Distribution of the Difference of Uniformly Random Elements In the search to decision reduction of 'On Ideal Lattices and Learning with Errors over Rings', the authors implicitly use the fact that the difference of distinct, uniformly random elements of a (finite) field is again uniformly random (Lemma 5.9). This is again used in the reductions of 'Worst-case to Average-case Reductions for Module Lattices', for Module Learning with Errors.

Firstly, why is this true?
Secondly, what happens when you relax the algebraic conditions on the sample space? For example, in an integral domain, one might not have inverses at all; is there a different sufficient and necessary condition for the difference of distinct, uniformly random elements in an integral domain to be again uniformly random? What about simple modules? Euclidean domains?
Firstly, why is this true?

This is easy to see, if we consider the finite field as a finite group with the addition operation (and ignore the multiplication operation)
If we consider the value $X - Y$, where $X, Y$ are uniformly and independently distributed elements of the group, then there are $n^2$ equiprobable $X, Y$ pairs possible (each with a probability $n^{-2}$.
For any group value $Z$, we have $Z = X - Y$ for $n$ possible $X, Y$ pairs (if we select any value for $X$, we see that the unique value of $Y$ that them sum to $Z$ is $X-Z$); each possible pair has probability $n^{-2}$, and so the probability for the sum to be $Z$ is $n^{-1}$.  This holds for all group members, and so the sum is equidistributed.

Secondly, what happens when you relax the algebraic conditions on the sample space?

As long as the relaxation still leaves us with a ring (that is, the addition operation is still a group), the above argument still holds.