Section 7.1 Overview of Proof of Dirichlet’s Theorem
I will add to this overview as I continue to work through the proof.
Subsection 7.1.1 Repeating the proof of Mertens’s theorems
To some extent the proof is a reprise of the proof of Mertens’s theorems:
- We start with estimating
by using a comparison between the sum and integral of a monotone function. - Next, we rewrite this as a sum involving the Von Mangoldt function,
by rewriting the previous sum as a double sum indexed over lattice points under a hyperbola. - This in turn lets us estimate
where we can control the error term thanks to Chebyshev’s estimate for the Chebyshev function. - We check that the terms
with powers have a convergent sum, so they contribute a bounded amount to the total, in other words an amount. Dropping these terms leaves the sum and only makes an difference.
(For Mertens’s theorems our next step was to use Riemann-Stieltjes integration and integration by parts to change the sum to However we will not take this step in our study of Dirichlet’s theorem.)
Our plan is basically to repeat all the same ideas, with the added restriction of summing over a congruence class modulo How will we do this? The basic idea is to use complex roots of unity.
Subsection 7.1.2 Complex roots of unity
A complex th root of unity is a complex number (the Greek letter xi) such that There are such numbers, forming the vertices of a regular -gon inscribed in the unit circle. The th roots of unity are for They form a group under multiplication. This group is cyclic.
A fundamental fact about roots of unity concerns the sum of roots of unity, or of powers of a root of unity.
Proof.
Geometrically, the points are equally spaced around the origin, so their average or center of gravity is at the origin.
Remark 7.1.2.
Proposition 7.1.3.
LetProof.
Otherwise, suppose that Let be the order of the smallest positive integer such that The assumption means
Note that must be a divisor of (This follows from Lagrange’s theorem in group theory. Alternatively, if with then Since then as well. The fact that contradicts that is the smallest positive integer such that unless, of course, is not positive. And this must be the case: so divides )
Then the values are th roots of unity, and they are pairwise distinct (why? exercise). So they are in fact all of the th roots of unity. By the previous proposition,
Finally, the full sum is simply equal to the sum of the first terms, repeated times. This is because so that each term in the sum is simply a repetition of the term steps earlier. Therefore the full sum is times zero, which is simply zero.
Remark 7.1.4.
So, our strategy for Dirichlet’s theorem is to consider sums
where, for each is a root of unity chosen in some way. Specifically, we will choose them so that if then but otherwise, the will cycle among all the roots of unity. Then we will add together all of these sums, with the result that the terms with will add up, but the other terms will cancel out (add up to zero).
Example 7.1.5.
We will have to figure out how to choose the roots of unity in general in order to get the cancellation we seek. And we will have to estimate the order of growth of these sums involving the roots of unity. This is our plan!