Algorithms and Statistics for Additive Polynomials Mark Giesbrecht - - PowerPoint PPT Presentation
Algorithms and Statistics for Additive Polynomials Mark Giesbrecht with Joachim von zur Gathen and Konstantin Ziegler Symbolic Computation Group Cheriton School of Computer Science University of Waterloo Waterloo, Ontario, Canada November
1/29
Algorithms and Statistics for Additive Polynomials
Mark Giesbrecht with Joachim von zur Gathen and Konstantin Ziegler
Symbolic Computation Group Cheriton School of Computer Science University of Waterloo Waterloo, Ontario, Canada
November 28, 2013
2/29
Polynomial Composition and Decomposition
Functional Composition
Let g, h ∈ F[x], for a field F. Compose g, h as functions f (x) = g(h(x)) = g ◦ h Generally non-distributive operation (not always, as we’ll see!):
g(h1(x) + h2(x)) g(h1(x)) + g(h2(x))
2/29
Polynomial Composition and Decomposition
Functional Composition
Let g, h ∈ F[x], for a field F. Compose g, h as functions f (x) = g(h(x)) = g ◦ h Generally non-distributive operation (not always, as we’ll see!):
g(h1(x) + h2(x)) g(h1(x)) + g(h2(x)) Decomposition
Given f ∈ F[x], can it be decomposed? Do there exist g, h ∈ F[x] such that f = g ◦ h?
f = x 4 − 2x 3 + 8x 2 − 7x + 5 f = g ◦ h g = x 2 + 3x − 5 h = x 2 − x − 2
3/29
Tame and Wild Decomposition
Let F be a field of characteristic p and f ∈ F[x] monic of degree d. Normalize f , g, h to monic and original: h(0) = 0
f is tame if p ∤ d f is wild if p | d
Traditionally this describes the ramification of F(x) over F(f (x)).
3/29
Tame and Wild Decomposition
Let F be a field of characteristic p and f ∈ F[x] monic of degree d. Normalize f , g, h to monic and original: h(0) = 0
f is tame if p ∤ d f is wild if p | d
Traditionally this describes the ramification of F(x) over F(f (x)).
Tame decomposition
Ritt (1922) describes all tame decompositions and “ambiguities”. For a fixed s, there are either 0 or 1 monic h ∈ F[x] of degree
s with h(0) = 0 such that f (x) = g(h(x)).
See von zur Gathen (2013) for complete decompositions.
3/29
Tame and Wild Decomposition
Let F be a field of characteristic p and f ∈ F[x] monic of degree d. Normalize f , g, h to monic and original: h(0) = 0
f is tame if p ∤ d f is wild if p | d
Traditionally this describes the ramification of F(x) over F(f (x)).
Wild decomposition
Life is much more difficult (G, 1988) For a finite field F of characteristic p, there are
f ∈ F[x] of degree d with > dλ log d monic, original, h ∈ F[x]
where λ = (6 log p)−1.
3/29
Tame and Wild Decomposition
Let F be a field of characteristic p and f ∈ F[x] monic of degree d. Normalize f , g, h to monic and original: h(0) = 0
f is tame if p ∤ d f is wild if p | d
Traditionally this describes the ramification of F(x) over F(f (x)).
Wild decomposition
On the bright side, there are at most (d − 1)/(s − 1) indecomposable monic, orginal h ∈ F[x] of degree s such that
f (x) = g(h(x)).
(Von zur Gathen, G, Ziegler, 2010)
4/29
Additive Polynomials
Additive or linearized polynomials are those such that
f (x + y) = f (x) + f (y)
Non-linear additive polynomials only exist in F[x] if F has prime characteristic p, and have the form
f = a0x + a1x p + a2x p2 + · · · + anx pn ∈ F[x].
4/29
Additive Polynomials
Additive or linearized polynomials are those such that
f (x + y) = f (x) + f (y)
Non-linear additive polynomials only exist in F[x] if F has prime characteristic p, and have the form
f = a0x + a1x p + a2x p2 + · · · + anx pn ∈ F[x]. Example
Let F125 = F5[θ]/(θ3 + θ + 1).
f = x 25 + (3θ2 + 4θ + 2)x 5 + (3θ2 + 4θ + 2)x
is an additive polynomial, and
f = (x 5 + (θ2 + θ + 4)x) ◦ (x 5 + 3θx) = (x 5 + (2θ2 + 4θ + 2)x) ◦ (x 5 + (θ2 + 2θ)x)
5/29
Ore’s Legacy
In 1932-4, Oystein Ore wrote four seminal papers for finite fields, differential algebra, and computer algebra
1
reine angew. Math., v. 168, pp. 233-252, 1932.
2
Mathematics", v. 34, no. 22, pp. 480–508, 1933.
3
Soc., v. 35, pp. 559-584, 1933.
4
[1,2] form the basis for modern computational theory of LODEs (Ore_algebra,OreTools) [3,4] have had great influence on theory of finite fields
6/29
Ore Polynomials in Computational Algebra
Additive polynomials are employed in Error correcting codes HFE and other cryptosystems Mathematical constructions in algebraic function fields General fun and parlour tricks. Despite their large (exponential) degrees we will see that we can compute very efficiently with them.
7/29
Ore Polynomials and Additive Polynomials
Let q = pe for prime p and integer e.
Fq the finite field with q elements.
Additive polynomials over Fq:
Fq[x; p] =
0in
aix pi ∈ Fq[x]
composition(◦), with x p ◦ ax = apx p.
7/29
Ore Polynomials and Additive Polynomials
Let q = pe for prime p and integer e.
Fq the finite field with q elements.
Additive polynomials over Fq:
Fq[x; p] =
0in
aix pi ∈ Fq[x]
composition(◦), with x p ◦ ax = apx p. Ore polynomials over Fq:
Fq[x; σp] =
0in
aix i ∈ Fq[x]
xa = σp(a)x σp(a) = ap is the Frobenius automorphism of Fq/Fp
7/29
Ore Polynomials and Additive Polynomials
Let q = pe for prime p and integer e.
Fq the finite field with q elements.
Additive polynomials over Fq:
Fq[x; p] =
0in
aix pi ∈ Fq[x]
composition(◦), with x p ◦ ax = apx p. Ore polynomials over Fq:
Fq[x; σp] =
0in
aix i ∈ Fq[x]
xa = σp(a)x σp(a) = ap is the Frobenius automorphism of Fq/Fp
8/29
The Geometry of Additive Polynomials
Assume f ∈ Fq[x; p] squarefree of degree pn
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The Geometry of Additive Polynomials
Assume f ∈ Fq[x; p] squarefree of degree pn
f squarefree ⇐⇒ f ′ = a0 0
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The Geometry of Additive Polynomials
Assume f ∈ Fq[x; p] squarefree of degree pn
f squarefree ⇐⇒ f ′ = a0 0
Roots Vf ⊆ Fq of f form Fp-vector space of dimension n.
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The Geometry of Additive Polynomials
Assume f ∈ Fq[x; p] squarefree of degree pn
f squarefree ⇐⇒ f ′ = a0 0
Roots Vf ⊆ Fq of f form Fp-vector space of dimension n. If W an Fp-subspace of Vf , and
h ∈ Fq[x] has roots exactly W
then h ∈ Fq[x; p] and ∃g ∈ Fq[x; p] such that f = g ◦ h.
8/29
The Geometry of Additive Polynomials
Assume f ∈ Fq[x; p] squarefree of degree pn
f squarefree ⇐⇒ f ′ = a0 0
Roots Vf ⊆ Fq of f form Fp-vector space of dimension n. If W an Fp-subspace of Vf , and
h ∈ Fq[x] has roots exactly W
then h ∈ Fq[x; p] and ∃g ∈ Fq[x; p] such that f = g ◦ h. Decomposing additive polynomials ≡ finding subspaces of Vf
8/29
The Geometry of Additive Polynomials
Assume f ∈ Fq[x; p] squarefree of degree pn
f squarefree ⇐⇒ f ′ = a0 0
Roots Vf ⊆ Fq of f form Fp-vector space of dimension n. If W an Fp-subspace of Vf , and
h ∈ Fq[x] has roots exactly W
then h ∈ Fq[x; p] and ∃g ∈ Fq[x; p] such that f = g ◦ h. Decomposing additive polynomials ≡ finding subspaces of Vf Let σq(a) = aq, the q-Frobenius automorphism. If W is also σq-invariant, then h ∈ Fq[x; p] Decomposing additive polynomial over Fq[x]
≡ finding σq-invariant subspace of Vf
9/29
The Geometry of Additive Polynomials (2)
Example
Again let F125 = F5[θ]/(θ3 + θ + 1), and
f = x 25 + (3θ2 + 4θ + 2)x 5 + (3θ2 + 4θ + 2)x
Then
µ = RootOf
σq = 3 3 2 3
Probably not the best way to work with additive polynomials...
10/29
Right Composition Factors as Eigenvectors of σq
Given f ∈ Fq[x; p], find #
10/29
Right Composition Factors as Eigenvectors of σq
Given f ∈ Fq[x; p], find #
= number of 1-dimensional σq-invariant subspaces of Vf = number of eigenvectors of σq
Remember, σq : Vf → Vf is a Fp-linear map
σq acts like an n × n matrix over Fp
10/29
Right Composition Factors as Eigenvectors of σq
Given f ∈ Fq[x; p], find #
= number of 1-dimensional σq-invariant subspaces of Vf = number of eigenvectors of σq
Remember, σq : Vf → Vf is a Fp-linear map
σq acts like an n × n matrix over Fp
New questions: How many eigenvectors can an n × n matrix over Fq have? How can we compute this?
11/29
Right Composition Factors as Eigenvectors of σq (2)
How many eigenvectors can a matrix have? Look at the (rational) Jordan form in Fn×n
p
Example: degree p2 (n = 2): the number of ways of decomposing
f = x p2 + a1x p + a0x = (x p + b0x) ◦ (x p + c0x)
Put σq in rational Jordan form; there are only four possibilities:
σq ∼
α 1 β
1 λ
λ µ
λ λ
Here λ, µ, α, β ∈ F∗
p, λ µ and y2 − βy − α ∈ Fp[y] is irreducible.
11/29
Right Composition Factors as Eigenvectors of σq (2)
How many eigenvectors can a matrix have? Look at the (rational) Jordan form in Fn×n
p
Example: degree p2 (n = 2): the number of ways of decomposing
f = x p2 + a1x p + a0x = (x p + b0x) ◦ (x p + c0x)
Put σq in rational Jordan form; there are only four possibilities:
σq ∼
α 1 β
1 λ
λ µ
λ λ
Here λ, µ, α, β ∈ F∗
p, λ µ and y2 − βy − α ∈ Fp[y] is irreducible.
11/29
Right Composition Factors as Eigenvectors of σq (2)
How many eigenvectors can a matrix have? Look at the (rational) Jordan form in Fn×n
p
Example: degree p2 (n = 2): the number of ways of decomposing
f = x p2 + a1x p + a0x = (x p + b0x) ◦ (x p + c0x)
Put σq in rational Jordan form; there are only four possibilities:
σq ∼
α 1 β
1 λ
λ µ
λ λ
1
Here λ, µ, α, β ∈ F∗
p, λ µ and y2 − βy − α ∈ Fp[y] is irreducible.
11/29
Right Composition Factors as Eigenvectors of σq (2)
How many eigenvectors can a matrix have? Look at the (rational) Jordan form in Fn×n
p
Example: degree p2 (n = 2): the number of ways of decomposing
f = x p2 + a1x p + a0x = (x p + b0x) ◦ (x p + c0x)
Put σq in rational Jordan form; there are only four possibilities:
σq ∼
α 1 β
1 λ
λ µ
λ λ
1 2
Here λ, µ, α, β ∈ F∗
p, λ µ and y2 − βy − α ∈ Fp[y] is irreducible.
11/29
Right Composition Factors as Eigenvectors of σq (2)
How many eigenvectors can a matrix have? Look at the (rational) Jordan form in Fn×n
p
Example: degree p2 (n = 2): the number of ways of decomposing
f = x p2 + a1x p + a0x = (x p + b0x) ◦ (x p + c0x)
Put σq in rational Jordan form; there are only four possibilities:
σq ∼
α 1 β
1 λ
λ µ
λ λ
1 2 p + 1
Here λ, µ, α, β ∈ F∗
p, λ µ and y2 − βy − α ∈ Fp[y] is irreducible.
An f ∈ Fq[x; σ] of degree p2 can have only 0, 1, 2, or p + 1 right composition factors of degree p.
12/29
Right Composition Factors as Eigenvectors of σq (3)
Example: degree p3 (n = 3): the number of ways of decomposing
f = x p3 + a2x p2 + a1x p + a0x = (x p2 + b1x p + b0x) ◦ (x p + c0x)
σq ∼
λ λ λ
λ 1 λ λ
λ 1 λ 1 λ
λ 1 λ µ
λ µ
λ
12/29
Right Composition Factors as Eigenvectors of σq (3)
Example: degree p3 (n = 3): the number of ways of decomposing
f = x p3 + a2x p2 + a1x p + a0x = (x p2 + b1x p + b0x) ◦ (x p + c0x)
σq ∼
λ λ λ
λ 1 λ λ
λ 1 λ 1 λ
λ 1 λ µ
λ µ
λ
3 1
12/29
Right Composition Factors as Eigenvectors of σq (3)
Example: degree p3 (n = 3): the number of ways of decomposing
f = x p3 + a2x p2 + a1x p + a0x = (x p2 + b1x p + b0x) ◦ (x p + c0x)
σq ∼
λ λ λ
λ 1 λ λ
λ 1 λ 1 λ
λ 1 λ µ
p + 1 1 2 λ λ µ
λ
3 1
12/29
Right Composition Factors as Eigenvectors of σq (3)
Example: degree p3 (n = 3): the number of ways of decomposing
f = x p3 + a2x p2 + a1x p + a0x = (x p2 + b1x p + b0x) ◦ (x p + c0x)
σq ∼
λ λ λ
λ 1 λ λ
λ 1 λ 1 λ
λ 1 λ µ
p + 1 1 2 λ λ µ
λ
3 1
An f ∈ Fq[x; σ] of degree p3 can have only
0, 1, 2, 3, p + 1, p + 2, or p2 + p + 1
right composition factors of degree p.
13/29
General categorization of number of composition factors
How many composition factors of degree p can an additive polynomial of degree pn have? Sn is the set of possible numbers:
S0 = {0} S1 = {0, 1} S2 = {0, 1, 2, p + 1} S3 = {0, 1, 2, 3, p + 1, p + 2, p2 + p + 1} S4 = {0, 1, 2, 3, 4, 2p + 2, p2 + p + 2, p3 + p2 + p + 1}
. . . . . . In general #Sn =
0kn P(k), where P(k) is the number of
additive partitions of k.
14/29
Efficient Counting of Composition Factors
Roots of f ∈ Fq[x; p] of degree pn may be in an extension field of high degree (O(pO(n2))). Can’t really compute directly with Vf . Want algorithms which take time poly in n log p (not pn)
14/29
Efficient Counting of Composition Factors
Roots of f ∈ Fq[x; p] of degree pn may be in an extension field of high degree (O(pO(n2))). Can’t really compute directly with Vf . Want algorithms which take time poly in n log p (not pn)
Look at the ring structure of Fq[x; p] Fq[x; p] is a (non-commutative) ring under the + and ◦
14/29
Efficient Counting of Composition Factors
Roots of f ∈ Fq[x; p] of degree pn may be in an extension field of high degree (O(pO(n2))). Can’t really compute directly with Vf . Want algorithms which take time poly in n log p (not pn)
Look at the ring structure of Fq[x; p] Fq[x; p] is a (non-commutative) ring under the + and ◦
Left (and right) Euclidean ring: LCLM and GCRD operations. No unique factorization (but Jordan-Hölder and Krull-Schmidt give a lot of structure to factorizations) Fast algorithms for +, ◦, lclm and gcrd (time O(n3 log2 q)).
14/29
Efficient Counting of Composition Factors
Roots of f ∈ Fq[x; p] of degree pn may be in an extension field of high degree (O(pO(n2))). Can’t really compute directly with Vf . Want algorithms which take time poly in n log p (not pn)
Example (F125[x; 5] again – a left Euclidean ring) f = x 25 + (3θ2 + 4θ + 2)x 5 + (3θ2 + 4θ + 2)x g = x 25 + (3θ2 + θ + 3)x 5 + (4θ2 + 2θ + 2)x f + g = 2x 25 + (3θ2 + 2θ + 3)x 5 + (4θ2 + 3θ + 2)x f ◦ g = x 625 + (4θ2 + 2)x 125 + · · · + (2θ2 + 3θ + 1)x
lclm(f , g) = x 125 + (θ2+ 3θ + 1)x 25 + (2θ2 + 3)x 5 + (2θ2 + 2θ + 3)x gcrd(f , g) = x 5 + 3θx
15/29
The Centre of It All
The centre of Fq[x; p] is also very useful:
centre(Fq[x; p]) = Fp[x; q] =
15/29
The Centre of It All
The centre of Fq[x; p] is also very useful:
centre(Fq[x; p]) = Fp[x; q] =
the usual (commutative) polynomials!
αix qi →
αiyi
for a0, . . . , an ∈ Fp
15/29
The Centre of It All
The centre of Fq[x; p] is also very useful:
centre(Fq[x; p]) = Fp[x; q] =
the usual (commutative) polynomials!
αix qi →
αiyi
for a0, . . . , an ∈ Fp A cool trick Given any f ∈ Fq[x; p] we can find a left multiple in the center with
O(n3 log2 q) operations in Fq.
15/29
The Centre of It All
The centre of Fq[x; p] is also very useful:
centre(Fq[x; p]) = Fp[x; q] =
the usual (commutative) polynomials!
αix qi →
αiyi
for a0, . . . , an ∈ Fp A cool trick Given any f ∈ Fq[x; p] we can find a left multiple in the center with
O(n3 log2 q) operations in Fq. f = x 25 + (3θ2 + 4θ + 2)x 5 + (3θ2 + 4θ + 2)x ∈ Fq[x; 5] f ∗ = x 1252 + 4x 125 + 3x ∈ Fp[x; 125] f ∗ is the minimal central left multiple (mclm) of f
16/29
The Centre of It All (2)
Basis of the factoring algorithm in G (1992, 1998): Factor the minimal central left multiple and take GCRDs:
f = x 25 + (3θ2 + 4θ + 2)x 5 + (3θ2 + 4θ + 2)x ∈ Fq[x; 5] f ∗ = x 1252 + 4x 125 + 3x ∈ Fp[x; 125] → y2 + 4y + 3 = (y + 1)(y + 3) f ∗ = (x 125 + x)
= (x 125 + 3x) ◦ (x 125 + x)
gcrd(f , f1) = x 5 + (θ2 + 2θ)x gcrd(f , f2) = x 5 + 3θx
Can’t completely decompose with this technique...
17/29
Decomposition in Fq[x; p]
Theorem (G 1992, 1998)
Given f =
0in aix pi ∈ Fq[x], can find g, h ∈ Fq[x], if they
exist, such that f = g ◦ h. Requires expected time O(n4 log2 q)
17/29
Decomposition in Fq[x; p]
Theorem (G 1992, 1998)
Given f =
0in aix pi ∈ Fq[x], can find g, h ∈ Fq[x], if they
exist, such that f = g ◦ h. Requires expected time O(n4 log2 q)
Hardest when minimal central left multiple is irreducible in Fp[y]. Construct a finite algebra A from f , called the eigenring; show that zero-divisors in A yields composition factors of f . Show how to find zero divisors in a finite algebra quickly. Build very explicit Krull-Schmidt and Jordan-Hölder like decompositions, which show structure of all decompositions
18/29
Enter the Eigenring
Decompose an algebra over Fq associated with f . Definitions Idealizer: If ⊆ R largest subring in which Rf a two-sided ideal
If = {u ∈ R : fu ∈ Rf }
Eigenring: Ef = If /Rf an associative algebra over Fq Theorems Ef has zero divisors uv = 0 iff f is decomposable. Zero divisors “split” f : gcrd(f , v) 1. Ef has orthogonal idempotents v 2 = 1,w 2 = 1 with
v + w = 1 iff f = lclm(f1, f2), with gcrd(f1, f2) = 1
Can find eigenring with O˜(n3) operations in Fq
19/29
Detour: Decomposing Associative Algebras over Fq
Describe associative algebra A by a Fq-basis a1, . . . , aℓ ∈ Fm×m
q
A = a1, . . . , aℓ ⊆ Fm×m
q
How do we Find zero-divisors or certify there are none If A semisimple, decompose A = A1 ⊕ · · · ⊕ Ar, for Ai simple If A is simple, find the explicit isomorphism with Fs×s
qη
Friedl & Rónyai (1985) show how to do all this in polynomial time (up to factoring polynomials in Fp[x]) G & Eberly (2000, 2004): “nearly optimal” O(m3 log m + m2ℓ)
Las Vegas for semisimple algebras over Fq Monte Carlo for general algebras over Fq
20/29
Detour: Decomposing Associative Algebras over Fq
Density Theorems Let A = a1, . . . , aℓ ⊆ Fn×n
q
be an associative algebra over Fq,
a randomly chosen from A, and f = minpoly(a) ∈ Fq[x].
For any A over Fq with zero divisors, Prob
If f = f1f2 then f1(a)f2(a) = 0, so f1(a), f2(a) zero divisors When A is a field, Prob{deg f = n} 1/4.
21/29
Central Multiples and Frobenius Automorphisms
Theorem (von zur Gathen, G, and Ziegler 2010) f ∈ Fq[x; p] squarefree of degree pn with roots Vf σq : Vf → Vf the Frobenius automorphism. f ∗ ∈ Fp[x; q] be the minimal central left multiple of f . f ∗ =
0in αix qi
f + =
0in αiyi is min poly of σq.
21/29
Central Multiples and Frobenius Automorphisms
Theorem (von zur Gathen, G, and Ziegler 2010) f ∈ Fq[x; p] squarefree of degree pn with roots Vf σq : Vf → Vf the Frobenius automorphism. f ∗ ∈ Fp[x; q] be the minimal central left multiple of f . f ∗ =
0in αix qi
f + =
0in αiyi is min poly of σq.
Can find the minimal polynomial of σq quickly Can compute the complete rational Jordan form of σq Given f ∈ Fq[x; p] of degree pn, we can compute the number
22/29
Back to our example in F125[x; 5]
f = x 25 + (3θ2 + 4θ + 2)x 5 + (3θ2 + 4θ + 2)x ∈ Fq[x; 5] f ∗ = x 1252 + 4x 125 + 3x ∈ Fp[x; 125] = (x 125 − 4x) ◦ (x 125 − 2x)
So σq ∼
4 2
σq has two eigenvectors f has two right factors of degree 5 h1 = x 5+θ2x+2θx, h2 = x 5+3θx
23/29
Subadditive/Projective Polynomials
Subadditive polynomials: Cohen (1990), Abhyankar (1997):
Ψ =
aix (pi−1)/(p−1) ∈ Fq[x] for b 0
Numerous applications: strong Davenport pairs, difference sets, cryptographically secure sequences, error-correcting codes... Bluher (2004) showed that x p+1 + ax + b has either 0, 1, 2, or
p + 1 roots in Fq. This looks familiar!
23/29
Subadditive/Projective Polynomials
Subadditive polynomials: Cohen (1990), Abhyankar (1997):
Ψ =
aix (pi−1)/(p−1) ∈ Fq[x] for b 0
Numerous applications: strong Davenport pairs, difference sets, cryptographically secure sequences, error-correcting codes... Bluher (2004) showed that x p+1 + ax + b has either 0, 1, 2, or
p + 1 roots in Fq. This looks familiar! Lemma Ψ has a root c ∈ Fq ⇐⇒ aix pi = g ◦ (x p − cx) for g ∈ Fq[x; p]
23/29
Subadditive/Projective Polynomials
Subadditive polynomials: Cohen (1990), Abhyankar (1997):
Ψ =
aix (pi−1)/(p−1) ∈ Fq[x] for b 0
Numerous applications: strong Davenport pairs, difference sets, cryptographically secure sequences, error-correcting codes... Bluher (2004) showed that x p+1 + ax + b has either 0, 1, 2, or
p + 1 roots in Fq. This looks familiar! Lemma Ψ has a root c ∈ Fq ⇐⇒ aix pi = g ◦ (x p − cx) for g ∈ Fq[x; p] Theorem
We can compute the number of roots in Fq of a subadditive
Ψ ∈ Fq[x] with O(n3 log2 q) operations in Fq
(even though deg Ψ ≈ pn−1).
24/29
Inverse Problem – degree p2 case
For each possible number of right components, how many additive polynomials of degree n have that many right components? Equivalently: how many subadditive polynomials in Fq[x] have each possible number of roots in Fq?
24/29
Inverse Problem – degree p2 case
For each possible number of right components, how many additive polynomials of degree n have that many right components? Equivalently: how many subadditive polynomials in Fq[x] have each possible number of roots in Fq?
Bluher (2004): For f = x p2 + a1x p + a0x ∈ Fq[x; p] (a0 0)
Right components # additive polynomials of degree p2
with that many right components
p2−p 2
· q2−1
p−1
1
(p−1) · q2−q p2−p
2
(p−1)(p−2) 2
· (q−1)2
(p−1)2
p + 1
(p−1) · (q−1)(q−p) p(p−1)2(p+1)
24/29
Inverse Problem – degree p2 case
For each possible number of right components, how many additive polynomials of degree n have that many right components? Equivalently: how many subadditive polynomials in Fq[x] have each possible number of roots in Fq?
Bluher (2004): For f = x p2 + a1x p + a0x ∈ Fq[x; p] (a0 0)
Right components # additive polynomials of degree p2 Get an elementary proof, algorithm for enumeration
with that many right components
p2−p 2
· q2−1
p−1
1
(p−1) · q2−q p2−p
2
(p−1)(p−2) 2
· (q−1)2
(p−1)2
p + 1
(p−1) · (q−1)(q−p) p(p−1)2(p+1)
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Inverse Problem – the degree p3 case
Von zur Gathen & G (2011): degree p3 in Fq[x; p]
Right components Number of f ∈ Fq[x; p] of degree p3
with specified number of right components
p3−p 3
· q3−1
p3−1
1
(p−1) · p2−p 2
· q−1
p−1 · q2−1 p2−1 + (p−1) · q3−q2 p3−p2
2
(p−1)(p−2) · q2−q p2−p · q−1 p−1
3
(p−1)(p−2)(p−3) 6
· (q−1)3
(p−1)3
p + 1
(p−1) · q2−q p2−p · q−p p−1 · 1 p2
p + 2
(p−1)(p−2) · q−1 p−1 · (q−1)(q−p) p(p−1)2(p+1)
p2 + p + 1
(p−1) · (q−1)(q−p)(q−p2) (p3−1)(p3−p)(p3−p2)
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Indecomposable Additive Polynomials
Indecomposable additive polynomials f ∈ Fq[x; p]
σq has a single, irreducible Jordan block f ∗ ∈ Fp[y] is irreducible of degree n
Let Np(n) =
µ(n/k)pk =
Number of indecomposable additive polynomials is then
qn − 1 pn − 1 · Np(n) ≈ qn n
Gives a very compact proof of a theorem of Odoni (1999) A random additive polynomial of degree pn in Fq[x; p] will be indecomposable with probability about 1/n Randomized polynomial-time algorithm for generating.
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Maximizing collisions
For f ∈ Fq[x; r] of degree pn, the number of distinct right components of degree p is at most (pn − 1)/(p − 1). Goal: Generate f ∈ Fq[x; r] with this maximal number Let Vf ⊆ Fq and σq : Vf → Vf Components maximized when σ = c · Id for some c ∈ Fp Happens when minpoly(σ) = y − c
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Maximizing collisions
For f ∈ Fq[x; r] of degree pn, the number of distinct right components of degree p is at most (pn − 1)/(p − 1). Goal: Generate f ∈ Fq[x; r] with this maximal number Let Vf ⊆ Fq and σq : Vf → Vf Components maximized when σ = c · Id for some c ∈ Fp Happens when minpoly(σ) = y − c Algorithm: Find right components of x q − cx of degree pn
all have (pn − 1)/(p − 1) distinct right components of degree p
Cost is O(n4) operations in Fq
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How many maximal collisions?
Count the number of n-dimensional subspaces of Vf ∗ :
S(q, p, n) = (q − 1)(q − p) · · · (q − pn−1) (pn − 1)(pn − p) · · · (pn − pn−1)
assuming pn−1 < q. There are p − 1 non-zero values for c ∈ Fp in f ∗ = x q − cx Total number of "maximal collision polynomials" is thus:
(p − 1) · S(q, p, n)
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Open Questions
Inverse theory for number of right factors of degree p of any polynomial in Fq[x; p] Automatically generate inverse formulas Compute number of right factors of any given degree of a polynomial in Fq[x; σ] Resolve conjecture: how many decompositions possible for a general polynomial?