Chapter 7 Flashcards
strategy to solve x³+y³=z³
- factorise x³+y³ in a ring ℤ[ω] = {a+bω|a,b ϵℤ}. ω = e^2πi/3
- Find a machine and a formula via the method of decent
Lemma. if a solution to x³+y³=z³ exists then…
Suppose that there exists a non-trivial integer solution to x³+y³=z³. Then there exists a non-trivial solution in which
(i) x and y are coprime in ℤ
(ii) 3|z but 3 is not a factor of x and y
Idea to prove that if there exists a non-trivial integer solution to x³+y³=z³ then there exists a non-trivial solution in which x and y are coprime
- Let d be a HCF of x,y
- Then we get that d|z (d³|x³,y³,z³)
- divide the eqn by d
- x/d and y/d are coprime
Idea to prove that if there exists a non-trivial integer solution to x³+y³=z³ then there exists a non-trivial solution in which 3|z and 3 is not a factor of x and y
- calculate cubes modulo 9 for a CSR modulo 9
- no choice of signs gives us ±1 + ±1 ≡ ±1 mod 9.
- so at least one of x³,y³,z³≡ 0 mod 9. hence is a multiple of 3.
- if 3|x replace (x,y,z) with (-x,y,-z) so 3| the new z. Do the same with y.
α* where α= a+bω
α*=(a-b)-bω
N(α) where α= a+bω
N(α) = a² + b² -ab
Units of ℤ[ω]
±1, ±ω, ±ω²
Theorem. ℤ[ω] is a..
The ring ℤ[ω] is a ER and hence a UFD
Theorem. product associate to a cube of a member of ℤ[ω].
Let α,β,γ∈ℤ[ω], any two of which are coprime. Suppose that
αβγ is associate to a cube of a member of ℤ[ω]
Then α,β,γ are each associate to cubes of members of ℤ[ω]
λ =
λ = 1 - ω
λ² is associate to 3 proof
N(λ)=3.
3 is is irreducible in ℤ so λ is irreducible and hence prime in ℤ[ω] .
λ²=-3ω=3(-ω)
since -ω is a unit λ² is associate to 3
Lemma. CSR mod λ
{0,1,2} is a complete set of residues mod λ in ℤ[ω]
idea of proof for showing {0,1,2} is a CSR modulo λ in ℤ[ω]
- Let α = a + bω then α ≡ a + b mod λ
- since {0,1,2} is a CSR mod 3 in ℤ then there exists r∈{0,1,2} s.t. a+b=r+3A (a,b are integers)
- Now λ|3 so a + b ≡ r mod λ ≡ α
Now show no two distinct members of {0,1,2} are congruent mod λ - taking norms of 0,1,2 we see that no two of them are congruent modulo λ
Lemma . integer congruent to a cube
Let α ∈ ℤ[ ω ]. Then there is an integer n such that
α³ ≡n mod λ³
idea to prove there is an integer n such that
α³ ≡n mod λ³
- α = x + ωy and ω = 1 − λ so α=a+bλ
- calculate α³
- recall that λ²|3 thus α³≡a³modλ³
- put n=a³
Lemma u is equal to the cube of a unit
Let u ∈ ℤ[ω] be a unit and let α ∈ ℤ[ω]. Suppose that
u≡α³modλ³. Then u is equal to the cube of a unit.
idea to prove u is equal to the cube of a unit
- by lemma u≡n mod λ³
- so λ³|u-n. Now λ² is associate to 3 so 3|u-n i.e. u≡nmod3
- take cases for each of the units to check if u-n/3 ∈ ℤ[ω]. This shows the only units which hold are ±1. so just set v=u=u³=v³
formula for fermats last theorem
ordλ(Z)
Theorem C
Suppose we have the following
(i) x³+y³=uz³
(ii) x,y,z ∈ ℤ[ω]{0} and u is a unit in ℤ[ω]
(iii) x,y are coprime in ℤ[ω]
(iv) λ|z
consider the numbers
A={x+y,x+ωy,x+ω²y}
Then,
(a) Each member of A is a multiple of λ
(b) λ is a HCF of any two distinct members of A
(c) Exactly one member of A is a multiple of λ²
Factorisation of x³+y³
(x+y)(x+ωy)(x+ω²y)
idea of proof for (a) Each member of A is a multiple of λ (Theorem C)
- calculate the differences of elements of A. The differences are all multiples of λ so either every element of A is a multiple of λ or no element of A is a multiple of λ.
- show λ|(x+y)(x+ωy)(x+ω²y) since (x+y)(x+ωy)(x+ω²y) = x³+y³=uz³ λ|z
- since λ is irreducible it is prime so λ is a factor of at least one element of A => λ is a factor of all the elements of A
idea of proof for (b) λ is a HCF of any two distinct members of A (Theorem C)
- Let d be a HCF of two elements of A
- by (a) we have λ|d
- also from the differences and since ω, ω² are units then d|λy so e|y and e is a common factor of two members of A
- But then e|x so as x and y are coprime we deduce that e is a unit. Thus d is associate to λ (i.e. λ is a HCF)
idea of proof for (c) Exactly one member of A is a multiple of λ² (Theorem C)
- Let B = {x+y /λ, x+ωy /λ , x+ω²y/λ }
- find the differences of elements in B (y,ωy, ω²y)
- λ|z and x,y are coprime so λ∤ y. Also λ∤ ωy and λ∤ ω²y. Thus no two members of B are congruent.
- we know {0,1,2} is a CSR. Since B has 3 members we deduce that exactly one member of B is congruent to 0,1,2 respectively. In particular, exactly one member of B is a multiple of λ. Hence exactly, one member of A is a multiple of λ²
Theorem B (the machine for FLT)
Consider the equations
(i) x³+y³=uz³
(ii) x,y,z ∈ ℤ[ω]{0} and u is a unit in ℤ[ω]
(iii) x,y are coprime in ℤ[ω]
(iv) λ|z, λ∤y, λ∤x
If (x,y,z,u) is a solution to the above then there exists another solution (X,Y,Z,U) in which ordλ(Z) = ordλ(z)-1
Idea of proof of theorem B
- recall from theorem C that λ is a HCF of any pair of the factors (x+y)(x+ωy)(x+ω²y) = x³+y³=uz³ and exactly one of the factors is a multiple of λ²
- suppose that λ²|x+ω²y
- divide (x+y)(x+ωy)(x+ω²y) = uz³ by λ³
- by theorem each of these new factors is associate to a cube
- Note ω²+ω+1=0 take a comb of the new factors to remove x,ys to get 0 = ωαX³+ ω²βY³ = ωγZ³
- divide through by the unit ωα to get X³ + δY³ = εZ³
- Now argue δ is the cube of a unit. Find a multiplicative inverse (Y and λ³ are coprime). So there exists a unit δ₀ s.t. δ=δ₀³ so X³ + (δ₀Y)³ = εZ³. We claim that (X,δ₀Y,Z,ε) is a solution.
- check all the hypotheses hold for this solution.
- Prove ordλ(Z) = ordλ(z)-1 by multiplying the factors and taking ords.
How to show Y and λ³ are coprime (for proof of theorem B when finding multiplicative inverse)
Since λ²|x+ω²y then from theorem C we know that λ²∤x+ωy so λ∤x+ωy/λ so λ∤Y. since λ is irreducible it follows that λ³ and Y are coprime.
