Rambles about arithmetic functions
Okay so arithmetic functions (let’s call them A) are functions f from the positive integers ℕ^+ to the complex numbers ℂ. Under the obvious coordinate wise addition they form an abelian group, and we can expand this to a ring with the Dirichlet convolution defined as
You can look up the proof if you’d like, but the question I have is what kind of ring this defines, so the rest of this post is some rambles about that.
It’s pretty clear from the second sum up there that f ∗ g = g ∗ f, so A is commutative.
Moving one rung up the these-rings-are-nicer-to-deal-with list, let’s try to prove that A is an integral domain; that is to say, no two nonzero functions have the zero function as their convolution.
Proof. Let f,g ∈ A be arithmetic functions not equal to the zero function 0. We then have to positive integers n,m ∈ ℕ^+ such that n is the smallest number such that f(n) ≠ 0 and m the smallest such that g(m) ≠ 0.
Now say that a,b are such that ab = nm. Then either a = n and b = m or a < n or b < m. This means that either f(a)g(b) = f(n)g(m) or f(a)g(b) = 0. It follows that
So f ∗ g ≠ 0, and we conclude that A is an integral domain, QED.
What fucks me up about this is that there is therefore an entire ass field that you can embed this structure in (this is a consequence of being an integral domain). What does it look like?? I want to know???
Oh yeah, probably should mention that A itself isn’t a field. The identity element for convolution is the function e that maps everything to 0, except for 1, which it maps onto 1. The group of units of A (those functions that have an inverse w.r.t. convolution) consists of the functions u such that u(1) ≠ 0 (multiplicative functions are a subgroup of these). Therefore any function that does map 1 onto 0 is not invertible, so A is not a field.
This just leaves more questions though. Is A a principle ideal domain? I doubt it, and if not, is it a unique factorisation domain? This post is too long already so I’ll explore this later on lol.
