Kovacic Picard-Vessiot Theory
Transcript of Kovacic Picard-Vessiot Theory
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Picard-Vessiot theory, algebraic groups andgroup schemes
Jerald J. KovacicDepartment of Mathematics
The City College of The City University of New YorkNew York, NY 10031email: [email protected]
URL: mysite.verizon.net/jkovacic
September 30, 2005
Abstract
We start with the classical definition of Picard-Vessiot extension.We show that the Galois group is an algebraic subgroup of GL(n).Next we introduce the notion of Picard-Vessiot ring and describe theGalois group as spec of a certain subring of a tensor product. We shallalso show existence and uniqueness of Picard-Vessiot extensions, usingproperties of the tensor product. Finally we hint at an extension ofthe Picard-Vessiot theory by looking at the example of the Weierstra-function.
We use only the most elementary properties of tensor products,
spec, etc. We will define these notions and develop what we need. Noprior knowledge is assumed.
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1 Introduction
Throughout this talk we fix an ordinary -field Fof characteristic 0 and withalgebraically closed field of constants
C = F
If you want, you may assume that F= C(x) is the field of rational functionsof a single complex variable.
I usually use the prefix - instead of the word differential. Thus I speak of
-rings and -fields, -ideals, etc.
2 Classical Picard-Vessiot theory
We consider a linear homogeneous -equation
L(y) = y(n) + an1y(n1) + + a0y = 0
Definition 2.1. By a fundamental system of solutions ofL(y) = 0, we meana set 1, . . . , n of elements of some -extension field G ofF such that
1. L(i) = 0,
2. 1, . . . , n are linearly independent over C.
We usually write = (1, . . . n) for the row vector.
Definition 2.2. By a Picard-Vessiot extension for L we mean a -field Gcontaining F such that
1. G = F = C,
2. G = F1, . . . , n where 1, . . . , n is a fundamental system of solutionsof L(y) = 0.
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Definition 2.3. Suppose that G is a Picard-Vessiot extension. Then the
group of -automorphisms ofG over F,
G(G/F) = - Aut(G/F)
is called the Galois group ofG over F.
Proposition 2.4. Suppose that G is a Picard-Vessiot extension. If G(G/F) then there is an invertible matrix c() with constant coefficients suchthat
= c().
The mappingc : G(G/F)
GL(n)
is an injective homomorphism.
Proof. The easiest way to see this is to look at the Wronskian matrix.
W =
1 n1 n...
...
(n1)1 (n1)n
Because 1, . . . , n are linearly independent over C the Wronskian is invert-
ible.
A simple computation shows that
W =
1 n1 n...
...
(n)1 (n)n
=
0 1 0 . . . . . . . 0... 0 1
...... 0
. . ....
.... . . 1 0
0 . . . . . . . . . 0 1a0a1. . . . . .an2an1
1 n1 n...
...
(n1)1 (n1)n
i.e.
WW1 = A =
0 1 0 . . . . . . . 0... 0 1...
... 0. . .
......
. . . 1 00 . . . . . . . . . 0 1a0a1. . . . . .an2an1
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The matrix A is called the companion matrix for L.
Differentiatec() = W1W
and you get 0, so c() GLC(n) = GL(C). The first row of W is so
W = W c() implies that = c() .
Suppose that , G(G/F). Then
c() = W1(W c()) = W1(W)c() = c()c() ,
because c() has constant coordinates and therefore is left fixed by . There-fore c is a homomorphism of groups. c is injective since G = F = F(W).
3 Algebraic subgroups of GL(n)
Here we take a classical point of view, later on we shall be more modernand use group schemes.
We start by putting a topology on affine m-space
Am = Cm
Definition 3.1. A subset X ofAm is Zariski closed if there exists a finiteset of polynomials in m variables
f1, . . . , f r C[X1, . . . , X m]
such that X is the zero set of f1 = = fr = 0, i.e.
X = {(a1, . . . , am) Cm | f1(a1, . . . , am) = = fr(a1, . . . , am) = 0} .
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We can drop the adjective finite in the definition. Indeed X being the zero
set of a collection fi (i I) of polynomials is equivalent to saying that X isthe zero set of the entire ideal
a = ((fi)iI)
and even the radical ideala = {f | fe a for some e N}
By the Hilbert Basis Theorem this ideal is generated by a finite number ofpolynomials.
Theorem 3.2. (Hilbert Nullstellensatz) There is a bijection between closedsubsets ofAm and radical ideals ofC[X1, . . . , X m].
Now lets put a topology on GL(n), the set of invertible n n matrices withcoefficients in C. We do this by embedding GL(n) into An
2+1:
c11 c1n...
...cn1 cnn
(c11, . . . , c1n, . . . , cn1, . . . , cnn, 1/ det cij) An2+1 .
The image is closed, it is the zero set of
det(Xij)Y = 1
where Y is the (n2 + 1)st coordinate.
Definition 3.3. A subset X GL(n) is Zariski closed is if it closed in thesubset topology as a subset ofAn
2+1.
Definition 3.4. A linear algebraic group is a closed subgroup of GL(n) forsome n.
4 The Galois group of a Picard-Vessiot ex-
tension
In this section G is a Picard-Vessiot extension ofF.
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Proposition 4.1. The image of
c : G(G/F) GL(n)is an algebraic subgroup of GL(n).
Proof. Let y1, . . . , yn be -indeterminates over F. This means that
y1, . . . , yn, y
1, . . . , y
n, y
1 , . . . y
n, . . .
is an infinite family of indeterminates over F. We use vector notation andwrite y = (y1, . . . , yn). Then
F{y} = F[y, y, . . . ]is a polynomial ring in an infinite number of variables. There is a homomor-phism over F, called the substitution homomorphism, defined by
: F{y} F{}yi iyi i
...
Evidently, it is a -homomorphism. Let p be its kernel
0 E p E F{y} E F{} E 0
For C GL(n) we let C be the substitution homomorphismC: F{y} F{y}
y yCThis means
yi j yjCji .Lemma 4.2. C is in the image of c : G(G/F) GL(n) if and only if
Cp p and C1p p(This is equivalent to Cp = p.)
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Proof. Suppose that C = c() for some
G(G/F). We have both
= C and Cy = yC
We have the commutative diagram:
0 E p E F{y} E F{} E 0
cC
c
0 E p E F{y} E F{} E 0
To show that Cp p, we chase the diagram. If a p then a = 0 so0 = (a) = (Ca)
which implies thatCa ker = p .
We have shown thatCp p .
For the other inclusion, use 1.
Now suppose that Cp = p. Then there is a -homomorphism
0 E p E F{y} E F{} E 0
cC
cC
c
0 E p E F{y} E F{} E 0In fact is defined by
a = (CA), where A = a (a F{}, A F{y}) .Since y = , we have the matrix equation
= (C(y)) = (yC) = C
We can see that is bijective by diagram chasing. Tberefore extends to a-automorphism of the field of quotients
: F = G G
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So
G(G/F) and since = C,
c() = C .
We think ofp as a vector space over F and choose a basis A for it. We alsoextend A to a basis B ofF{y} over F, so that A B.Lemma 4.3. There exist polynomials
Qbc F[X11, . . . , X nn] b, c Bwith the property that for every C GL(n) and b B,
C(b) =dB
Qbc(C) d .
Proof. We first examine how C acts on F{y}. Let M be the set of monomials,thus an element M ofM is a power product
M =r
k=1y
(ek)ik
fk
of the yi and their derivatives. Since the coordinates of C are constants,
C(y(e)i ) =
k
y(e)k Cki .
The right hand side is linear combination of the y(e)i with coefficients that
are coordinates of C. If we apply C to a monomial we will get productof these right hand sides which is a linear combination of monomials withcoefficients that are polynomials over Z in the coordinates of C. I.e. there
exist polynomials
PMN Z[X11, . . . , X nn] M, N Msuch that
CM =NM
PMN(C)N .
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Because B and M are both bases ofF
{y
}over F, we can express each element
of B as a linear combination over F of monomials, and, conversely, everymonomial as a linear combination over F of elements ofB. It follows thatthere exist polynomials
Qbc F[X11, . . . , X nn] b, c Bwith the property that for every C GL(n) and b B,
C(b) =dB
Qbc(C) d .
Lemma 4.4. There is a (possibly infinite) family of polynomials
Ri F[X11, . . . , X nn, Y] i Isuch that C GL(n) is in the image of c if and only if
Ri(C,1
detC) = 0, i I
Proof. We know, by Lemma 4.2, that C GL(n) is in the image of c if andonly if
Cp p and C1p p .Recall that A is a basis ofp over F and, by the previous lemma,
C(a) =b
Qab(C) b
so Cp p if and only ifQab(C) = 0 for every a A, b B, b / A .
Similarly C1p p if and only if
Qab(C
1) = 0 for every a A, b B, b / A .
Of course, the coordinates of C1 are 1detC
times polynomials in the coordi-nates ofC. Thus there exist polynomials
Rab F[X11, . . . , X nn, Y]
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such that
Rab(C, 1detC) = Qab(C
1)
To conclude the proof of the theorem we need to find polynomials as above,except that the coefficients should be in C not F.
Choose a basis ofF over C. We then can write
Ri =
Si
whereSi C[X11, . . . , X nn, Y] .
If Ri(C,1
detC) = 0 then
0 =
Si(C,1
detC)
Because the elements of are linearly independent over C, we must have
Si(C,1
detC) = 0 for all
It follows from the previous lemma that C GL(n) is in the image of c ifand only if
Si(C,1
detC) = 0 for all i I and
This proves the theorem.
5 Matrix equations
Starting with a linear homogeneous -equation (a scalar -equation) we chosea fundamental system of solutions 1, . . . , n and formed the Wronskian
1 n...
...
(n1)1 (n1)n
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We discovered that
W
= AWwhere
A =
0 1 0 . . . . . . . 0... 0 1
...... 0
. . ....
.... . . 1 0
0 . . . . . . . . . 0 1a0a1. . . . . .an2an1
was a matrix with coefficients in F.
We can also start with a matrix equation
Y = AY
where A MatF(n) is any matrix with coordinates in F, and look for asolution matrix Z that is invertible. The matrix Z is called a fundamentalsolution matrix for the matrix -equation.
6 The Picard-Vessiot ring
Let A MatF(n) be a given n n matrix with coefficients in F.Definition 6.1. By a Picard-Vessiot ring for A we mean an integral domainR such that
1. (qfR) = F = C,
2. R = F[Z, Z1] where ZZ1 = A MatF(n).
Item 2 could also be written R = F[Z, 1detZ
]. There is a popular abuse of
notation that writes R = F[Z,1det ].
Proposition 6.2. If G = F = F(W) is a Picard-Vessiot extension, asbefore, thenR = F[W, W1] is a Picard-Vessiot ring.
Conversly, ifR is a Picard-Vessiot ring then G = qfR is a Picard-Vessiotextension.
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IfF contains a non-constant this is a consequence of the cyclic vector the-
orem. IfF= C it must (and can be) proven by a different method.
Definition 6.3. By the Galois group ofR over F, denoted G(R/F) we meanthe group of a -automorphisms ofR over F.
Proposition 6.4. IfG = qfR, then G(R/F) = G(G/F).
7 Differential simple rings
Definition 7.1. Let R be a -ring. We say that R is -simple ifR has noproper non-trivial -ideal.
In algebra (not -algebra) a simple (commutative) ring R is uninteresting.Indeed (0) is a maximal ideal and the quotient
R/(0) = R
is a field, i.e. R is a field. But in -algebra the concept is very important.
Example 7.2. Let R = C[x] where x
= 1 i s -simple. If a R is anon-zero -ideal then it contains a non-zero polynomial. Choose a non-zeropolynomial P(x) in a having smallest degree. But P a has smaller degree,so P = 0. But that makes P C so 1 a.
Note that (0) is a maximal -ideal (there is no proper -ideal containing it)but is not a maximal ideal.
More generally, ifR is any -ring and m a maximal -ideal ofR then R/mis -simple. It is a field if and only ifm is a maximal ideal.
Proposition 7.3. LetR be a Picard-Vessiot ring. ThenR is -simple.
Proposition 7.4. Suppose thatR is a -simple ring containingF. Then
1. R is an integral domain, and
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2. (qfR) = C.
This suggests a way of creating Picard-Vessiot rings.
Theorem 7.5. LetA MatF(n). Then there exists a Picard-Vessiot ringRfor A.
Proof. Let y = (yij) be a family of n2 -indeterminates over F and let
S = F{y}[ 1det y
]
(The derivation on F{y} extends to S by the quotient rule.) We want to finda maximal -ideal ofS that contains the -ideal
a = [y Ay]
We can do this, using Zorns Lemma, as long as a is proper, i.e. no power ofdet y is in a. But this is certainly true since every element ofa has order atleast 1 and det y has order 0.
Let m be a maximal -ideal ofS that contains a and set
R = S/m
R is -simple so it is a domain and (qfR) = C. If Z is the image of thematrix y in R then
Z = AZ
since a is contained the kernel ofS R.Corollary 7.6. Given a linear homogeneous-equation L(y) = 0 there existsa Picard-Vessiot extension for L.
8 Example where [y A] is not maximal
The example I gave at the seminar was wrong. I had forgotten that thecontaining ring is F{}[ 1
det y].
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Let F= C(ex) and A = 1 (a 1
1 matrix). The we must look at the ideal
[y y] F{y}[1y
] = F{y, 1y}
I had asserted that [yy] [y], which is indeed true, but not relevant, since1 [y]. However
[y y] [y ex] .Indeed
y y = (y ex) (y ex)Also [y ex] is a maximal -ideal (even a maximal ideal) since it is the kernelof the substitution homomorphism
F{y, 1y} F{ex, ex} = C(ex) = F
9 Tensor products
Let R and S be -rings that both contain F. We are interested in the tensorproduct
R
FS
This is a -ring. The easiest way to describe it uses vector space bases.
Let {xi}, (i I), be a vector space basis ofR over F and {yj}, (j J), bea basis ofS over F. Consider the set of all pairs (xi, yj) and the set RFSof all formal finite sums
i.j
aij(xi, yj) where aij F
RFS is a vector space over Fwith basis (xi, yi).
If x =
i aixi R and y =
j bjyj S we write
x y =i
j
aibj(xi, yj) RFS
We have
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1. (x + x)
y = x
y + x
y
2. x (y + y) = x y + x y3. a(x y) = ax y = x ay
One defines multiplication so that
(x y)(x y) = xx yyand shows that R F S is a ring. Finally we define a derivation with theproperty
(x
y) = x
y + x
y
and we get a -ring.
We have two canonical mappings
: R RFSa a 1
and
: S RFSa 1 a
10 CR C
Be careful. Tensor products are usually much worse than the rings youstarted with. For example
CR Cis not a field, in fact it is not even an integral domain! Indeed
(i 1 + 1 i)(i 1 1 i) = 1 1 i i + i i 1 1 = 0
In fact CR C C C. Every element ofCR C has the formx = a(1 1) + b(i 1) + c(1 i) + d(i i)
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We define : C
R C
C2 by
(x) =
(a + d) + (b c)i, (a d) + (b + c)iIt is straightforward to check that is an isomorphism or rings.
11 Uniqueness of a Picard-Vessiot ring
Suppose that R and S are both Picard-Vessiot rings for the matrix A. Say
R = F[Z, Z1
] S = F[W, W1
]
whereZZ1 = A = WW1
If Z and W were in some common ring extension T of both R and S, then
W = ZC
for some matrix of constants, C T. We can easily find a common ringextension of both R and S, namely
RFSAnd we can find one whose ring of constants is C
T= (RFS)/mwhere m is a maximal -ideal ofRFS. (It is -simple!)Proposition 11.1. Suppose thatm is a maximal -ideal ofRFS and let
: RFS (RFS)/m = Tbe the canonical homomorphism. Then
: R RFS T
and
: S RFS T
are isomophisms.
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Proof. The kernel of is a proper -ideal ofR. Because R is -simple, this
ideal must be (0), so is injective.
Since Z = AZ and W = AW and A has coefficients in F,
C = (1 W)(Z 1)1
is a matrix of constants and hence has coordinates in C. Therefore
(1 W) = C(Z 1) = (CZ 1)and
(1 S) (R 1)
orT= (R S) (R 1) = ((R)) .
The following proposition says that a Picard-Vessiot ring for A is uniqueup to -isomorphism. It follows that a Picard-Vessiot extension for a linearhomogeneous -equation is also unique up to a -isomorphism.
Theorem 11.2. R andS are-isomorphic.
Proof. Both R and S are -isomorphic to T.
12 RFR
We continue to assume that R is a Picard-Vessiot ring. Here we are interestedin the -ring
RFRand in particular relating it to the Galois group G(R/F).
Let G(R/F). Define a mapping : RFR R
by(a b) = ab.
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Proposition 12.1. If
G(R/F) then the kernel of is a maximal-ideal
m.
Proof.(RFR)/m R
Because R is -simple, m is a maximal -ideal.
Proposition 12.2. Let G(R/F). Thenm is generated as an ideal by
a 1 1 a a R .
Proof. If a R then
(a 1 1 a) = a a = 0
so a 1 1 a m.
Now suppose that
x =i
ai bi m so thati
aibi = 0
Then
x =i
(ai 1)(1 bi bi 1) +i
aibi 1
= i
(ai 1)(bi 1 1 bi)
With this we can prove the converse of Proposition 12.1.
Theorem 12.3. Letm be a maximal -ideal ofR FR. Then there exists G(G/F) such thatm = m.
Proof. SetT= (RFR)/m, : RFR T
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By Proposition 11.1 the mappings
: R T and : R T
are isomorphisms.
Define : R R by = ( )1 ( )
i.e., so thatR
E R
c c
RFR E T ' RFRcommutes.
Let a R, then But
(a 1 1 a) = ((a)) (a) = ( )(a) ( )(a) = 0
Thereforea
1
1
a
ker = m
By Proposition 12.2m m
But m is a maximal -ideal, therefore
m = m .
We have shown that G(R/F) is in bijective correspondence (i.e. can be
identified with the set of maximal -ideals ofRFR, i.e.G(R/F) max diffspec(RFR) .
We have diffspec but we want spec.
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13 C
R C bis
C is a Galois extension of R with Galois group {id, }, being complexconjugation. It turns out that C R C has precisely two prime ideals andthey are both maximal. The first is generated by
a 1 1 a a C
which corresponds the the identity automorphism, and the other generatedby
a
1
1
a a
C
which corresponds to the automorphism . Thus
max spec(CR C) = spec(CR C)
is a finite scheme, which is in fact a group scheme and is isomorphic to theGalois group.
14 The constants ofRFR
Definition 14.1. LetK = (RFR)
Remember that R = C, so we might expect K to be rather small (maybeK = C). This is very far from the truth.
Example 14.2. Let F= C(x) and let
Z = (ex) GL(1)
Note thatZ = Z, so A = 1 .
The Picard-Vessiot ring is
R = F[ex, ex] .
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Then
(ex e
x)
= ex e
x + ex (e
x) = 0so
c = ex ex KExample 14.3. Now let
Z =
1 log x0 1
Here
Z =
0 1
x
0 0
=
0 1x0 0
1 log x0 1
= AZ
andR = F[log x]
Letc = log x 1 1 log x
then
c =1
x 1
1
1
x
= 0
so K.
If M, N MatR(n) we defineM N MatRFR(n)
be the matrix whose ijth coordinate is
(M N)ij =k
Mik Nkj
Proposition 14.4.Suppose that
R=F
[Z, Z
1
]. Then = Z Z1
is a matrix of constants.
Theorem 14.5.
K = C[, 1]
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15 Ideals in R
CK
Definition 15.1. LetI(K)
denote the set of ideals ofK and
I(RCK)
the set of -ideals ofRCK.
Suppose that ao is an ideal ofK, then
RC aois a -ideal ofRCK. This gives a mapping
: I(K) I(RCK)
Ifa RCK is a -ideal then
{c K | 1 c a}
is an ideal ofK and we have a mapping
: I(RCK) I(K)
Theorem 15.2. The mappings and are bijective and inverse to eachother.
The mappings and are order-preserving, so we get a bijection betweenmaximal ideals ofK and maximal -ideals ofRCK.
16 RFR RC K
This is one of the most important theorems of Picard-Vessiot rings.
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Recall that
K = (RFR)so, in particular, K RFR. We have a homomorphism
: RCK RFRgiven by
r k (r 1) kTheorem 16.1. RFR RCK
Proof. We consider
: RCK RFRThe kernel is a -ideal ofRCK. By Theorem 15.2, there is an ideal ao Kwith
R ao = ker But restricted to 1 K is injective, so ao = 0. Therefore is injective.
For surjectivity we need to show that 1 R R 1)[K]. But1 Z = (Z 1)(Z1 Z) = (Z 1) (R 1)[K]
17 specK
Theorem 17.1. IfR is a Picard-Vessiot ring and
K = (RFR)
then
G(R/F)
max diffspec(RFR)
max diffspec(RCK) max specK
Proof. The first line is Theorem 12.3, the second line is Theorem 16.1 andthe last is Theorem 15.2.
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18 Zariski topology on specK
X = specK is the set of prime ideals ofK. Ifa K is a radical ideal, thenwe define
V(a) = {p K | a p}Note that V is order-reversing:
a b = V(a) V(b)
Also
V((1)) = V((0)) = X
V(a b) = V(a) V(b)V(i
ai) =i
V(ai)
We put a topology on X, called the Zariski topology, by defining the closedsets to be sets of the form V(a) for some radical ideal a ofK.
By a closed point p of X we mean a point (prime ideal) such that
V(p) = {p} .
Thus the closed points are precisely the maximal ideals, i.e. the set of closedpoints is what I previously called max specK.
We can do the same thing for -rings R. Thus diffspecR is the set of prime-ideals, if a is a radical -ideal ofR then V(a) is defined similarly. Andwe get a topology, called the Kolchin topology. The set of closed points ismax diffspecR.
Beware Despite the similarity of definitions of diffspecR and specK, thereare vast differences in the theory.
I want to describe max specK a little further. We know that
K = C[, 1det
]
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where = Z1
Z
MatK(n). Let X = (Xij) be indeterminates over C
and Y another indeterminate. Then
: C[X, Y] KX Y 1
det
We have an exact sequence
0 E a E C[X, Y] E K E 0
where a is the kernel of .
An element of max specK comes from a maximal ideal ofC[X, Y] that con-tains a and conversely, i.e.
max specK {m C[X, Y] | m is a maximal ideal that contains a}
If c GL(n) is a zero ofa then
m = (X
c, (det c)Y
1)
is a maximal ideal containing a. The converse is also true - this is the weakHilbert Nullstellensatz. Therefore the set of maximal ideals containing a,max specK, is the zero set ofa.
19 Affine scheme and morphisms
Theorem 19.1. LetR and S beC-algebras. An algebra homomorphism
: R S
induces a scheme morphism
a : spec S spec R
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Conversely, a scheme morphism
f: spec S spec Rinduces an algebra homomorphism
f# : R SThere is a bijection
Mor(spec S, spec R) Hom(R, S)
Note that the arrows get reversed.
Theorem 19.2. LetR and S beC-algebras. Then
spec R spec S = spec(R C S)
20 Group scheme
A group in the category of sets is well-known. But a group in the category
of schemes is somewhat different. It is NOT a group in the category of sets.In category theory one deals with objects and arrows. Here too. We writeG = specK and C = specC. All products are over C, i.e. = C.Definition 20.1. G = specK is a group scheme if there are mappings
m : G G G, (multiplication)e : C G, (identity)i : G G, (inverse)
such that the following diagrams commute.
G G G midG E G G
cidGm
cm
G G m E G
(associativity)
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G C idGe E G G
cm
GidG E G
Tm
C G eidG E G G
(identity)
G G(idG,i)
mdd
dG E C
e E G
(i,idG)
dd
dm
G G
(inverse)
21 Hopf algebra
We can translate the group scheme mappings into algebra homomorphisms.
Definition 21.1. K is a Hopf algebra if there are mappings
: K KK, (comultiplication)I: K C, (coidentity)S: K K, (coinverse or antipode)
such that the following diagrams commute.
K E KK
c
cidK
KK idK E KKK
(coassociativity)
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KK idKI E K CT
KidK E K
c
KK IidK E CK
(coidentity)
KKidKS
dd
ds
K'
CI' K
SidKd
dds
KK
(antipode)
Theorem 21.2. K is a Hopf algebra if and only ifspecK is a group scheme.
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22 Sweedler coring
There is a natural structure of coring (which I will not define) on R FRdefined by
: RFR (RFR) R (RFR)a b a 1 1 b
I: RFR Fa
b
ab
S: RFR RFRa b b a
This looks like a Hopf algebra but, in fact, is not quite.
Proposition 22.1. The mappings above restrict to
: K KCK
I
:K
C
S: K KTheorem 22.2. K together with , I and S is a Hopf algebra.
Theorem 22.3. specK is a group scheme.
23 Matrices return
We can compute the comultiplication on K. Recall that
K = C[,1
det ], = Z1 Z .
so() = Z1 F1 R 1 FZ = Z1 FZR Z1 FZR
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because Z has coordinates in R. Thus
() = C i.e.
(ij) =k
ik C kj
which is matrix multiplication.
AlsoI() = I(Z1 Z) = Z1Z = 1 GLR(n)
so
I
() = 1 GLC(n)Finally
S() = S(Z1 FZ) = ZFZ1 = (Z1 Z)1 = 1
24 The Weierstra -function
Up to now we have dealt only with Picard-Vessiot extensions. The Galois
group is a subgroup of GL(n), in particular, it is affine. There is a moregeneral theory, the theory of strongly normal extensions. Here we examineone simple example. We use classical language of algebraic geometry.
Start with projective 2-space P2 = P2(C). This is the set of equivalence classof triples
(a,b,c) C3 (a,b,c) = 0 ,modulo the equivalence relation
(a,b,c) (a,b,c) C, = 0 .
The equivalence class of (a,b,c) is denoted [a,b,c].
Recall that a polynomial P C[X, Y , Z] is homogeneous if every term hasthe same degree. A subset S P2 is closed (in the Zariski topology) if it isthe set of all zeros of a finite set of homogeneous polynomials
f1, . . . , f r C[X, Y , Z]
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We define E
P2 to be the elliptic curve, the zero set of the single homoge-
neous polynomialY2Z 4X3 + g2XZ2 + g3Z3
where g2, g3 C and the discriminant g32 27g23 is not 0.
If [a,b,c] E and c = 0 then
[a,b,c] = [a
c,
b
c, 1] = [x,y, 1], y2 = 4x3 g2x g3 .
If c = 0 then it follows that a = 0. We get the single point [0, 1, 0] which wedenote by
.
We can interpret the equation y2 = 4x3 g2x g3 as defining a Riemannsurface. It has genus 1. We can integrate on this surface and the integral isdefined up to homotopy (which we call periods).
Theorem 24.1. (Abel) Given P1, P2 E there is a uniqueP3 E such thatP1
dt
s+
P2
dt
s=
P3
dt
s(mod periods)
Here t is a dummy variable and s2 = 4t3
g2t
g3.
This puts an addition on E and makes it an algebraic group.
It turns out that[x,y, 1] = [x, y, 1]
Suppose that [x1, y1, 1] and [x2, y2, 1] are in E and x1 = x2. Then
[x1, y1, 1] + [x2, y2, 1] =
[
(x1 + x2) +
14
y2 y1x2 x1
2
,
y2 y1x2 x1
x3
y1x2 y2x1
x2 x1, 1]
Weierstra inverted the integral to define (x):
x =
(x)
dt
s.
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so that
2 = 43 g2 g3 .
In general, we simply define to be a solution of this -equation.
Definition 24.2. A -field extension G ofF is said to be Weierstrassian if
1. G = F = C,
2. G = F where 2 = 43 g2 g3.
Compute
2 = 122 g2 to get = 62 12g2 ,therefore
G = F = F(, ) .
Let G(G/F) be the group of all -automorphisms ofG over F. If G(G/F)then
2 = 43 g2 g3
We may think of [,
, 1] as an element of E(G
), the elliptic curve withcoordinates in G. (Recall E had coordinates in C.) The above equationshows that [, , 1] is also an element of E(G). So we can subtract thesepoints.
Assume that = and let[,, 1] = [, , 1] [, , 1] = [,, 1] + [, , 1] .
From the formulas above we have:
= ( + ) + 14
2
We claim that is a constant. First compute
+
= 2( )
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and then
= + 12
+
2( ) = 0Because
2 = 43 g2 g3 , is also a constant.
By assumption, G = F = C so [,, 1] E.Theorem 24.3. There is an mapping
G(G
/F
) Egiven by
[, , 1] [, , 1]It is injective and the image is an algebraic subgroup of E.
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