Intermediate Field

Twin prime seems to be related to Galois theory.

Galois described field extension and intermediate field.
F=coefficient, K=field extension, M=intermediate field. G=Galois group.

Therefore

F⊂M⊂K

G(K/M)

G(K/F)

∴

G(K/M)⊂G(K/F)

There is cubic equation.
x^3+ax+b=(x-α)(x-β)(x-γ)=0
α+β+γ=0, αβ+βγ+αγ=a, αβγ=-b

G	e	f1	f2	g1	g2	g3
α	α	β	γ	β	α	γ
β	β	γ	α	α	γ	β
γ	γ	α	β	γ	β	α

e⊂H⊂G

H1=(e,g1)
H2=(e,g2)
H3=(e,g3)
H=(e,f1,f2)

H	e	f1	f2
α	α	β	γ
β	β	γ	α
γ	γ	α	β

I define

F⊂M⊂K

and

G(K/M)⊂G(K/F)

Therefore, G=F and H=M.

K⊂M⊂F

This is upside-down.

∴

K(G(M))=M, G(K(H))=H

Homomorphic

I describe symmetry which is based on Galois theory.
I think that it must be related to prime numbers.

Quadratic equation is like a mirror. There is the formula.











f(x)=0 have two solutions, α and β.


∴α+β=-a

As you know, f(α)=f(β)=0.

If f(α)=β and f(β)=α, f(f(α))=α and f(β)=α.

I define α+β=-a.

Therefore

f(α+β)=f(-a)
f(α)+f(β)=β+α=-a
β+f(β)=-a
∴f(β)=-a-β=α+β-β=α.

There is symmetry.

Integration

When you count prime numbers in each 1000000, there are less prime numbers. This is known as power law.

I have found the pattern of prime numbers.

Шn+1=Шn^2+2(д-1)Шn

This is absolute convergence.
Therefore, it would be replaced as f(z).
I can pick up all prime numbers infinitely, so, in this case, f(z) must be the sum of prime numbers in each 100.


Z=Z(T)=x(T)+iy(T)
T is a parameter.

f(Z)=f(Z(T))

C is the curve of this graph, and you can divide it by N.

T0=Ta,T1,T2,.....,TN=Tb
Z0=Za,Z1,Z2,.....,ZN=Zb



ΔZn≡Zn-Zn-1, ΔTn≡Tn-Tn-1



Z=Z(T) and ΔZn≡Zn-Zn-1


∴

Riemann surface

This figure is piling squares which are prime numbers.


The pattern includes complex plane, so I think that I can describe Riemann surface.



θ=2nπ　(n=0,±1,±2........) and arg Z=θ+2nπ (-π＜θ≦π).
Therefore, arg Z change according to n.

n=0 →　-π＜arg Z≦π
n=1 →　π＜arg Z≦3π
n=5 →　9π＜arg Z≦11π
n=-5 → -11π＜arg Z≦-9π

∴(2n-1)π＜arg Z≦(2n+1)π

Complex planes Z depend on n, and these are piled infinitely.

－Π＜θ≦Π

I define how to expand circles in complex plane.




Then, this is apparent.

(n=0,±1,±2.......)

Therefore



－Π＜θ≦Π

Z keep circling because of Z=X+iY

Maclaurin's expansion

It is impossible to find all prime numbers, so Maclaurin's expansion is a good example to show convergence.

If x had a certain value, this mathematical formula would be convergence.
Then, you see complex number Z.



This complex plane don't include point at infinity, so it must be absolute convergence.



Z=X0 means expansion of the circle, and X0 is a radius of each circles.

Absolute Convergence

We tend to think that infinity must be the realm of philosophy. Our body can’t stand eternity. It is just the ideal. Riemann’s zeta function must be almost correct, but nobody can prove this difficulty because of infinity. You can see the goal, but it moves forward with you. You never reach it.
Therefore, we need to know absolute convergence to ignore infinity, but it includes complex plane.



In order to prove absolute convergence, you should accept Z=Z0 and |Z-a|＜|Z0-a|.
M must be very huge values.


(n=0.1,2....)







Then you see that |Z0-a| is the radius of the circle.
∴ |Z-a|＜|Z0-a|

This means expansion because of Σ, but it must be absolute convergence.

Coastline

It is difficult to accept that a coastline would be infinite. We can't see infinity on the map. This must be a mathematical trick.


D＞1
D is fractal dimensions.

As you can see that a coastline is very complicated,　it is difficult to measure the line. It depends on dividers. If you had high quality dividers, the line would be longer because you divide the line smaller and smaller.

ε is a length of a divider which you measure.
n(ε) is the sum of ε.

L(ε) is an approximate length of the coastline.
∴ L(ε)=εn(ε)



F is the length of the coastline.

If F=100(meter),D=1.2, and ε=3(meter),
n(ε)=26.75805.
Therefore, you need about 26.76 small lines to measure 100(meter) of the coastline.

L(ε)=80.27416
This is far from 100.

If ε=2(meter),
n(ε)=43.52753
L(ε)=87.05506.
This is close to 100.

However, If ε=0.1,
n(ε)=1584.893
L(ε)=158.4893.
This is over 100.

If ε=0.0001,
n(ε)=6309573.445
L(ε)=630.9573445.
This is far from 100.

Therefore, you see that you can't measure the coastline because of innovation of dividers.
If D=1, F is equal to L(ε) so there is no approximation.
However, in this case, D is more than 1.
This is chaos.