that the induced map \( \textrm{Frac}\, (k[x, y]/(y^2 − x^2(x + 1))) \to k(t) \) is injective. However, since \( \frac{y}{x} \mapsto t \) under the induced map, it is surjectiv as well and thus an isomorphism of fraction fields.

In a similar fashion, since \( k[t] \) is a UFD it is integrally closed by Exercise 5.4.F; then by a similar universal proprty argument to that of Exercise 9.7.A we know that there is an induced injective morphism \( \widetilde{A} \hookrightarrow k[t]\). However, this map is also necessarily surjective since \( \widetilde{A} \) must contain the element \( y/x \) (as \( u^2 - (x + 1) \) is a monic polynomial with elements in \( A \) which vanished at \( y/x \) ) and therefore the previous surjection \( y/x \mapsto t \) holds.

Proof:

Similar to the tactic for the nodal curve, we argue that a cubic curve should meet a line at three points (counted with multiplicity) and most lines through the origin should meet our cuspidal cubic \( y^2 = x^3 \) with multiplicity two at the origin. Therefore, by setting \( t = y/x \) we get that \( t^2 = y^2/x^2 = x^3/x^2 = x \). Similarly, \( t = y/x \) implies that \( y = tx = t^3 \). Thus, we get a map \( A = k[x, y] / (y^2 - x^3) \to k[t] \) given by \( (x, y) \mapto (t^2, t^3) \). Using a similar argument to before, it is easy to see that this map is injective since any element \( f \in k[x, y] / (y^2 - x^3) \) may be factored \( f(x, y) = (y^2 - x^3)a(x, y) + b(x, y)y + c(x, y) \) — after applying the map we see that the terms \( b(x, y)y \) and \( c(x, y) \) must have different degrees in \( t \). Again, we must have that the integral closure \( \widetilde{A} \) contains \( y / x \) as it is a root of the monic polynomial \( u^2 - x \), and thus the map \( \widetilde{A} \to k[t] \) is an isomorphism. That is to say we may describe the integral closure of \( \spec A \) as \( \spec k[t^2, t^3] \).

Proof:

The approach becomes slightly different, since a quartic meets a line in 4 points counted by multiplicity; any line through the origin intersects the tacnode \( \spec k[x, y] / (y^2 - x^4) \) at the origin with multiplicity two, so there must be two other points (contrary to the previous two exercises). Notice however, that the tacnode itself is not irreducible as \( \operatorname{char}\, k \neq 2 \) implies \( (y^2 - x^4) = (y - x^2)(y + x^2) \). By Exercise 9.7.D, the normalization of \( \spec k[x, y] / (y^2 - x^4) \) is the disjoint union of the normalization of each irreducible component. Since \( k[x, y] / (y - x^2) \cong k[x] \) which is a UFD, its easy to see that the normalization is simply the disjoint union \( \A^1_k \coprod \A^1_k \).

Proof:

We use a similar approach to that of the hint in Exercise 9.7.E: letting \( A = \Z[15 i] \), we know that since \( i \) is integral over \( A \) and \( \Z \subset \Z[15 i] \) we have \( \Z[i] \subset \widetilde{A} \). However, since \( \Z[15 i] \subset \Z[i]\) and \( \Z[i]\) is integrally closed from the hint, we have \( \widetilde{A} = \Z[15 i] \). Notice that \( \spec A \cong \spec \Z[i] \) precisely when \( 15 \) is invertible, i.e. away from \( [(3)], [(5)] \).

Proof:

As usual, we begin with the construction affine-locally and then glue. Much as we say \( A \) is normal if it is integrally closed in \( K(A) \), for any field extension \( K(A) \hookrightarrow L \) we say that \( A \) is normal over \( L \) if it is integrally closed over \( L \). Much as the set of elements of \( K(A) \) that are integral over \( K(A) \) forms a subring \(\widetilde{A} \subset K(A) \), we may define the integral closure of \(A\) in \( L \) by an identical argument. In particular, notice that every element of \( L \) is the root of a polynomial with coefficients in \( K(A) \) — by clearing denominators, we have that every element of \( L \) is the root of a polynomial with coefficients in \( A \). Suppose \( \alpha \in L \) is the root of \( a_n x^n + \dots + a_0 \) so that \( a_n \alpha^n + \dots + a_0 = 0\). Then \( \alpha \) is integral over \( A[1/a_n] \) and by multiplying through by \( a_n^{n-1} \) we can see that \( a_n \alpha \) is integral over \( A \). Thus, if we let \( (\widetilde{A})_L \) denote the integral closure of \( A \) in \( L \), it follows that \( K((\widetilde{A})_L) = L \). Therefore, the injection \( A \hookrightarrow (\widetilde{A})_L \) induces a dominant morphism \( \spec (\widetilde{A})_L \to \spec A \). Universality is the same as in Exercise 9.7.A, and we can show that the cocycle condition is satisfied in the same way as Exercise 9.7.B.

Proof:

Let \( A \) denote the integral closure of \( \Z \) in \( \Q(i) = \Q[i] \); clearly we have \( \Z \subset A \subset \Q[i] \). Now clearly \( A \) must contain \(i\) since \( x^2 + 1 \) is a monic polynomial with coefficents in \( \Z \) and roots in \( \Q[i] \). In fact, this is precisely the integral closure of \( \Z \) in \( \Q[i] \).

Proof:

Any element of \( \Q[\sqrt{n}] \) may be written \( a + b \sqrt{n} \) for some \( a, b \in \Q \); for such an element to be an algebraic integer, it must be the root of a polynomial with integer coefficients lying in \( \Q[\sqrt{n}] \). In other words, \( \Z[\sqrt{n}] \). When \( n \equiv 1 \mod 4 \) we should obtain \( \Z[ \frac{1 + \sqrt{n}}{2}] \), when \( n \equiv 2 \mod 4 \) we obtain the same result as \( n \equiv 3 \mod 4 \) and lastly when \( n \equiv 0 \mod 4 \) we must reduce since \( \Q(\sqrt{d}) = \Q( \sqrt{ a^2 d } ) \) for any integer \( a \).

Proof:

1. Like part (b), the intention here is for the field extension to be \( k(x, y) \) not \( k(y) \). Now any element \(f \in k[x, y] / (x^2 + x - y^2) = A \) can be written of the form \( f(x, y) = (x^2 + x - y^2)a(x, y) + b(x)\cdot y + c(x) \). Modding out by \( (x^2 + x - y^2) \), it is easy to see that any element of \( A \) is integral over \( k[x] \) in \( k(x, y) / (x^2 + x - y^2) = K(A) \) and thus \( k[x, y] / (x^2 + x - y^2) \subset (\widetilde{k[x]})_{K(A)} \). However, since \( A \) itself is normal by Exercise 5.4.H(a) (since \( x^2 + x = x(x +1) \) has no repeated factors in \( k[x]\)) we have that \( A \) is in fact the normalization.

2. Recall from §4.4.6 (likely already well-understood), the projective line is taken by gluing the subschemes \( D(t) \cong \spec k[t, \frac{1}{t}] = k[t]_t \) and \( D(u) \cong \spec k[u, \frac{1}{u}] = k[u]_u \) and gluing via the isomorphism of rings \( u \mapsto \frac{1}{t} \). Now since localization and integral closure commute, localization and normalization must commute as well (over affines) so that we get a diagram

   $$ \begin{CD} \Big( k[u, z] / (u^2 + u - z^2)\Big)_u @. \Big( k[t, y] / (t^2 + t - y^2) \Big)_t \\ @AAA @AAA \\ k[u]_u @>>{u \mapsto 1/t}> k[t]_t \end{CD} $$

   It is worth noting that while \( y \) satisfies \( y^{-1} = \frac{y}{t} - \frac{y}{t+1} \) in the field extension \( k(t,y) / (t^2 + t - y^2) \), we do not get this in the integral closure since \( 1 + t \) is not invertible. However, since \( \frac{y}{x} \) is a root of \( \zeta^2 - (1 + \frac{1}{t}) \) in \( k[t, \frac{1}{t}, \zeta] \), we get that \( \frac{y}{x} \) is in the integral closure of \( k[t]_t \) in the field extension. Since \( \frac{1}{t} \) is \( \frac{y^2}{t^2} - 1 \), we may simply express the integral closure as \( k[t, \frac{y}{t}] / (t^2 + t - y^2)\). To make the above diagram commute, we must still have \( u \mapsto \frac{1}{t} \); therefore, it suffices to determine the image of \( \frac{z}{u} \). Some algebra shows that we must have \( \frac{z^2}{u^2} \) maps to \( t^2 ( \frac{1}{t^2} + \frac{1}{t} ) = t \cdot \frac{y^2}{t^2} \). However this is not a transition function since going back to \( k[u, z/u] / (u^2 + u - z^2) \) is not the identity — so I'm a bit stuck here.

Proof:

1. First note that \( \nu \) is affine since it is, by construction, locally of the form \( \spec \widetilde{A} \to \spec A \). Since finite morphisms may be checked affine locally and \( X \) is quasicompact (finite type = quasicompact + locally finite type), we may simply assume \( \nu \) is of the form \( \nu : \spec \widetilde{A} \to \spec A \). But then as \( K(A) \) is trivially a finite extension over itself and \( X \) is finite type over a Noetherian ring making it noetherian, we have that \( \widetilde{A} \) is finitely-generated as an \(A\)-module. Therefore, \( \nu \) is finite.

2. For the first part, normality implies integral by definition so Theorem 9.3.7 still applies by looking affine locally. The second statement is a slight adjustment of part (a) to the case where we have a finite extension that is not necessarily the trivial extension \( K(A) / K(A) \); by taking the separable closure of \( K(A) \) in \( L \) we obtain the result.

Proof:

This follows from the next exercise together with Proposition 6.5.5.

Proof:

By definition, we know that the normalization map is a dominant morphism which gives a bijection of irreducible components. In particular, this is equivalent to the fact that on all open affines \( \spec A \subset X \), the morphism \(\iota : A \hookrightarrow \widetilde{A} \) is an injection. By restricting to the image of \( \iota \) (which is a dense open subset), we obtain an isomorphism of rings which in turn induces an isomorphism of affine schemes. By gluing, we obtain the intended isomorphism.

Proof:

Recall that dominant morphisms induce injections (in the opposite direction) of function fields so that birational maps give isomorphisms of function fields. It suffices to check that \( \rho \) extends to an isomorphism on affine open sets; let \( \spec A \subset X \) and consider \( \rho^{-1}(\spec A) = \spec B \) (since finite morphisms are by definition affine). Since \(X\) is normal, we must have that \(B\) is integral over \(A\). Since \( K(A) \) is integral over \( A \) and \( K(A) = K(B) \), we have that \( B \) is integrally closed and thus \( B = A \)