Compute the kernel of the group hom $\Omega : \Bbb{Q}^{\times} \to \Bbb{Z}^+$.

The $\Omega$ function is the counting function that returns precisely the number of primes $\Omega(n)$ (including multiplicity) that divide a natural number $n \in \Bbb{N}$. For example $\Omega(6) = 2, \Omega(8) = 3$, etc. It is known and easily seen to be completely multiplicative on $\Bbb{N}$ ie. $\Omega(ab) = \Omega(a) + \Omega(b)$ for all $a, b \in \Bbb{N}$.

Extend the definition to all of $\Bbb{Z}\setminus 0$ by defining $\Omega(-n) := \Omega(n)$ for all $n \gt 0$. Now extend the definition to all of $\Bbb{Q}^{\times} = \Bbb{Q}\setminus 0$ by defining $\Omega(a/b) = \Omega(a) - \Omega(b)$. Then what you have is a group homomorphism from the multiplicative rationals onto (surjective) $\Bbb{Z}^+$:

Let $$ a/b, c/d \in \Bbb{Q} $$

Then $$\Omega(\dfrac{c}{d} \dfrac{a}{b}) = \Omega(\dfrac{ca}{db}) = \\\Omega(ca) - \Omega(db) = \\ \Omega(c) + \Omega(a) - (\Omega(d) + \Omega(b)) =\\ \Omega(c) - \Omega(d) + \Omega(a) - \Omega(b) = \\ \Omega(c/d) + \Omega(a/b)$$

It is well-defined since if $\dfrac{a}{b} = \dfrac{a'}{b'}$, then $ab' = a' b$ so that $$\Omega(a) + \Omega(b') = \\ \Omega(ab') = \Omega(a'b) = \\ \Omega(a') + \Omega(b) \implies \\ \Omega(a) - \Omega(b) = \Omega(a') - \Omega(b') \implies \\ \Omega(\dfrac{a}{b}) = \Omega(\dfrac{a'}{b'})$$.

Since we have a surjective group homomorphism $\Omega: \Bbb{Q}^{\times} \to \Bbb{Z}^+$. Was wondering how we could more explicitly compute the kernel which is :

$$ \ker \Omega = \{ a/b \in \Bbb{Q}^{\times}: \Omega(a) = \Omega(b) \} $$

For example $p/q \in \ker \Omega$ for all $\pm$ primes $p, q\in \Bbb{Z}$. Thus isn't it a weird or "exotic" normal subgroup of $\Bbb{Q}^{\times}$? Does it have a name?

By the first isomorphism theorem for groups, $\Bbb{Q}^{\times}/\ker \Omega \simeq \Bbb{Z}^+$.

Question 2. Can we extend $\Omega$ to $\Bbb{Q}(i)$ easily?

Solution 1:

If I am not mistaken, your map is essentially the degree map from the Picard group of $\mathbb{Q}$, defined in a multiplicative way (this is not a problem). The kernel is the the subgroup of degree zero "divisors". In this is indeed the case, the theory is well developped, and can be extended to all number fields (among which $\mathbb{Q}(i)$) as Arakelov geometry (see for example this, although it may not be the most elementary). It is close (but not the same) as the ideal class group of a number field, where instead of quotienting by the subgroup of degree $0$ divisors, you quotient by the subgroup of principal ideals. Note that the class group of $\mathbb{Q}$ is trivial so it is not very interesting.