Some addenda to the Casas-Alvero conjecture
Wouter Castryck (KU Leuven)
joint work with Robert Laterveer and Myriam Ounaïes (Univ. Strasbourg)
Arithmétique en Plat Pays (Univ. Lille-1) December 10, 2012
Consider a polynomial
) (
...
)
(x a0xd a1xd1 ad1xad ai C f
. )!
1 (
! )
(
...
) 1 (
) (
1 0
) 1 (
1 2
1 1
0 )
1 (
a d
x a d x
f
a x
a d
x da x
f
d
d d
d
along with its derivatives
Suppose that each derivative has a root in common with , i.e.f (x) .
0 ) ( )
( :
} 1 ,...,
1
{ ( )
i d c C f c f i c
Conjecture (E. Casas-Alvero, 2001): These common roots are all the same, and hence
c d
x a x
f ( ) 0( ) for some .cC
Part I:
Introduction Part II:
p-Adic methods Part III:
Gröbner bases
Normalization
The relative position of the roots of a polynomial and those of its derivatives does not change under transformations of the form
. 0 ,
),
( )
(x af bx c a b
f
Geometrically: rotations, translations, stretchings of the complex plane.
) 1 )(
1 (
)
(x x x3 x2 f
)
)(
1
( x
f f (2)(x) )
)(
4
( x
f f (5)(x)
)
)(
3
( x
f Example:
) ( )
(x af x
f (x) af (e x)
ff ((xx)) afaf ((rereiixx) c)
f i
Corollary: If there exists a counterexample to the Casas-Alvero conjecture, then there exists a counterexample of the form
).
...
)(
1 (
)
(x x2 x xd3 a1xd4 ad3 f
Degree 4.
Example: CA conjecture in low degree
) (
)
)(
1 (
)
(x x2 x xa aC f
6 6
24 )
)(
3
( x x a
f
a x
a x
x
f (2)( ) 12 2 (6 6) 2 Degree 3.
) 1 (
)
(x x2 x f
2 6
)
)(
2
( x x
f
} 3 , 3 / 1 , {1
a
} 2 / 3 , 3 / 2 , 0
{
a
Degree 5.
) ,
( )
)(
)(
1 (
)
(x x2 x xa xb a bC f
) (
6 )
1 (
24 60
)
( 2
) 3
( x x a b x ab a b
f
ab x
b a ab x
b a x
x
f (2)( ) 20 3 12( 1) 2 6( ) 2 )
1 (
24 120
)
)(
4
( x x a b
f
Feasible up to degree ±8. (Diaz-Toca, Gonzalez-Vega, 2006)
Their insolvability can be checked using Gröbner bases (see later).
Plugging in
yields 4³ systems of non-linear equations in the variables
b a , , 1 , 0
. ,b a
Rolle: If all roots of are contained in a line, then the roots of are contained in the intervals separated by the roots of .
The Gauss-Lucas theorem
) (x
f f (1)(x)
)
)(
1
( x
f ) (x f )
(x f
Gauss-Lucas: The roots of are inside the convex hull of the roots of .f (1)(x) f (x) A weak generalization is:
Explicit formula: if are the roots of and is a root of different from all ‘s, then
ad
a1,..., f (x)
a f (1)(x) ai
1 . 1
1
2 1
2
d
i i
d
i
i i
a a
a a a a
A Casas-Alvero counterexample has a root of multiplicity at least 2.
Thus we have the following possible configurations of the roots:
Alternative proof in degree 4
↯
↯
↯ ↯
A similar proof can be given in degree 5.
Difficult problem.
Refining Gauss-Lucas?
Jensen, 1913: Suppose that the roots of are distributed symmetrically about a line. Then the roots of are contained in the union of the symmetry axis and the Jensen disks.
) (x f
)
)(
1
( x
f
Conjecture (Sendov, 1958): Suppose that all roots of can be caught in a closed disk of radius . Then if one centers a closed disk of radius around a root of , it will always contain a root of .
) (x f )
(x
f f (1)(x)
R R
Open for degree > 8.
Extremal case: . xd 1
Known constraints:
Casas-Alvero: current state of the art
Number of distinct roots > 4.
Number of recycled roots > 2. If is a pair of distinct roots of a polynomial
, then there is at least one such that neither nor (Draisma, de Jong, Knopper).
} , {a1 a2 )
(x
f i f (i)(a1) 0 f (i)(a2) 0
k
k d p
p
d , 2 (Graf von Bothmer, Labs, Schicho, van de Woestijne)
Degree > 19. The Casas-Alvero conjecture is true in degrees
k k k
p d
p d
p d
7
4 3
2 p
7 , 5 ,
3 p
with 366 exceptional primes, the largest one being
249847120216983926479165256672374830117371749836786068968700949838499096141806825287856933123954724798488422551659890912229726792102063
if if
12
d (—, Laterveer, Ounaïes)
(same idea) ( “ )
By contradiction.
Normalize: , leading coefficient Write
Number of recycled roots > 2
Draisma, de Jong, Knopper, 2010: If is a pair of distinct roots of a
polynomial , then there is at least one such that neither nor .
} , {a1 a2 )
(x
f i
0 ) ( 2
)
( a
f i
0 ) ( 1
)
( a
f i
1
,
0 2
1 a
a 1.
).
( ...
)
(x xd a1xd1 ad1x ai C f
1 )
1
( (x) d!x (d 1)!a
f d
0 ) 0
)(
1
(d
f or f (d1)(1) 0
R
a12
1 2
) 2
( ( 1)! ( 2)!
! 2 ) !
( d x d a x d a
x
f d
0 ) 0
)(
2
(d
f or f (d2)(1) 0 a2 R and so on... so is a real polynomial. f (x)
Now is a linear function, so with graph eitherx f (d1)(x)
or
So is either strictly positive or strictly negative on the interval
)
)(
1
( x
f d
0,1.Proof:
But then is strictly increasing or decreasing on x f (d2)(x)
0,1.So again strictly positive or strictly negative...
But then is strictly increasing or decreasing on And so on...
One eventually finds that is strictly in- or decreasing on : impossible.
)
)(
3
( x
f
x d
0,1.) (x f
x
0,1Corollary (to the previous result): The Gauss-Lucas hull of a counterexample to the Casas-Alvero conjecture contains at least 2 roots in its interior.
Number of distinct roots > 4
Only roots that need to be recycled: and (some of) the interior roots.
Let be a counterexample to the Casas-Alvero conjecture.
Let be a root of that is located on the boundary of the Gauss- Lucas hull with maximal multiplicity.
) (x f
) (x f a
a
a
Thus: at least 4 roots in total. If exactly 4, then these are contained in a line.
Using Rolle, one can exclude the latter case (omitted here, nothing deep).
So by Draisma et al.: at least 2 interior roots.
Proof:
Part I:
Introduction Part II:
p-Adic methods Part III:
Gröbner bases
Biggest breakthrough thus far: Graf von Bothmer, Labs, Schicho and van de Woestijne, 2007.
Their proof was rewritten in more elementary language by Draisma and de Jong in 2010, using p-adic valuations.
p-adic valuations
largest exponent such thatvp(x)
) 0
p( v
e pe | x. Over the integers.
N Z)
p( Note: v
) ( )
( )
/
(x y v x v y
vp p p
Over the rationals.
Z Q)
p( Note: v
) ( )
( )
(ab v a v b
vp p p and )) ( ), ( min(
)
(a b v a v b
vp p p
(equality if ) vp(a) vp(b)
Over the complex numbers. Possible to extend to (via ).
vp C
Q C)
p( Here: v
Far from unique!
Qp
C
Valuations of binomial coefficients
)!
(
!
! r p r
p r
p
0 if
r vp p
1
r vp p
0, p ,r
1,2,..., 1
. p
if r
Special case of: Legendre: ,
1
) ( )
( )
(
p
n s r
s r
n s r
vp n p p p
where is the sum of the p-adic digits of . sp(x) x
r
vp n the number of carries when adding to in base . r nr p So:
Proof of the Casas-Alvero conjecture in degree p
By contradiction. Let
) (
... 1 2
) 1
( 1 1 2 2 1 C
p p p ap x ai
p x p
p a x
p a x
x f
be a counterexample.
Using a normalization of the form , we may assume:f (x) pep f (pex)
( ) ( ) 0
0min vp z f z
j k p k j
k p
k a x
k p j
k p
k x p
k a
p
)!
(
)!
) (
(
j k p kx k a
p k
p j
k p
k
p
)!
(
!
! )!
(
)!
( p k j
kx j a
p k
p j
k p
j
p
)!
(
!
! )!
(
)!
( Derivative of a term:
j k p kx k a
j p j
p
p
)!
(
!
j p j
p j
p j
p
j a
j p
j x p
j a x p
j a x p
x p f
j p
2 ...
) 1
! ( )!
( 2
2 1
1 )
(
So:
evaluate in a common root :z vp(a1) 0
1 )
1 (
1 ) 1
! (
!
1 f x x a
p
p
:
1
p For j
:
2
p For j
2 1
2 )
2 (
2 2 1
) 2
! (
!
2 f x x a x a
p
p
similarly: vp(a2) 0 and so on: all ‘s non-negativevp(ai)
Now: evaluate in a root with :z vp(z) 0
z p a
z p p a
z p a
z
z p a
z p p a
z p a z
z f
p p
p p
p p
p p
1 2
2 1
1
1 2
2 1
1
... 1 2
1
... 1 2
) 1 ( 0
↯
Degrees p
k, 2p
k, 3p
k, ...
: still true that as soon as .
pk 0
r v p
k p
z p a
z p p a
z p a
z k
k k
k
p k
k p
k p
k p
1 2
2 1
1 ... 1
1 2
pk
r
0
still gives the
desired contradiction.
: still true that as soon as , unless when . pk
2 2 0
r v p
k
p 0 r 2pk r pk
z p a
z p p a
z p p a
z p a
z k
k k k
k k
p k
k p
p k
k p
k p
k p
1 2 2
2 2 1
2 1 2
1 2
... 2 ... 2
2 2 1
2
would still give the desired contradiction if we would know that vp(apk) 0. But this we know!
p b1 0mod
Idea: reduce mod to f (x) p
. 2 mod
)
( 2 1 1 a p
p b p
x b
x k
k k
p k
p k
p
with
Then satisfies CA conditions over x2 b1x Fp Following the same line of thinking:
Graf von Bothmer et al.: Let be a prime number. If the Casas-Alvero
conjecture holds in degree over , then it holds in degree over , for each non-negative integer .
Fp
p
n npk
k
C Counterexample:
0 2
) (
2 3
) (
) 1 (
) (
) 2 (
2 2
) 1 (
2
x x
f
x x x
x f
x x x
f
If , the Casas-Alvero conjecture is true over any field.n 2
If , the Casas-Alvero conjecture is true over any field of characteristic not 2.n 3
The list of bad ‘s can be computed using a Gröbner basis computation and integer factorization. Feasible up to .n 7
p
2 1 1
2 1
1 ... 1
2 1 1
1 a x
p x p
p a p x
xp p p p
x1
Constraints in degree p + 1
Same normalization: let
) (
1
... 1 2
1 1
) 1
( 1 1 2 1 1 2 C
p p p ap x ai
p x p
p a x
p a x
x f
be a counterexample, where we may assume . min
vp(z ) f (z) 0
0As before, we obtain that all ‘s are non-negative.vp(ai) In addition:
( ) 0
1 ... 1
2 1 1
1 2
1 1
2 1
1
a z v z
p z p
p a z
z p a
p p
p p
p
implies .vp(a1) 0
We may assume (via ).
Note that
) (
)!
1 1 (
! 1 )!
1 (
)
( 1 1
)
( p a p x a
p x p
x
f p
so is a root of . a1 f (x) a1 1 f (x) (a1)(p1) f (a1x) ) (
1
... 1 2
1 1
) 1
( 1 2 1 1 2 C
p p p ap x ai
p x p
p a p x
x x
f 0 ) 1
(
f
xp p
p x
x 1
2 1 1
2 1
... 1 2
1 a x
p x p
p a
pxp p p
1
pxp
1
pxp pxp
coefficients with vp 0
) 0 ) ( (
2
2
2
b xp vp b
) 0) (
(
bp vp bp
0 )
...
)(
1 (
)
(x x xp pxp 1 b2xp 2 bp
f
1 x
1
p
p px
x
2 2
b xp
bp
0 ) ( i
p b v
This implies that all other roots have valuation > 0. In particular, 1 is a simple root.vp(ai) 0 Now
1 2
1
2 1
... 1 2
1 1
... 1 2
1 1
1 1 ) 1 (
0
p ap
p a p
p p p a
a p p
f p
so for at least one . vp(ai) 0 2 i p1
, 2 ...
) 1 )! (
1 (
! 2
2 1
) 1 (
i i
i i
i
p a
i x i
i a i x
x x
p f
i
But as before
so for such an , the common root of and should be 1.i f (x) f (p1i)(x) Summarizing:
The root of must be recycled at least once more (not by the first derivative).
)
)(
( x
f p
Pushing the argument further, one can show that this root must be recycled at least twice more.
Moreover: severe constraints on the indices for which has a common root with the last derivative.
This will prove useful in the next part.
p i
i r
...
2 1 f (ij)(x)
Part I:
Introduction Part II:
p-Adic methods Part III:
Gröbner bases
Proving the unsolvability of a linear system
Given a linear system of equations:
. 0 ...
:
, 0 ...
:
, 0 ...
:
3 3 2
2 1
1
2 2
3 23 2
22 1
21 2
1 1
3 13 2
12 1
11 1
m n
mn m
m m
m
n n
n n
b x
a x
a x
a x a V
b x
a x
a x
a x
a V
b x
a x
a x
a x a V
How do we verify that it has no solutions?
Method: Gaussian elimination.
... until one finds the equation 1 0. ) 0
(
C
cV c
Vi i
)
, (
c i j cV
V
Vi i j C throw away equations 0 0
operations are invertible, so solution set does not change eliminate coefficient
of , then of , ... x1 x2
Proving the unsolvability of a non-linear system
Given a system of polynomial equations:
. 0 ) ,..., ,
( :
, 0 ) ,..., ,
( :
0 ) ,..., ,
( :
2 1
2 1 2 2
2 1 1 1
n m
m
n n
x x
x f V
x x
x f V
x x
x f V
How do we verify that it has no solutions?
Method: Buchberger’s algorithm.
... until one finds the equation 1 0. ) 0
(
C
cV c
Vi i
) (
i j V
m V
m
Vi ij i ji j may create new solutions...
Works by Hilbert’s Basis Theorem and Hilbert’s Nullstellensatz.
Worst-case complexity: horrendous.
eliminate coefficients of monomials in lexicographical order (for instance).
1 2
:
3 2
3 :
2 2 4
2 1 2
2 1 5
2 1 2
3 1 1
x x
x V
x x x
x x
x V
2 2 2 1
3
2 3
2x V x V throw away equations 0 0.
) (
i j V
m V
mij i ji j add
reduce these ‘modulo’ the former system
operations are invertible, so solution set does not change
Back to Casas-Alvero
We will stick to , the smallest case not covered by the d 12 p-adic method.
).
( ) )(
)(
1 (
)
(x x2 x x a2 x a3 x a10
f
Hypothetical counterexample:
We set and . a0 0 a1 1
. 0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) (
, 0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) (
7 ) 11 ( 2
) 10 ( 0
) 9 ( 7
) 8 ( 4
) 7 (
3 ) 6 ( 1
) 5 ( 3
) 4 ( 2
) 3 ( 1
) 2 (
a f
a f
a f
a f
a f
a f
a f
a f
a f
a f
One reason for to be a counterexample could be: f (x)
system of 10 equations in 9 variables, that can be verified using Buchberger.
Problems:
Each system would take a lifetime on a computer with an astronomical amount of memory.
There are 1110 such systems.
5 5
We call (1,2,3,1,3,4,7,0,2,7) a scenario for this counterexample.
If there are no counterexamples corresponding to (1,2,3,1,3,4,7,0,2,7), then there are no counterexamples corresponding to (1,2,3,1,3,4,5,0,2,5), and vice versa.
It suffices to consider the scenarios for which .(s1,s2,...,s10) si max
s1,...,si1
11110 systems 678 570 systems
29 392 systems (applying the stuff to ).p1 p 11
Hybrid representation
) )(
)(
)(
1 (
)
(x x2 x x a2 x a3 x a5
f
Suppose we wish to exclude counterexamples
that match with the scenario (1,2,0,3,1,4,5,0,2,2).
) )(
)(
)(
)(
(x a6 x a7 xa8 x a9 x a10
. 0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) (
, 0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) ( ,
0 ) (
2 ) 11 ( 2
) 10 ( 0
) 9 ( 5
) 8 ( 4
) 7 (
1 ) 6 ( 3
) 5 ( 0
) 4 ( 2
) 3 ( 1
) 2 (
a f
a f
a f
a f
a f
a f
a f
a f
a f
a f
System:
Note that the variables are never being plugged in. a6,a7,a8,a9,a10
) (x5 b1x4 b2x3 b3x2 b4x b5
We use an according hybrid representation for .f (x) Advantage: the variables appear
linearly in all equations of the system.
They can be get rid of using Gaussian elimination.
5 4 3 2
1,b ,b ,b ,b b
So: the more unused variables, the easier the system becomes!
By the stuff: the number of unused variables is at least 2.
In fact, the lower the number of unused variables, the more restrictive the condition becomes.
1 p
1 p
# unused time per case memory # scenarios # scenarios (with p+1 stuff)
0 impossible 1 0
1 impossible 55 0
2 3 weeks 90 GB 1155 5
3 20 hours 10 GB 11880 146
4 15 minutes 1.4 GB 63987 1469
5 10 seconds 0.2 GB 179487 6298
6 0.01 seconds 0.1 GB 246730 11586
7 negligible negligible 145750 8172
8 negligible negligible 28501 1668
9 negligible negligible 1023 48
10 negligible negligible 1 0
Some details that were suppressed
Computations are done modulo a small prime p to avoid coefficient inflation (some theoretical justification needed).
Instead of Buchberger’s algorithm, we used Faugères F4 method.
Instead of the lexicographical order, we used grevlex.
and now...?
It seems unlikely that the p-adic method will ever lead to a complete proof.
A larger unexploited area seems to lie at the analytic side of the story (so far, only algebrists have attacked this problem).
Without new insights, pushing the Gröbner basis method to the next open case ( ) is utopic. And in any case, Gröbner bases can only check one at a time, so they will never lead to a complete proof.
20
d d
WARNING: This is an addictive problem.
