Hilbert’s Problem Solving

David Hilbert was a most concrete, intuitive mathematician who invented, and very consciously used, a principle: namely, if you want to solve a problem first strip the problem of everything that is not essential. Simplify it, specialize it as much as you can without sacrificing its core. Thus it becomes simple, as simple as it can be made, without losing any of its punch, and then you solve it. The generalization is a triviality which you don’t have to pay too much attention to.


Lord of the Ring

Lord of the “Ring”:
The term Ring first introduced by David Hilbert (1862-1943) for Z and Polynomial.
The fully abstract axiomatic theory of commutative rings by his student Emmy Noether in her paper “Ideal Theory in Rings” @1921.

eg. 3 Classical Rings:
1. Matrices over Field
2. Integer Z
3. Polynomial over Field.

Ring Confusions
Assume all Rings with 1 for * operation.

Ring has operation + forms an Abelian group, operation * forms a semi group (Close, Associative).

1) Ever ask why must be Abelian + group ?
Apply Distributive Axioms below:
(a+b).(1+1) = a.(1+1) + b.(1+1)
= a + a + b + b …[1]

(a+b).(1+1) = (a+b).1 + (a+b).1
= a + b + a + b …[2]

a + (a + b) + b = a + (b + a) +b
=> a + b = b + a

Therefore, + must be Abelian in order for Ring’s * to comply with distributive axiom wrt +.

2). Subring
Z/6Z ={0,1,2,3,4,5}
3.4=0 => 3, 4 zero divisor

has subrings: {0,2,4},{0,3}

3). Identity 1 and Units of Ring

Z/6Z has identity 1
but 2 subrings do not have 1 as identity.
subrings {0,2,4}:
4.4=4 => identity is 4
4 is also a unit.

Units: Ring R with 1.
∀a ∈ R ∃b ∈ R s.t.
a.b=b.a = 1
=> a is unit
and b its inverse a^-1

Z/6Z: identity for * is 1
5.5 = 1
5 is Unit besides 1 which is also unit. (1.1=1)

Prime Secret: ζ(s)

Riemann intuitively found the Zeta Function ζ(s), but couldn’t prove it. Computer ‘tested’ it correct up to billion numbers.


Or equivalently (see note *)

\frac {1}{\zeta(s)} =(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})(1-\frac{1}{5^{s}})(1-\frac{1}{p^{s}})\dots

ζ(1) = Harmonic series (Pythagorean music notes) -> diverge to infinity
(See note #)

ζ(2) = Π²/6 [Euler]

ζ(3) = not Rational number.

1. The Riemann Hypothesis:
All non-trivial zeros of the zeta function have real part one-half.

ie ζ(s)= 0 where s= ½ + bi

Trivial zeroes are s= {- even Z}:
s(-2) = 0 =s(-4) =s(-6) =s(-8)…

You might ask why Re(s)=1/2 has to do with Prime number ?

There is another Prime Number Theorem (PNT) conjectured by Gauss and proved by Hadamard and Poussin:

π(Ν) ~ N / log N
ε = π(Ν) – N / log N
The error ε hides in the Riemann Zeta Function’s non-trivial zeroes, which all lie on the Critical line = 1/2 :

All non-trivial zeroes of ζ(s) are in Complex number between ]0,1[ along real line x=1/2

2. David Hilbert:

If I were to awaken after 500 yrs, my 1st question would be: Has Riemann been proven?’

It will be proven in future by a young man. ‘uncorrupted’ by today’s math.

Note (*):

\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots = \sum \frac {1}{n^{s}} …[1]

\frac {1}{2^{s}}\zeta(s) =  \frac{1}{2^{s}}(1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots)

\frac {1}{2^{s}}\zeta(s) =  \frac {1}{2^{s}}+ \frac{1}{4^{s}} + \frac{1}{6^{s}} + \frac{1}{8^s} +\dots … [2]


(1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{3^{s}} + \frac{1}{5^{s}} + \dots + \frac{1}{p^{s}} +\dots

\text {Repeat with} (1-\frac{1}{3^s}) \text { both sides:}

(1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{5^{s}} + \frac{1}{7^{s}} + \dots + \frac{1}{p^{s}} +\dots


(1- \frac{1}{p^{s}}) \dots (1- \frac{1}{5^{s}})(1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1


\zeta(s) = \prod \frac {1}  {1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}

Note #:
\zeta(s) = \prod \frac {1}  {1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}

Let s=1
RHS: Harmonic series diverge to infinity
\prod \frac {1}{1- \frac{1}{p}}= \prod \frac{p}{p-1}
Diverge to infinity => there are infinitely many primes p

English: Zero-free region for the Riemann_zeta...

English: Zero-free region for the Riemann_zeta_function (Photo credit: Wikipedia)