Download Basic Algebraic Geometry

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1.1
1.1.1
Some algebraic geometry
Basic de…nitions.
Zariski and standard topology.
Let F be a …eld (almost always R or C). Then we say that a subset X F n
is Zariski closed if there is a set of polynomials S
F [x1 ; :::; xn ] such that
n
n
X = F (S) = fx 2 F js(x) = 0; s 2 Sg. One checks that the set of Zariski
closed sets, Z, in F n satis…es the axioms for the closed sets in a topology.
That is
F n and ; 2 Z.
If T
Z then \Y 2T Y 2 Z.
If Y1 ; :::; Ym 2 Z then [j Yj 2 Z.
The Hilbert basis theorem (e.g. A.1.2 [GW]) implies that we can take the
sets S to be …nite.
If X is a subset of F n then the Z-topology on X is the subspace topology
corresponding to the Zariski topology.
If F = R or C then we can also endow F n with the standard
metric
Pn
2
2
topology corresponding to d(x; y) = kx yk where kuk =
i=1 jui j for
u = (u1 ; :::; un ). We will call this topology the standard topology (in the
literature it is also called the classical, Hausdor¤ or Euclidean topology). Our
…rst task will be to study relations be tween these very di¤erent topologies.
Examples. n = 1: Then if Y 2 Z then Y = F or Y is …nite.
n = 2:A Zariski closed subset is a …nite union of plane curves or all of
2
F : Thus the Zariski topology on F 2 is not the product topology.
1.1.2
Noetherian topologies.
The Zariski topology on a closed subset, X, of F n is an example of a
Noetherian topology. That is if Y1
Y2
:::
Ym
::: is a decreasing
sequence of closed subsets of X then there exists N such that if i; j
N
then Yi = Yj (this is a direct interpretation of the Hilbert basis theorem).
If X is a topological space that be cannot written X = Y [ Z with Y
and Z closed and both are proper the X is said to be irreducible. Clearly an
irreducible space is connected but the converse is not true.
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Exercise. What are the irreducible Hausdor¤ topological spaces?
Lemma 1 Let X be a Noetherian topological space. Then X is a …nite union
of irreducible closed subspaces. If X = X1 [ X2 [
[ Xm with Xi closed
and irreducible and if Xi * Xj for i 6= j then the Xi are unique up to order.
For a proof see A.1.12 [GW].The decomposition X = X1 [ X2 [
[ Xm
with Xi closed and irreducible and Xi * Xj for i 6= j is called the irredundant
decomposition of X into irreducible components. Each of the Xi is called an
irreducible component.
1.1.3
Nullstellensatz.
If X is a subset of F n then we set IX = ff 2 F [x1 ; :::; xn ]jfjX = 0g. If I
F [x1 ; :::; xn ] is an ideal then we set F n (I) = fx 2 F n jf (x) = 0; f 2 Ig.
Lemma 2 A closed subset, X, of F n is irreducible if and only if IX is a
prime ideal (i.e. F [x1 ; :::; xn ]jX is an integral domain).
We now recall the Hilbert Nullstellensatz.
Theorem 3 Let F be algebraically closed and let I be an ideal in F [x1 ; :::; xn ].
Then
1. F n (I) =p; if and only if I = F [x1 ; :::; xn ]:
2. IX(I) = I:
We will give a proof of 1. in this theorem for the case of F = C since the
method of proof will be consistent with the techniques of the later material
to be presented. That 1. implies 2. is the trick of Rabinowitz which we defer
to the references.
We prove that if I is a proper ideal then Cn (I) 6= ;. Let M be a maximal
proper ideal with I M. Then k = C[x1 ; :::; xn ]=M is a …eld containing C.
1
n
Since the image of the monomials xm
xm
span k as a vector space over
1
n
C we see that dimC k is at most countable. Since C is algebraically closed, if
k 6= C there must be t 2 k that is transcendental over C: We assert that the
elements f t 1 a ja 2 Cg are independent over C. Indeed, if
m
X
i=1
ci
t
ai
2
=0
with a1 ; :::; am 2 C and distinct and c1 ; :::; cm in C then multiplying by
m
Y
(t
ai )
i=1
we …nd that
m
X
j=1
cj
Y
(t
ai ) = 0:
i6=j
This de…nes a polynomial satis…edY
by t that is trivial if and only if all of the
cj are 0 (evaluate at ai and get ci (ai aj )) Since C is not countable this
j6=i
is a contradiction. We have proved that k is one dimensional over C with
basis 1 + M. Thus xi + M = zi 1 + M with zi 2 C. We have shown that
xi zi 2 M for all i = 1; :::; n. Since the ideal Ifzg is maximal we see that
M = Ifzg and since I M we have z 2 Cn (I).
1.2
1.2.1
A¢ ne varieties and dimension.
A¢ ne varieties.
Let X; Y be sets and let f : X ! Y be a mapping. If g is a function from Y
to F then we set f g = g f which is a function on X.
In this section we will assume that F is algebraically closed (you can
assume it is C). An a¢ ne variety over F is a pair (X; R) with X a topological
space and R and algebra over F of continuous functions from X to F (F is
endowed with the Zariski topology) such that there exists a Z-closed subset
Z in F n for some n and a homeomorphism, f , of X onto Z such that f :
F [x1 ; :::; xn ]jjZ ! R is an algebra isomorphism. A morphism a¢ ne varieties
(X; R) and (Y; S) is a map f : X ! Y such that f S R. An isomorphism
is a morphism that is one to one and onto and f is an isomorphism of
algebras.
We note that if (X; R) is an a¢ ne variety then R must be a …nitely
generated algebra over F and if a 2 R is such that ak = 0 for some k > 0 then
a = 0. That is, the only nilpotent element in R is 0. The converse is also true.
Let R be a …nitely generated algebra over F without nilpotents. Let r1 ; :::; rm
generate R as an algebra over F . Then we have an algebra homomorphism
: F [x1 ; :::; xm ] ! R de…ned by (xi ) = ri . Let I = ker and set Z =
F n (I). Then since R = F [x1 ; :::; xm ]=I and R has no nilpotents we must
3
p
have I = I. Thus I = IZ (by the nullstellensatz). Thus R is isomorphic
with O(Z) = F [x1 ; :::; xm ]jZ . Actually more is true. The topological space
X is also completely determined by R. Indeed, the nullstellensatz implies
that the maximal ideals in O(Z) are the deals Ifzg + I for z 2 Z. Thus
the set Z can be identi…ed with the set of maximal ideals in O(Z). The
topology is determined as follows. If Y is a closed subset of Z and J is the
ideal of elements in O(Z) such that Y = fz 2 Zjg(z) = 0; g 2 J g then
Y + fz 2 ZjJ
Ifzg g for I an ideal of R. Returning to R we may de…ne
MR to be the set of maximal proper ideals endowed with the topology that
has as its closed sets the sets MR (I) = fMj maximal with I
Mg. We
have just seen that if we use the fact that R is isomorphic with O(Z) and
the maximal ideals in O(Z) are the images of the ideals Ifzg for z 2 Z. This
implies that if M 2 MR then R=M = F 1 + M. Thus if r 2 R then we can
de…ne r(M) = c if r = c1 + M. We therefore have made R into an algebra
of F valued functions on MR . The bottom line is that all of the information
is in the algebra R.
Recall that if F = C and if Y
Cn is Z-closed then we also have the
standard (metric) topology on Y . If (X; R) is an a¢ ne variety and if (X; R)
is isomorphic with (Y; C[x1 ; :::; xn ]jY ) and (V; C[x1 ; :::; xm ]jV ) with Y Z-closed
in Cn and V Z-closed in Cm . Then we have f : X ! Y and g : X ! V
homeomorphisms in the Z-topology such that f : C[x1 ; :::; xn ]jY ! R and
g : C[x1 ; :::; xm ]jV ! R algebra isomorphisms. This g f 1 : Y ! V is
given by the map = ( 1 ; :::; m ) with j = (g f 1 )(xjjV ) 2 C[x1 ; :::; xn ]jY .
Thus g f 1 is continuous in the standard metric topology. This implies
that endowed with the standard topology Y and V are homeomorphic. We
therefore see that this implies that (X; R) also has a natural metric topology
which we will also call standard.
We will say that an irreducible a¢ ne variety (X; R) is normal if R is
integrally closed in K(X) the quotient …eld of R.
Theorem 4 If (X; R) is an irreducible a¢ ne variety then if S is the integral
closure of R in K(X) then S is …nitely generated over F and (MS ; S) is a
normal a¢ ne variety.
For a proof see [S] Theorem II.5.4. We will call the variety (MS ; S) with
the map M ! z with z 2 X such that M \ R = Mz the normalization of
(X; R) one can show that this map is …nite (in particular …nite to one).
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1.2.2
Dimension.
Let (X; R) be an irreducible a¢ ne variety over an algebraically closed …eld F .
Then R is an integral domain. We denote by K(X) the quotient …eld of R.
Then K(X) is a …eld extension of F:If (X; R) is isomorphic with a Z-closed
subset of F n then the …eld K(X) is isomorphic is generated as a …eld by at
most n generators. We de…ne the dimension of X to be the transcendence
degree of K(X) over F .
Example. (F n ; F [x1 ; :::; xn ]) has dimension n since K(F n ) = F (x1 ; :::; xn )
the …eld of rational functions in n variables. If X
F n is Z-closed then
dim X n.
This is a precise de…nition but is hard to compute. We will now discuss
an e¤ective way to calculate the dimension of an a¢ ne variety.
Let R be an algebra over F then a …ltration of R is an increasing sequence
F0 R
F1 R
:::
Fm R
::: of subspaces of R such that
L Fi R F j R
Fi+j R If F is a …ltration of R then we de…ne GrFR =
j 0 Fj R=Fj 1 R
(F 1 R = f0g) with multiplication
(a + Fi 1 R)(b + Fj 1 R) = ab + Fi+j 1 R
for a 2 Fi R and b 2 Fj R. If we set GrF j R = Fj R=Fj 1 R then GrFR is a
graded algebra over F . Here, a graded
over F is an algebra S such
L algebra
j
that as a vector space over F , S = j 0 S such that S i S j S i+j .
We will say that a …ltration, F, of R is good if dim GrF 1 R < 1 and it
generates GrFR as an algebra over F .
Example. Let X
F n be Z-closed with R = F [x1 ; :::; xn ]jX : Then we set
Fj R equal to the span of the restrictions xajX with a = (a1 ; :::; an ) 2 Nn and
xa = xa11
xann . Then GrFR is generated by the images of the xi in GrF 1 R.
Theorem 5 Let R be an algebra over F with a good …ltration F then there
exists, N 2 N and a polynomial hF (q) such that dim Fj R = hF (j) for j N .
Furthermore, if F and G are good …ltrations of R then deg hF = deg hG .
Examples.
1. Let X = F n with R = F [x1 ; :::; xn ] with the …ltration, F T; given by
n
degree as in the example above. Then dim Fj R = n+j
= jn! + lower in j.
n
2. Let f 2 F [x1 ; :::; xn ] be a non-constant polynomial. Let
R = F [x1 ; :::; xn ]=f F [x1 ; :::; xn ]
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on R we …lter by degree in the xi . Then if deg f = m > 0 we have
0 ! Fk j F [x1 ; :::; xn ] ! Fk F [x1 ; :::; xn ] ! Fk R ! 0
is exact with the map from …ltered degree k
by f . Thus
dim Fk R =
n+k
n
n+k
n
j
=j
j to k given by multiplication
kn 1
+ lower degree in k:
(n 1)!
Theorem 6 Let (X; R) be an a¢ ne variety over F then if F is a good …ltration of R then deg hF is the maximum of the dimensions of the irreducible
components of X.
We will denote by DimR the degree of hF for F a good …ltration of R.
More important for computations is
Theorem 7 Let I be an ideal in F [x1 ; :::; xn ] then
p
DimF [x1 ; :::; xn ]=I = DimF [x1 ; :::; xn ]= I:
This implies that we can compute the dimension of a variety from a
de…ning set of equations.
We also note that Theorem 6 tells us that we should (and do) de…ne
dim X = DimR = max dim Z for Z an irreducible component of X.
We record here two results that we will need later.
Theorem 8 If X is an irreducible a¢ ne variety and if Y is a closed subvariety then dim Y
dim X with equality if an only if Y = X:
Theorem 9 Let (X; R) be an irreducible a¢ ne variety of dimension n. Then
if f is a non-constant element of R then every irreducible component of
X(f ) = fx 2 Xjf (x) = 0g is of dimension n 1:
For a proof see Theorem I.6.5 p. 74 in [S].
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1.2.3
Tangent space.
If f 2 F [x1 ; :::; xn ] then we de…ne for
p = (p1 ; :::; pn ) 2 F n ; dfp : F n ! F
by
dfp (z1 ; :::; zn ) =
Here
@f
(p)
@xi
X @f
(p)zi :
@xi
is the formal derivative. We have the Taylor formula
f (p + x) = f (p) + dfp (x) +
X
c x =
j j 2
f (p) + dfp (x) +
X
xi gi (x)
i
here for = ( 1 ; :::; n ), x = x1 1
xnn and gi is an appropriate choice of
polynomial which can choose so that gi (0) = 0. If F = R or C this coincides
with the usual derivative. We set Tp (X) = fv 2 F n jdfp (v) = 0; f 2 IX g.
We will now de…ne this space in terms of the algebra R = F [x1 ; :::; xn ]jX . If
p 2 X then we set Mp = ff 2 Rjf (p) = 0g. Then the Zariski cotangent
space to X at p is the vector space Mp =M2p . We will now de…ne a natural
pairing between this space and Tp (X). If f 2 R and f (p) = 0 then we
choose g 2 F [x1 ; :::; xn ] such that gjX = f . If z 2 Tp (X) then we assert that
dgp (z) depends only on f and not on the choice of g. Indeed, if hjX = f
then g h 2 IX . Thus d(g h)p (z) = 0. Now if f 2 M2p then f is a sum of
elements of the form u1 u2 with ui 2 Mp . If hijX = ui then using
d(h1 h2 )p = h1 (p)d(h2 )p + h2 (p)d(h1 )p = 0
we see that we do indeed have a natural (independent of choices) pairing of
Mp =M2p and Tp (X). Furthermore, if f (p) = 0 and dfp = 0 then f +IX 2 M2p
(see Taylor’s formula above).
We thus have an intrinsic de…nition of Tp (X) = (Mp =M2p ) :
Theorem 10 Let X be an irreducible a¢ ne variety. If p 2 X then dim Tp (X)
dim X. Furthermore, dim X = minp2X dim Tp (X).
For this c.f. [GW] A.3.2. We have
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Theorem 11 Let X be an irreducible a¢ ne variety then the set of p 2 X
such that dim Tp (X) = dim X is Z-open.
We can see this as follows. We assume that X is Z-closed in F n . Let
n d
d = dim X. Then if f = (f1 ; :::; fn d ) 2 IX
then we set Uf equal to the
subset of X consisting of those points, p, such that some (n d) (n d)
minor of
@fi
(p)
@xj
is non-zero. One sees that each Uf is Z-open in X (perhaps empty) and the
union of the Uf is the set of smooth elements.
Let X be an irreducible a¢ ne variety. We will call a point smooth if
dim Tp (X) = dim X.
In our application of these ideas we will use (the case n = 1) of the
following theorem.
Theorem 12 If (X; R) is a normal a¢ ne variety then the set of singular
points (i.e. not smooth points) is Z-closed and of dimension at most n 2:
For a proof of this see [S], Theorem II.5.3 p.117. This result implies that
since dim ; = 1 we have the normalization of a curve is smooth (all points
are smooth).
We will need to extend the concept of tangent space to non-irreducible
varieties, For this we will need some new concepts.
1.2.4
A¢ ne varieties revisited.
We take F to be algebraically closed. Let (X; R) be an a¢ ne variety. Let
2 R be non-constant. Let Xf g = fx 2 Xj (x) 6= 0g. If (X; R) is
isomorphic with the a¢ ne algebraic set Y = F n (S) then R is isomorphic
with O(Y ) = F [x1 ; :::; xn ]jY under the map f : X ! Y (i.e. f is a
Z-homeomorphism and f : O(Y ) ! R is an algebra isomorphism). Let
= . We consider the
2 F [x1 ; :::; xn ] be such that if = jY then f
subset
1
Z = f(x;
)jx 2 Yf g g
(x)
of F n+1 . We look upon F [x1 ; :::; xn ] as the subalgebra of F [x1 ; :::; xn+1 ]
spanned by the monomials that don’t involve xn+1 . We note that if h(x1 ; :::; xn+1 ) =
8
xn+1 (x1 ; :::; xn )
1 then
Z = F n+1 (S [ fhg):
One checks that O(Z) is the localization of O(Y ) by the multiplicative set
2
f1; ; ; :::g. Thus if R( ) is the corresponding localization of R then we have
(Xf g ; R( ) ) is an a¢ ne variety.
Now let U be Z-open in Y_ : Then U = Y
W \ Y with W Z-closed in
n
F . That is W = F n (S) with Sa …nite set of polynomials. This implies that
U = [ 2S Yf g . This implies that the a¢ ne subvarieties (Xf g ; R( ) ) de…ne a
basis for the open sets in the Z-topology.
If U X is open then we de…ne OX (U ) to be the functions from U to
F such that for each p 2 U there is a 2 R such that (p) 6= 0, Xf g U
and jX g 2 R( ) . To make sure that this makes sense we need to know
Theorem 13 Let (X; R) be an a¢ ne variety. Then OX (X) = R.
1.2.5
The tangent space revisited.
Let (X; R) be an a¢ ne variety and let for each Z-open subset U X, OX (U )
be as in the previous section. If p 2 X and if U and V are Z-open in X and
p 2 U \ V then we say that 2 OX (U ) and 2 OX (V )are equivalent if
there exists W a Z-open subset with p 2 W
U \ V such that jW = jW .
We de…ne OX;p to be the set of equivalence classes under this relation.
If 2 OX (U ) for some Z-open U with p 2 U let [ ]p be the corresponding
equivalence class. Then one checks that we can de…ne an algebra structure
on OX;p by de…ning [ ]p [ ]p = [ jW ]p [ jW ]p for 2 OX (U ) and 2 OX (V )
and p 2 W U \ V .
This algebra is called the local ring of X at p. One can de…ne this ring
to be the localization of R at Mp (see section 1.4). If [ ]p 2 OX;p then the
value of (p) depends only on [ ]p and we use the notation (p) for the value
at p of 2 OX;p . This de…nes a homomorphism of OX;p onto F . We will
use the notation mX;p for the kernel of this homomorphism. Then it is easily
seen that OX;p is a local ring with maximal ideal mX;p (see section 1.4). We
consider the map p : Mp =M2p ! mX;p =m2X;p by ! [ ]p + m2X;p .
Theorem 14 If X is irreducible and if p 2 X then
9
p
is a bijection.
This leads to a de…nition of Zariski cotangent space. Tp (X) = mX;p =m2X;p .
As before we say that p 2 X is smooth if dim Tp (X) = dim X. If X =
0
[m
j=1 Xj is its decomposition into irreducible components and if X = [p2Xj Xj
then OX;p = OX 0 ;p and the cotangent space depends only on this closed subvariety.
1.2.6
The formal algebra at a smooth point.
Let (X; R) be an a¢ ne variety over F and algebraically closed …eld. We
assume that p 2 X is a smooth point. Let u1 ; :::; ud 2 mX;p be such that if
ui = ui + m2X;p form a basis of mX;p =m2X;p . If = (a1 ; :::; d ) is a multiindex
then we set (as usual) u = u1 1
ud d . Let pk : R ! R=mkX;p :
Lemma 15 u1 ; ::; ud are algebraically independent over F .
Lemma 16 In the notation above the set fpl u j0
OX;p =mlX:p over F .
j j < lgis a basis of
Corollary 17 In the notation above
lim OX;p =mkX;p = F [[x1 ; :::; xd ]]
the formal power series in the indeterminates x1 ; :::; xd .
Theorem 18 Let (X; R) be an a¢ ne variety over F of dimension d and let
p be a smooth point of X. Let X = X1 [
[ Xm be the decomposition
of X into irreducible components and let X 0 = [p2Xj Xj (the union of the
irreducible components containing X). Then X 0 is irreducible.
Proof. We may assume that X = X 0 . Thus we need only show that if p is a
smooth point of X and if every irreducible component of X contains p then
X is irreducible. We …rst note that the map OX;p ! lim OX;p =mkX;p given
by a 7 ! fa + mkX;p g is injective. Indeed, if a 7! 0 then a 2 mkX;p for all k
and thus Corollary 29 in the appendix (section 1.4) implies that a = 0. This
implies that OX;p is an integral domain. We assert that this implies that R is
an integral domain. Indeed, we assert that the map a 7 ! a=1 injects R into
OX;p . To see this we note that a=1 = 0 implies that there exists 2 R such
that (p) 6= 0 and a = 0. This implies that if Y is an irreducible component
of X then ajY = 0. Since (p) 6= 0 the set fy 2 Y j (y) 6= 0g is Z-dense in
Y and thus ajY = 0. We have assumed that every irreducible component of
X contains p sot a = 0:This implies that Rinjects into an integral domain so
Xis irreducible.
10
1.2.7
Relations between the standard and Zariski topologies.
We now look at the case when F = C.
We look upon C with the standard topology as R2 and Cn as R2n . If f
function from a set S to C we will write f (s) = fR (s) + ifI (s) with fR and fI
real valued functions on S: If > 0 we set B n = fx 2 Cn j kxk < gthought
of as the ball in R2n . We will now prove
Theorem 19 Let X be Z-closed in Cn and irreducible be of dimension d. Let
p 2 X be a smooth point then there exists > 0 and a C 1 map : B d ! Cn
such that (0) = p, (B d ) X is an open neighborhood of p in the standard
topology and : B d ! (B d ) is a homeomorphism in the standard topology.
Proof. Let f1 ; :::; fm generate IX . After possible relabeling we may assume that (df1 )p ; :::; (dfn d )p are linearly independent elements of (Cn ) . Let
ui = (fi )R and un d+i = (fi )I then we note that (dui )p = ((dfi )p )R and
(dun d+i )p = ((dfi )p )I where duj is the standard calculus di¤erential on R2n .
This implies that (du1 )p ; :::; (du2n 2d )p are linearly independent at p. Now after relabeling we may assume that the restrictions of x1 p1 ; :::; xd pd
to X form a basis of Mp modulo M2p . Set u2n 2d+i = (xi
pi )R and
2n
u2n d+i = (xi pi )I . Then (du1 )p ; :::; (du2n )p form a basis of (R ) . The
real analytic inverse function theorem implies that there exists an open standard neighborhood U of p in R2n such that if = (u1 ; :::; u2n ) then (U ) is
open and : U ! (U ) is a (real analytic) di¤eomorphism. Let > 0 be
so small that B n d B d
(U ). Then
1
(0
B d ) = Cn (f1 ; :::; fn d ) \
1
(B n
d
B d ):
1
We set (z) =
(0; z) for z 2 B d :
Let Z = Cn (f1 ; :::; fn d ) and let Z = [ri=1 Zi be the decomposition of
Z into irreducible components. Since X is irreducible we may assume that
X Z1 . On the other hand since f1 ; :::; fn d 2 IZ1 we see that dim Tp (Z1 )
d = dim X. Thus Z1 = X:
Let W be the Z-closure of (B d ). Then W C(f1 ; :::; fn d ). We also note
that xi
de…nes an analytic function on B d for i = 1; :::; n. This implies that
de…nes an injective homomorphism of O(W ) into the analytic functions
on B d . Since this algebra is an integral domain this implies that O(W ) is an
integral domain and hence that W is irreducible. Furthermore the functions
xi pi , i = 1; ::; d are algebraically independent on (B d )and hence on W .
11
thus the dimension of W is at least d. Since p 2 W we see that dim W = d
(the tangent space at p is at most of dimension d). Thus W is an irreducible
component of Y containing p. Since p is smooth we must have Z = X the
unique irreducible component of Y containing p.
In our application of these results we will use (the case n = 1) of the
following theorem.
In general the standard metric topology (which we will denote as the Stopology) is much …ner than the Z-topology on an a¢ ne variety (they are
equal if and only if the variety is …nite). However, we will show that the
S-closure of a Z-open set in an a¢ ne variety is the Z-closure.
Theorem 20 Let X be Z-closed in Cn and irreducible and let U 6= ; be
Z-open in X. Then the S-closure of U is X.
Proof. We prove this result by induction on dim X. If dim X = 1 then
X U is …nite. Let p 2 X U . Let f : Y ! X be the normalization
of X then f is surjective and continuous in the S-topology. Theorem 13
implies that Y is smooth. Let q 2 f 1 (p) as in Theorem 12 choose > 0
and : B 1 ! (B 1 ) be a homomorphism in the S-topology onto an S-open
subset of Y with (0) = q. By taking possibly smaller we may assume
that (B 1 ) \ f 1 (X U ) = fqg. Choose a sequence zj 2 B 1 f0g such that
limj!0 zj = 0. Then xj = f ( (zj )) 2 U and in the S-topology limj!1 xj = p.
Now assume that the result has been proved for all X of dimension n 1.
Assume dim X = n + 1: Let p 2 X U . Let Y = X U = Y1 [
[ Yr
be its decomposition into irreducible components and assume that Y1 ; :::; Ys
are exactly the ones that are not equal to fpg (usually s = r). Let pi 2 Yi
be such that pi 6= p. Let
O(X) be such that (p) = 0 and (pj ) 6= 0
for j = 1; :::; s. Set Z = fx 2 Xj (x) = 0g. Let Z = Z1 [
[ Zt
be the decomposition of Z into irreducible components. Assume that some
Zj X U . Then since X is irreducible and U is non-empty we see that the
Yi are all of dimension at most n. Since Zj is irreducible it must be contained
in one of the Yj . On the other hand dim Zk = n for all k. Thus we must have
Zj = Yi for some i:But pi 2
= Z. This contradiction implies Zi * X U . Let
Zj be such that p 2 Zj . Then Zj has dimension n and U \ Zj 6= ;. Thus the
inductive hypothesis implies that p is in the S-closure of U \ Zj hence of U .
Exercises.
1. Why didn’t we start the induction from dimension 0(where the result
is trivial)?
12
2. Prove that if X is an a¢ ne variety and U is a Z-open subset of X then
the Z-closure of U is the same as the S-closure.
1.3
1.3.1
Projective and quasiprojective varieties.
Projective algebraic sets.
We de…ne projective n-space to be the set of all one dimensional subspaces
of F n+1 and denote it Pn (F ) or Pn if F is understood. If L 2 Pn then L = F v
with v 2 L f0g. Thus we can look upon Pn as the set F n+1 f0g modulo
the equivalence relation v
w if there exists c 2 F = F f0g such that
n+1
cw = v. If (x0 ; x1 ; :::; xn ) 2 F
f0g then we denote the corresponding line
by [x0 ; x1 ; :::; xn ] and x0 ; x1 ; :::; xn are called the homogeneous coordinates of
the point in Pn .
Let f 2 F [x0 ; x1 ; :::; xn ] then f is said to be homogeneous of degree m
if f (cx) = cm f (x) for x 2 F n+1 and c 2 F . Assuming that F is in…nite
this condition is equivalent to saying that f is an F -linear combination of
monomials x = x0 0 x1 1
xnn with j j = 0 + ::: + n = m. We denote by
m
F [x0 ; :::; xn ] the space of polynomials that are homogeneous of degree m.
We note that if x 2 F n+1 f0g and if f 2 F m [x0 ; :::; xn ] then whether or not
f (x) = 0 depends only on [x]. De…ne the closed sets in the Zariski topology
on Pn to be the sets of the form
Pn (S) = f[x] 2 Pn jf (x) = 0; f 2 Sg
and S is a set of homogeneous polynomials. One checks that with these sets
satisfy the conditions necessary to be the closed sets of a topology.
Of particular interest are the Z-open sets Pnj = Pn Pn (xj ):Let [x] 2 Pnj
(that is xj 6= 0 ) then
[x] = [
x0
xj 1
xj+1
xn
1
x] = [ ; :::;
; 1;
; :::; ]:
xj
xj
xj
xj
xj
We therefore see that we have a bijective mapping from F n to Pnj ;
by
j : (a1 ; :::; an ) 7! [a1 ; :::; aj ; 1; aj+1 ; :::; an ]:
j,
given
Suppose that f 2 F [x1 ; :::; xn ] is of degree m. Then the j-th homogenization
of f is the polynomial obtained as follows:
13
Let f (x1 ; :::; xn ) =
P
j j m
fj (xo ; :::; xn ) =
c x . Write
X
m j j
c xj
xo 1
+1
j
xj j 1 xj+1
xn n :
j j m
Then fj is homogeneous of degree m in the variables x0 ; :::; xn and j fj = f .
Thus if we endow Pnj with the subspace Zariski topology then j : F n ! Pnj
de…nes a Z-homeomorphism.
If f 2 F m [x0 ; :::; xn ] then de…ne Ri f (x1 ; :::; xn ) = f (x1 ; :::; xj ; 1; xj+1 ; :::; xn ).
m
Then Ri is a linear bijection between F m [x0 ; :::; xn ] and m
j=0 F [x1 ; :::; xn ] =
Fm [x1 ; :::; xn ].
Let X be a Z-closed subset of Pn . Then we have an open covering X =
[ni=0 Pni \ X: Let i be as above. If X = Pn (S) then i 1 (Pni \ X) = F n (Ri S).
1.3.2
Sheaves of functions.
Let X be a topological space then a sheaf of functions on X is an assignment
F : U ! F(U ) with F(U ) a subalgebra of the algebra of all functions from
U to F satisfying the following properties:
1. If V
U then F(U )jV
F(V ):
2. If fV g 2I is an open covering of U then if
2 F(V ) is given such
that if x 2 V \ V then (x) = (x) then there exists 2 F(U ) such that
for all .
jV =
Examples.
1. It is easily seen that if (X; R) is an a¢ ne variety then OX is a sheaf
of functions on X. We call OX the structure sheaf of X.
2. If F is a sheaf of functions on X and if U is an open subset of X then
we de…ne FjU to be the assignment V ! F(V ) for V open in U (hence in X).
3. If X is a topological space and if F = R or C then the assignment of
the assignment U ! C(U; F ) (the continuous functions from U to F ) is a
sheaf or functions.
4. If F = R then the assignment U open in Rn to C 1 (U ) (the functions
from U to R such that all partial derivatives of all orders are continuous) is
a sheaf of functions denoted CR1n . Similarly,U ! C ! (U ) the assignment to
U the real analytic functions from U to R.
We consider the category of all pairs (X; F) with X a topological space
and F a sheaf of functions on X. We will call this the category of spaces
14
with structure. With morphisms between (X; F) and (Y; G) continuos maps
f : X ! Y such that f G(U ) F(f 1 U ).
Example.4. The category of smooth manifolds is the full subcategory of
spaces with structures (X; F) where X is a Hausdor¤ space such that for
each p 2 X there is an open neighborhood of p, U , such that (U; FjU ) is
isomorphic with (V; CR1n jV ) for V an open set of Rn for some …xed n. We will
1
denote the sheaf F by CX
and call n the dimension of X.
We are now ready to give a de…nition of algebraic variety.
1.3.3
Algebraic varieties.
The example in the previous subsection indicates how one should de…ne an
algebraic variety. We de…ne a pre-variety to be a topological space with a
structure (X; OX ) such that for every p 2 X there exists an open neighborhood, U , of p in X and an a¢ ne variety (Y; OY ) such that (U; OXjU ) is
isomorphic with (Y; OY ).
Examples.
1.We have seen that the structure that we attached to an a¢ ne variety
(X; R) which we denoted as (X; OX ) is a pre-variety.
2. Let X be a Z-closed subset of Pn . Set Xi = X \ Pni and let i be as
in 1.3.1. Then in 1.3.1 we observed that i 1 (Xi ) is Z-closed in F n . We pull
back the structure as in example on Yi = i 1 (Xi ). Then getting sheaves of
functions OXi . If U
Xi \ Xj then OXi jU = OXj jU . We de…ne for U
X,
Z-open OX (U ) to be the functions on U such that jU \Xi 2 OXi (U \ Xi ).
Then (X; OX ) is a pre-variety.
3. If U is open in X a Z-closed subset in Pn then we take OU to be the
restriction of OX to U and (U; OU ) is a pre-variety.
We now come to the notion of algebraic variety. For this we need to
de…ne products of pre-varieties. If X
F n and Y
F m are Z-closed sets
n
m
m+n
then we note that X Y is Z-closed in F
F =F
(as a topological
space–recall that this is not the product topology). To see this we take F n
to be F n 0 and F m to be 0 F m then if X = F n (S) and Y = F m (T )
then we look at f 2 S as a function of x1 ; :::; xn and g 2 T as a function of
xn+1 ; :::; xn+m then X Y = F n+m (S [ T ).
15
This de…nes a product of a¢ ne varieties. One can do this intrinsically by
de…ning (X; R) (Y; S) to be (X Y; R F S). Here (
)(x; y) = (x) (y).
If (X; OX ) and (Y; OY ) are pre-varieties then we can de…ne the pre-variety
(X Y; OX Y ) as follows: If U X and V
Y are open and (U; OXjU ) and
(V; OY jV ) are isomorphic to a¢ ne varieties then take OX Y jU V to be OU V
as in 1.3.2 Example 2. This de…nes a product (here the topology on U V
is the topology of the product of a¢ ne varieties de…ned as above). One can
prove that this is the categorical product in the category of pre-varieties.
We can …nally de…ne an algebraic variety. Let (X; OX ) be a pre-variety
then it is an algebraic variety over F if (X) = f(x; x)jx 2 Xg is a closed
set in X. This condition is usually called separable. We have (c.f. Lemma
A.4.2 [GW])
Theorem 21 Let X be Z-open in the Z-closed subset, Y , in Pn . Then the
corresponding pre-variety (as in example 3 above) is an algebraic variety.
A variety as in the above theorem will be called quasi-projective. An algebraic variety isomorphic with Z-closed subset of Pn will be called a projective
variety.
Theorem 22 If (X; OX ) and (Y; OY ) are quasi-projective then so is (X
Y; OX Y ).
For this it is enough to show that Pn Pm is isomorphic with a Z-closed
subset of PN for some N . To prove this we consider the map (F n+1 f0g)
(F m+1 f0g) ! F n+1 F m+1 f0g given by
(x; y) 7 ! x
y:
If we take the standard bases ej of F n+1 and F m+1 then we can identify
F n+1 F m+1 with F (m+1)(n+1) using the basis ei ej for 0
i
n and
0
j
m. One checks that this de…nes a morphism of (F n+1 f0g)
(F m+1 f0g) ! F n+1 F m+1 f0g:Also, [x y] = [x0 y 0 ] with (x; y) 2
(F n+1 f0g) (F m+1 f0g) if and only if [x] = [x0 ] and [y] = [y 0 ]. Further,
one sees that if zij is the coordinate of z 2 F n+1 F m+1 then the condition
that [z] is in the image of this map is that zij zkl = zkj zil = zil zkj . These are
homogeneous equations of degree 2. Thus the image is closed. We have thus
embedded Pn Pm as a closed subset of Pnm+n+m . This de…nes the topology
on the corresponding product of algebraic varieties.
16
Exercise. Prove the assertions in the argument above (hit if z = x y then
zij = xi yj ). Also show that the images of the Pni Pm
j are open in the image
and isomorphic to F n+m .
The following theorem is Theorem A.4.8 in [GW].
Theorem 23 If (X; OX ) is an irreducible projective variety then OX (X)consists
of the constants.
1.3.4
Local properties of algebraic varieties.
It is almost counter-intuitive that the assertion that for X a topological space
the assertion that Y
X is closed is a local property. That is if for each
p 2 X there exists U X open with U \ Y closed in U then Y is closed in
X. To see this we note that this local condition implies that there is an open
covering of X, fU g 2I such that U \ Y is closed in U . This implies that
U
U \ Y is open in U hence in X. But [ 2I (U
U \Y )=X Y.
One notes that the topology that we have put on an algebraic variety is
Noetherian so algebraic varieties have a unique (up to order) decomposition
into irreducible algebraic varieties.
Since any two non-empty open sets in an irreducible topological space
must intersect we see that we can de…ne the dimension of an irreducible
algebraic set to be the dimension of any open non-empty a¢ ne subvariety.
We de…ne the dimension of an algebraic variety to be the maximum of the
dimensions of the irreducible components.
We will say that a point in an irreducible algebraic variety is smooth if
it is smooth in an a¢ ne open set containing it. Thus the set of all smooth
points is open in the variety.
We note that if F = C then a variety has in addition a standard metric
topology. To distinguish the two topologies we will use the terms Z-topology
and S-topology.
Theorem 19 now implies
Theorem 24 If (X; OX ) is an irreducible d-dimensional algebraic variety
over C then the set of smooth points endowed with the S-topology is a C 1
manifold of dimension 2d.
In addition using the fact that closedness is a local property we have
Theorem 25 Let (X; OX ) be an algebraic variety over C. If U is Z-open in
X then the S-closure of U is the same as the Z-closure.
17
In our study of algebraic groups we will be using the following theorem
the local version is e.g. Theorem A.2.7 in [GW].
Theorem 26 Let (X; OX ) and (Y; OY ) be non-empty irreducible algebraic
varieties. Then if f : X ! Y is a morphism such that the image of X is
dense in Y (here we are using the Z-topology if F = C) and if U is open and
non-empty in X then f (U ) contains an open non-empty subset of Y:
Corollary 27 Let (X; OX ) and (Y; OY ) be non-empty irreducible algebraic
varieties over C. Then if f : X ! Y is a morphism such that the image of
X is Z-dense in Y then it is S-dense.
1.4
1.4.1
Appendix. Some local algebra.
The Artin Rees Lemma.
Let R be a Noetherian algebra over a …eld F . In this context we recall the
Artin Rees lemma.
Lemma 28 Let I be an ideal in R. Let M be a …nitely generated module
over R and let N be an R-submodule of M . Then there exists l 2 N such
that I l+j M \ N = I j (I l M \ N ) for all j 2 N:
Proof. Let t be an indeterminate and consider the subalgebra
R = R + tI + t2 I 2 + :::
of R[t];this algebra is sometimes called the Rees algebra of R with respect to
I. If I is generated by r1 ; :::; rm (i.e. T = Rr1 + ::: + Rim ) then R is the
algebra over R generated by tr1 ; :::; trm . Thus R is Noetherian. We consider
the R -module
M = M + tIM + t2 I 2 M + ::::
If M is generated as a module by m1 ; :::; mr as an R-module then so is M
as an R module. Let
N1 = N + tIM \ N + t2 I(IM \ N ) + ::: + tr+1 I r (IM \ N ) + :::
N2 = N + tIM \ N + t2 (I 2 M \ N ) + ::: + tr+2 I r (I 2 M \ N ) + :::
18
and in general
Nm = N + tIM \ N + t2 (I 2 M \ N ) + ::: + tm (I m M \ N )+
tm+1 I(I m M \ N ) + ::: + tm+j I j (I m M \ N ) + :::
Then N1
N2
::: and all are R -submodules. Thus there is an index l
such that Nl+j = Nl for all j. Comparing coe¢ cients of t yields the lemma.
Corollary 29 Let M be a maximal proper ideal in R. If M is an R-module
then \k 1 Mk M = 0.
Proof. Let N = \k 1 Mk M . Let l be as in the Artin Rees lemma. Then
N = Ml+j M \ N = Mj (Ml M \ N ) = Mj N:
Thus N = MN . Now Nakayama’s lemma (below) implies that N = 0:
We now recall Nakayama’s Lemma.
Lemma 30 Let R be a commutative ring with identity and let I be an ideal
in R such that 1 + a is invertible for all a 2 I (e.g. I is maximal). Then if
M is a …nitely generated R-module such that IM = M then M = 0.
Proof.
Let M = Rm1 + ::: + Rmr . Then there exist aij 2 I such that
Pr
j=1 aij mj = mi . Thus we have
r
X
(
ij
aij )mj = 0:
j=1
If we apply Cramer’s rule we have det(I
[aij ])mk = 0 for all k. But
det(I [aij ]) = 1 + a with a 2 I. Thus mk = 0 all k.
1.4.2
Localization.
Let R be a …nitely generated algebra over F . Let S
R be closed under multiplication containing 1 and doesn’t contain 0we will call S a multiplicative subset. Then we de…ne R(S) to be the set R S modulo the
equivalence relation (r; s) is equivalent with (t; u) if there exists v 2 S such
that v(ru ts) = 0. We observe that if (r; s) is equivalent with (r0 ; s0 ) and
(t; u) is equivalent with (t0 ; u0 ) then (rt; su) is equivalent with ((r0 t0 ; s0 u0 ) and
(ru + ts; su) is equivalent with (r0 u0 + t0 s0 ; s0 u0 ) we can thus de…ne an algebra
structure on R(S) . We write the equivalence class of (r; s) as r=s. We have
an algebra homomorphism of R to R(S) given by r ! r=1.
19
Lemma 31 If R and S are as above then R(S) is Noetherian.
Let M be an ideal in R such that S = R M is multiplicative. This
condition is equivalent to the primality of M. Then we set RM = R(S) .
Then RM has the unique maximal ideal m =MR(S) . This is the de…nition
of a local ring.
Let R be a local ring with maximal ideal M. Let jkl : R=Ml ! R=Ml :for
l k be the natural projection. We de…ne
lim R=Mk
to be the set of sequences fak g with ak 2 R=Mk and jkl al = ak for l
k.
Then this ring with addition and multiplication given by component wise
operations is called the completion of R and denoted R.
20