Download ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let

ABSTRACT NONSINGULAR CURVES
Affine Varieties
Notation. Let k be a field, such as the rational numbers Q or the complex
numbers C.
n
We call affine n-space the collection A (k) of points P = a1 , a2 , . . . , an with coordinates ai ∈ k. For example, A1 (k) = k. Similarly, we
call projective n-space the collecn
tion P (k) of equivalence
classes
: · · · : an : a0 , where we say two nonzero points
a1 n+1
a1 , . . . , an , a0 , b1 , . . . , bn , b0 ∈ A (k) are equivalent if ai = λ bi for some nonzero
b1
a1
= . We can
λ ∈ k. For example, a1 : a0 = b1 : b0 ∈ P1 (k) formally if and only if
a0 b 0
always embed An (k) ,→ Pn (k) via the map a1 , . . . , an 7→ a1 : · · · : an : 1 . We will be
interested in subsets of these spaces.
Denote k[x1 , x2 , . . . , xn ] as the collection of polynomials in n variables with coefficients
in k. Given a collection of polynomials {f1 , f2 , . . . , fm }, we denote the ideal
hf1 , f2 , . . . , fm i = g1 f1 + g2 f2 + · · · + gm fm g1 , g2 , . . . , gm ∈ k[x1 , x2 , . . . , xn ] .
Recall that ℘ = hf1 , f2 , . . . , fm i is a prime ideal if and only if the quotient ring O =
k[x1 , x2 , . . . , xn ]/℘ is an integral domain. Whenever this is the case, we call an affine
variety a set in the form
n
X = P ∈ A (k) f1 (P ) = f2 (P ) = · · · = fm (P ) = 0 .
I claim that any f ∈ O is actually a function f : X 7→ P1 (k). To see why, say P
that f =
h1 + ℘ = h2 + ℘ for some polynomials
h
,
h
.
Since
h
−
h
∈
℘,
we
have
h
−
h
=
1
2
1
2
1
2
α gαfα ,
P
so that h1 (P ) − h2 (P ) = α gα (P ) fα (P ) = 0. Hence f (P ) = h1 (P ) : 1 = h2 (P ) : 1 is
a well-defined element. We are motivated by the following philosophy, usually attributed to
Alexander Grothendieck:
Instead of studying a variety X, study its ring of functions O.
Zariski Topology. We wish to place a topology on an affine variety X. We say that a
subset V ⊆ X is closed if there exists a subset T ⊆ k[x1 , x2 , . . . , xn ] such that
V = Z(T ) = P ∈ X f (P ) = 0 for all f ∈ T .
T may be taken to be an ideal containing ℘, but this is not crucial. We say that a subset
U ⊆ X is open if U = X − V is the compliment of some closed set V . I claim that:
• Both ∅ and X are open sets.
S
• If {Uα } is a arbitrary collection of open sets that
T α Uα is also open.
• If {Uα } is a finite collection of open sets, then α Uα is also open.
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Let me explain why. It is easy to see that ∅ = X − Z {0} and X = X − Z {1} . Say that
each Uα = X − Vα for some closed set Vα = Z(Tα ). De Morgan’s Laws imply that
!
[
\
[
Uα = X −
Vα = X − Z
Tα ,
α
α
α
\
[
Y
!
Uα = X −
α
Vα = X − Z
α
Tα
.
α
This topology is called the Zariski topology, named after Oscar Zariski.
Relation with Complex Analysis. Here are how the definitions we’ve studied before
match up with the ones gives thus far:
Complex Analysis
Commutative Algebra
C, the ambient space
D, region in Cn
Ω, open set
f : D → C, analytic function
S, sheaf of analytic functions
A1 (k), affine space
X, affine variety in Pn (k)
U = X − Z(T ), Zariski open set
f : X → P1 (k), rational function
O, ring of rational functions
Example. We seek a working definition for the types of maps allowed between varieties. To
this end, we consider a simple
√ example.
Let k be a field such that −1 6∈ k. Let f1 (x, y) = x2 + y 2 − 1 be a polynomial from the
ring k[x, y], and consider the following affine variety:
2
2
2
X = (x, y) ∈ A (k) x + y − 1 = 0 .
This is simply the unit circle. There is a morphism ψ : P1 (k) → X defined by
2
a1 − a20 2 a1 a0
,
a1 : a0 7→
.
a21 + a20 a21 + a20
Note that this map is well-defined and only involves ratios of polynomials. We will show that
this map is birational by exhibiting another morphism ϕ : X → P1 (k) such that ψ ◦ ϕ = idX
and ϕ ◦ ψ = idY a maps also involving ratios of polynomials.
To this end, consider the following subsets of X:
U1 = (x, y) ∈ X (x, y) 6= (−1, 0) = X − Z {x + 1} ,
U2 = (x, y) ∈ X (x, y) 6= (+1, 0) = X − Z {x − 1} .
These are open sets such that X = U1 ∪ U2 , so they define an open cover of the unit circle.
Define now the following maps:
)
(
ϕ1 : U1 → P1 (k)
x, y 7→ 1 + x : y
which send
ϕ2 : U2 → P1 (k)
x, y →
7
y :1−x
2
Using the formal relation
2
a1
y
a1 − a20 2 a1 a0
1+x
,
=
=
⇐⇒
(x, y) =
x +y =1
⇐⇒
y
a0
1−x
a21 + a20 a21 + a20
we immediately see that ϕ1 U1 ∩U2 = ϕ2 U1 ∩U2 . In particular, the pairs hU1 , ϕ1 i ' hU2 , ϕ2 i are
equivalent, so they define a morphism
(
1
+
x
:
y
if (x, y) ∈ U1 , and
ϕ : X → P1 (k)
which sends
(x, y) 7→
y : 1 − x if (x, y) ∈ U2 .
2
2
It is easy to verify that ψ ◦ ϕ = idX and ϕ ◦ ψ = idY . Note how we glued two maps together
to find a morphism.
Abstract Nonsingular Varieties
Points as Primes. Recall that Grothendieck’s philosophy is to replace X with O whenever
possible. But what exactly is the relationship between them?
Assume now that k is algebraically closed. Fix a prime ideal ℘ = hf1 , f2 , . . . , fm i in
k[x1 , x2 , . . . , xn ]; then O = k[x1 , x2 , . . . , xn ]/℘ is an integral domain. Continue to denote
X as the collection of P ∈ An (k) such that f (P ) = 0 for all f ∈ ℘.
Proposition 1. There is a one-to-one correspondence between points P = a1 , a2 , . . . , an
on X and maximal ideals mP = hx1 − a1 , x2 − a2 , . . . , xn − an i in mSpec O.
Proof. Fix a point P = a1 , a2 , . . . , an on X. We show that m = hx1 −a1 , x2 −a2 , . . . , xn −
an i is a maximal ideal. Recall that an ideal m ⊆ O is maximal if and only if O/m is a field.
Since O/m ' k[a1 , . . . , an ] = k, we see that m is indeed maximal.
Conversely, fix a maximal ideal m in mSpec O. We show that mP = hx1 − a1 , x2 −
a2 , . . . , xn − an i for some point P = a1 , a2 , . . . , an on X. Since O/m is a finite extension
of k and k is algebraically closed, we have a map
∼
O −−−→ O/m −−−→ k.
Let ai ∈ k be the image of xi ∈ O, and consider the ldeal mP = hx1 −a1 , x2 −a2 , . . . , xn −an i.
Since each polynomial xi − ai is in the kernel of this map, we must have mP ⊆ m ⊆ O. But
mP is a maximal ideal, so m = mP .
Abstract Varieties. To recap, there is an injective map X ' mSpec O ,→ Spec O, for
some integral domain O, which is defined by P 7→ mP . Even though is map not surjective,
for any integral domain O we say that Spec O is a abstract variety. Note that X ' mSpec O
is really an affine variety, although we will abuse notation and call Spec O an affine variety
as well.
Example. Let O = Z. The only prime ideals in O are m = p Z and {0} for any rational
prime p ∈ Z, because
(
Fp for m = p Z, and
O/m '
Z for m = {0}.
In particular, mSpec O ' {2, 3, 5, . . . , p, . . . } consists of the rational primes and Spec O =
{0} ∪ mSpec O. Hence the affine variety mSpec O has points which are the rational primes,
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yet the abstract variety Spec O is strictly larger. These are varieties – but we cannot express
the points as zeroes of some polynomial!
Zariski’s Notion of Nonginsgularity. Let’s continue to denote X as the collection of
P ∈ An (k) such that f (P ) = 0 for all f ∈ ℘. We say that X is a nonsingular affine variety if
we can define a tangent at every point P ∈ X. We can make this rigorous using derivatives
as follows: Since X is the collection of zeroes for the set ℘ = hf1 , f2 , . . . , fm i, we define the
Jacobian of X at a point P to be the m × n matrix


∂f1
∂f1
∂f1
 ∂x1 (P ) ∂x2 (P ) · · · ∂xn (P ) 


..
..
..
..
.
J f1 , f2 , . . . , fm (P ) = 
.
.
.
.


 ∂fm

∂fm
∂fm
(P )
(P ) · · ·
(P )
∂x1
∂x2
∂xn
Since ℘ is a prime ideal, the number of polynomials cannot be greater than the number of
variables: m ≤ n. The rank of this matrix should be as large as possible, so we say that X
is a nonsingular affine variety of rankk J f1 , f2 , . . . , fm (P ) = m for every P ∈ X.
Points as DVRs. Keeping Grothendieck’s philosophy in mind, we wish to translate this
idea into one involving the ring O. We give some motivation: Say that X
is the collection
of P ∈ An (k) such that f (P ) = 0 for all f ∈ ℘. If P = a1 , a2 , . . . , an is a point on X,
then any f in O = k[x1 , x2 , . . . , xn ]/℘ has a Taylor series expansion
n
X
∂f
f (x1 , x2 , . . . , xn ) = f (P ) +
(x1 − a1 )
∂xi
i=1
+
n X
n
X
1
i=1 j=1
∂ 2f
(xi − ai ) (xj − aj ) + · · ·
2 ∂xi ∂xj
Modulo the maximal ideal mP = hx1 − a1 , x2 − a2 , . . . , xn − an i, we have f (Q + P ) ≡ f (P )
mod mP and f (Q + P ) ≡ f (P ) + ∇f (P ) · Q mod m2P for any Q ∈ X. This means
• f mod mP is the same as computing f (P ).
• f mod m2P is the same as computing ∇f (P ).
For now, assume that n = 2 and m = 1, i.e., O = k[x, y]/ f1 . The following proposition
explains then importance of mP /m2P in associating P ∈ X with a discrete valuation ring. It
is a restatement of Proposition 9.2 on pages 94-95 in Atiyah-Macdonald.
Theorem 2. For each P ∈ X, the following are equivalent:
i. X is nonsingular
at P , i.e., ∇f1 (P ) 6= (0, 0).
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ii. dimk mP /mP = 1.
iii. mP OP is a principal ideal.
iv. OP , the localization of O at mP , is a discrete valuation ring.
v. OP is integrally closed.
Proof. (i) ⇐⇒ (ii). We make a general observation, following Zariski. Say that we have
O = k[x1 , x2 , . . . , xn ]/℘, as before. For each point P ∈ X, we define the Jacobian to be the
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m × n matrix

∂f1
∂f1
∂f1
 ∂x1 (P ) ∂x2 (P ) · · · ∂xn (P ) 


..
..
..
..
.
J f1 , f2 , . . . , fm (P ) = 
.
.
.
.



 ∂fm
∂fm
∂fm
(P )
(P ) · · ·
(P )
∂x1
∂x2
∂xn
Hence multiplication by this matrix induces a short exact sequence

{0} −−−→ TP (X) −−−→ An (k) −−−→ Am (k)
where TP (X) is the tangent space of
X at P . We have a perfect (i.e., bilinear and nondegenerate) pairing TP (X) × mP /m2P → k defined by Q, f 7→ ∇f (P )· Q. (For this reason,
2
we call mP /m2P the cotangent space of X at P .) Hence
dimk mP /mP = dimk TP (X). But
dimk TP (X) = n − m if and only if J f1 , f2 , . . . , fm (P ) has rank m. The case of interest
follows with n = 2 and m = 1.
(ii) ⇐⇒ (iii). We will abuse notation and think of mP as an ideal of OP . As mP is a
maximal ideal, Nakayama’s Lemma states that we can find t ∈ mP where t ∈
/ m2P . Consider
the injective map O/mP → mP /m2P defined by x 7→ t x.Clearly this is surjective if and only
if mP = t OP is principal. Recall now that dimk O/mP = 1.
(iii) =⇒ (iv). Say that mP = t OP as a principal ideal. In order to show that OP is a
discrete valuation ring, it suffices to show that any nonzero x ∈ OP is in the form x = tm y
for some m ∈ Z and y ∈ OP× . Consider the radical of the ideal generated by x:
p
(x) = y ∈ OP y n ∈ x OP for some nonnegative integer n .
p
As OP has a unique nonzero prime ideal, we must have (x) = m. But then there is largest
nonnegative integer m such that tm−1 ∈
/ x OP yet tm ∈ x OP . Hence y = x/tm ∈ OP but
y∈
/ mP .
(iv) =⇒ (v). Say that OP is a discrete valuation ring. Say that x ∈ K is a root of
a polynomial equation xn + a1 xn−1 + · · · + an = 0 for some ai ∈ OP . Assume by way
of contradiction that x ∈
/ OP . Then vP (x) < 0, so that vP (1/x) > 0, hence y = 1/x isan
element of OP . Upon dividing by xn−1 we have the relation x = − a1 +a2 y +· · ·+an y n−1 ∈
OP . This contradiction shows that OP is indeed integrally closed.
(v) =⇒ (iii). Say that OP is integrally closed. We must construct anpelement t ∈ OP
such that mP = t OP . Fix a nonzero x ∈ mP . By considering the radical (x) and noting
that mP is a finitely generated OP -module, we see that there exists some m ∈ Z such that
m−1
mm
6⊆ x OP . Choose y ∈ mm−1
such that y ∈
/ x OP , and let t = x/y be
P ⊆ x OP yet mP
P
an element in K. Consider the module (1/t) mP ⊆ OP ; we will show equality. As y ∈
/ OP ,
we have 1/t ∈
/ OP , so that 1/t is not integral over OP . Then (1/t) mP cannot be a finitely
generated OP -module, we have (1/t) mP 6⊆ mP . As there is an element of (1/t) mP which is
not in mP , we must have equality: (1/t) mP = OP . Hence mP = t OP as desired.
Examples. Take f1 (x, y) = y 2 − x3 + x, and P = (0, 0).
Then ∇f1 (P ) =2 (1, 0) is nonzero.
2
2
2
The maximal ideal is mP = x, y and mP = x , x y, y , so that x = x·x +(−1)·y 2 ∈ m2P .
Hence mP /m2P is a 1-dimensional k-vector space spanned by y alone. Now take f1 (x, y) =
y 2 −x3 , and P = (0, 0). Then ∇f1 (P ) = (0, 0) is zero. The quotient mP /m2P is a 2-dimensional
k-vector space spanned by both x and y.
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Dedekind Domains. Here is an application to the rings one studies in Algebraic Number
Theory.
Corollary 3. The following are equivalent:
i. X is a nonsingular
curve, i.e., ∇f1 (P ) 6= (0, 0) for all points P ∈ X.
ii. dimk m/m2 = 1 for all maximal ideals m of O.
iii. O is a Dedekind domain.
Proof. (i) ⇐⇒ (ii). Recall that the map X → mSpec O defined by P 7→ mP is a bijection.
Now use the previous theorem. (ii) ⇐⇒ (iii). A Dedekind domain is a Noetherian integral
domain of dimension 1 that is integrally closed. Using the previous theorem, it suffices show
that the localization Om is integrally closed for each maximal ideal m if and only if O is
integrally closed. But this is clear. (If not, consult Theorem 5.13 on page 63 of AtiyahMacdonald.)
Abstract Nonsingular Curves. Let O be a Noetherian integral domain with Krull dimension dim O = 1. We say that X = Spec O is an abstract curve. Note that X is nonsingular
precisely when O is a Dedekind domain. For instance, when O = Z, we think of X = Spec Z
as a nonsingular curve!
Abstract Nonsingular Varieties. This can be generalized considerably. Let O be a
Noetherian integral domain. As O is an integral domain, we have O ⊆ K; as O is Noetherian,
the Krull dimension, dim O, is finite. We say that the abstract variety X = Spec O is
nonsingular if O is integrally closed. It is not hard to show that dimk m/m2 = dim O for
any maximal ideal m of O; we define this as the dimension of X.
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