Spring Week 1 and 2

MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
˙
CIHAN
BAHRAN
4.3.1(†). Let X be a flat scheme of finite type over a Dedekind scheme S. Let
s ∈ S be a closed point and x ∈ Xs . Let t denote a generator of the maximal ideal
of OS,s . We suppose that X is regular at x. Show that Xs is regular at x if and
only if t ∈ mx r m2x , where mx is the maximal ideal of OX,x .
Let’s first observe that X is locally Noetherian: Given x ∈ X, let V be an affine open
subset neighborhood of s = f (x) in Y . Now pick U to be an affine neighborhood of x
contained in f −1 (V ). As S is Dedekind, OS (V ) is a Noetherian ring and since f is of
finite type, OX (U ) is a finitely generated OS (V )-algebra. Thus OX (U ) is Noetherian.
Now we can apply Theorem 3.12 to get
dim OXs ,x = dim OX,x − dim OS,s .
Because S is Dedekind, dim OS,s is either 0 or 1. Suppose dim OS,s = 1. Note that
• X is regular at x, so dim OX,x = dimk(x) mx /m2x ,
• Since k(s) = OS,s /(t), we have OXs ,x ∼
= OX,x ⊗OS,s k(s) ∼
= OX,x /tOX,x . The
residue field of OXs ,x is k(x).
• As a k(x)-vector space,
mx /tOX,x ∼
= mx /(m2x + tOX,x )
(mx /tOX,x )2
Therefore
Ä
ä
dim OXs ,x = dimk(x) mx /m2x − 1
hence Xs is regular at x if and only if
Ä
ä
Ä
dimk(x) mx /m2x − 1 = dimk(x) mx /(m2x + tOX,x )
if and only if
Ä
ä
ä
dimk(x) (m2x + tOX,x )/m2x = 1
if and only id t ∈
/ m2x (t is surely in mx because OS,s → OX,x is a local homomorphism).
4.3.2. Let f : X → Y be a morphism of finite type. Show that f is flat if and
only if for every point y ∈ Y and for every closed point x of Xy , f is flat at x (Use
Proposition 1.2.13).
Remark. Let f : X → Y be a morphism of finite type with Y locally Noetherian.
We can show that the set of points x ∈ X such that f is flat at x is open in X, see
[41], IV.11.3.1.
We first assume that the claim is true when Y is affine. Let {Vλ }λ∈Λ be an affine open
covering of Y and write Uλ = f −1 (Vλ ).
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MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
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Assume f is affine. Then f |Uλ : Uλ → Vλ is flat for every λ ∈ Λ. Now fix (arbitrary)
y in Y and a closed point x ∈ Xy . Now y belongs Vλ0 for some λ0 ∈ Λ. Let U := Uλ0
and V := Vλ0 . Uy can be identified topologically with the open subspace Xy ∩ U of Xy .
And since x is closed in Xy , x is closed in Uy . So by our assumption f |U is flat at x,
#
#
which means (f |U )#
x : OV,y → OU,x is flat. But (f |U )x = fx so f is flat at x.
Conversely, assume that for every point y ∈ Y and for every closed point x of Xy , f is
flat at x. Fix λ ∈ Λ and let y ∈ V := Vλ . Write U := Uλ . If x is a closed point of Uy ,
then writing {x} for the closure of x in Xy we have
{x} ∩ Uy = {x}
How to use finite type??????
4.3.3. Let f : X → Y be a morphism of locally Noetherian schemes. Show that
dim X ≤ dim Y if f is injective, and dim X ≥ dim Y if f is flat and surjective.
We prove an easy lemma first.
Lemma 1. Let X be a scheme. Then dim X = sup{dim OX,x : x ∈ X}.
Proof. Let {Uλ }λ∈Λ be an affine open covering of X. Then for every λ ∈ Λ we have
OUλ ,x = OX,x so
sup{dim OX,x : x ∈ X} = sup{sup{dim OUλ ,x : x ∈ Uλ } : λ ∈ Λ}
and by Remark 2.5.6 we have
dim X = sup{dim Uλ : λ ∈ Λ} .
Thus we may reduce to the case X = Spec A is affine. And in this case we have
dim X = dim A
= sup{dim Ap : p ∈ Spec A}
= sup{dim OX,x : x ∈ X} .
Assume f : X → Y is injective. Fix x ∈ X and write y = f (x). Since f is injective,
the fiber Xy is a singleton as a set and hence dim OXy ,x = 0. Thus by Theorem 3.12
we have
dim OX,x ≤ dim OY,y ≤ dim Y .
Taking the supremum over x ∈ X we get dim X ≤ dim Y .
Now assume f : X → Y is flat and surjective. Fix y ∈ Y and let x ∈ X such that
f (x) = y. Then by Theorem 3.12 we have
dim OX,x − dim OY,y = dim OXy ,x ≥ 0
so
dim OY,y ≤ dim OX,x ≤ dim X .
Now taking the supremum over y ∈ Y we get dim Y ≤ dim X.
MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
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4.3.4(†). Let k be a field. Let A be the polynomial ring k[x, y]. Show, by computing
the dimension of the fibers, that the canonical morphism Proj A[T0 , T1 ]/(yT0 −
xT1 ) → Spec A is not flat.
4.3.5. Let A be a reduced ring, and let a ∈ A be a non-zero divisor. Let us consider
a ring B containing A[T ] and an element s ∈ B such that sn = aT for some n ≥ 1.
Show that the sub algebra A[T ][s] of B is free (and hence flat) over A[T ] (consider a
linear relation between 1, · · · , sn−1 and show that the coefficients are zero in Ap [T ]
for every minimal prime ideal p of A).
4.3.7. Let f : X → Y be a morphism of schemes. Let X1 , X2 be two closed
subschemes of X, flat over Y . We suppose that there exists a dense open subscheme
V of Y such that X1 ∩ f −1 (V ) and X2 ∩ f −1 (V ) are equal as closed subschemes
of f −1 (V ). Let us suppose that Y is integral. We want to show that X1 = X2 as
schemes.
(a) Show that one can suppose that X = Spec B, Y = Spec A, and V are affine.
(b) Let I, J be the ideals of B defining X1 and X2 . Tensoring the exact sequence
0 → A → OY (V ) by B/I, show that I = ker(B → OX (f −1 (V ))/(I)).
Deduce from this that I = J.
4.3.8(†). Let f : X → Y be a morphism of locally Noetherian schemes. Let Z be
a closed subscheme of X. We suppose that there exists a point y ∈ Y such that
Zy = Xy as schemes.
(a) Show that if Z is flat over Y at z ∈ Xy , then Z equals to X in a neighborhood
of z (use Exercise 1.2.14 and Nakayama’s lemma)
(b) Show that if Z is flat over Y and if f is moreover of finite type (resp.
proper) over Y , then there exists an open neighborhood V 3 y such that
Z ∩ f −1 (V ) → f −1 (V ) is an open immersion (resp. an isomorphism) (use
Exercise 3.2.5).
4.3.9(†). Let f : X → Y be a flat morphism of finite type with X, Y Noetherian.
We will show that f is open.
(a) Use Exercise 3.2.17 to show that f (X) contains a non-empty open subset
V of Y .
(b) By considering the morphism X ×Y (Y r V ) → Y r V , show that f (X)r V
contains a non-empty open subset of Y r V .
(c) Show that Y r f (X) is closed in Y by using the fact that the topological
space Y is Noetherian; that is, every descending chain of closed subsets in
Y is stationary.
If X is empty, then there is nothing to show. So we may assume X 6= ∅.
(a) Pick some x ∈ X. Let V be an affine neighborhood of f (x) in Y and let U be an
affine neighborhood of x in X such that f (U ) ⊆ V . Then the restriction f |U : U → V
is flat of finite type with U, V Noetherian and U 6= ∅. And it suffices to show that
f |U (U ) = f (U ) contains a non-empty open subset of V . Therefore we reduce to the
case where X = Spec B and Y = Spec A are affine with A, B nonzero Noetherian rings.
MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
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And f = Spec ϕ where ϕ : A → B is a ring homomorphism which makes B a finitely
generated and flat A-algebra.
We claim that f (X) contains a generic point of Y . For the sake of contradiction, suppose
not. Then let p1 , . . . , pr be the minimal primes of A (we know Noetherian rings have
finitely many minimal primes) and write η1 , . . . , ηr for the corresponding generic points
in Y . Then Xηj = ∅ for every j = 1, . . . , r. Now writing N for the nil-radical of A,
Q
we have an injective ring homomorphism A/N ,→ rj=1 A/pj . Tensoring by the flat
A-algebra B, we get an injection
B/N B ,→ ⊕r (B ⊗A A/pj ) ,→ ⊕r (B ⊗A k(ηj )) ∼
= ⊕r OX (Xη ) = 0
j=1
j=1
j=1
ηj
j
which yields B = N B hence B = 0, a contradiction.
So WLOG, η1 ∈ f (X). Consider the open subset W = Y r {η2 , . . . , ηr } of Y . Since η1
is a generic point we have η1 ∈
/ {ηj } for every j 6= 1 and hence
η1 ∈
/
r
[
{ηj } =
j=2
r
[
{ηj } = {η2 , . . . , ηr }
j=2
which means η1 ∈ W . Thus U := f −1 (W ) 6= ∅.
Since W is a non-empty open subset of the irreducible component {η1 } of Y , it is
irreducible. The restriction f |U : U → W is flat, so by Lemma 3.7 it is dominant. f |U
is also of finite type, hence by Exercise 3,2.17 f |U (U ) = f (U ) contains a dense open
subset V of W . Here
• V is non-empty because W is non-empty,
• V ⊆ f (U ) ⊆ f (X),
as desired.
(b) and (c) Suppose V ( f (X). So the collection of nonempty open subsets of Y that
are strictly contained in f (X) is nonempty. Since Y is Noetherian, this collection has
a maximal element, which we may assume to be V .
The closed subset Y r V of Y has a unique reduced closed subscheme structure, so we
can form the fiber square
q
T := X ×Y Y r V
p
X
f
/
YrV
/
Y.
The properties flat and finite type are stable under base change, so q is flat and of finite
type. Also being a closed subscheme of a Noetherian scheme, Y r V is Noetherian. Since
q is of finite type, T is locally Noetherian. Covering Y r V by finitely many affine open
subschemes, since q is quasi-compact their inverse images form a finite open covering
of T by quasi-compact schemes; hence T is quasi-compact. Thus T is Noetherian. Also
Y r V 6= ∅ by our assumption.
So by (a), q(T ) contains a non-empty open subset of Y r V . We claim that q(T ) =
f (X)r V . Certainly by the commutativity of the fiber square we have f (T ) ⊆ f (X)r V .
Conversely, let x ∈ X such that y = f (x) ∈
/ V . By Exercise 3.1.7, the set
{z ∈ T : p(z) = x, q(z) = y}
is non-empty. Thus there exists z ∈ T such that q(z) ∈ f (X)r V .
MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
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So f (X)r V contains a non-empty open subset of Y r V . This open subset is necessarily
of the form (Y r V ) ∩ W = W r V for some open subset W of Y . Now since
∅ 6= W r V ⊆ f (X)r V
we have V ( V ∪ W ⊆ f (X). By the maximality of V , the open subset V ∪ W should
be equal to f (X). So f (X) is open.
In fact it follows from here that f is open because if U is an open subset of X, then U
is a Noetherian scheme and f |U : U → Y is flat and of finite-type, so by above f (U ) is
open.
4.3.10(†). Let k be a field of characteristic char(k) 6= 2. Let Y = Spec k[u, v]/(v 2 −
u2 (u + 1)), and let f : X → Y be the normalization morphism. Show that f is
unramified surjective, but not ´etale.
Let A = k[u, v]/(v 2 − u2 (u + 1)). Since the polynomial v 2 − u2 (u + 1) has degree 2
when considered as an element of k[u][v] and has no roots, it is irreducible, and hence
generates a prime ideal because k[u, v] is a UFD. Thus A is an integral domain.
Let’s write x and y for the images of u and v in A, respectively. Clearly x 6= 0 and
y/x ∈ Frac(A) r A. Yet (y/x)2 = x + 1 so y/x is integral over A. So A is not normal.
Therefore f is not flat by Example 3.5. So f is not ´etale.
Unramified???????????????????????
4.3.11. Let f1 : X1 → Y, f2 : X2 → Y be morphisms of locally Noetherian schemes
of finite type. Let us suppose that for every y ∈ Y , there exists i = 1 or 2 such
that fi is unramified (resp. ´etale; resp. smooth) at all points of fi−1 (y). Show that
X1 ×Y X2 → Y is unramified (resp. ´etale; resp. smooth).
4.3.12(†). Let f : X → Y be a flat morphism of algebraic variety. We suppose
that Xy is irreducible for every closed point y ∈ Y and that Y is irreducible. Show
that X is irreducible.
For the sake of contradiction, suppose X is not irreducible. Then there are non-empty
open subsets U, V of X such that U ∩ V = ∅.
Note that being algebraic varieties, X and Y are Noetherian schemes. Also the structure
morphisms X → Spec k and Y → Spec k are of finite type by definition and moreover
they are quasi-compact since X, Y are Noetherian. Hence f , which is compatible with
the structure morphisms is of finite type by Proposition 3.2.4.
Thus f is open by Exercise 4.3.9. Hence f (U ) and f (V ) are non-empty open subsets
of Y , so they must intersect since Y is irreducible. Since the set of closed points of
an algebraic variety forms a dense subset, there exists a closed point y of Y such that
y ∈ f (U ) ∩ f (V ).
As topological spaces, we may identify Xy with f −1 (y). So the open subsets U ∩ Xy and
V ∩ Xy of Xy are non-empty. But U ∩ V = ∅, this contradicts Xy being irreducible.
4.3.14. Let us keep the notation and hypotheses of Lemma 3.16. Let J = ann(b).
(a) Show that bB is flat over A.
MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
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(b) Using the exact sequence 0 → J → B → B, where the last map is multiplication by b, show that J ⊗A A/mA = 0. Deduce from this that J = 0 and
that b does not divide zero in B.
4.3.15(†). Let X, Y be integral schemes, separated, and of finite type over a locally
Noetherian scheme S. Let f : X Y be a birational map. Let T → S be a faithfully
flat morphism such that XT and YT are integral. Suppose that fT : XT
YT is
defined everywhere and that Y is affine. We want to show that f is also defined
everywhere. See Exercise 5.2.14 for the Y arbitrary case.
(a) Show that we can suppose S, T are affine (and even local).
(b) Let us identify the function fields K(X) and K(Y ). Show that f is defined
everywhere if and only if OY (Y ) ⊆ OX (X).
(c) Let M = (OX (X) + OY (Y ))/OX (X). Show that M ⊗OS (S) OT (T ) = 0.
Deduce from this that M = 0 and that f is defined everywhere.
(a) Let us assume the claim is true when S is affine and deduce the general case. For
the general case, let {Si } be an affine open covering of S. For each i, Yi and Ti be the
inverse images of Si under the morphisms Y → S and T → S, respectively. Also let
{Yij }j be an affine open covering of {Yi } for each i.
By the construction of the fiber products (see pages 79-80), the schemes
{Yij ×Si Ti }i,j = {Yij ×S Ti } = {(Yij )Ti }i,j
glue together to give YT . Let Xij = fT−1 (Yij )?????????????????
(b) Write S = Spec A and T = Spec B where A is a Noetherian local ring and B is a local
ring. Also let R = OY (Y ) so we may write Y = Spec R since Y is affine. By restricting
f , we may assume it maps an open subset U of X isomorphically onto a principal
open set D(a) of Y . That is, we have a ring isomorphism ϕ : Ra → OX (U ). And f
is defined everywhere if and only if this ring isomorphism restricts to an isomorphism
R → OX (X), in other words if and only if ϕ(R) ⊆ OX (X) (note that OX (X) injects
in OX (U ) since X is integral). Since identifying the function fields makes ϕ into the
identity map, we are done.
(c) Write C = OX (X), so together with our notation above, M = (C + R)/C. Clearly
R ⊆ C iff M = 0. fT iso????
4.3.16. Let B be a semi-local ring, that is, having only a finite number of maximal
ˆ be the formal completion of B for the topology defined
ideals m1 , . . . , mn . Let B
ˆ ∼
ˆi , where B
ˆi is the formal completion of Bm , for the
∩i mi . Show that B
= ⊕i B
topology defined by its maximal ideal.
4.3.17. Let A be a Noetherian local ring, and let B be a finite A-algebra with
c , where ˆ denotes the formal
maximal ideals m1 , . . . , mn . Show that B ⊗A Aˆ ∼
= ⊕i B
mi
completion for the topology defined by the maximal ideal.
4.3.18(†). Let X be a quasi-projective scheme over a locally Noetherian scheme
S. Let G be a finite group acting on X. Hence the quotient Y := X/G exists
(Exercise 3.3.23).
(a) Let G act on an A-algebra B, and C be a flat A-algebra. Then G acts on
B ⊗A C by the identity on the second component. Let us denote the A-linear
MATH 8254 ALGEBRAIC GEOMETRY WEEK 1 AND 2
map g − idB : B → B by φg . Show that ker(φ ⊗ idC ) = ker(φg ) ⊗A C and
that
ker(φg ⊗ idC ) ∩ ker(φg0 ⊗ idC ) = (ker(φg ) ∩ ker(φg0 )) ⊗A C
in B ⊗A C. Deduce from this that (B ⊗A C)G = B G ⊗A C. Show that the
quotient morphism f : X → Y commutes with flat base change.
(b) Let us fix a point x ∈ X. Let us set D := {g ∈ G : gx = x}. This is the
decomposition group of x. Show that D acts canonically on OX,x as well as
ˆX,x . Let y = f (x). Show that (O
ˆX,x )D = O
ˆY,y
on the formal completion O
D
(use Exercise 3.17 and (a)). Show that OY,y = (OX,x ) if G = D.
(c) Show that the morphism f : X → Y factors into X → X/D and X/D → Y .
Let z be the image of x in X/D. Show that X/D → Y is ´etale at z.
4.3.19(†).
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