Linnik's Theorem Flashcards

0
Q
State linnik's Theorem.
Define N(α,Τ:χ)
A

There exists effectively computable constants c1 & c2 such that whenever (a,q)=1 there exists a prime pΞa mod q with pα,
|γ|<T}

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1
Q

What is the generalised Riemann hypothesis? If this were true how would this give us an asymptotic formula for the sum over all p<x with pΞa mod q of: Σlogp ?

A

That the completed L-functions Λ(s,χ) is zero only at points where Re(s)=1/2
Plugging this bound for the sum over zeroes in our proof of PNT in AP’s and hooding T=sqr(x) would give the asymptotic formula for the sum as:
x/φ(q) + O(sqr(x)•(logx)^2)

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2
Q

What is Bombieri’s theorem?

What is a theorem of this kind called?

A

There is a constant c>0 such that for all T and 1/2=, since they show there are relatively few zeros in some region to the right of the half (we hope there should be none)

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3
Q

Use Bombieri’s theorem to show that if there exists a primitive real character χ’ mod q’ st L(s,χ’) has a real exceptional β’>1-δ/logT, then we get a great bound for real part of all other zeros with |γ|<T

A

Specifically, if the above conditions hold then:
β=s theorem gives a certain sum less than 1, but this sum counts exactly the number of zeros not satisfying the above conditions, so none do.

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4
Q

How might we go about proving Linnik’s theorem?

A

We can show Bombieri’s theorem implies Linnik’s, so we should just try to prove this. This is still very tricky, but manageable

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5
Q

Let χmod q be induced by χ’mod q’ and Re(s)>1/2. Then show that:
-L’/L(s,χ)=-L’/L(s,χ’) + O(logq)

A

Looking at the difference of these two as sums, we see this is easy

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6
Q

If we have χmodq, q=<T, then what can we say about the logarithm of L(σ+iv,χ) and the logarithmic derivative when σ is larger than 1-c/logT, for a c we choose specifically?

A

Well if σ>σ0=1+c/logT then it is easy to show that the first is less than loglogT+O(1) and the second is O(logT).
So assume that it is no between these values. By rewriting the logarithmic derivative here as the difference between two L functions we can calculate it as a sum of zeros and bound those sum to give the same bound. Then the logarithm is bounded by an integral + logarithm of the L function at σ0+iv and as the integral is O(1) we are done.
In fact, these bounds are valid for ζ and we can remove the restriction 2<|v| if there is no exceptional zero of L.

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7
Q

How can we use derivatives of the L function to detect zeros of L?

A

We can show that if L zeros at some point then there will be large derivatives near here. Formula are messy and I still have to go over the proofs! If there are multiple zeros we have to look at multiple derivatives to assure a large one. This is all set out with lots of precision in notes

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8
Q

What is Turans power sum method?

How do we use it to effectively give a lower bound for the derivatives of F’/F at particular points?

A

If we have n complex numbers, and some K>n, then there is some
K+1<K+n such that the modulus of the sum of each of these n numbers to the power k is larger than the maximum modulus of each one, divided by 32, all to the power k.
Proof is long
Use the expression for this as a sum over poles. Is then quite obvious.

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