Multiplicative sharing of 1 Let R be a ring.

Is there a known/standard way for N parties to perform some exchanges, such that at the end each party holds an element $r_i\in R^\ast$ and $r_1r_2\dots r_N=1$ (and each party $i$ does not know $r_j$ for each $j\neq i$)?

(Or equivalently, share 0 additively?)
If the ring is multiplicatively commutative, the obvious way to do this is:


Have each pair $a, b$ create a secure connection, and jointly select two random invertible elements $r_{ab}$ and $r_{ba}$ with $r_{ab}r_{ba} = 1$ (that is, they're inverses of each other).
Then, $$r_a = \prod_{b \ne a} r_{ab}$$


This is secure (in that any group of parties know nothing about the shares that everyone else has, except the product of their shares), however it does require $O(N^2)$ communication.

If we reduce the requirement to "no single party can deduce anything about any other party, then we can reduce the communication to $O(N)$ (by having $a$ exchange with $a+1$ and $N$ exchange with $1$).