Automorphisms of the Affine Line over Nonreduced Rings

Automorphisms of the
Affine Line over
Nonreduced Rings
University of New Mexico Geometry Seminar
Taylor Dupuy
Wednesday, February 13, 13
The Affine Line
Algebra
ring
S
Geometry
scheme
S[T ]
A1S = Spec(S[T ])
S ! S[T ]
A1S ! Spec(S)
affine line over S
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Spec(S)
Automorphisms of
Affine Line
1
AS
1
AS
Spec(S)
Schemes
1
Aut(AS )
Rings
AutS (S[T ])
op
group of polynomials invertible under composition
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Affine Linear Subgroup
1
AS
AL1 (S) ⇢ Aut(A1S )
1
AS
(T ) = a + bT
a2S
b 2 S⇥
Spec(S)
Group Law
(
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1 (T )
1
= a 1 + b1 T ,
2 )(T )
2 (T )
= a 2 + b2 T
= a 1 + b1 a 2 + b1 b2 T
Automorphisms over
Domains
Theorem.
S a domain
=) Aut(A1S ) = AL1 (S)
proof.
S[T ] = S[ (T )] =) deg( (T ))  1
Automorphisms of the affine line over domains
are really really really boring.
Wednesday, February 13, 13
Non-Boring
Automorphisms
S = Z/p2 (Ring with nilpotents!)
(T ) = T + pT
1
(T ) = T
100
mod p
2
pT 100 mod p2
(T ) 2 Aut(A1Z/p2 )
Iterates have bounded degree:
Has finite order:
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deg(
(T ) has order p
n
(T ))  100
MAIN POINTS
• Univariate polynomials under composition
n
Z/p
have finite order (over
)
• Iterates of a univariate polynomial under
composition have bounded degree.
• Univariate polynomials under composition
are really algebraic groups! (over Z/p )
n
• Univariate polynomials automorphism
groups are solvable!
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Examples
polynomial
1+T
1+T
2 4
T +2 T
T + 23 T 4
T + 22 T 10 + 23 T 5
T 2 + 22 T 10
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order
2
16
4
2
8
8
coefficient ring
Z/23
Z/24
4
Z/2
Z/24
Z/24
Z/24
More Examples
2
2
T + pT + p T
4
mod p
p = 5, order = 125
p = 7, order = 343
p = 11, order = 1331
This is the typical case
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(r = 4)
r
pr
1
Even More Examples
Aut(A1Z/54 )
order25
80T 10 + 350T 9 + 620T 8 + 300T 7 + 180T 6 + 145T 5 + 560T 4 + 525T 3
+265T 2 + 571T + 191,
order125
555T 10 + 400T 9 + 605T 8 + 305T 7 + 435T 6 + 470T 5 + 250T 4 + 490T 3
+515T 2 + 346T + 356,
order500
230T 10 + 405T 9 + 335T 8 + 410T 7 + 205T 6 + 325T 5 + 620T 4 + 195T 3
+10T 2 + 62T + 160,
order625
370T 10 + 70T 9 + 75T 8 + 65T 7 + 385T 6 + 450T 5 + 200T 4 + 560T 3
+395T 2 + 606T + 487,
order125
390T 15 + 330T 14 + 300T 13 + 290T 12 + 220T 11 + 230T 10 + 580T 9 + 220T 8
+575T 7 + 430T 6 + 600T 5 + 365T 4 + 230T 3 + 395 ⇤ T 2 + T + 285
Wednesday, February 13, 13
Our Setup
We Study: Aut(A1S ) = AutS (S[T ])op
Where: S = R/q
(non reduced!)
qR = hqi prime
n
Rings we are thinking about:
R=Z
q=p
R = F [t]
q=t
coord ring of affine
R=
scheme
q=p
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(wittfinitesimal)
(infinitesimals)
corresp.
to
(
)
O(U ⇥
1
AZ )
Subgroups: Abelian
ones!
gr,s ⇢ Aut(A1R/qr ) , s
important feature
q s · q s ⌘ 0 mod q r
r/2
reduction map
gr,s :=
1
ker(Aut(AR/qr )
(T ) = T + q s f (T ) 2 gs,r,
!
1
Aut(AR/qs )
f (T ) 2 R/q
r s
[T ]
Should be viewed as q-adically close to
the identity! Like a Lie algebra!
Group Law:
(T + ps f (T )) (T + ps g(T )) = T + ps (f (T ) + g(T ))
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Subgroups: Bounded
Degree
Defn/Proposition
8m
1
e
Ad (n, R, q) ⇢ Aut(AR/qn )
2, deg(
mod q m )  d2m
2
Corollaries
1
2
Aut(A
1) Every iterate of
R/q n ) has bounded
degree.
1
2
Aut(A
2) Every
Z/pn ) has finite order.
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Corollaries
1
1) Every iterate of 2 Aut(AR/qn ) has bounded
degree.
1
2) Every 2 Aut(AZ/pn ) has finite order.
If
The Point:
2 Aut(A1R/qn ) and deg( ) = d then
ed (n, R, q)
2A
Remark:
The explicit bound on the degree using this method is
super shitty.
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8m
ed (R, q) ⇢ Aut(A1 n )
A
R/q
n
2, deg(
mod q m )  d2m
2
example: n = 2
d = whatever
2
e
Ad (2, R, q) polynomials mod q of degree less than d
(T ) = a0 + a1 T + qf (T )
e(T ) = ã0 + ã1 T + q f˜(T )
ordT (f ), ordT (f˜)
2
composing these polynomials gives
( e(T ))
⌘
=
a0 + a1 [ã0 + ã1 T + q f˜(T )] + qf (ã + ãT )
a0 + a1 ã0 + (a1 ã1 )T
+q(a1 f˜(T ) + f (ã0 + ã1 T ))
which shows the set is closed under composition.
Wednesday, February 13, 13
next case: Aed (3, R, q)
ordT (f )
2
2
ordT (g)
(T ) = a0 + a1 T + qf (T ) + q g(T ) mod q
deg(
deg(
3
3
2
mod q )  d
3
mod q )  2d
Want to show when we compose two of these guys
we get one back. Look at:
deg(
deg(
2
mod q )
3
mod q )
composing gives
( e(T ))
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=
deg(f
deg(g
a0 + a1 e(T )
mod q),
mod q), deg(f
2
mod q ),
+q[f (ã0 + ã1 T ) + qf (ã0 + ã1 T )f˜(T )]
+q 2 g(ã0 + ã1 T )
0
information
d
deg(
deg(
2
mod q )
3
mod q )
deg(f
deg(g
computation
mod q),
mod q), deg(f
2
mod q ),
( e(T ))
=
2d
a0 + a1 e(T )
+q[f (ã0 + ã1 T ) + qf 0 (ã0 + ã1 T )f˜(T )]
+q 2 g(ã0 + ã1 T )
• (deg f (ã0 + ã1 T ) mod q 2 )  2d,
• (deg f 0 (ã0 + ã1 T )f˜(T ) mod q)  (d
1) + d
• (deg g(ã0 + ã1 T ) mod q)  2d,
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deg( ( e(T )))  2d
Algebraic Groups
• Algebraic varieties where group laws are
given by polynomial expressions.
• Example: matrix groups like the general
linear group
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As Algebraic Groups!
Theorem.
There exist G/Fp finite dimensional such that
G(Fp ) ⇠
= An (Z, p).
There exist G/Fp infinite dimensional such that
1
G(Fp ) ⇠
Aut(A
=
Z/pn ).
Wednesday, February 13, 13
Algebraicity Idea:
Apply (Greenberg Transform = p-Jet Functors)!
n
Gr (X) = Jpn (X) mod p
Key Property:
Grn (X)(Fp ) = X(Z/pn+1 )
defined over Fp
higher dimension
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mixed characteristic
Solvability
Theorem.
The groups Aut(A1Rn ) and An (R, q)
are solvable.
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Solvable Groups
The collection of solvable groups is built inductively:
base defn: If a group is abelian then it is solvable.
inductive part: a group is solvable when one of the
following is true
1) It is the the extension of an abelian group by a
solvable group.
2) It is the extension of a solvable group by an
abelian group.
extension of H by V
1!V !E!H!1
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Example
Claim: AL1 (S) is solvable.
proof.
⇥
AL1 (S) ⇠
S
o
S
=
1 ! S ! S o S⇥ ! S⇥ ! 1
Wednesday, February 13, 13
Subgroups: Newton
Polygonish condition
The collection of invertible polynomials of the form
2
2
4
(T ) ⌘ a0 + a1 T + qa2 T + q a3 T + · · · + q
d 1
form a subgroup
d
ad T 2 R/q [T ]
And (R, q) ⇢ Aut(A1R/qd )
Coefficients are increasingly divisible by q
proof is similar to the computations we did before.
Wednesday, February 13, 13
d
Idea Behind Solvability
Use the groups
gr,s = ker(Aut(A1R/qr ) ! Aut(A1R/qs ))
s
r/2
or their variants to build up the groups we want.
Example. A2 (R, q) is solvable.
proof is by induction.
Wednesday, February 13, 13
Expected Degree
Bounds
The Theorem we stated earlier is far from optimal for
automorphisms of quotients of the integers.
Here is the expected better bound (for a typical
element):
2 Aut(A1Z/pr )
deg(
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n
mod pr )  (deg( )
1)(r
1) + 1