When I was thinking about this question: International Zhautykov Olympiad 2019 problem 5 I learned that, when $0 < a,b$ are integers that divide $n >0$ and $d$ is their $\gcd$, then if for a divisor $m \mid n$, $K_m$ denotes the kernel of the reduction map $(\mathbb Z/n)^\times \to (\mathbb Z/m)^\times$, we have $K_aK_b = K_d$.
Note that Bézout's theorem (mod $n$) says precisely that the same is true for the reduction maps $(\mathbb Z/n, +)\to (\mathbb Z/m, +)$.
The proof I have for the multiplicative version uses the Chinese remainder theorem: do the case where $n$ is a prime power first, then glue the solutions for a given element of $K_d$ using the CRT. Slightly more generally, it shows that any PID $R$ satisfies the following:
If $0 \neq N \subset A, B$ are ideals of $R$, then $$1+(A+B) = (1+A)(1+B) + N$$
I'm a bit dissatisfied with the proof for PID's, and would like to understand the equality more conceptually. So my question is: Which other commutative rings $R$ satisfy this property? Is there a characterization?
The identity $1+(A+B) = (1+A)(1+B) + N$ is local, so it holds for every ring that is locally a PID; in particular, any Dedekind domain.
The below alternative proof that $K_a K_b = K_d$ generalizes to the ring of integers of a number field.
Let $l = \mathrm{lcm}(a, b)$ and embed $K_l$ diagonally in $K_a \times K_b$. The map $$(K_a \times K_b) / K_l \to K_d : (x, y) \mapsto xy^{-1}$$ is injective. We have $|K_a| = \phi(n) / \phi(a)$, and the identity $$\phi(a) \phi(b) = \phi(d) \phi(l)$$ implies that $|(K_a \times K_b) / K_l| = |K_d|$. Thus the above map is surjective.