Vakil Solutions: Chapter 7

☆ A number of solutions are adaped from Howard Nuer's own solutions to his Technion 2019 Algebraic Geometry course. I will later go through and accredit the exercises where his approach is used.

Section 7.1: An example of a reasonable class of morphisms: Open embeddings

Exercise 7.1.A:

Verify that the class of open embeddings satisfies properties (i) and (ii) of §7.1.1.

Proof:

To see that the class of open embeddings is local on the target, first suppose that $π : X \to Y$ is an open embedding and $V \subset Y$ is open. If $π^{- 1} (V) = \emptyset$ then the claim is trivial. Otherwise, $π^{- 1} (V)$ is nonempty and open in $X$ . By definition $π$ factors as $(X, O_{X}) \tilde{⟶} (U, O_{Y} |_{U}) ↪ (Y, O_{Y})$ , so that the restriction map factors as $(π^{- 1} (V), O_{X} |_{π^{- 1} (V)}) \tilde{⟶} (U \cap V, O_{Y} |_{U \cap V}) ↪ (Y, O_{Y})$ . Since $U \cap V$ is open, the restriction map is therefore an open embedding.

Simialrly, assume $π : X \to Y$ is a morphism, ${V_{i}}$ is an open cover of $Y$ , and for each $i$ we have $π_{i} : π^{- 1} (V_{i}) \to V_{i}$ is an open immersion. Since the $V_{i}$ cover $Y$ , $π^{- 1} (V_{i})$ form an open cover of $X$ . Moreover, each $π_{i}$ factor through some isomorphism $ρ_{i} : (π^{- 1} (V_{i}), O_{X} |_{π^{- 1} (V_{i})}) \tilde{⟶} (\tilde{V_{i}}, O_{Y} |_{\tilde{V_{i}}})$ (where $\tilde{V_{i}} \subset V_{i}$ open subset since $π$ need not be a surjection). By our usual sheaf axioms, the $O_{X} |_{U_{i}}$ glue to $O_{X}$ and the $O_{Y} |_{\tilde{V_{i}}}$ glue to $O_{Y} |_{Im (π)}$ . However, since $Im (π) = ⋃_{i} \tilde{V_{i}}$ these glue to an open subscheme of $(Y, O_{Y})$ so that $π$ is an open immersion.

To see that property (ii) holds, let $π : X \to Y$ and $ρ : Y \to Z$ be open embeddings. Then $π$ factors through an isomorphism $(X, O_{X}) \tilde{⟶} (U, O_{Y} |_{U})$ to an open subset $U \subset Y$ . By (i), the restriction $ρ |_{U}$ is also an open embedding and thus factors through an isomorphism $(U, O_{Y} |_{U}) \tilde{⟶} (V, O_{Z} |_{V}) ↪ (Z, O_{Z})$ ( where $V \subset Z$ open subset). Since the composition of isomorphisms is an isomorphism, our map $ρ \circ π$ factors through the iso $(X, O_{X}) \tilde{⟶} (V, O_{X} |_{V})$ and is thus an open embedding.

\begin{matrix} ◼ \end{matrix}

Exercise 7.1.B:

Verify that the class of open embeddings satisfies property (iii) of §7.1.1. More specifically: suppose

ι : U \to Z

is an open embedding and

ρ : Y \to Z

is any morphism. Show that

U \times_{Z} Y

exists and

U \times_{Z} Y \to Y

is an open embedding. (Hint: I’ll even tell you what

U \times_{Z} Y

is:

(ρ^{- 1} (U), O_{Y} |_{ρ^{- 1} (U)})

. In particular, if

U ↪ Z

and

V ↪ Z

are open embeddings then

U \times_{Z} V ≅ U \cap V

Proof:

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

\begin{matrix} (ρ^{- 1} (U), O_{Y} |_{ρ^{- 1} (U)}) & \overset{}{\to} & (Y, O_{Y}) \\ ρ |_{ρ^{- 1} (U)} ↓ & ↓ ρ \\ (U, O_{Z} |_{U}) & \to_{ι}^{} & (Z, O_{Z}) \end{matrix}

Must simply be the inclusion $j : (ρ^{- 1} (U), O_{Y} |_{ρ^{- 1} (U)}) ↪ (Y, O_{Y})$ . To see that the universal property is satisfied, let $(W, O_{W})$ be another scheme along with morphisms $α : W \to U$ and $β : W \to Y$ such that $ι \circ α = ρ \circ β$ . Since $Im α \subset U$ we must also clearly have $Im (ρ \circ β) \subset U$ so that $Im (β) \subset ρ^{- 1} (U)$ . But then the unique map $(W, O_{W}) \to (ρ^{- 1} (U), O_{Y} |_{ρ^{- 1} (U)})$ must simply be $β$ itself. Therefore the fiber product exists.

The inclusion $j : (ρ^{- 1} (U), O_{Y} |_{ρ^{- 1} (U)}) ↪ (Y, O_{Y})$ is clearly an open embedding, as $ρ^{- 1} (U) \subset Y$ is open by virtue of $π$ being continuous.

\begin{matrix} ◼ \end{matrix}

Exercise 7.1.C:

Suppose

π : X \to Y

is an open embedding. Show that if

Y

is locally Noetherian, then

X

is too. Show that if

Y

is Noetherian, then

X

is too. However, show that if

Y

is quasicompact,

X

need not be. (Hint: let

Y

be affine but not Noetherian, see Exercise 3.6.G(b).)

Proof:

Since $π : X \to Y$ an open embedding, it factors through some isomorphism $ρ : (X, O_{X}) \tilde{⟶} (U, O_{Y} |_{U})$ . First suppose $Y$ is locally Noetherian such that $Y$ can be covered by $V_{i} = Spec B_{i}$ where each $B_{i}$ is Noetherian. Then clearly $U$ can be covered by such sets as well, say $U \cap V_{i} = Spec (B_{i})_{q_{i}}$ — notice $B_{i}$ Noetherian implies $(B_{i})_{q_{i}}$ Noetherian by §3.6.16. Since $ρ$ is an isomorphis, we may pull such a cover back to $(X, O_{X})$ and be guaranteed that our rings are still Noetherian (since an isomorphism of schemes induces an isomorphism of sheaves). The proof showing $Y$ Noetherian implies $X$ Noetherian is identical.

To see that $Y$ quasicompact does not imply $X$ quasicompact, we may use the example of Exercise 3.6.G(b) almost verbatim. If we consider $A = Spec k [x_{1}, x_{2}, \dots]$ along with the maximal ideal $m = (x_{1}, x_{2}, \dots)$ , then $Spec A$ is clearly quasicompact as it is affine. However, we showed in the linked exercise that the compliment of $V (m)$ was not quasicompact, so we simply let $π$ denote the inclusion of $V (m)^{c}$ into $Spec A$ .

\begin{matrix} ◼ \end{matrix}

Exercise 7.1.D:

Show that the notion of "open embedding" is not local on the source.

Proof:

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

\begin{matrix} ◼ \end{matrix}

Algebraic interlude: Lying Over and Nakayama

Exercise 7.2.A:

Show that if

ϕ : B \to A

is a ring morphism,

(b_{1}, \dots, b_{n}) = 1

B

, and

B_{b_{i}} \to A_{ϕ (b_{i})}

is integral for all

i

, then

ϕ

is integral. Hint: replace

B

ϕ (B)

to reduce to the case where

B

is a subring of

A

. Suppose

a \in A

. Show that

b_{i}^{t} a^{m} \in B + B a + B a^{2} + \dots + B a^{m - 1}

for some

t

and

m

independent of

i

. Use a "partition of unity" argument as in the proof of Theorem 4.1.2 to show that

a^{m} \in B + B a + B a^{2} + \cdot \cdot \cdot + B a^{m - 1}

Proof:

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

a^{m_{i}} + \frac{β_{i, m_{i} - 1}}{b_{i}^{t_{i, m_{i} - 1}}} a^{m_{i} - 1} + \dots + \frac{β_{i, 0}}{b_{i}^{t_{i, 0}}} = 0

Letting $t_{i}$ be the maximum of all $t_{i, j}$ , we have that $ b^{t_i} a^{m_i + \beta_{i, m_i - 1} a^{m_i-1} + \dots + \beta_{i, 0} = 0 $ — in other words, $b^{t_{i}} a^{m_{i}} \in B + B a + \dots + B a^{m_{i} - 1}$ . By taking $m$ to be the maximum of all $m_{i}$ and $t$ the maximum of all $t_{i}$ , we have $b_{i}^{t} a^{m} \in B + B a + \dots + B a^{m - 1}$ for each $i$ . Similar to Theorem 4.1.2, we know that as $Spec B = ⋃ D (f_{i}) = ⋃ D (f_{i}^{t})$ , so $(b_{1}, \dots, b_{n}) = 1$ implies $(b_{1}^{t}, \dots, b_{n}^{t}) = 1$ . Therefore, we may find $r_{i} \in B$ such that $1 = r_{i} b_{1}^{t} + \dots + r_{n} b_{n}^{t}$ and thus

a^{m} = a^{m} \sum r_{i} b_{i}^{t} \in \sum r_{i} (B + B a + \dots + B a^{m - 1}) = (B + B a + \dots + B a^{m - 1})

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.B:

Show that the property of a homomorphism $ϕ : B \to A$ being integral is always preserved by localization and quotient of $B$ , and quotient of $A$ , but not localization of $A$ . More precisely: suppose $ϕ$ is integral. Show that the induced maps $T^{- 1} B \to ϕ (T)^{- 1} A$ , $B / J \to A / ϕ (J) A$ , and $B \to A / I$ are integral (where $T$ is a multiplicative subset of $B$ , $J$ is an ideal of $B$ , and $I$ is an ideal of $A$ ), but $B \to S^{- 1} A$ need not be integral (where $S$ is a multiplicative subset of $A$ ). (Hint for the latter: show that $k [t] \to k [t]$ is an integral homomorphism, but $k [t] \to k [t]_{(t)}$ is not.)
Show that the property of $ϕ$ being an integral extension is preserved by localization of $B$ , but not localization or quotient of $A$ . (Hint for the latter: $k [t] \to k [t]$ is an integral extension, but $k [t] \to k [t] / (t)$ is not.)
In fact the property of $ϕ$ being an integral extension (as opposed to integral homomorphism) is not preserved by taking quotients of $B$ either. (Let $B = k [x, y] / (y^{2})$ and $A = k [x, y, z] / (z^{2}, x z - y)$ . Then $B$ injects into $A$ , but $B / (x)$ doesn’t inject into $A / (x)$ .) But it is in some cases. Suppose $ϕ : B \to A$ is an integral extension, and $J \subset B$ is the restriction of an ideal $I \subset A$ . (Side Remark: you can show that this holds if $J$ is prime.) Show that the induced map $B / J \to A / J A$ is an integral extension. (Hint: show that the composition $B / J \to A / J A \to A / I$ is an injection.)

Proof:

Let $ϕ : B \to A$ be integral and first consider the case that $T$ is a multiplicative subset ( so that $ϕ (T)$ is as well ). Let $\overset{―}{a} = \frac{a}{ϕ (t)} \in ϕ (T)^{- 1} A$ be arbitrary. Since $a \in A$ and $ϕ : B \to A$ is integral, we can find some $n > 0$ and $b_{i} \in B$ such that $a^{n} + ϕ (b_{n - 1}) a^{n - 1} + \dots + ϕ (b_{0}) = 0$ . Dividing the equation by $ϕ (t)^{n}$ gives us
$\begin{aligned} 0 & = \frac{a^{n}}{ϕ (t)^{n}} + \frac{ϕ (b_{n - 1})}{ϕ (t)} \frac{a^{n - 1}}{ϕ (t)^{n - 1}} + \dots + \frac{ϕ (b_{0})}{ϕ (t)^{n}} \\ = {(\frac{a}{ϕ (t)})}^{n} + ϕ (\frac{b_{n - 1}}{ϕ (t)}) {(\frac{a}{ϕ (t)})}^{n - 1} + \dots + ϕ (\frac{b_{0}}{t^{n}}) \end{aligned}$
So that $\frac{a}{ϕ (t)}$ integral over $T^{- 1} B$

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

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

To see, however, that integrality is not preserved by localization of $A$ we follow the example given: the identity $k [t] \to k [t]$ is trivially integral, however the induced localization map $k [t] ↪ k [t]_{(t)}$ given by $f (t) \mapsto \frac{f (t)}{1}$ is certainly not integral. To see this, notice that by taking $a = \frac{1}{1 + t}$ , for any $f_{n - 1} (t), \dots, f_{0} (t)$ we have
$\frac{1}{(1 + t)^{n}} + \frac{(f_{n - 1} (t))^{n - 1}}{1} \frac{1}{(1 + t)^{n - 1}} + \dots + \frac{f_{0} (t)}{1} \neq 0$
since we cannot cancel out the first $\frac{1}{(1 + t)^{n}}$ term.
This follows from part (a) since if $ϕ : B \to A$ is integral, then $ϕ (T) : T^{- 1} B \to ϕ (T)^{- 1} A$ is integral (additionally by algebra if $ϕ$ is injective then $ϕ (T)$ is injective ). Additionally, we have already seen that integrality need not hold for localization of $A$ — to see that integral extensions need not be preserved follows from the simple fact that $ϕ : B \to A$ injective does not imply $B \to A \to A / I$ is injective.
We know from part (a) again that $B / J \to A / J A$ is integral so it suffices to show it is injective. Since $J = I \cap B \subset I$ , $B / J \to A / I$ factors through $A / J A$ . Since $R^{'} \to R^{''}$ and $R \to R^{'} \to R^{''}$ injections imply $R \to R^{'}$ is an injection (easy proof by contradiction), the result follows.

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.C:

Show that if

C \to B

and

B \to A

are both integral homomorphisms, then so is their composition.

Proof:

Let $a \in A$ be arbitrary; since the map $B \to A$ is integral, $a$ is contained in a subalgebra $R \subset A$ with is finitely generated as a $B$ -module, say by $b_{1}, \dots, b_{n}$ . Since the map $C \to B$ is integral, each $b_{i}$ is contained in some subalgebra $S_{i} \subset B$ that is finitely generated as a $C$ -module. Then $\prod S_{i}$ is also finitely generated so that $R$ must also be a finitely generated $C$ -module. Since $a \in A$ was arbitrary the result follows.

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.D:

Suppose

ϕ : B \to A

is a ring morphism. Show that the elements of

A

integral over

B

form a subalgebra of

A

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.E:

Show that the special case where

A

is a field translates to: if

B \subset A

is a subring with

A

integral over

B

, then

B

is a field. Prove this. (Hint: you must show that all nonzero elements in

B

have inverses in

B

. Here is the start: If

b \in B

, then

1 / b \in A

, and this satisfies some integral equation over

B

Proof:

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

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

\frac{1}{b^{n}} + β_{n - 1} \frac{1}{b^{n - 1}} + \dots + β_{0} = 0

Multiplying through by $b^{n}$ , we may rewrite the equation as

1 = - β_{n - 1} b - \dots - β_{0} b^{n} = b (- β_{n - 1} - \dots - β_{0} b^{n - 1})

Thus, $b^{- 1} = - β_{n - 1} - \dots - β_{0} b^{n - 1}$ so that $B$ is also a field.

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.F:

Suppose $ϕ : B \to A$ is an integral homomorphism (not necessarily an integral extension). Show that if $q_{1} \subset q_{2} \subset \dots \subset q_{n}$ is a chain of prime ideals in $B$ and $p_{1} \subset \dots \subset p_{m}$ is a chain of prime ideals of $A$ such that $p_{i}$ "lies over" $q_{i}$ (and $1 \leq m < n$ then the second chain can be extended to $p_{1} \subset p_{2} \subset \dots \subset p_{n}$ so that this remains true. ) (Hint: reduce to the case m = 1, n = 2; reduce to the case where $q_{1} = (0)$ and $p_{1} = (0)$ ; use the Lying Over Theorem 7.2.5.)
Draw a picture of this theorem (akin to Figure 7.1).

Proof:

By induction we only consider the case that $m = 1$ and $n = 2$ ; that is to say, suppose that $q_{1} \subset q_{2}$ is a chain of primes in $B$ and $p_{1} \subset A$ with $q_{1} = p_{1} \cap B$ (we may assume without loss of generality that $B$ is a subring of $A$ ). Then the induced map $\overset{―}{ϕ} : B / q_{1} \to A / p_{1}$ is clearly integral, since for any $\overset{―}{a} \in A / p_{1}$ and representative $a$ we may choose $b_{i}$ such that $a^{n} + b_{n - 1} a^{n - 1} + \dots + b_{0} = 0$ and reduce mod $p_{1}$ . Moreover, the map must also be injective since if $\overset{―}{b_{1}} = \overset{―}{b_{2}}$ in $A / p_{1}$ for some $b_{1}, b_{2} \in B$ then $b_{1} - b_{2} \in p_{1} \cap B = q_{1}$ so that they are indeed equivalent in $B / q_{1}$ . Therefore, the map is an integral extension so by the Lying Over theorem there exists some $\overset{―}{p_{2}} \subset A / p_{1}$ lying over $q_{2} / q_{1} \subset B / q_{1}$ . If we let $π : A \to A / p_{1}$ denote the canonical projection, then $p_{2} := π^{- 1} (\overset{―}{p_{2}})$ is the desired prime lying over $q_{2}$ .
Take $B = Q [x, y]$ , $A = Q [x, y, z] / (z^{2} + 1)$ and consider $(0) \subset (x) \subset (x, y)$ in $B$ and $(z) \subset (x, z)$ in $A$ .

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.G:

Suppose

A

is a ring, and

I

is an ideal of

A

contained in all maximal ideals. Suppose

M

is a finitely generated

A

-module, and

N \subset M

is a submodule. If

N / I N \to M / I M

is surjective, then

M = N

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.H:

Suppose

(A, m)

is a local ring. Suppose

M

is a finitely generated

A

-module, and

f_{1}, \dots, f_{n} \in M

, with (the images of)

f_{1}, \dots, f_{n}

generating

M / m M

. Then

f_{1}, \dots, f_{n}

generate

M

. (In particular, taking

M = m

, if we have generators of

m / m^{2}

, they also generate

m

Proof:

Take $N = ⟨ f_{1}, \dots, f_{n} ⟩ \subset M$ . By assumption, $N / m N \to M / m M$ is surjective ( since $N \to N / m M$ is surjective and the composition $N \to M / m M$ is surjective by assumption ) we have by the previous exercise that $M = N$ .

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.I:

Recall that a

B

-module

N

is said to be faithful if the only element of

B

acting on

N

by the identity is 1 (or equivalently, if the only element of

B

acting as the 0-map on

N

is 0). Suppose

S

is a subring of a ring A, and

r

∈ A. Suppose there is a faithful

S [r]

-module

M

that is finitely generated as an

S

-module. Show that

r

is integral over

S

. (Hint: change a few words in the proof of version 1 of Nakayama, Lemma 7.2.8.)

Proof:

Let $m_{1}, \dots, m_{n}$ be a generating set of $M$ as an $S$ -module and suppose $r m_{i} = \sum a_{i j} m_{j}$ for some $a_{i j} \in M$ . Taking $Z = (a_{i j})$ this implies that

(r I_{n \times n} - Z) (\begin{matrix} m_{1} / / ⋮ / / m_{n} \end{matrix}) = 0

Multiplying through by $adj (r I_{n \times n} - Z)$ implies that

det (r I_{n \times n} - Z) (\begin{matrix} m_{1} / / ⋮ / / m_{n} \end{matrix}) = 0

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.2.J:

Suppose

A

is an integral domain, and

\tilde{A}

is the integral closure of

A

K (A)

, i.e., those elements of

K (A)

integral over

A

, which form a subalgebra by Exercise 7.2.D. Show that

\tilde{A}

is integrally closed in

K (\tilde{A}) = K (A)

Proof:

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

\begin{matrix} ◼ \end{matrix}

A gazillion finiteness conditions on morphisms

Exercise 7.3.A:

Show that the composition of two quasicompact morphisms is quasicompact. (It is also true — but not easy — that the composition of two quasiseparated morphisms is quasiseparated. This is not impossible to show directly, but will in any case follow easily once we understand it in a more sophisticated way, see Proposition 10.1.13(b).)

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.B:

Show that any morphism from a Noetherian scheme is quasicompact.
Show that any morphism from a quasiseparated scheme is quasiseparated. Thus by Exercise 5.3.A, any morphism from a locally Noetherian scheme is quasiseparated. Thus readers working only with locally Noetherian schemes may take quasiseparatedness as a standing hypothesis.

Proof:

Let $ϕ : X \to Y$ be a morphism of schemes with $X$ Noetherian and suppose $U \subset Y$ is affine. Since $ϕ$ is continuous we have that $ϕ^{- 1} (U) \subset X$ is open. By Exercise 3.6.T since $X$ is Noetherian as a topological space we have that $ϕ^{- 1} (U)$ is open.
Let $ϕ : X \to Y$ be a morphism of schemes with $X$ quasiseparated and suppose $U \subset Y$ is affine. Then $ϕ^{- 1} (U)$ is open; we wish to show that it is in addition quasiseparated. Let $W_{1}, W_{2}$ be affine open subsets of $ϕ^{- 1} (U)$ ; then as open affines of $X$ , we know that $W_{1} \cap W_{2}$ is a finite union of open affine sets (by Exercise 5.1.F).

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.C:

(quasicompactness is affine-local on the target) Show that a morphism $π : X \to Y$ is quasicompact if there is a cover of $Y$ by affine open sets $U_{i}$ such that $π^{- 1} (U_{i})$ is quasicompact.
(quasiseparatedness is affine-local on the target) Show that a morphism $π : X \to Y$ is quasiseparated if there is a cover of $Y$ by affine open sets $U_{i}$ such that $π^{- 1} (U_{i})$ is quasiseparated.

Proof:

Suppose that $V \subset Y$ is an affine open subset; for each $i$ we have $V \cap U_{i}$ is a distinguished open of $U_{i}$ and is therefore also affine. Since $π |_{π^{- 1} (U_{i}})$ is quasicompact (as a morphism), we have that $π^{- 1} (U_{i} \cap V)$ is also quasicompact and thus a finite union of open affines. Taking the union over all $i$ is still a finite union of open affines so that $π : X \to Y$ is quasicompact.
In a similar fashion, let $V \subset Y$ be an affine open subset so that $V \cap U_{i}$ is a distinguished open (and thus affine) for each $i$ . Since $π |_{π^{- 1} (U_{i})}$ is quasiseparated for each $i$ , we have that $π^{- 1} (U_{i} \cap V)$ is quasicompact. Thus, $π^{- 1} (V)$ is the finite union of quasicompact sets and thus quasicompact.

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.D:

Show that affine morphisms are quasicompact and quasiseparated. (Hint for the second: Exercise 5.1.G.)

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.E:

What is the natural map

Γ (X, O_{X})_{s} \to Γ (X_{s}, O_{X})

of the Qcqs Lemma 7.3.5? (Hint: the universal property of localization, Exercise 1.3.D.)

Proof:

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

\frac{f}{s^{n}} \mapsto f \cdot (s^{- 1})^{n}

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.F:

Suppose

Z

is a closed subset of an affine scheme

Spec A

locally cut out by one equation. (In other words,

Spec A

can be covered by smaller open sets, and on each such set

Z

is cut out by one equation.) Show that the complement

Y

Z

is affine. (This is clear if

Z

is globally cut out by one equation

f

, even set-theoretically; then

Y = Spec A_{f}

. However,

Z

is not always of this form, see §19.11.10.)

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.G:

Show that a morphism

π : X \to Y

is finite if there is a cover of

Y

by affine open sets

Spec A

such that

π^{- 1} (Spec A)

is the spectrum of a finite

A

-algebra.

Proof:

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

For each $i$ , suppose $g_{i, 1}, \dots, g_{i, n_{i}}$ is a generating set of $B_{π^{♯} f_{i}}$ (as an $A_{f_{i}}$ -module) for some $g_{i, j} \in B$ . Taking large enough $N$ , we may define a map $ϕ : B^{\oplus N} \to A$ such that at each $f_{i}$ we have $(coker ϕ)_{f_{i}} = 0$ . Then for any $x \in coker ϕ$ there exist $r_{i} \geq 0$ such that $f_{i}^{r_{i}} x = 0$ . Taking $r = \prod r$ , notice that since $(f_{1}, \dots, f_{n}) = 1$ we also have $(f_{1}^{r}, \dots, f_{n}^{r}) = 1$

x = 1 \cdot x = \sum c_{i} f_{i}^{r} x = 0

Therefore, $coker ϕ = 0$ so that $B$ is a finitely generated $A$ -algebra.

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.H:

Show that if

X \to Spec k

is a finite morphism, then

X

is a finite union of points with the discrete topology, each point with residue field a finite extension of

k

, see Figure 7.5. (An example is

Spec (F_{8} \times F_{4} [x, y] / (x^{2}, y^{4}) \times F_{4} [t] / (t^{9}) \times F_{2}) \to Spec F_{2}

.) Do not just quote some fancy theorem! Possible approach: By Exercise 3.2.G, any integral domain which is a finite

k

-algebra must be a field. If

X = Spec A

, show that every prime

p

A

is maximal. Show that the irreducible components of

Spec A

are closed points. Show

Spec A

is discrete and hence finite. Show that the residue fields K(A/p) of A are finite extensions of k. (See Exercise 7.4.D for an extension to quasifinite morphisms.)

Proof:

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

A / (m_{1} \cdot m_{2} \dots m_{n}) ≅ A / m_{1} \times A / m_{2} \times \dots \times A / m_{n}

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.I:

Show that the composition of two finite morphisms is also finite.

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.J:

R

is an

A

-algebra,define a graded ring

S_{∙}

S_{0} = A

,and

S_{n} = R

for

n > 0

. (What is the multiplicative structure?\) Hint: you know how to multiply elements of

R

together, and how to multiply elements of

A

with elements of

R

.) Describe an isomorphism

Proj S_{∙} ≅ Spec R

. Show that if

R

is a finite

A

-algebra (finitely generated as an

A

-module) then

S_{∙}

is a finitely generated graded ring over

A

, and hence that

Spec R

is a projective

A

-scheme (§4.5.9).

Proof:

The $Z$ -grading on $S_{∙}$ occurs in a natural way; $S_{0} = A$ acts on each $S_{n} = R$ as $R$ is an $A$ -algebra. Moreover, for $a \in R = S_{i}$ and $b \in R = S_{j}$ we declare $a b \in S_{i + j}$ as this is required to make this a graded $A$ -algebra (one should keep in mind that $S_{∙}$ looks similar to $R [x]$ , so as $Proj R [x] ≅ Spec R$ we hope to find a bijection $S_{∙} \leftrightarrow R [x]$ — see Exercise 6.4.A). Notice from our $A$ -structure on $S_{∙}$ , it is ambiguous how to embed any element $f \in R$ into $S_{∙}$ as $[f]_{n} = [f]_{1} [1_{R}]_{1}^{n + 1}$ ; however, as we only care about prime ideals, this allows us to construct a well-defined bijection $Spec (S_{[f]_{1}})_{0} \to Spec R$ . For each $f \in R$ , we define $R \to (S_{[f]_{1}})_{0}$ by

r \mapsto \frac{[r f]_{1}}{[f]_{1}}

In fact, this gives us an isomorphism $R ≅ (S_{[1]_{1}})_{0}$ in quite an obvious way. Notice that if $p \in Proj S_{∙}$ with $p \notin D_{+} ([1]_{1})$ then $p$ must contain $S_{+}$ since for any $[f]_{n} = [f]_{n - 1} [1]_{n - 1}$ . Thus, $Proj S_{∙} = D_{+} ([1]_{1})$

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.K:

Show that finite morphisms have finite fibers. (This is a useful exercise, because you will have to figure out how to get at points in a fiber of a morphism: given

π : X \to Y

, and

q \in Y

, what are the points of

π^{- 1} (q)

? This will be easier to do once we discuss fibers in greater detail, see Remark 9.3.4, but it will be enlightening to do it now.) Hint: if

X = Spec A

and

Y = Spec B

are both affine, and

q = [q]

, then we can throw out everything in

B

outside

q

by modding out by

q

; show that the preimage is

Spec (A / π^{♯} q A)

. Then you have reduced to the case where

Y

is the

s p e c

of an integral domain

B

, and

[q] = [(0)]

is the generic point. We can throw out the rest of the points of

B

by localizing at

(0)

. Show that the preimage is

Spec

A

localized at

π^{♯} (B ∖ {0})

. Show that the condition of finiteness is preserved by the constructions you have done, and thus reduce the problem to Exercise 7.3.H.

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.L:

Show that the open embedding

A_{C}^{1} - {0} \to A_{C}^{1}

has finite fibers and is affine, but is not finite.

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.M:

Prove that integral morphisms are closed, i.e., that the image of closed subsets are closed. (Hence finite morphisms are closed. A second proof will be given in §8.2.5.) Hint: Reduce to the affine case. If

π^{♯} : B \to A

is a ring map, inducing the integral morphism

π : Spec A \to Spec B

, then suppose

I \subset A

cuts out a closed set of

Spec A

, and

J = (π^{♯})^{- 1} (I)

, then note that

B / J \subset A / I

, and apply the Lying Over Theorem 7.2.5 here.

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.N:

Suppose

B \to A

is integral. Show that for any ring homomorphism

B \to C

, the induced map

C \to A \otimes_{B} C

is integral. (Hint: We wish to show that any

\sum_{i = 1}^{n} a_{i} \otimes c_{i} \in A \otimes_{B} C

is integral over

C

. Use the fact that each of the finitely many

a_{i}

are integral over

B

, and then Exercise 7.2.D.) Once we know what "base change" is, this will imply that the property of integrality of a morphism is preserved by base change, Exercise 9.4.B(e).

Proof:

Let $ϕ : B \to A$ and $ψ : B \to C$ be our ring morphisms giving $A$ and $C$ the structure of $B$ -algebras. Recall that $A \otimes_{B} C$ comes with the $B$ -algebra structure induced by the commutative diagram

\begin{matrix} B & \overset{ϕ}{\to} & A \\ ψ ↓ & ↓ {id}_{A} \otimes 1_{C} \\ C & \to_{1_{A} \otimes {id}_{C}}^{} & A \otimes_{B} C \end{matrix}

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

\begin{aligned} 0 = 0 \otimes c^{n} & = (a^{n} \otimes c^{n}) + (ϕ (b_{n - 1}) a^{n - 1} \otimes c^{n}) + \dots + (ϕ (b_{0}) \otimes c^{n}) \\ = (a^{n} \otimes c^{n}) + (ϕ (b_{n - 1}) \otimes c) (a^{n - 1} \otimes c^{n - 1}) + \dots + (ϕ (b_{0}) \otimes c^{n}) (1 \otimes 1) \\ = (a^{n} \otimes c^{n}) + (1 \otimes ψ (b_{n - 1}) c) (a^{n - 1} \otimes c^{n - 1}) + \dots + (1 \otimes ψ (b_{0}) c^{n}) (1 \otimes 1) \end{aligned}

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.O:

Show that a morphism

π : X \to Y

is locally of finite type if there is a cover of

Y

by affine open sets

Spec B_{i}

such that

π^{- 1} (Spec B_{i})

is locally finite type over

B_{i}

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.P:

(easier) Show that finite morphisms are of finite type.
Show that a morphism is finite if and only if it is integral and of finite type.

Proof:

This is trivial by definition since if $π : X \to Y$ is a finite morphism and $Spec A \subset Y$ , then $π^{- 1} (Spec A) = Spec B$ with $B$ a finite $A$ -algebra and thus finitely generated $A$ -algebra.
The forward direction follows from the previous part (and finite algebras are integral). Conversely, suppose $π : X \to Y$ is integral and of finite-type. Since integral maps are affine, we may restrict to the case $Y = Spec B$ and $X = Spec A$ so that $A$ is integral and finite type over $B$ . To see that $A$ is a finite $B$ -algebra, let $x_{1}, \dots, x_{n}$ be the generators of $A$ (as a $B$ -algebra). Now $x_{1}$ is integral over $B$ so $B [x_{1}]$ is a $B$ -subalgebra finite over $B$ by Lemma 7.2.1. Similarly, $x_{2}$ is integral over $B [x_{1}]$ so by induction the result follows.

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.Q:

Show that every open embedding is locally of finite type, and hence that every quasicompact open embedding is of finite type. Show that every open embedding into a locally Noetherian scheme is of finite type.
Show that the composition of two morphisms locally of finite type is locally of finite type. (Hence as the composition of two quasicompact morphisms is quasicompact, Easy Exercise 7.3.A, the composition of two morphisms of finite type is of finite type.)
Suppose $π : X \to Y$ is locally of finite type, and $Y$ is locally Noetherian. Show that $X$ is also locally Noetherian. If $π : X \to Y$ is a morphism of finite type, and $Y$ is Noetherian, show that $X$ is Noetherian.

Proof:

This follows since if we restrict to $X = Spec A$ and $U \to Spec A$ with $U = ⋃ D (f_{i})$ then each $A_{f_{i}}$ is finitely generated by $f_{i}^{- 1}$ over $A$ . If $X$ is locally Noetherian, then we may modify the above assumption $X = Spec A$ to the case that $A$ is a Noetherian ring. By Exercise 3.6.T, every open subset of a Noetherian topological space is quasicompact; thus, $U \to X$ is a quasicompact morphism and therefore of finite type.
Suppose $ϕ : X \to Y$ and $ψ : Y \to Z$ are locally of finite type and let $U = Spec A$ be an affine open. Using the equivalent definition of locally of finite type, $ψ^{- 1} (Spec A)$ may be covered by affine opens $Spec B_{i}$ such that each $B_{i}$ is a finitely generated $A$ -algebra. Similarly for each $i$ , $ϕ^{- 1} (Spec B_{i})$ may be covered by affine opens $Spec C_{i j}$ such that $C_{i j}$ is a finitely generated $B_{i}$ -algebra. Since each $B_{i}$ is a finitely generated $A$ -algebra, we get that $C_{i j}$ is a finitely generated $A$ -algebra.
Since $Y$ is locally Noetherian, it may be covered by open affines $Spec A$ where each $A$ is Noetherian. Since $π : X \to Y$ is locally of finite type, $π^{- 1} (Spec A)$ may be covered by $Spec B_{i}$ where $B_{i}$ is a finitely generated $A$ -algebra. By Exercise 3.6.X, since $B_{i}$ is finitely-generated as a module and $A$ is Noetherian, $B_{i}$ must be Noetherian as well. By running over our affine cover $Spec A$ , we may cover $X$ by $Spec B_{i}$ with $B_{i}$ Noetherian so that $X$ is locally Noetherian.

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.R:

Suppose

p

is prime and

r \in Z^{+}

. Let

q = p^{r}

, and

k = F_{q}

. Define

ϕ : k [x_{1}, \dots, x_{n}] \to k [x_{1}, \dots, x_{n}]

ϕ (x_{i}) = x_{i}^{p}

for each

i

, and let

F : A_{k}^{n} \to A_{k}^{n}

be the map of schemes corresponding to

ϕ

Show that $F^{r}$ is the identity on the level of sets, but is not the identity morphism.
Show that $F$ is a bijection, but is not an isomorphism of schemes.
If $K = {\overset{―}{F}}_{p}$ , show that the morphism $F : A_{K}^{n} \to A_{K}^{n}$ of $K$ -schemes corresponding to $x_{i} \mapsto x_{i}^{p}$ is a bijection, but no power of $F$ is the identity on the level of sets.

Proof:

Recall that $F^{r}$ as a map of sets is simply the map $Spec k [x_{1}, \dots, x_{n}] ∋ p \mapsto (ϕ^{r})^{- 1} (p)$ . Now if $ϕ^{r} (f) = f^{p^{r}} \in p$ then $f \in p$ (since $p$ is prime) so that $p = (ϕ^{r})^{- 1} (p)$ — therefore, the map of sets is indeed a bijection.

Now by construction, $F_{q}$ is the splitting field of $X^{p^{r}} - X$ over $F_{p}$ so that every element $a \in k$ satisfies $a^{p^{r}} = a$ . However, this need not hold for elements $f \in k [x_{1}, \dots, x_{n}] - k$ (for example, $x_{1}$ is free so $ϕ^{r} (x_{i}) = x_{i}^{p^{r}} \neq x_{i}$ ).
On the level of sets, $F$ is clearly a bijection since $F^{- 1} = F^{r - 1}$ . However, $F$ cannot be an isomorphism since $F^{*} = ϕ : k [x_{1}, \dots, x_{n}] \to k [x_{1}, \dots, x_{n}]$ is not surjective (pick $x^{s}$ with $s$ coprime to $p$ )
If we consider $S$ to be the set of all polynomials of the form $x^{p^{n}} - x$ for some $n > 0$ then $K = \overset{―}{F_{p}}$ is the splitting field of $S$ (a common Galois theory exercise). To see that $F$ is a bijection requires a little more work than before, as it need not hold that $ϕ (f)$ is a power of $f$ for any $f \in K [x_{1}, \dots, x_{n}]$ . However, if we write $f = \sum_{ν} a_{ν} x^{ν}$ for some multi-index $ν$ , then for each $ν$ there exists some $r_{ν}$ such that $a^{p^{r_{ν}}} = a$ and thus $ϕ^{r_{ν}} (a_{ν} x^{ν}) = a_{ν} (x^{ν})^{p^{r_{ν}}} = (a_{ν} x^{ν})^{p^{r_{ν}}}$ . Taking $r = \prod_{ν} r_{ν}$ , we have that $ϕ^{r} (f) = f^{p^{r}}$ . Now any prime $p \in K [x_{1}, \dots, x_{n}]$ is finitely-generated (as $K [x_{1}, \dots, x_{n}]$ is obviously Noetherian) so that we may apply this argument to a finite set of generators of $p$ giving us some $s >> 0$ with $(ϕ^{s})^{- 1} (p) = p$ . Thus, our map $A_{K}^{n} \to A_{K}^{n}$ must be a bijection. However, our integer $s$ depends on each point $[p] \in A_{K}^{n}$ so that no $s$ works universally. That is to say if we choose some arbitrary $r^{'} > 0$ and consider the map $F^{r^{'}} (x_{1} - a) = x_{1}^{p^{r^{'}}} - a$ then we could simply choose $a \in K$ such that $a$ is not a root of $X^{p^{r^{'}}} - X$ .

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.S:

Suppose

X

is a scheme over

F_{p}

. Explain how to define (without choice) an endomorphism

F : X \to X

such that for each affine open subset

Spec A \subset X

F

corresponds to the map

A \to A

given by

f \mapsto f^{p}

for all

f \in A

.(The morphism

F

is called the absolute Frobenius morphism.)

Proof:

If we let ${Spec A_{i}}$ be any affine cover, then by definition each $A_{i}$ is a $F_{p}$ -algebra so we get a map $O_{X} (Spec A_{i}) \to O_{X} (Spec A_{i})$ given by $f \mapsto f^{p}$ , which by the previous exercise induces the identity on points of $Spec A_{i}$ . Then these maps clearly glue together since our open sets glue.

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.T:

Show that the notion of "locally of finite presentation" is affine-local on the target.

Proof:

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

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.U:

Show that if

π : X \to Y

is locally of finite presentation, then for any affine open subscheme

Spec B

Y

and any affine open subscheme

Spec A

X

with

π (Spec A) \subset Spec B

A

is a finitely presented

B

-algebra. In particular, the notion of "locally of finite presentation" is affine-local on the source.

Proof:

This ammounts to showing that if $(f_{1}, \dots, f_{n}) = A$ and $A_{f_{i}}$ a finitely-presented $B$ -algebra for each $i$ , then $A$ is a finitely-presented $B$ -algebra. I will outsource this proof to [Stacks 00EO].

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.V:

Show that open embeddings are locally finitely presented.

Proof:

Let $ι : U ↪ X$ be our open embedding. Indeed we may reduce to the case $X = Spec A$ and consider the identity morphism $id : X \to X$ . Taking $U = D (f_{i}) = Spec A_{f_{i}} \subset X$ , we simply apply the previous exercise.

\begin{matrix} ◼ \end{matrix}

Exercise 7.3.W:

Show that the composition of two locally finitely presented morphisms is locally finitely presented. (Then once we show that the composition of two quasiseparated morphisms is quasiseparated in Proposition 10.1.13(b), we will know that the composition of two finitely presented morphisms is finitely presented — recall that the composition of two quasicompact morphisms is quasicompact, by Easy Exercise 7.3.A.)

Proof:

It suffices to reduce to the affine case, so that we have ring maps $ϕ : A \to B$ and $ψ : B \to C$ where $B$ is a finitely presented $A$ -algebra and $C$ is a finitely presented $B$ algebra. Thus, there exist integers $m, n > 0$ and surjections $π : A [x_{1}, \dots, x_{n}] \to B$ and $ρ : B [y_{1}, \dots, y_{m}] \to C$ such that $\ker (π)$ and $\ker (ρ)$ are finitely-generated ideals of $A [x_{1}, \dots, x_{n}]$ and $B [y_{1}, \dots, y_{m}]$ respectively. Then $π$ may be extended to a morphism $\tilde{π} : A [x_{1}, \dots, x_{n}, y_{1}, \dots, y_{m}] \to B [y_{1}, \dots, y_{m}]$ in the natural way, so that $\ker (\tilde{π}) = \ker (π) [y_{1}, \dots, y_{m}]$ is finitely generated in $A [x_{1}, \dots, x_{n}, y_{1}, \dots, y_{m}]$ . By taking compositions, we see that $ρ \circ \tilde{π}$ is clearly a surjection, so that $C$ is a finitely-generated $A$ -algebra. To see that it is finitely presented, we must show that $\ker (ρ \circ \tilde{π})$ is finitely generated.

Suppose $x \in \ker (ρ \circ \tilde{π})$ , or in other words $\tilde{π} (x) \in \ker ρ$ . Since $\ker ρ$ is assumed to be finitely generated in $B [y_{1}, \dots, y_{m}]$ (say by some $v_{1}, \dots, v_{k}$ ) we have that $\tilde{π} (x) = \sum g_{i} v_{i}$ for $g_{i} \in B [y_{1}, \dots, y_{m}]$ . Since $\tilde{π}$ surjects onto $B [y_{1}, \dots, y_{m}]$ , we can find some $f_{i j}, w_{i j} \in A [x_{1}, \dots, x_{n}, y_{1}, \dots, y_{m}]$ with $\tilde{π} (\sum_{i, j} f_{i j} w_{i j}) = \sum_{i} g_{i} v_{i}$ . By linearity, this implies $\tilde{π} (x - \sum_{i, j} f_{i j} w_{i j}) = 0$ or $x - \sum_{i, j} f_{i j} w_{i j} \in \ker \tilde{π}$ . Since this is finitely generated by $\ker π$ and $y_{1}, \dots, y_{m}$ , we must have that $x$ is generated by $\ker (π), y_{1}, \dots, y_{m}$ and $\ker ρ$ .

\begin{matrix} ◼ \end{matrix}

Images of morphisms: Chevalley’s Theorem and elimination theory

Exercise 7.4.A:

Recall that a subset of a topological space X is locally closed if it is the intersection of an open subset and a closed subset. (Equivalently, it is an open subset of a closed subset, or a closed subset of an open subset. We will later have trouble extending this to open and closed and locally closed subschemes, see Exercise 8.1.M.) Show that a subset of a Noetherian topological space

X

is constructible if and only if it is the finite disjoint union of locally closed subsets. As a consequence, if

X \to Y

is a continuous map of Noetherian topological spaces, then the preimage of a constructible set is a constructible set. (Important remark: the only reason for the hypothesis of the topological space in question being Noetherian is because this is the only setting in which we have defined constructible sets. An extension of the notion of constructibility to more general topological spaces is mentioned in Exercise 9.3.I.)

Proof:

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

Now suppose we let $F^{'}$ denote the collection of subsets of $X$ that can be written as a finite disjoint union of locally closed subsets; by the previous paragraph we have $F^{'} \subset F$ . To show that $F^{'} = F$ , it suffices to show that $F^{'}$ also satisfies properties (i), (ii), and (iii). Clearly $F^{'}$ contains all open sets $U \subset X$ since $U \cap X = U$ and $X$ is closed. To see that $F^{'}$ is closed under intersections, notice that if $\underset{i}{∐} Z_{i} \cap U_{i}, \underset{j}{∐} Z_{j}^{'} \cap U_{j}^{'} \in F^{'}$ , then the intersection is $\underset{i, j}{∐} (Z_{i} \cap Z_{j}^{'}) \cap (U_{i} \cap U_{j}^{'})$ which is also clearly an element of $F^{'}$ . To prove that $F^{'}$ is closed under compliments, we let $F_{n}^{'}$ denote the collection of subsets of $X$ which can be expressed as a disjoint union of exactly $n$ locally-closed sets. First notice that for any $Y \in F_{1}^{'}$ we have $Y = Z \cap U$ . Then $Y^{c} = Z^{c} \cup U^{c} = (U^{c} \cap X) ∐ (Z^{c} \cap U)$ which is clearly in $F^{'}$ . Proceeding inductively, suppose that compliments in $F_{i}^{'}$ are contained in $F^{'}$ for $i < n$ . If we let $Y \in F_{n}^{'}$ then we can write $Y = Y_{n - 1} ∐ Y_{1}$ for some $Y_{n - 1} \in F_{n - 1}^{'}$ and $Y_{1} \in F_{1}$ . Then $Y^{c} = Y_{n - 1}^{c} \cap Y_{1}^{c}$ ; by inductive hypothesis both $Y_{n - 1}^{c} \in F^{'}$ and $Y_{1}^{c} \in F^{'}$ and we have shown $F^{'}$ is closed under intersection. Since $F^{'} = ⋃_{n} F_{n}^{'}$ , this completes the proof.

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.B:

Show that the generic point of

A_{k}^{1}

does not form a constructible subset of

A_{k}^{1}

(where

k

is a field).

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.C:

Show that a constructible subset of a Noetherian scheme is closed if and only if it is "stable under specialization". More precisely, if $Z$ is a constructible subset of a Noetherian scheme $X$ , then $Z$ is closed if and only if for every pair of points $y_{1}$ and $y_{2}$ with $y_{1} \in y_{2}$ , if $y_{2} \in Z$ , then $y_{1} \in Z$ . Hint for the "if" implication: show that $Z$ can be written as $∐_{i = 1}^{n} U_{i} \cap Z_{i}$ where $U_{i} \subset X$ is open and $Z_{i} \subset X$ is closed. Show that $Z$ can be written as $∐_{i = 1}^{n} U_{i} \cap Z_{i}$ (with possibly different $n$ , $U_{i}$ , $Z_{i}$ ) where each $Z_{i}$ is irreducible and meets $U_{i}$ . Now use &qupt;stability under specialization" and the generic point of $Z_{i}$ to show that $Z_{i} \subset Z$ for all $i$ , so $Z = ⋃ Z_{i}$ .
Show that a constructible subset of a Noetherian scheme is open if and only if it is "stable under generization". (Hint: this follows in one line from (a).)

Proof:

For the "only if" direction, suppose that $Z$ is closed and let $y_{1} \in \overset{―}{{y_{2}}}$ with $y_{2}$ in $Z$ . Since $Z$ is closed, $\overset{―}{{y_{2}}} \subset Z$ so that $y_{1} \in Z$ as well. Conversely, suppose $Z$ is constructible and stable under specialization; as $Z$ is constructible, we may write $Z = ∐_{i = 1}^{n} U_{i} \cap Z_{i}$ by Exercise 7.4.A with each $U_{i}$ open and $Z_{i}$ closed. As $X$ is Noetherian, we know that each $Z_{i}$ may be written as $Z_{i} = Z_{i, 1} \cup \dots \cup Z_{i, n_{1}}$ for $Z_{i, j}$ irreducible components of $Z$ (by Proposition 3.6.15). Now by using an inductive argument, if $Z_{i} = Z_{i, 1} \cup Z_{i, 2}$ then we may write $Z_{i} \cap U_{i} = (Z_{i, 1} \cap (Z_{i, 2}^{c} \cap U)) \cup (Z_{1} \cap U)$ — however, $Z_{i, 1}$ now may not be reducible in the Noetherian subspace $(Z_{i, 2}^{c} \cap U)$ , so we must again break it into finitely many irreducible components (as a helpful example, take $U$ to be an open neighborhood of the origin and consider $Z = V (x y)$ — by removing $V (x)$ we have split $V (y)$ into two irreducible components). After finitely many steps, this allows us to write $Z = ∐ Z_{i} \cap U_{i}$ where each $Z_{i}$ is irreducible. Let $η_{i}$ denote the unique generic point of $Z_{i}$ - since ${η_{i}}$ nontrivially intersects any open set intersecting $Z_{i}$ , we know $η_{i} \in U_{i}$ so that $η_{i} \in Z$ . Thus, for any $x \in ⋃_{i = 1}^{n} Z_{i}$ we know that $x \in \overset{―}{{η_{j}}}$ for some $η_{j}$ ; as we just showed $η_{j} \in Z$ this implies that $x \in Z$ since $Z$ is stable under specialization. In other words, $⋃_{i = 1}^{n} Z_{i} \subset Z = ∐_{i = 1}^{n} Z_{i} \cap U_{i}$ . However since $U_{i} \cap Z_{i} \subset Z_{i}$ for all $i$ we get the reverse containment so that $Z$ is closed.
Much like the previous part, it is immediate that open sets are stable under generization: if $Z$ is open and $y_{1} \in \overset{―}{{y_{2}}}$ with $y_{1} \in Z$ , then by definition every open neighborhood of $y_{1}$ nontrivially intersects ${y_{2}}$ . Since $Z$ is such an open neighborhood, $y_{2} \in Z$ . For the reverse direction, if $Z$ is a constructible set stable under generization, then $Z^{c}$ is a constructible set stable under specialization and therefore closed. Thus $Z$ is open.

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.D:

Suppose

π : X \to Spec k

is a quasifinite morphism. Show that

π

is finite. (Hint: deal first with the affine case,

X = Spec A

, where

A

is finitely generated over

k

. Suppose

A

contains an element

x

that is not algebraic over

k

, i.e., we have an inclusion

k [x] ↪ A

. Exercise 7.3.H may help.)

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.E:

Assume Chevalley’s Theorem 7.4.2. Show that a morphism of affine

k

-varieties

π : X \to Y

is surjective if and only if it is surjective on closed points (i.e., if every closed point of

Y

is the image of a closed point of

X

). (Once we define varieties in general, in Definition 10.1.7, you will see that your argument works without change with the adjective "affine" removed.)

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.F:

Show that

B

itself satisfies

(†)

Proof:

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

x_{j} = \frac{a_{1, j}}{b_{i, j}} x_{1} / 1 + \dots + \frac{a_{n, j}}{b_{n, j}} x_{n} / 1

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.G:

Reduce the proof of Lemma 7.4.4 to the following statement: if

A

is a finitely-generated

B

-algebra satisfying

(†)

, then

A [T]

does too. (Hint: induct on the number of generators of

A

as a

B

-algebra.)

Proof:

If $A$ is a finitely generated $B$ -algebra, then it is isomorphic to $B [T_{1}, \dots, T_{n}] / I$ for some $n \geq 0$ and ideal $I \subset B [T_{1}, \dots, T_{n}]$ . Thus, a finitely generated $A$ -module $M$ is a finitely-generated $B [T_{1}, \dots, T_{n}]$ -module with $I M = 0$ . By assuming the statement in the exercise, one may use induction to show that $M$ as a $B [T_{1}, \dots, T_{n}]$ -module must have some corresponding $f \in B$ such that $M_{f}$ is a free $B_{f}$ -module.

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.H:

Show that multiplication by

T

induces a surjection

ψ_{n} : M_{n} / M_{n - 1} \to M_{n + 1} / M_{n}

Proof:

By inductive construction of our $M_{n}$ , we can simply express our $A$ -submodules as

M_{n} = ⟨ S ⟩ + T ⟨ S ⟩ + \dots + T^{n} ⟨ S ⟩

Then for any $r > 0$ , we have that $M_{i} / M_{i - 1} ≅ T^{i} ⟨ S ⟩ S$ , from which it is obvious that multiplication by $T$ gives a surjection $ψ_{n} : M_{n} / M_{n - 1} \to M_{n + 1} / M_{n}$ .

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.I:

Show that for

n ≫ 0

ψ_{n}

is an isomorphism. Hint: use the ascending chain condition on

M_{1}

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.J:

Show that there is a nonzero

f \in B

such that

(M_{i + 1} / M_{i})_{f}

is free as a

B_{f}

-module, for all

i

. Hint: as

i

varies,

M_{i + 1} / M_{i}

passes through only finitely many isomorphism classes.

Proof:

Since we are assuming $A$ satisfies $(†)$ , each $M_{i + 1} / M_{i}$ is a finitely-generated $A$ -module so that there exists some nonzero $f_{i} \in B$ such that $(M_{i + 1} / M_{i})_{f_{i}}$ is a free $B_{f_{i}}$ -module. Since there are only finitely many isomorphism classes by the previous exercise, we may take $f = f_{1} f_{2} \dots f_{n}$ for sufficiently large $n ≫ 0$ so that the claim follows.

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.K:

Suppose

M

is a

B

-module that is an increasing union of submodules

M_{i}

, with

M_{0} = 0

, and that every

M_{i + 1} / M_{i}

is free. Show that

M

is free. Hint: first construct compatible isomorphisms

ϕ_{n} : ⨁_{i = 0}^{n - 1} M_{i + 1} / M_{i} \to M_{n}

by induction on

n

. Then show that the colimit

ϕ := \lim_{\leftarrow} ϕ_{n} : ⨁_{n = 0}^{\infty} M_{i + 1} / M_{i} \to M

is an isomorphism. Side Remark: More generally, your argument will show that if the

M_{i + 1} / M_{i}

are all projective modules (to be defined in §23.2.1), then

M

is (non-naturally) isomorphic to their direct sum.

Proof:

We first wish to construct the compatible isomorphisms $ϕ_{n} : ⨁_{i = 0}^{n - 1} M_{i + 1} / M_{i} \to M_{n}$ . The base case $n = 1$ is obvious since $M_{0} = 0$ . Now suppose $ϕ_{j}$ is an isomorphism for $j \leq n$ , and consider the short exact sequence

0 \to M_{n} \to M_{n + 1} / M_{n} \to M_{n + 1} \to 0

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

ϕ := \lim_{\leftarrow} ϕ_{n} : ⨁_{n = 0}^{\infty} M_{i + 1} / M_{i} \to M

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.L:

Suppose

π : X \to Y

is a finite type morphism of Noetherian schemes, and

Y

is irreducible. Show that there is a dense open subset

U

Y

such that the image of

π

either contains

U

or else does not meet

U

. (Hint: suppose

π : Spec A \to Spec B

is such a morphism. Then by the Generic Freeness Lemma 7.4.4, there is a nonzero

f \in B

such that

A_{f}

is a free

B_{f}

-module. It must have zero rank or positive rank. In the first case, show that the image of

π

does not meet

D (f) \subset Spec B

. In the second case, show that the image of

π

contains

D (f)

Proof:

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

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.M:

Show that to prove Chevalley’s Theorem, it suffices to prove that if

π : X \to Y

is a finite type morphism of Noetherian schemes, the image of

π

is constructible.

Proof:

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.N:

Reduce further to the case where

Y

is affine, say

Y = Spec B

. Reduce further to the case where

X

is affine

Proof:

Since $Y$ is a Noetherian scheme, it is quasicompact by Exercise 3.6.T; thus we may take some finite affine cover $Spec B_{i}$ for $Y$ . But then

π (X) = ⋃_{i = 1}^{n} π (π^{- 1} (Spec B_{i}))

Since the finite union of constructible subsets is constructible, it suffices to consider $π |_{π^{- 1} (Spec B_{i})}$ --- in other words, simply consider $Y = Spec B$ .

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

\begin{matrix} ◼ \end{matrix}

Exercise 7.4.O:

Complete the proof of Chevalley’s Theorem 7.4.2, by making the above argument precise.

Exercise 7.4.P:

Fix an algebraically closed field

\overset{―}{k}

. Suppose

f_{1}, \dots, f_{p}, g_{1}, \dots, g_{q} \in \overset{―}{k} [W_{1}, \dots, W_{m}, X_{1}, \dots, X_{n}]

are given. Show that there is a (Zariski)-constructible subset

Y

{\overset{―}{k}}^{m}

such that

f_{1} (w_{1}, \dots, w_{m}, X_{1}, \dots, X_{n}) = \dots = f_{p} (w_{1}, \dots, w_{m}, X_{1}, \dots, X_{n}) = 0

and

g_{1} (w_{1}, \dots, w_{m}, X_{1}, \dots, X_{n}) \neq 0 \dots g_{q} (w_{1}, \dots, w_{m}, X_{1}, \dots, X_{n}) \neq 0

has a solution

(X_{1}, \dots, X_{n}) = (x_{1}, \dots, x_{n}) \in {\overset{―}{k}}^{n}

if and only if

(w_{1}, \dots, w_{m}) \in Y

Proof:

Following the hint, let $A = k [W_{1}, \dots, W_{m}, X_{1}, \dots, X_{n}]$ so that $V (f_{1}, \dots, f_{p}) = Spec A / (f_{1}, \dots, f_{p}) ↪ Spec A = A^{n + m}$ is finite type, as $A / (f_{1}, \dots, f_{n})$ is finitely-generated over $A$ . Now for each $D (g_{i})$ , since $A_{f_{i}}$ is a finitely-generated $A$ -algebra, $D (g_{i}) ↪ V (f_{1}, \dots, f_{p})$ is of finite type. Since there are finitely many $g_{i}$ , we get that the morphism $D (g_{1}, \dots, g_{q}) ↪ V (f_{1}, \dots, f_{m})$ is of finite type and thus the composition. Thus, the map $D (g_{1}, \dots, g_{q}) ↪ A^{m + n}$ is of finite type so by Chevalley's theorem, it has a constructible image. Since $\overset{―}{k} [W_{1}, \dots, W_{m}, X_{1}, \dots, X_{n}]$ is finitely generated over $\overset{―}{k} [W_{1}, \dots, W_{m}]$ , the morphism $A^{m + n} \to A^{m}$ is of finite-type. Applying the Nullstellensatz cleans things up a bit.

\begin{matrix} ◼ \end{matrix}

Thanks for reading! 😁

Solutions to The Rising Sea (Vakil)

Chapter 7

Section 7.1: An example of a reasonable class of morphisms: Open embeddings

Exercise 7.1.A:

Proof:

Exercise 7.1.B:

Proof:

Exercise 7.1.C:

Proof:

Exercise 7.1.D:

Proof:

Algebraic interlude: Lying Over and Nakayama

Exercise 7.2.A:

Proof:

Exercise 7.2.B:

Proof:

Exercise 7.2.C:

Proof:

Exercise 7.2.D:

Proof:

Exercise 7.2.E:

Proof:

Exercise 7.2.F:

Proof:

Exercise 7.2.G:

Proof:

Exercise 7.2.H:

Proof:

Exercise 7.2.I:

Proof:

Exercise 7.2.J:

Proof:

A gazillion finiteness conditions on morphisms

Exercise 7.3.A:

Proof:

Exercise 7.3.B:

Proof:

Exercise 7.3.C:

Proof:

Exercise 7.3.D:

Proof:

Exercise 7.3.E:

Proof:

Exercise 7.3.F:

Proof:

Exercise 7.3.G:

Proof:

Exercise 7.3.H:

Proof:

Exercise 7.3.I:

Proof:

Exercise 7.3.J:

Proof:

Exercise 7.3.K:

Proof:

Exercise 7.3.L:

Proof:

Exercise 7.3.M:

Proof:

Exercise 7.3.N:

Proof:

Exercise 7.3.O:

Proof:

Exercise 7.3.P:

Proof:

Exercise 7.3.Q:

Proof:

Exercise 7.3.R:

Proof:

Exercise 7.3.S:

Proof:

Exercise 7.3.T:

Proof:

Exercise 7.3.U:

Proof:

Exercise 7.3.V:

Proof:

Exercise 7.3.W:

Proof:

Images of morphisms: Chevalley’s Theorem and elimination theory