Solutions to The Rising Sea (Vakil)

Proof:

We have already proved the case of \( n = 0 \), so consider \( n > 0 \). We still have that \( \OO(n) \) is trivial on \( U_0 = D(x_0) = \spec k[x_{1/0}] \) and \( U_1 = D(x_1) = \spec k[x_{0/1}] \), so that the data of a global section is a polynomial \( f(x_{1/0}) \in k[x_{1/0}] \) and a polynomial \( g(x_{0/1}) \in k[x_{0/1}] \) such that the two agree up to transition function. On \( U_1 \), this means

If we write \( f(t) = a_r t^r + \dots + a_1 t + a_0 \) and \( g(t) = b_s t^s + \dots + b_1 t + b_0 \) then the above equation tells us

As \( g \) is a polynomial with positive degree terms, all of our negative degree coefficients on the left must vanish. Since we must also have our top degree terms are equal, \( n = s \) and thus our section is uniquely determined by \( a_i = b_{n - i} \) for \( 0 \leq i \leq n \) giving us \( \Gamma(\P^1, \OO(n)) = n + 1 \)

In the case that \( n \) is strictly negative, \( (\star) \) tells us that the function on the left with only strictly negative degree terms is equal to the function on the right with only non-negative degree terms which is absurd — thus there can be no nontrivial global sections.

Proof:

Using the previous exercise and proceeding by contrapositive, if \( \OO(m) \cong \OO(n) \) for both \( m, n \geq 0 \) then they must be isomorphic on all open sets and thus at the level of global sections, so that \( m + 1 = n +1 \). If one of \( m \) or \( n \) is nonnegative and the other is negative, the two clearly cannot be isomorphic again by the previous exercise. In the case that both \( m, n \) are strictly negative with \( m \leq n \), consider tensoring by \( \OO(-m) \) and sheafifying: \( (\OO(m) \otimes \OO(-m))_{sh} = \OO \) which has 1 dimension of global sections, while \( \OO(n) \otimes \OO(-m) = \OO(n - m)\) which has \( n - m + 1 \) dimensions of global sections. since \( n \neq m\) this implies the two cannot be isomorphic.

Proof:

The idea is similar to before, just with more tedious combinatorics. Let \( U_i = D(x_i) = \spec k[x_{0/i}, \dots, x_{m/i}] / (x_{i/i} - 1) \) continue to denote our affine open cover. Then the data of a global section is the data of polynomials \( f_i (x_{0/i}, \dots, x_{m/i}) \) with

In general we have the formula \( x_{r/i} \cdot x_{i/j} = x_{r/j} \), but since \( x_{r/r} = 1 \) on our open intersections we should always have one variable inverting; in other words

Suppose our polynomial \( f_i \) has highest degree \(d_i\) and \( f_j \) has highest degree \(d_j\). Moreover, if we write

and

(you should only be using \( m \) index variables since we are on an affine patch). Then \( \star \) above tells us that the right hand side should simplify to

Since we require this be a polynomial with no negative degree terms, we see that we must have \( c_{1, \alpha_1, \dots, \alpha_m} = 0 \) when \( \sum_k \alpha_k > n\). Therefore, we restrict our attention to \( \sum \alpha_k \leq n \) and in particular each exponent \( \alpha_k \leq n \). Since the two sides are equal, we must have \( d_i = d_j \) and thus the \( c_{i, \alpha_1, \dots, \alpha_m} \) are a permutation of the \( c_{j, \alpha_1', \dots, \alpha_m'} \) and our sections are uniquely determined by the \( c_{i, \alpha_1, \dots, \alpha_m} \) corresponding to their respective monomials. Thus, \( \Gamma( \P^m_k, \OO(n) ) \) should have dimension equal to the number of monomials with degree less than \(n\) which is precisely \( {n + m \choose m} \) (as explained in Exercise 8.2.K)

Proof:

Let \( \mathscr{L} \) be an invertible sheaf on \( \P^1_k \). By Exercise 13.2.C we know that since \( \mathscr{L} \) is locally free, it is trivial on both distinguished open subsets \( D(x_0) \) and \( D(x_1) \). Thus, \( \mathscr{L} \) is uniquely determined by its transition function on \( D(x_0x_1) = \spec k[x_0, x_1]_{x_0x_1} = \spec k[x_{0/1}, x_{1/0}] = \spec [y, y^{-1}] \). Any such function can be written as \( \frac{p(y)}{y^n} \); by possibly factoring out some power of \( y \) from \( p(y) \), we may assume without loss of generality that \( p(y) \) has nonzero constant term. Now in order for this to be invertible, we must have \( p(y) \) is invertible in \( k[y, y^{-1}] \) which implies that \( p(y) \) is in fact constant \( a \in k \). Thus, our transition function is of the form \(a\left( \frac{x_0}{x_1} \right)^n \) so that scaling sections over \( D(x_0) \) and \( D(x_1) \) by \(a\) we get \( \mathscr{L} = \OO(n) \).

Proof:

1. At any closed point \( [(x - a)] \) we have that the local ring \( k[x]_{(x - a)} \) clearly has maximal ideal generated by \( (x - a) \). In particular, when \( (x - a) \neq (x + 1) \) or \( (x) \) we get that \( x^3 / (x+1) \) is invertible and thus has valuation \( 0 \). However, by definition of the usual valuation over the local ring \( k[x]_{(x-a)} \), when \( (x - a) = (x) \) we have that \( 1/ (x +1) \) is invertible so that the valuation at \( (x) \) is 3. A similar argument show that \( x^3 \) is invertible when \( (x+1) \) is our closed point so that the valuation is \( -1 \). Thus, \( \div x^3 / (x+1) = 3[(x)] - [(x+1)] \) as expected.

2. By the discussion in §14.1, we know that global sections of \( \P^1 \) correspond to homogeneous polynomials of degree 1 in \( x \) and \(y \) — now \( x^2 / (x + y) \) is not a global section of course because it is not defined at the point \( [1 , -1] \); however, on the dense open set \( \spec k[x, y]_{(x + y)} \subset \proj k[x, y]\) this gives us a well defined rational function of degree (in the sense of our graded ring) \( 2 - 1 = 1 \). (This idea is made clear by fiat in Hartshorne where if \( \OO_{\proj S} \) corresponds to some graded ring, one defines \( \OO_{\proj S}(n) \) to be the sheaf associated to \( S(n) \)). By a similar idea to part (a), we should obtain \( \div x^2 / (x + y) = 2 [1 , 0] - [1 , -1] \).

Proof:

More generally if \( X \) has irreducible components \( X_i \) with generic points \( \eta_i \), we may define \( \OO_X(D) \) by

Proof:

Following the hint, we may assume without loss of generality that \( X = \spec A \) since the construction is affine local, and fix \( f \in A \). Then by definition we have

Now for any nonzero \( s \in K(X) \) we have \( \div\vert_{D(f)} s = \sum_{Y \cap D(f) \neq \emptyset} \textrm{val}_{Y \cap D(f)}(s) [Y \cap D(f)] \). Then on each \( Y \cap D(f) \), we certainly have that \( f \) is an invertible section as \( \frac{1}{f} \) has no poles and thus we obtain a morphism \( \Gamma(\spec A, \OO_X(D))_f \to \Gamma(\spec A_f, \OO_X(D)) \) (as usual we should always have this map). To produce a map in the other direction, suppose that \( s \in \Gamma(\spec A_f, \OO_X(D)) \). By definition, \( \div \vert_{\spec A_f}\,s + D\vert_{\spec A_f} \geq 0 \) so that \( \div\,s \) can only have negative coefficients on \( V(f) \). But then

When \( Y \cap V(f) \neq 0 \) we have \( \textrm{val}_Y(f) \) is strictly positive, so by choosing \( N \) large enough we have \( \div (f^N s) + D \geq 0 \) on all of \( \spec A \) giving the desired isomorphism.

Proof:

1. If \( \frac{f(x)}{g(x)} \in K(\A^1_k) = k(x) \) is a global section, then \( \div \frac{f(x)}{g(x)} \) is of the form

   $$ a_1[p_1] + \dots + a_n[p_n] - b_1[q_1] - \dots - b_n[q_n] $$

   where \( a_i, b_i \geq 0 \) and the \( p_i, q_i \) are closed points on the affine line (since \( k[x] \) is a PID) corresponding to the zeros of \( f(x) \) and \( g(x) \), respectively. Since \( \div \frac{f(x)}{g(x)} −2[(x)] + [(x − 1)] + [(x − 2)] \geq 0 \), we must necessarily have that \( a_i[p_i] = r[(x)] \) for some \( r \geq 2 \) and either (i) all \( b_j = 0 \) or (ii) some \( [q_j] = [(x - 1)] \) or \( [q_j] = [(x- 2)] \), in which case we may have the corresponding \( b_j = -1 \). In the first case \( g(x) \) is constant and \( x^2 \) is clearly a \( k[x] \)-multiple of \( x^2 / (x - 1)(x- 2) \), so \( \frac{f(x)}{g(x)} \) is a \(k[x] \)-multiple of \(x^2\). In the latter case it is clearly a \( k[x] \)-multiple of \( x^2 / (x - 1)(x-2) \).

2. By part (a), multiplication by \( \frac{(x - 1)(x - 2)}{x^2} \) is an isomorphism.

Proof:

1. We follow the idea of this StackExchange answer which seems to pin the idea that Ravi is talking about with the base.

   For the first part of the hint, we wish to show that those open sets \( U \) with \( \OO(\div s)\vert_U \cong \OO_U \) form a basis for the Zariski topology. Now since \( \mathscr{L} \) is a line bundle, it is trivial on a sufficently small affine open cover, so we may simply take the usual base of the Zariski topology subordinate to our cover — therefore, if we are able to show that \( \OO(\div s) \) is trivial on any open subset \( U \) where \( \mathscr{L} \) is trivial then we are done.

   If we suppose \( U \) is a trivializing subset of \( \mathscr{L} \) with isomorphism \( \phi_U : \mathscr{L}\vert_U \to \OO_X\vert_U \), then \( f = \phi_U(s) \) is a rational section with \( \div s \vert_U = \div f \vert_U \). Then we obtain a map \( \OO_U(\div s) \to \OO_U \) given by multiplication by \( f \) which is well defined since for any \( t \in \OO_U(\div s) \)

   $$ \div (f \cdot t)\vert_U = \div f\vert_U + \div t\vert_U = \div s\vert_U + \div t\vert_U \geq 0 $$

   The key idea is that since \( X \) is normal, the divisor of \( f t \) being effective implies that \( ft \) is regular. The same idea shows that division by \(f\) is well-defined and gives the desired inverse (as indicated in the hint).

   As described in §13.3, this uniquely extends to define a morphism of sheaves over all of \(X\).

2. As indicated in the hint, this is already true by construction. Let \( 1 \in K(X)^\times \) denote the unit section on the dense open subset \( X \backslash \textrm{Supp}(\div s) \); as any open subset \( U \) necessarily intersects this dense open (and in particular the ones in the base described above), our isomorphisms \( \phi_U \) above are well-defined and certainly map \( 1 \mapsto s \) by construction.

Proof:

By the discussion of §14.1, we may identify \( \OO(m) = \OO(1)^{\otimes m} \) so that \( s^m \in \OO(m) \) corresponds to \( x_0^m \) (recall from said discussion that we may think of \( \OO(m) \) as the homogeneous degree \(m\) polynomials). By the previous exercise, we know that \( \OO(D) \cong \OO(1) \) with \( \sigma : 1 \mapsto s \). Taking the \( m^{th} \) tensor product, we see that \( \sigma \otimes \dots \otimes \sigma : \OO(D)^{\otimes m} \cong \OO(m) \) takes the canonical section \(1\) to \( s^s \). Since \( \P^n_k \) is normal, Proposition 14.2.1 tells us that \( \div \) is an injective homomorphism (taking tensor products to sums) so that \( mD \) uniquely corresponds to \( (\OO(m), s^{\otimes m}) \). Applying the previous exercise again we get \( \OO(mD) \cong \OO(m) \).

Proof:

1. Following this approach given on StackExchange, we may use the fact that \( \OO_X(D) \) is invertible to find a trivialization \( \phi_U : \OO_U \to \OO_X(D)\vert_U \) on some open subset \( U \subset X \). Since \( \phi_U \) is an isomorphism, \( \OO_X(D)\vert_U \) is generated by \( \tau = \phi(1) \). From Exercise 14.2.E we know that this map should locally look like multiplication by \( \tau \), so that the inverse image of \( 1 \in \OO_X(D)\vert_U \) should be \( \tau^{-1} \). We wish to show that \( \div_{\OO_U} \tau + D\vert_U = 0\); if we suppose to the contrary that the support is non-empty, then we may find some principal divisor \( W = \div f \) with a negative coefficient (corresponding to a pole of \(f\)) so that \( f \) is not a regular function on \( U \) — assuming without loss of generality (by restricting to a smaller open subset) \(f \) has no other poles in \( U \), \( \div (f \cdot\tau) + D\vert_U \geq 0 \) by construction, which contradicts the fact that \( \tau \) generates \( \OO_X(D)\vert_U \). Thus, we have shown that \( \div_{\OO_U}(\tau) = -D\vert_U \) so that \( \div_{\OO_X(D)}(1) = \div_{\OO_U}(\tau^{-1}) = D\vert_U \)

2. Using the notation of part (a) above, we simply have \( D\vert_U = \div_{\OO_U} (\tau^{-1}) \) proving the claim.

Proof:

1. Let \( D = V(x, z) \) be our codimension 1 subvariety. If \( D \) were locally principle, we could find an open subset \( U \) of the origin such that \( D\vert_U = \div s \) for some \( s \). However, by considering the stalk at the origin we may follow the same argument of that in 12.1.3 / 12.1.5 to see that \( (x ,z) \) is not a principal ideal of \( (k[x, y, z] / (xy - z^2))_{(x, y, z)} \) proving the claim.

2. The local ring \( (k[x, y, z]/(xy - z^2))_{(x, z)} \) is regular in codimension 1 and the maximal ideal corresponding to \( (x, z) \) is generated by \( z \) since \( y \) is necessarily a unit and thus \( x = z^2 / y \). But then the valuation of \(x\) is by definition \(2\) so that \( \div x = 2D \) (if one wanted to be more precise, we would have to show there are no other divisors \( D^\prime \) such that \( x \) has a nontrival order of vanishing).

Proof:

Following the hint, since \( X \) is factorial it is therefore normal so that irreducible components are connected components. Therefore, it suffices to consider only the irreducible components of \( D \) as they are all disconnected — in this vain, let \( D = [Y] \) for some \( Y \subset X \) closed subvariety regular of codimension 1. By the previous exercises, we know that \( \OO(D) \) will be invertible iff \( D \) is locally principal iff we can find some affine open cover such that on each \( U\) in the cover, \( [Y \cap U] \) is principal. In fact, from the discussion of §14.2.2, by Algebraic Hartog's lemma we have \( \OO(D) \) is isomorphic to \( \OO \) away from the support of \(D\) so we only need to consider some affine cover of \( Y \).

Fixing an arbitrary \( p \in Y \), by factorality we know that \( \OO_{X, p} \) is a UFD so that by Lemma 11.1.6 the prime \( \qq \) of height 1 corresponding to \(Y\) is principal. If we let \( f \) be an element of \( K(X) \) that restricts to a generator of \( \qq \), then as \( X \) is Noetherian \( f \) has only finitely many zeroes and poles \( y_1, \dots, y_s \). Now as \( \qq = (f) \) corresponds to \( Y \) in the local ring \( \OO_{X, p} \), \( [Y] \) is the only term of \( \div f \) containing \( p \) with a positive coefficent

Proof:

1. Since \( \mathscr{L} \) is invertible, on a sufficiently small affine neighborhood \( U \) there should be an isomorphism \( \phi_U : \OO\vert_U \to \mathscr{L}\vert_U \) inducing an isomorphism \( \mathscr{L}(D)\vert_U \cong \OO(D)\vert_U \); in this case the map \( \phi\vert_U \) should take rational sections \( \OO_{X, \eta} \) to rational sections \( \mathscr{L}_\eta \) (where \( \eta \) is the unique generic point of \( X \)).

2. Since \( D_2 \) is locally principle, we may restrict to some open subset \( U \) such that \( D_1\vert_U = \div s_1\vert_U \) for some rational function \( s_1 \in K(X) \). Then we consider the map \( \phi_U : \OO_U(D_1\vert_U)(\div s) \to \OO_U(D_1\vert_U + D_2\vert_U ) \) given by \( t \mapsto ts_1 \). Since \( \div t\vert_U + D_2\vert_U \geq 0 \), we have

   $$ \div s_1 t\vert_U + D_2\vert_U = \div t\vert_U + \div s_1\vert_U + D_2 \vert_U = \div t\vert_U + (D_1 + D_2)\vert_U \geq 0 $$

   (notice we may interpret rational sections of \( \OO(D_1) \) as rational functions in \( K(X) \) in the obvious way) so that the map is well defined. By a similar argument to that of Exercise 14.2.E, this is locally an isomorphism on \( U \) and thus induces an isomorphism on sheaves.

Proof:

Suppose \( f \in k[x_0, \dots, x_n] \) is a degree \( d \) polynomial and \( Y = V_+(f) \). Similar to the discussion of 14.2.9 we have an exact sequence

where the first map sends \( 1 \) to \( [Y] \); as one may expect, \( [Y] \) maps to \( \OO(d) \) under the inclusion \( \pic X \hookrightarrow \operatorname{Cl} X \). Since \( \pic \P^n \) is generated by \( \OO(1) \), we in fact get an exact sequence

where the first map is multiplication by \( d \); thus \( \pic (\P^n - Y) \cong \Z / (d) \).

Proof:

Using the Veronese map \( \nu_d : \P^n \to \P^N \), we may identify \( Y \) with the intersection of the image of \( \P^n \) with a hyperplane \( H \). But then \( \P^n - Y\) corresponds to the image of \( \P^n \) in \( \A^N \cong \P^N - H \) which is a closed subvariety and thus itself affine. It is immediate that \( D(x_i) = \P^n - H_i \); as these form the usual open cover for projective space, they also clearly form an distinguished open cover for \( \P^n - Y \). Now \( \P^n - H_i \cong \A^n \) and thus we have \( \pic \P^n - H_i = 0 \) for all \( i \). Using the excision sequence, we have

where the first map sends \(1\) to \( [Y] \). Since the last term is surrounded by trivial groups, it itself must be trivial.

Proof:

If we suppose that \( H_i \) were in fact cut out by a single equation when restricted to \( \P^n - Y\), then \( [H_i] \) would be trivial in the class group, so that by considering the excision sequence again

the usual forst map sending \( 1 \) to \( [H_i] \) would be the zero map, so that \( \operatorname{Cl}(\P^n - Y) \cong \operatorname{Cl}(\P^n - Y - H_i) \). However, this contradicts the results from the previous exercises showing that the left hand side is torsion while the right hand side is trivial.

Proof:

Following this answer by Qi Zhu, we first notice that \( k[x,y] / (x^2 + y^2 - 1) \cong (k[x, y]_{(x^2 + y^2)})_0 \) so that \( \spec \R[x, y] / (x^2 + y^2 - 1) \cong \P^1_{\R} / V_+(x^2 + y^2) \). Over \( \R \), this describes a degree \(2\) hypersurface; by Exercise 14.2.L it has non-trivial class group, so \( \spec A \) cannot be a UFD. However, when \( k = \C \) we have that \( V(x^2 + y^2) \) is the union of two hyperplanes, so that \( A \otimes_\R \C \) is a UFD.

Proof:

The isomorphism \(X − L− M \cong \A^2\) should somewhat come for free, since \( \P^1 \) minus any point is isomorphic to \( \A^1 \) (and in particular, \( \{\infty\} \) only makes sense once an affine open is chosen) and \( \A^1_k \times_k \A^1_k \cong \A^2_k \). Let \(\spec k[x, y] \), \( \spec k[x, 1/y] \), \( \spec k [1/x, y] \) and \( \spec k[1/x, 1/y] \) denote the usual affine cover of \( \P^1 \times \P^1 \)

If \( Y \) is any curve distinct from \( L \) and \( M \), we have that on any affine open subset — in particular on \( \spec k[x,y] \) — \( Y \) is defined by some polynomial \( f \in k[x, y] \). Let \( a \) be the degree of \(f\) in \(x\) and \( b \) the degree of \(f\) in \(y\). Then

so that quotienting by principal divisors we get \( [Y] = a[L] + b[M] \) giving our surjection \( \Z \oplus \Z \to \textrm{Cl}(X)\). This map must be injective since there are no rational functions \( f \in k(x, y) \) such that \( \div(f) = aL + bM \) as a divisor on \(X\) unless \(a = b = 0\).

Proof:

Since \( \pic \P^2 \cong \Z \) we can already know \( \P^2 \not\cong \P^1 \times \P^1 \). To see that the two varieties are not isomorphic, notice that \( \P^1 \times \P^1 \cong \proj k[x, y ,z , w ] / (xw - yz) \) so if we let \( \P^2 \cong \proj k[x, y, z] \). Then on \( D_x \) we have \( k[ y, z ,w ] / (w - yz) \cong k[y, z] \).

Proof:

Following the account from Hartshorne (II.6.5.2), we can use the excision exact sequence

(where the first map sends 1 to \(Z\)) to see that since the local ring of \(Z\) at its generic point is generated by \( Z \), \( x = 0 \) implies \( z^2 = 0 \) so that \( \div (x) = 2Z \). Then on \( D(x) \cap X \), we have \( y = z^2 / x \) so that \( D(x) \cap X \cong \spec k[x, x^{-1}, z] \) — since \( k[x, x^{-1}, z] \) is a UFD, \( \textrm{Cl}(X - Z) = 0 \) and thus \( \textrm{Cl}(X) \) is generated by \( [Z] \). In particular, we showed that the kernel is \( (2) \) so that \( \textrm{Cl}(X) \cong \Z / 2\Z \). From the diagram 14.2.7.1, we may think of \( \pic X \) as the group of locally principal divisors modulo the principal divisors ( in the next section we will see that this is just the same thing as Cartier divisors modulo principal divisors). By Exercise 14.2.H, we know that \( Z \) is not locally principal. As \( \pic X \hookrightarrow \textrm{Cl}(X) \) where the latter is generated by \( Z \) we obtain \( \pic X = 0 \).

Proof:

We follow the logic of Yuchen Liu's answer to this problem.

Suppose to the contrary that \( n[Z] \) is locally principal for some \( n > 1 \), and let \( U_i = \spec A_i \) be an open cover together with \( f_i \in K(A_i)^\times \) such that \( n[Z \cap U_i] = \div f_i \). Since \( n[Z \cap U_i] \) is effective, the \( f_i\) are in fact regular by Algebraic Hartog's lemma so that \( f_i \in A_i \) and thus \( Z \cap U_i \subset V(f_i) \) for all \( i \). By Krull's principal ideal theorem, every irreducible component of \( V(f_i)\) is exactly codimension 1 (i.e. Weil) so that \( \div f = n[Z \cap U_i] \) implies \( V(f_i) = Z \cap U_i \). Therefore, \( U_i \cap (X - Z) = D_{A_i}(f_i) \) is an affine scheme for all \(i\).

Now consider the Weil divisor \( Y \subset X \) cut out by \( y = z = 0 \). It is easy to see that both \( Y \) and \( Z \) are themselves isomorphic to \( \A^2 \). If \( X - Z \) were affine, then \( Y - Z \) would be a closed subvariety of an affine variety and thus affine. But \( Y - Z \) is isomorphic to the affine plane minus the origin which we know is not affine by §4.4.1. Thus, \( X - Z \) is locally affine but itself not affine, which contradicts Proposition §7.3.4 (affine-ness of a morphism is affine local on the target) in the following way: the immersion \( \phi : X - Z \hookrightarrow X \) pulls back each \( U_i \) to \( D(f_i) \) so it is affine, but \( \phi^{-1}(X) \) is not affine.

Proof:

As described in the hint, we know by Exercise 5.4.F and Lemma 11.1.6 that \( A \) is integrally closed and \( \operatorname{Cl}\spec A = 0\). For the reverse direction, Lemma 11.3.5 tells us that since \( A \) is Noetherian, it suffices to show that all codimension 1 primes are principal. Moreover, Exercise 11.1.A tells us there is an inclusion preserving bijection between prime ideals \(\pp \) of height 1 in \( A \) and codimension 1 closed subvarieties of \(\spec A\). Following the proof of Hartshorne, let \( \pp \) be a prime ideal of height 1 and let \(Y \subset \spec A\) denote the corresponding divisor. By assumption, \( Y \) is principal so we can find some \( f \in K(A)^\times \) with \( (f) = Y \) — as \( \nu_Y(f) = 1 \) this necessarily implies that \( f \in A_\pp \) with \( f \) a generator / uniformizer of the maximal ideal \( \pp A_\pp \). If \( \pp^\prime \subset A \) is any other prime ideal, then \( \nu_Y(\pp^\prime) = 0 \) so that \( f \in A_{\pp^\prime}\). By Algebraic Hartog's lemma, this implies \( f \in A \) so that \( f \in A \cap \pp A_\pp = \pp \). To see that \( f \) generates \( \pp \), let \( g \) be any other element so that \( \nu_Y(g) \geq 1 \) and \( \nu_Z(g) \geq 0\) for any other divisor \( Z \neq Y \). Then \( \nu_Z(g/f) \geq 0 \) for all Weil divisors \( Z \) so that \( g / f \in A_{\pp^\prime}\) for all \( \pp^\prime \) of height 1 and thus \( g/f \in A \) by Algebraic Hartog's lemma again — in other words, \( g \in f\cdot A \)

Proof:

Since\( (x) \) is principal we know that it is height 1 and thus corresponds to a divisor \( [(x)] \in \operatorname{Cl} \spec A \). Using the excision exact sequence above, we have an inclusion of the subgroup generated by \( [(x)] \) into \( \operatorname{Cl} \spec A \). Since \( A_x \) is assumed to be a UFD, we know by the previous exercise that \( \textrm{Cl}\spec A_x = 0 \). Thus, \( \operatorname{Cl} \spec A \) is generated by \( [(x)] \) — however, since \( [(x)] \) is clearly principal, we get that \( \operatorname{Cl} \spec A = 0 \).

By the previous exercise, if we can show that \( A \) is integrally closed then we are done. We know that \( A[1/x] \) is integrally closed as it is a UFD; now suppose that

has some root \( T \in K(A) \) for \( a_i \in A \). Since the \( a_i \) can be interpreted as elements of \(A_x\), we have by integral closedness of \(A_x\) that \( T \in A_x \), i.e. \( T = r/x^m \) for some \( r \in A \) and \( m \geq 0 \) where \( r \) and \( x^m \) are coprime. If \( m > 0 \), this implies \( r \notin (x) \) and \( r \) is a root of

which means

since \( (x) \) is prime this implies \( r \in (x) \) — a contradiction.

Proof:

By considering \( x = a \) from the previous problem, we have that \( A_a = k[a, \frac{1}{a}, b, x_1, \dots, x_n] / (ab - x^2_1 - \dots - x^2_\ell)\). Since

we may eliminate \( b \) so that the ring is isomorphic to \( k[a, \frac{1}{a}, x_1, \dots, x_n] \) which is a UFD. By the previous problem, \(A\) itself must be a UFD.

Proof:

Continuing to let \( \mathscr{I} \) denote the ideal sheaf of \( D \) with \( \mathscr{I}\vert_{\spec A} = I \), since \( I \cong A \) as \( A \)-modules we know that \( I = (a) \) for some non-zero divisor \( a \in A \) (as our isomorphism is determined by the image of the generator of \(A\), i.e. \( 1_A \)). If we suppose that there is some other nonzero divisor \( a^\prime \in A\) with \( I = (a^\prime) \), then basic algebra tells us that \( (a) = (a^\prime) \) implies \( a = ua^\prime \) for some invertible element \( u \in A \).

Proof:

By definition, if \(D\) is an effective Cartier divisor we can find a sufficiently small affine open cover \( U_i = \spec A_i \) such that \( D\vert_{U_i} = V(f_i) \) for some nonzero divisor \( f_i \) (i.e. \(D\) is locally principal and regular). In this case, the short exact sequence of sheaves

is locally given by

Now on the affine open cover \( U_i \), Important Definition 14.2.2 describes \( \OO(D) \) as the sheaf consisting of rational functions \( t \in K(X)^\times \) such that if \( t\vert_{U_i} = t_i \in A_i \), then \( t_i f_i \in A_i \) (by Algebraic Hartog's Lemma). Thus, \( \OO(D) \) is \( \OO_X \)-module which is locally generated by \( \frac{1}{f_i} \) on the cover \( U_i \), which is precisely the definition of \( \mathscr{I}_D^\vee \).

From the discussion preceding this exercise, we know that the section \( s_D \) canonically corresponds to the map \( \OO_X \to \mathscr{I}_D^\vee \). Explicitly, we know that \( \mathscr{I}_D^\vee \) is the invertible sheaf locally generated on \( U_i \) by \( \frac{1}{f_i} \) (in this case it makes sense that the support of \( s_D \) is defined away from the support of \( D \), as it is the vanishing locus of \( f_i \)). Then it should follow from Exercise 14.2.G that the divisor corresponding to \( s_D \) is itself \( D \).

Proof:

By tensoring \( 0 \to \OO_X \to \mathscr{L} \) with \( \mathscr{L}^{-1} \), we obtain a morphism \( \mathscr{H}om( \mathscr{L}, \OO ) \to \OO\) given by \( \phi \mapsto \phi \circ s \). Then the fact that \( s \) is not locally a zerodivisor is the same as saying this map is injective on stalks, i.e. injective everywhere. In particular, on any trivializing open affine \( U = \spec A \), if we let \( a \) be the image of \( s \), then \( V(s)\vert_U \) corresponds to the principle closed subscheme \( \spec A / (a) \). In particular, this tells us that \( V(s) \) is by definition an effective Cartier divisor.

For any \( f \in \Gamma(U, \OO(\textrm{div}\,s)) \) we have a map \( f \mapsto fs \) which can be easily checked to be an isomorphism.