Solutions to The Rising Sea (Vakil)

Proof:

The hint effectively gives us the equality: if \( \dim A = n \) (w.r.t Krull dimension) then we can find a sequence of nested primes of length \(n\):

By applying \( V(\cdot) \) from the bijection in Exercise 3.7.E we obtain a chain of length \( n \) of irreducible closed subvarieties, such that \( n \leq \dim \spec A \). To see that this is indeed the supremum, if we suppose to the contrary that there is a longer chain then we may simply use the bijection in the opposite direction to obtain a chain of prime ideals of length \( > n \) which would be impossible.

Proof:

We first prove the fact suggested in the hint. If we let \( U \) be an open subset of a topological space \(X\) and suppose \( Z \subset U \) is a closed irreducible subset, then by definition of the subspace topology there exists some closed \( W \subset X \) such that \( W \cap U = Z \). If \(W\) is reducible, say \( W = W_1 \cup W_2 \) then it must be the case that either \( W_1 \cap U = \emptyset \) or \( W_2 \cap U = \emptyset\) (as otherwise we could write \( (U \cap W_1) \cup (U \cap W_2) = Z \) ). By replacing \( W \) with \( W_i \), we may assume \( W \) is irreducible giving us the first direction.

In the other direction, assume \( W \subset X \) is irreducible and suppose to the contrary \( W \cap U = Z_1 \cup Z_2 \) for some \( Z_1, Z_2 \) closed in the subset topology of \( U \). Then there exist some closed \( W_1, W_2 \subset X \) such that \( W_i \cap U = Z_i \). Now \( W \cap (W_1 \cup W_2) \) is a closed subset of an irreducible set and thus irreducible. Thus, we must have \( W \cap W_1 = W \) or \( W \cap W_2 = W \) giving us either \( W \cap U = Z_1 \) or \( W \cap U = Z_2 \).

To now prove the claim of the exercise, suppose for the forward direction that \( X \) admits a cover of open affines \( \{ U_i \} \) of dimension at most \( n \) with equality achieved on some open subset, say \( U_k \). Then there exists a nested chain of closed irreducible subsets \( V_0 \subset \dots \subset V_n \) in \( U_k \). By the above bijection, for each \( V_j \) there exists some closed irreducible subsets \( \widetilde{V_j} \subset X \) intersecting \( U \). However, it is worth noting that these subsets are not automatically contained in one another; they may satisfy containment in \( U \) while diverging in different directions outside of \( U \). Therefore, we consider the chain

As closed subsets of irreducible sets are irreducible, inclusions are well behaved (as they are when restricted to \( U \) where the intersections become trivial), this gives us that \( n \leq \dim X \). If we assume to the contrary that there exists some chain of closed irreducible subsets \( X_0 \subset X_1 \subset \dots \subset X_m \) of length \( m > n \) in \(X\), then by assuming without loss of generality that \( X_0 \) corresponds to a point \(p\) we may take an affine neighborhood \( U\) of \( p\) and use the above bijection in the other direction. Since \( \overline{ U \cap X_i } = X_i \) (by irreducibility) we obtain a chain of lenth \( m \) in \( U \) contradicting our hypothesis — thus \( \dim X = n \).

The reverse direction follows by an almost identical argument to the previous paragraph.

Proof:

By Exercise 5.3.B we know that \( X \) (our scheme) has a finite number of irreducible components, where every point of \( X \) must lie in some irreducible component. However, as the dimension of \( X \) is the maximum of the dimension of these irreducible components, each component must have dimension \( 0 \) and thus be a point.

Proof:

Following the hint, it suffices to show that if \( \phi : k \to A \) is an integral extension then \( \dim A = 0 \). Let \( \pp \subset \mm \) be a chain of length \( 1 \) in \( A \) of prime ideals. Modding out by \( \pp \) we may assume \(A\) is a domain and \( (0) \subset \mm^\prime \) is a chain of length 1. Then for any non-zero \( a \in A \) we have that \( a \) is the root of some monic polynomial with coefficients in \( k \):

Notice that we must have \( s_0 \neq 0 \) as otherwise \( a \) would be a zerodivisor. But then as \( k \) is a field, \( s_0 \) is invertible and thus

so that \( a \) is a unit and thus \( \mm = A \), contradicting the fact that \( \mm \) is prime. Thus, \( \dim A = 0\).

Proof:

Let \( \pi^\sharp : B \to A \) denote the integral extension of rings. We first show that \( \dim \spec A \geq \dim \spec B \); following the hint, suppose \( n = \dim \spec B \) such that there exists a chain of prime ideals

As already mentioned, since \( \dim \spec A = \dim \spec A / \mathfrak{N}(A) \), we may assume without loss of generality \( \qq_0 = [(0)]\) so that we may naturally pick \( \pp_0 = [(0)] \) to be the prime ideal in \( A \) "lying over" \( \qq_0 \). Then by the Going-Up theorem, we may construct a nested chain of primes

with each \( \pp_i \) lying over \( \qq_i \). It is clear that no two \( \pp_i \) can be equal, as otherwise intersecting with \( B \) we would have that \( \qq_0 \subset \dots \subset \qq_n \) is not a strict chain.

Conversely, we may simply use the map \( \pi \) to map a chain of primes in \( A \) to a chain of primes in \(B\); to see that strict inequality is preserved, suppose to the contrary that \( \pi(\pp_i) = \qq_i = \qq_{i+1} = \pi(\pp_{i+1}) \). Then \( \pp_i \subset \pp_{i+1}\) is a chain of length one over the fibre \( \pi^{-1}( [\qq_i] ) \) so that \( \dim \pi^{-1}([\qq_i]) \geq 1 \) — a contradiction of our last exercise.

Proof:

The normalization \( \nu : \widetilde{X} \to X \) may be constructed affine locally on an affine cover \( \{ \spec A_i \} \) via the integral closure \( A_i \hookrightarrow \widetilde{A_i} \). By Exercise 11.1.B above, this tells us we may assume without loss of generality that \( \widetilde{X} = \spec \widetilde{A} \) and \( X = \spec A \). But then as \( \spec \widetilde{A} \to \spec A \) is an integral extension, the claim follows from the previous exercise.

Proof:

1. Following the hint, we may assume without loss of generality that \( A \) is an integral domain since \( \dim A = \dim A / \mathfrak{N}(A) \). Letting \( A^\prime := A \times_k K \), we know by Exercise 3.7.F that the irreducible components of \( \spec A^\prime \) are in correspondence with the minimal primes of \( A^\prime \). Letting \( \pp \subset A^\prime \) denote an arbitrary minimal prime, consider the kernel of the composition \( A \hookrightarrow A^\prime \twoheadrightarrow A^\prime / \pp \): if \( a \) is in the kernel then by abuse of notation we may assume \( a \in \pp\). By Remark 5.5.12 we know \( a \) is a zero-divisor; however since \( K \) is faithfully-flat, multiplication by any non-zero is injective so it must be the case that \( a = 0 \) and thus \( A \to A^\prime / \pp \) is injective. In particular, this implies that \( \spec A \) is contained inside the the irreducible component \( X^\prime \) corresponding to \( \pp \). Since \( K \) is algebraic over \( k \) and \( A \hookrightarrow A^\prime / \pp \) is injective, the latter map is an integral extension so by Exercise 11.1.E we have \( \dim X^\prime = \dim X \). Since the irreducible component \( X^\prime \) / minimal prime \( \pp \) was arbitrary, this holds true for all irreducible components — that is, \( X_K \) has pure dimension \( n \).

2. We may again assume \( A \) is an integral domain so that \( \spec A \) has one connected component. Notice that while \( \spec A^\prime = \spec A \otimes_k K \) may have more than one irreducible component (on account of \( k \) not necessarily being separably closed), all irreducible components must have dimension \( n \) by assumption. By picking a \( \pp \subset A^\prime \) lying over \( (0) \subset A \), we again must have that \( \dim \spec A \geq \dim \spec A^\prime / \pp \) by virtue of mapping any chain of length \( n \) down to \( A \) and showing no two primes above map to the same prime below by Exercise 11.1.D. The rest of the argument should be pretty much the same.

Proof:

It was never explained in 3.2.Q, but it is a fairly non-trivial fact that the prime ideals of \( \Z[x] \) are of the form \( (0), (p), (q(x)), \) or \( (p, q(x)) \) for \( p \) prime and \( q(x) \) irreducible. I will not attempt to give a proof since Arturo Magidin's answer on StackExchange covers this very well. From this, it is easy to see that the two longest possible chains are either

or

giving us \( \dim \Z[x] = 2 \) as desired.

Proof:

This should follow by a slight adjustment of Exercise 11.1.B: if \( \spec A \) is any open subset of \( X \) intersecting \( Y \), then an increasing chain of closed irreducible sets starting from \( Y \) corresponds to an increasing chain of closed irreducible sets (in \( \spec A \)) starting from \( Y \cap \spec A \). If we let \( \pp \) denote the minimal prime ideal corresponding to the irreducible component \( Y \cap \spec A \), then we know that \( \OO_{X,\eta} \cong A_{\pp} \) so the claim follows from the preceding paragraph in the text.

Proof:

By definition of codimension of \( \codim_X Y = m \), we know that there exists an increasing chain of closed irreducible subsets of \( X \)

Similarly, by definition of \( \dim Y = n \) we know that there is an increasing chain of closed irreducible subsets

Then

is a chain of closed irreducible subsets of length \( m + n \). Since the dimension of \(X\) is taken as the supremum over all such chains, we get \( m + n \leq \dim X \)

Proof:

1. Notice that since the \( m_i \) are increasing and their differences are increasing, we necessarily have \( \lim_{i \to \infty} m_i = \infty \) so that \( \bigcup_{i} \pp_i = ( x_{m_1 + 1}, x_{m_1 + 2}, \dots ) \) (i.e if we had chosen \( m_1 = 1\) then we would have \( \bigcup_i \pp_i = (x_2, x_3, \dots) \)). Thus, \( S \) consists of those polynomials that are not divisible by \( x_j \) for \( j > m_1 \). Since the constants are obviously in \( S \), we trivially have \(1 \in S\). Moreover, if \( f, g \) are not divisible by \( x_j \) for \( j > m_1 \) then clearly \( fg \) is not either.

2. Recall from Exercise 3.2.K that the prime ideals in \( S^{-1} A \) are precisely the primes \( \qq \) in \( A \) with \( \qq \cap S = \qq \cap (A - \bigcup_i \pp_i) = 0 \); thus, \( \spec S^{-1} A \) consists of those primes contained in \( (x_{m_1 + 1}, x_{m_1 + 2}, \dots) \). In this case, it should become obvious that we may construct a chain

   $$ (x_{m_i + 1}) \subset (x_{m_i + 1}, x_{m_i + 2}) \subset \dots \subset (x_{m_i + 1}, \dots, x_{m_{i+1}}) $$

Proof:

For the first part (a), the definition is clearly symmetric and reflexive. For transitivity, if \( F_1 \sim F_2 \) and \( F_2 \sim F_3 \), then \( F_1 F_2 \) is algebraic over \( F_1 \) and \( F_2 \) and \( F_2F_3 \) is algebraic over \( F_2 \) and \( F_3 \). Notice that as \( F_1 F_2 \) is algebraic over \( F_2 \) we have \( F_1 F_2 F_3 \) is algebraic over \( F_2 F_3 \). As the latter is algebraic over \( F_3 \), then as the composition of algebraic extensions is algebraic we have that \( F_1 F_2 F_3 \) is algebraic over \( F_3 \). By a similar argument, we must have it is algebraic over \( F_1 \). Since \( F_1 F_3 \) is the smallest field extension containing \( F_1 F_3 \), we then necessarily have that \( F_1 F_3 \) is algebraic over \( F_1, F_3 \) as required.

For part (b) Stacks has a nice proof of this

Proof:

Since a maximal ideal \( \mm \subset k[x_1, \dots, x_n] / I \) is a maximal ideal containing \( I \), we may assume without loss of generality that \( I = (0) \). Now \( A / \mm \) is clearly a field extension; however, as \( \operatorname{tr. deg.} A / \mm = 0 \) we have that each \( x_i \) is algebraic over \( k \). By taking the maximum of the degrees of such polynomials, we get that \( A / \mm \) is a finite extension of \( k \).

Proof:

Recall from Exercise 6.5.B that a dominant morphism corresponds to an inclusion of fields in the opposite direction: \( K(Y) \hookrightarrow K(X) \). Thus, we obtain \( \operatorname{tr.deg}(Y) \leq \operatorname{tr.deg}(X)\).

Proof:

We saw in Exercises 3.6.F that this scheme is indeed integral. To show it has dimension 2, we follow the hint and use Exercise 11.1.B to compute the dimension on an affine cover. First notice that for \( D(w) \), we have

In a similar manner,

Moreover, the equation \( wy - x^2 = 0 \) implies \( D(x) \subset D(w) \) and similarly \( xz - y^2 \) implies \( D(y) \subset D(z) \). It should be fairly obvious that the transcendence degree of \( K(A) \) (where \( A = k[x, w]_w \)) is 2.

Proof:

Following the hint and the proof of 11.2.7, we may reduce to the case that \( X = \spec A\) and \(Y = \spec B\) are affine (irreducible and thus integral) \( k \)-varieties by Exercise 11.1.B — let \( n = \dim X \) and \( m = \dim Y \). By Noether Normalization there exists surjective finite morphisms \( X \to \A^{n}_k \) and \( Y \to \A^{m}_k \). As finite morphisms and surjectivity are preserved by base-change by Exercise 9.4.B and Exercise 9.4.D, respectively, we know that \( X \times_k Y \to \A^{n}_k \times_k \A^{m}_k = \A^{n+m}_k \) is a surjective finite morphism. In the language of algebra, this tells us that \( K(A \otimes_k B) \) is a finite and thus integral extension of \( k[x_1, \dots, x_n, y_1, \dots, y_m] \) so that \( \dim X \times_k Y = n + m \) by Exercise 11.1.E

Proof:

Let \( A = k[x_1, \dots, x_n] / I \) and consider \( \pp \subset \qq \subset S^{-1}A \). Since any chain of prime ideals in \( S^{-1} A \) corresponds to a chain of prime ideals in \( A \) not meeting \(S\), we may simply lift to \( A \). In a similar manner, since a chain of prime ideals in \( A \) simply corresponds to a chain of prime ideals in \( k[x_1, \dots, x_n] \) contained in \( I \), it suffices to show that \( k[x_1, \dots, x_n] \) is catenary.

We proceed by induction on \( n \); in the case \( n = 1 \) we know that \( k[x] \) is a PID, so any prime ideal is maximal and thus the longest possible chain is \( (0) \subset \pp = \qq \). Next suppose that \( k[x_1, \dots, x_{n-1}] \) is catenary, and let

be a maximal chain. Notice we may assume \( (0) \) is the minimal prime since \( k[x_1, \dots, x_n] \) is a domain, and moreso we may assume that \( \pp_1 = (f) \) for some irreducible \( f \), as otherwise we may pick some \( f \in \pp_1 \), decompose into irreducible factors \( f_1^{\alpha_1} \dots f_r^{\alpha_r} \)(as \( k[x_1, \dots, x_n] \) is UFD) and obtain \( (f_1) \subset \pp_1 \). Similar to the proof of Noether's normalization, up to some change of variables \( x_i^\prime = x_i - a_i x_n \), we may assume without loss of generality that \( f = x_n^d + F_{d-1}(x_1, \dots, x_{n-1})x_n^{d-1} + \dots + F_0(x_1, \dots, x_{n-1}) \). Then \( k[x_1, \dots, x_n] / (f) \) is an integral extension of \( k[x_1, \dots, x_{n-1}] \). By reducing modulo \( (f) \) and intersecting with \( k[x_1, \dots, x_{n-1}] \), we have that

remains a strictly nested chain (possibly by using lying over theorem) so that by inductive hypothesis we necessarily have that it is a maximal chain of fixed length. In fact, for the polynomial ring \( k[x_1, \dots, x_{n-1}] \) we know that this must be of length \( n - 1 \) so that our original maximal chain is of length \( k = n \).

Proof:

Notice that if \( Z \) is an irredcible subset maximal among those smaller than \(X\) then we necessarily have \( \codim_X Z = 1 \); thus showing that \( \dim Z = \dim X - 1 \) is equivalent to showing \( \dim Z + \codim_X Z = \dim X \). By an inductive argument, taking any maximal chain \( Z_0 \subset Z_1 \subset \dots \subset Z_r \subset X \) where each is a closed irreducible subset maximimal among those smaller than the next, \( \dim Z_i = \dim Z_{i+1} - 1 \) would imply Theorem 11.2.9.

Proof:

By Exercise 11.1.B it suffices to work affine-locally, so we take \( X = \spec A \) with \( A \) a finite (and thus integral) extension of \( k[x_1, \dots, x_d] \). Since \( Z \) is a closed irreducible subset maximal among those smaller than \( X \) by the previous exercise, it corresponds to a height 1 prime \( (0) \subset \pp_1 \) in \(A\). Pulling back to \( k[x_1, \dots, x_d] \), let \( \qq_1 \) denote the corresponding prime. If we assume that \( \pi(Z) \) is a hypersurface, i.e. the Krull dimension of \( \qq_1 \) is \( d-1 \), then we may find a sequence of nested primes of length \( d - 1 \), which by the Going Up theorem would give us a nested sequence of primes in \( B \) of length \( d - 1 \) — thus \( \dim Z \geq d - 1 \). However, since \( Z \) is maximal among those closed irreducible subsets smaller than \(X\), we have \( \dim Z = d - 1 \) proving the theorem.

Proof:

Since \( p \) is a closed point, we have that it is its own generic point — therefore the second and third integers are the same by Exercise 11.1.I. Furthermore, we know the first and the third are related by definition of codimension: the codimension of \( p \) in \( X \) is the maximum over all chains of irreducible components starting at \( p \). However since \( p \) is a closed point we have that \( \dim p = 0\) and thus by theorem 11.2.9 we have \( \codim_p X = \dim X \).

Proof:

We follow the hint closely; taking \( X \) to be the incidence correspondence described above. To make this precise, let \( y_0, \dots, y_N\) denote the coordinates of \( \P^{ { d + 3 \choose 3} - 1} \) corresponding to the degree \( d \) homogeneous monomials in \( k[x, y, z, w] \). Since \( \mathbb{G}(1, 3) \) may be covered by Schubert cells isomorphic to \( \A^4 \) — for example, one Schubert cell could be lines of the form \( [x, y, ax + by, cx + dy] \) where \( a, b, c, d \) become our affine coordinates. Then by plugging the line \( [x, y, ax + by, cx + dy] \) into our degree \(d\) homogeneous polynomial, we obtain a different degree \(d\) polynomail now with coefficients in terms of the \( a, b, c, d \). For example, taking \( d = 4 \) and plugging \( [x, y, ax + by, cx + dy] \) into \( x^4 + y^4 + z^4 \) would give us

By replacing the degree 4 components with their respective \( y_i \), this gives a polynomial in the respective \( \P^{34} \times \A^4 \). Since this extends nicely to across each Schubert cell covering \( \mathbb{G}(1, 3) \), one obtains a closed subscheme of \( \P^{ 34} \times \mathbb{G}(1, 3) \). The same idea generalizes to other \( d > 3 \), though it would be tedious to generalize the argument.

It follows immediately from Exercise 11.1.B that \( \dim \mathbb{G}(1, 3) = 4\). By a similar argument to that above, notice that any line \( \ell \) lying on a degree-\(d\) hypersurface corresponds a system of \( d + 1 \) linear conditions (by a linear change of coordinates of our line, this should correspond to Exercise 8.2.J). Since \( \dim \P^{ { d + 3 \choose 3} - 1} \times_k \mathbb{G}(1, 3) = { d + 3 \choose 3} - 1 + 4 \) by Exercise 11.2.E; by intersecting with \( d+1 \) linear conditions, we reduce the dimension by \( (d+1) \) by Exercise 11.2.G. Finally since projection \( \P^{ { d + 3 \choose 3} - 1} \times \mathbb{G}(1, 3) \to \P^{ { d + 3 \choose 3} - 1} \) is a closed proper map (since properness is preserved under base change) and \( X \hookrightarrow \P^{ { d + 3 \choose 3} - 1}\) is a closed subscheme, the composition is proper telling us that the image lies in a proper closed subscheme. In particular, since \( 4 - (d + 1) < 0 \) for \( d > 3 \) we have that \( \dim X < \dim \P^{ { d + 3 \choose 3} - 1} \) and thus the general hypersurface does not contain a line.

Proof:

Let \( f \in k[x_0, \dots, x_n] = S_\bullet \) be irreducible homogeneous and \( V_+(f) = \proj k[x_0, \dots, x_n] / (f) \hookrightarrow \P^n_k \) denote the reduced and thus integral closed subscheme. By Theorem 11.2.9 it suffices to show that \( \codim_{V(f)} \P^n_k = 1 \). Clearly since \( V_+(f) \subset \P^n_k \) we have that the codimension is \( \geq 1 \). Now suppose to the contrary that \( V_+(f) \subset Y \subset \P^n_k \) for some irreducible closed \( Y \). By Exercise 8.2.C we may assume without loss of generality that \( Y = V_+(I) \) so that \( S_+ \subset I \subset (f) \). By choosing some nonzero \( g \in I \) we may further assume \( I = (g)\). Then as \( f \) is irreducible and \( k[x_0, \dots, x_n] \) is a UFD we may write \( g = \alpha f^r \cdot s_1^{r_1} \dots, s_k^{r_k} \). If not all \( s_i = 0 \) then we would have that \((g)\) is not prime contradicting the assumption that \( Y\) is irreducible — thus \( (g) = (f) \) so that \( \dim V_+(f) = n - 1 \).

Proof:

We follow This answer by Tom Oldfield on Math StackExchange. Let \( \dim A = n \) such that there is a maximal chain of primes

where we necessarily have \( \pp_n = \mm \) as the unique maximal prime ideal. It may be the case that \( f \in \pp_i \) for all \( i \), in which case \( f \) is a zerodivisor and thus \( \dim A / (f) \geq n \). Otherwise, we may assume without loss of generality that \( f \) is not a zero divisor, and find some \( k \) such that \( f \in \pp_k \backslash \pp_{k-1} \) and thus \( \pp_{k - 1} + (f) \subset \pp_k \). Now it is likely not the case that \( \pp_{k-1} + (f) \) is prime, but as \( A \) is Noetherian we may find some \( \qq_{k-1} \) minimal with respect to \( \pp_{k-1} + (f) \subset \qq_{k-1} \subset \pp_k \) (i.e. minimal among those primes containing \(f\)). Repeating this argument inductively for \( 0 \leq j\) <\( k - 1 \) (since \( f \notin \pp_{j} \)), we may find \( \qq_j \) minimal among primes \( \pp_j + (f) \subset \qq_j \subset \qq_{j+1} \). To see that \( \qq_j \neq \qq_{j+1} \), notice that if this were the case we would have that \( \pp_j \subset \pp_{j+1} \subset \qq_{j+1} \) would be a strict chain of primes — however since \( \qq_{j+1} = \qq_j \) is minimal among primes containing \(f\), it must have codimension 1 by Krull's Principal ideal theorem, a contradiction.

Proof:

1. Let \( S_\bullet := k[x_0, \dots, x_n] \) and \( S_+ = (x_0, \dots, x_n) \). Without loss of generality, we may write \( X = V(I) \) for some homogeneous ideal \( I \) and let \( H = V(f) \) denote our hypersurface for some homogeneous \(f \in S_\bullet \). Following the hint, let \( C(X) = \spec (S_\bullet / I) \) denote the affine cone of \( I \). Since both \( I \) and \((f)\) are homogeneous, we have \( I, (f) \subset S_+ \) and \( \spec (S_\bullet / (f)) \cap C(X) = V( I + (f) )\). It suffices to show that there exists another point \( [\pp] \neq [S_+]\) in the intersection. Similar to the previous exercise, since \( S_\bullet \) is Noetherian we can find some \( I + (f) \subset \pp \subset S_+\) minimal with respect to prime ideals containing \( I + (f) \) as follows: if we suppose to the contrary that \( S_+ \) is minimal with respect to such prime ideals, then by Krull's Principal Ideal Theorem applied to \( S_\bullet / I \), the irrelevant ideal \( S_+ = (x_0, \dots, x_n) \) has codimension at most 1 (over \( I \)). However, since \( V_+(I) \) is assumed to be dimension at least 1, we can find some chain of homogeneous prime ideals (contained in \( S_+ \) of length 1): \( I \subset \pp_0 \subset \pp_1 \subset S_+ \) — a clear contradiction. Therefore, there must exist some (minimal) prime \( I + (f) \subset \pp \subset S_+ \) which corresponds to a point in \( X \cap H \subset \P^n \)

2. Proceed by induction on \( r \); the base case \( r = 1 \) was considered in the previous problem. Assume the statement holds true for \( \leq r - 1 \), let \( X = V(I) \) be irreducible of dimension \( r \) and without loss of generality let \(Y = V(f_1, \dots, f_r) \) be generated by linear homogeneous functions such that \( Y \) is codimension \(r\). By again observing the affine cones, we have \( C(X) \cap C(Y) = V(I + (f_0, \dots, f_r)) \), with both ideals contained in the irrelevant ideal \( S_+ = (x_0, \dots, x_n)\). By inductive hypothesis, we may find a prime \( \pp \) minimal with respect to the property that \( I + (f_1, \dots, f_{r-1}) \subset \pp \subset (x_0, \dots, x_n) \). By a similar argument to that above, we may find a prime \( \qq \) minimal with respect to \( \pp + (f_r) \subset \qq \subset (x_0, \dots, x_n) \)

3. Following the hint, we proceed by induction on the number \( m \) of generic points. For the base case that \(X\) has a unique generic point (implying that \( X \) is an irreducible closed subset); intersecting with a generic hyperplane \( V(f_1) \) gives us a closed subscheme one dimension lower (i.e. of dimension \(r - 1\)), so intersecting with \( r \) hyperplanes should give us a closed subscheme of dimension \( 0\) (i.e. finitely many points). In the case that \( k \) is infinite, we may choose one last hyperplane that does not contain any of these points.

   For the inductive step, let \( p_1, \dots, p_{m+1} \) denote our generic points. By inductive assumption, we have a collection of \( r + 1 \) hypersurfaces \( f_{1, i}, \dots, f_{r+1, i} \) whose intersection does not vanish on \( p_1, \dots, \widehat{p_i}, \dots, p_{m+1} \). Then by taking \( g_j = \sum_i f_1 \dots \widehat{f_i}\dots f_{m+1} \) for \( 1 \leq j \leq r + 1 \), we obtain a collection of \( r+1 \) hypersurfaces that do not vanish on \( X \).

4. Similar to the base case of the previous part, we may choose some sufficiently general \( f_i \in X \cap V(f_{i-1}) \) that is not a zero-divisor, such that each intersection reduces the previous dimension by 1. Thus, the intersection of \( r \) general hyperplanes will give a closed subscheme of dimension \( 0 \), which in the case of \( \P^n_k \) is a finite number of points.

Proof:

Following the hint, we proceed by induction on \( n \): the base case \( n = 1 \) is trivial. Now suppose the claim is true for \( n \): namely let \( \pp_1, \dots, \pp_{n+1} \) be our prime ideals, \( I \not \subset \pp_i \), and assume for each \( 1 \leq i \leq n \) that there exists some \( g_i \in I \) with

That is to say, for each \( j \neq i \), we have that \( g_j \notin \pp_i \) so by primality of the \( \pp_j \) we have that

Then by taking \( f = f_1 + \dots + f_n + f_{n+1} \), we necessarily have that \( f \in I - \bigcup_{i=1}^{n+1} \pp_i \)

Proof:

Since \( X \) and \(Y\) are equidimensional and the result is local to the irreducible components of \( X \cap Y \), we assume without loss of generality that \( X, Y \) are integral varieties. Recall by Exericse 10.1.B that \( \A^n_k \) is separated so that \( \delta : \A^d_k \hookrightarrow \A^d_k \times_k \A^d_k \) is a closed embedding. It is fairly easy to see that the ideal of \( \Delta \subset \A^d_k \times_k \A^d_k \) is of the form \( (x_1 - y_1, \dots, x_d - y_d) \) where \( k[x_1, \dots, x_d] \) denotes the coordinate ring of the first term and \( k[y_1, \dots, y_d] \) the second. It is fairly easy to see that at any closed point \( p = [ (x_1 - a_1, \dots, x_n - a_n, y_1 - a_1, \dots, y_n - a_n) ] \in \Delta\), \( x_1 - y_1, \dots, x_d - y_d \) gives a regular sequence of \( \OO_{\A^d \times \A^d, p} \) so that \( \Delta \) is a regular embedding of codimension \(d\).

Now as \( X, Y \) are both closed subvarieties of \( A^d_k \), we obtain closed embeddings \( X, Y \hookrightarrow \A^d_k \) giving us \( X \times_k Y \) as a closed subvariety of \( A^d_k \times_k \A^d_k \). By our classical interpretation of the diagonal, \( \Delta \cap X \times_k Y \) consists of those points where the embeddings agree, which is to say the collection of points in our original \( \A^d_k \) where \( X\) and \(Y\) align. Now by Theorem 11.2.9, we know that \( \dim X = d - m \) and \( \dim Y = d - n \), so by Exercise 11.2.E we have that \( \dim X \times_k Y = 2d - (m + n) \). Now as \( \Delta \) is locally cut out by \( d \) equations, we obtain by Krull's Principal ideal theorem that \( \Delta \cap X \times_k Y \) is of dimension at least \( d - (m + n) \) in \( \Delta \). As \( \Delta \cong \A^d_k \), this again tells us that the codimension of \( X \cap Y \) is at most \( m + n \).

Proof:

Following the same general argument, we may reduce to the case that \( X \) and \(Y\) are irreducible and thus integral projective subvarieties. If we suppose \( X = V_+ (I) \) and \( Y = V_+(J) \), let \( C(X) = \spec k[x_0, \dots, x_n] / I \) and \( Y = \spec k[x_0, \dots, x_n] / J \) denote theire cones — by definition we still have that \( C(X) \) and \( C(Y) \) are of codimension \( d \) and \(e\) respectively in \( \A^{n+1} \). Then \( C(X) \cap C(Y) \) have codimension at most \( d + e \leq n \), so by Theorem 11.2.9 we know that \( C(X) \cap C(Y) \) has dimension at least \( n + 1 - (d + e) \geq 1 \). As \( (x_0, \dots, x_n) \) is contained in both \( C(X) \cap C(Y)\), this tells us that we can find a line \( \ell \) passing through the origin in \( C(X) \cap C(Y) \) corresponding to a point in \( X \cap Y \), as desired.

Proof:

For the sake of contrapositive, suppose \( f \) is not a unit so that \( A / (f) \) is non-trivial. As \( A / (f) \) is also Noetherian, it must have finitely many minimal prime ideals. Picking some \( \pp \subset A / (f) \) minimal, it corresponds to a prime ideal \( \pp^\prime \subset A\) minimal with respect to those containing \(f\). But then \( \pp^\prime \) has codimension either 0 or 1 by Krull's Principal Ideal Theorem.

Proof:

Suppose \( V(r_1, \dots, r_\ell) = Z_1 \cup \dots \cup Z_k \) where each \( Z_i \) is an irreducible component. Similar to the approach of the previous exercise, we may choose some functions \( f_i \in \pp_i \) vanishing on \( Z_i \) (where \( \pp_i \) denotes the corresponding minimal prime) in such a way that \( f_i \) does not vanish on \( Z_j \) for \( j \neq i \). By taking \( g_i = \prod_{j \neq i} f_i \), we have that localizing \( V(r_1, \dots, r_\ell) \) give us \( Z_i \). Since \( A_{g_i} \) is also a finitely generated domain over \( k \), we may assume without loss of generality that \( Z = V(r_1, \dots, r_\ell) \). Then by Theorem 11.2.9, we have that \( \codim_Z X = \dim A - \dim A / (r_1, \dots, r_\ell) \). Since \( A \) is a finitely generated domain, we may use 11.2.1 to show that \( \dim A / (r_1, \dots, r_\ell) = \mathrm{tr.deg.} K(A / (r_1, \dots, r_\ell)) \geq \mathrm{tr.deg} A - \ell \) (since each relation induces at most one algebraic dependence). Thus, \( \textrm{codim}_Z X \leq \ell \) as desired.

Proof:

1. Notice that the maximal ideal is necessarily dimension \( 0 \) as there is no increasing chain of prime ideals starting at \( \pp \) — alternatively, one can say that \( \mm \) has maximal height and thus the codimension of \( \mm \) is the dimension of \( A \). Since \( [\mm] \) is indeed the only irreducible component of \( V(r_1, \dots, r_\ell) \), Krull's Height theorem (and the previous exercise) tell us that \( \dim [\mm] \geq \dim A - \ell \) i.e. \( \dim A \leq \ell \).

2. Following the hint, we proceed by induction on \( d \). For the base case \( d = 1 \), we necessarily have that \( \mm \) is height 1 and thus principal (notice that \( \mm \) must contain an irreducible non-zerodivisor since \( \mathfrak{N}(A)\) is always height 0). Thus \( [\mm] = V(f) \) for some irreducible non-zerodivisor \(f\).

   For the inductive step, suppose the claim holds true for Notherian local rings of dimension strictly less than \(d\). Let \( \pp_1, \dots, \pp_n \) denote the minimal primes of \( A \) corresponding to the irreducible components of \( \spec A \) of dimension \( d \). Since \( A / \pp_i \) is also Noetherian, we may always find some minimal prime \( \qq_i \supset \pp_i \) such that \( \qq_i \) corresponds to an irreduicble closed subset of codimension 1 and dimension \( d - 1 \). Then as the \( \qq_j \) are dimension \( d - 1 \) we have that they are not contained in any \( \pp_i \) and thus by Prime Avoidance we may find some \( h_i \in \qq_i \backslash \bigcup_{j=1}^n \pp_j \) — that is, \( h_i \) vanishes on the codimension 1 \( [\qq_i] \) but not on any irreducible component \( [\pp_i] \). Then by taking \( g_d = \prod_{i=1}^n h_i \), we obtain a function that vanishes on all our codimension 1 components so that \( \dim A / (g_d) = \dim A - 1 \) (this follows from Exercise 11.3.B since it is not a zero-divisor / does not vanish on every associated prime \( \pp_i \)). Since \( A / (g_d) \) is still a local ring (by the Correspondence theorem the prime ideals of \( A / (g_d) \) are the prime ideals of \( A \) containing \( g_d \)) with unique maximal ideal \( \overline{\mm} = \phi(\mm) \) where \( \phi : A \to A / (g_d) \) is the natural surjection, there exist some \( \overline{g_1}, \dots, \overline{g_{d-1}} \) such that \( V(\overline{g_1}, \dots, \overline{g_{d-1}}) = [\overline{\mm}] \). Then pulling back along \( \phi \), we obtain prime ideals \( g_i \) Lying over the \( \overline{g_i}\) such that \( V(g_1, \dots, g_{d-1}, g_d) = [\mm] \).

Proof:

1. For the forward direction, suppose that \( f_1 = f_2 = \dots = f_d = 0 \) has a common zero in \( \P^{n-1} \), call it \( [a_1, \dots, a_n] \). Then by taking the line \( L = [ 1 - t, t a_1, \dots, ta_n ] \), we have that for any point in \( L \) corresponding to some \( t \in k \)

   $$ \begin{align} f(1 - t, ta_1, \dots, ta_n) &= f_d(ta_1, \dots, ta_n) + (1 - t) f_{d-1}(ta_1, \dots, ta_n) + \dots + (1 - t)^{d-1} f(ta_1, \dots, ta_n) \\&= t^d f_d(a_1, \dots, a_n) + (1 - t) t^{d-1}f_{d-1}(a_1, \dots, a_n) + \dots + (1 - t)^{d-1}t f(a_1, \dots, a_n) \\&= 0 \end{align} $$

   Thus, \( L \subset X \). Conversely, suppose \( L \subset \P^n\) is a line through \( p \) that is contained in \(X\). By Exercise 11.3.C(a) \( L \) must intersect the hyperplane \( H = V(x_0) \) at some point \( q = [0, b_1, \dots, b_n] \). Then we necessarily have that \( f_d(b_1, \dots, b_n) = 0 \); however we also know that \( L \) contains points \( [(1- t), tb_1, \dots, tb_n] \) for all \( t \in k \) so that by choosing \( 0 < t < 1 \) we have

   $$ 0 = (1 - t)t^{d - 1}f_{d-1}(b_1, \dots, b_n) + \dots + (1 - t)^{d-1}t f_1(b_1, \dots, b_n) $$

   Since each \( (1 - t)^i t^{d-i} \neq 0 \) and the \( f_d \) are of distinct degrees, we must have that \( f_i(b_1, \dots, b_n) = 0 \) giving us a common zero in \( \P^{n-1}\).

2. In a similar approach to that above, let \( p \in X \) and suppose \( p \in D(x_i) \) so that we may write \( p = [a_0, \dots, a_{i-1}, 1, a_{i+1}, \dots, a_n] \) and refactor \(f\) as

   $$ f(x_0, \dots, x_n) = \widetilde{f}_d(x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_n) + \dots + x_i^{d-1} \widetilde{f}_1(x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_n) $$

   Then similar to the part above, we may show that there being a line through \( p \) contained in \(X\) is equivalent to there being a common zero of \( \widetilde{f_1} = \dots = \widetilde{f_d} = 0 \). By Exercise 11.3.F, we have that \( d \leq n - 1 \) implies \( V(\widetilde{f}_1, \dots, \widetilde{f}_{d}) \subset \P^{n-1}\) is non-empty giving us the result.

3. To answer the hint, notice that on \( D(x_0) \) by choosing our original \( p = [1, 0, \dots, 0]\), we can find a surface with no lines by taking \( f_i = x_i^i \) — this cannot have a common point by part (a) since that would imply \( [0, \dots, 0] \in \P^{n-1} \). Then by part (a), there is no line in \( X \) passing through \( p \) iff \( f_1, \dots, f_d\) does not have a common solution in \( \P^{n-1} \). However, the general intersection of \( d > n - 1 \) hypersurfaces \( V(f_1, \dots, f_d) \) is trivial in \( \P^{n-1} \) (e.g. by Exercise 11.3.C(d) taking the intersection of \( n - 1 \) will in general give us a finite number of points, and finding another hypersurface that does not intersect these is an open condition) telling us that most hypersurfaces containing \( p \) do not have a line through \( p \).

Proof:

We will follow the well-written answer of Fang Hung-Chien on StackExchange. Since the codimension of a point only depends on the irreducible component it is contained in (assuming equidimensional), which can further be reduced to some affine set containing the minimal primes, we may assume without loss of generality that \( X = \spec A \) and \( Y = \spec B \) such that \( q = [\qq] \) for some \( \qq \subset B \) and \( p = [\pp] \) for some \( \pp \subset A \) with \( (\pi^\sharp )^{-1}(\pp) = \qq \). In addition, Exercise 11.2.I tells us we need only look at closed points \( p, q \). From §11.1.4 we know that \( \codim_X p := \dim A_\pp \) and from §9.3.2 we have that \( \pi^{-1}(q) \) corresponds to \( A \otimes_B \kappa(\qq) \). By letting \( \pp^\prime \) denote the image of \( \pp \) under \( A \hookrightarrow A \otimes_B \kappa(\qq) \), the problem reduces to showing

Since localization commutes with quotients (i.e. by exactness of localization), we obtain an isomorphism \( ( A \otimes_B \kappa(\qq) )_{\pp^\prime} \cong A_\pp / \qq A_\pp \).

Following the hint, if we let \( \qq = V(b_1, \dots, b_d) \) and \( \pp / \qq A = V(\overline{a_1}, \dots, \overline{a_e}) \) be our system of parameters. Let \( a_i \) be any lift of \( \overline{a_i} \) to \( A \). Then we can find some sufficiently large integer \( \ell \gg 0 \) such that

and further some \( r \gg \ell \) such that

That is to say, \( [\pp] = V(a_1, \dots, a_e, b_1, \dots, b_d) \), so by Krull's Height Theorem we obtain

Proof:

Let \( U = \spec B_g \) denote our distinguished open subset. Moreover if \( \pi^\sharp : B \to A \) denotes our ring map, then the assumption may be stated that the corresponding map \( B_g \to A_{\pi^\sharp(g)} \) factors as

Fixing \( q \in U \), corresponding to some prime \( \qq \subset B_g \), since we assumed that \( \pi \) is dominant we simply need to show that \( \pi^{-1}(q) \) has pure dimension \( m - n \). Since \( K(B_g) = K(B) \), we have that \( \dim B_g = \dim B \) and may thus consider \( \qq \in B \) so that \( \pi^{-1}(q) \) has scheme structure \( \spec A \otimes_B \kappa(q) \). As this is finitely generated (our varieties are finite type), any minimal prime \( \pp \subset A \otimes_B \kappa(q)\) will have dimension \( \dim( A \otimes_B \kappa(q) ) = \dim ( (A \otimes_B \kappa(q)) /\pp) \) so that \( \pi^{-1}(q) \) is equidimensional (this fact will hold in a slightly further generality, see [Stacks 00QK]). In fact, by dividing by any minimal prime we may assume without loss of generality that \( \pi^{-1}(q) \) is irreducible.

Since base-change preserves fibres and surjectivity, let \( \pi^{-1}(q) \to \A^{m - n}_{\kappa(q)} \to \spec \kappa(q) \) denote the composition where the first map remains finite and surjective. But then by Exercise 11.1.E and Noether Normalization, this implies \( \dim \pi^{-1}(q) = \dim A^{m-n}_{\kappa(q)} = m - n \).

Proof:

Let \( X = \bigcup_i Z_i \) denote the decomposition of \( X \) into finitely many irreducible components. Notice that for any \( x \in Z_i \), the fibre over \( \pi(x) \in Y \) must also lie completely in \( Z_i \) as otherwise the fibre would be irreducible. By Theorem 11.4.1, for each irreducible component \( Z_i \) there exists a non-empty open subset \( V \subset \pi(Z_i) \) where \( \dim \pi^{-1}(y) = \dim Z_i - \dim \pi(Z_i) \) for \( y \in V\). Notice however that since \( \pi(Z_i) \) is closed (as \( \pi \) is proper) and \( Y \) is irreducible, we must have \( \pi(Z_i) = Y \) (since otherwise we could take \( Y = \pi(Z_i) \cup Y \backslash U \)). But then for all \( x \in X \), the fibre of \( \pi(x) \) lies in \( Z_i \) so we can conclude \( X = Z_i \).

Proof:

If we can prove that dimension of fibres is upper-semicontinuous on irreducible varieties, then the dimension of \( \pi^{-1}( \pi(p) ) \) is in terms of the largest irreducible component of \( \pi^{-1}(\pi(p)) \), we may first reduce to the case that \( X \) is itself irreducible — note that this also implies that \( Y \) is irreducible.

Now for part (a) we may automatically assume that \( \pi \) is dominant since \( p \mapsto \dim \pi^{-1} ( \pi(p) ) \) only depends on the information of \( \pi(X) \). However, for part (b) if \( \pi \) is not dominant, then for any point \( y \notin \pi(X) \) we have \( \dim \pi^{-1}(y) = \dim \emptyset = - \infty \) so the preimage of any \( [-\infty, n) \) will always contain \( Y \backslash \pi(X) \). In other words, the dimension of fibres can only jump up once we reach \( \pi(X) \subset Y\).

Proof:

By the previous exercise we need only restrict our attention to \( \pi(X) \). Then for any \( q \in \pi(X) \), there exists some \( p \in X \) with \( \pi(p) = q \). Since \( \dim \pi^{-1}( \pi(p) ) = \dim \pi^{-1}(q)\) is upper semi continuous at \( p \) by part (a) the result follows.

Proof:

Let \( X = \spec A \). Since finite-ness is affine-local on the target, we may also assume \( Y = \spec B \) — both \( A, B \) are necessarily integral domains and \( X , Y\) are assumed to be integral varieties. Let \(\eta_Y = [(0)] \) denote the unique generic point of \( Y \). Moreover, by definition of generically finite we know that the preimage of \( \eta_Y \) is finite and non-empty, so that \( \pi \) is necessarily dominant and of finite-type. By Noether normalization and Zariski's theorem, \( K(B) \subset K(A) \) is a finite extension of fields. Let \( \{ a_i \}\) be a set of generators for \( A \) over \(B\), so that in \( K(A) \) each \( a_i \) satisfies some polynomial in \( K(B) \). Clearing denominators, we may assume that the \( a_i \) satisfy polynomials in \( B \) — by taking the product of the leading coefficients of these polynomials, call it \(b\), and localizing \( B_b \) and \( A_b \) we may assume these polynomials are in fact monic. Then \( A_b \) is integral and finitely-generated over \( B_b \) and thus a finitely-generated \( B_b \)-module.

Proof:

Since we are assuming \( \pi\vert_{U_i} \) is finite, Exercise 7.3.M tells us that \( \pi\vert_{U_i} \) is closed. Since there are finitely many \( U_i \), the image of any closed \( Z \subset X\) is the finite union of the closed \( \pi\vert_{U_i}(Z) \) which is therefore closed.

Proof:

Since \( X \) and \( Y \) are integral, they contain unique generic points. Recall that a set is dense iff it contains each generic point. Since \( \pi \) is assumed to be dominant and \( \eta_Y \in V_i \subset \pi(U_i) \), we know that each affine \( U_i \) contains the generic point \( \eta_X \). Thus, \( \pi (X \backslash U_1) \) cannot be all of \( Y \).