Let S be a regular R lattice in Kn. Although the ring is rather unusual, we can still demonstrate uniformity among the cells, and relate the lattice count to the covolume. You may want to review the related proof for integer lattices in real space.
Let x be a lattice point, that is, a point in Rn. Let's locate the cell that contains x, then pull x back to the base cell. Write x in terms of the basis of S. In other words, x is the sum of ciSi, where ci lies in K. If ci looks like an improper fraction, a quotient of polynomials, where the numerator has degree at least as large as the denominator, write it as a mixed fraction, a polynomial plus a proper fraction whose denominator exceeds the numerator. If ci is a laurent series, separate it into a power series and a linear combination of reciprocal powers of t. Do this for each coefficient, and separate the whole parts from the fractions. The polynomials or power series, analogous to integers, can be subtracted away. This moves x back to the base cell.
If x and y are distinct points in the base cell, subtract them to find a new vector. Verify that the difference between two proper fractions of polynomials gives another proper fraction of polynomials. The same holds for linear combinations of reciprocal powers of t. This is not an element of S, and y cannot be a translate of x.
The lattice points in the base cell of S correspond 1-1 with the points in every other cell, and represent the quotient module Rn/S.
If x is in the base cell, multiply by anything in F and the result is in the base cell. Thus Rn/S is an F vector space. This isn't a surprise really, since R and S are both F vector spaces. In this context, let the lattice count be the dimension of the vector space in the base cell, rather than the cardinality of its elements.
To get us started, let S be a rectangular box, where the ith vector runs along the ith coordinate. For instance, the first vector might equal 3t2+5t+1 in the first coordinate, and zero elsewhere. When translating x back to the base cell, divide the first coordinate by 3t2+5t+1 and take the remainder. Do this for all n coordinates, and x lies in the base cell.
Add up the degrees of the polynomials that define the rectangular box to find the dimension of the vector space inside. At the same time, arrange the vectors in a diagonal matrix and take the determinant. This is the product of the polynomials, and its degree is the sum of the individual degrees. The degree of the determinant equals the lattice count, the dimension of the vector space inside the base cell. We want to show this for any lattice S, not just a rectangular box.
Apply the second proof, which uses elementary row operations to move from any given matrix to a diagonal matrix. At each step, the determinant and the lattice count are left unchanged, or multiplied by the same factor. For example, multiply one of the vectors by t2. If x is in the original cell, represent it as a linear combination of basis vectors of S. All coefficients are between 0 and 1. The new cell has the same restrictions, except one of the coefficients can range from 0 to t2. Now x is in the new cell, but the same is true when the designated coefficient of x is multiplied by t, or t2, or any linear combination thereof. Each x now accounts for F2 elements in the new cell. Multiplying a vector by t2 multiplies the number of points in the cell by F2, and increases the dimension by 2. At the same time, multiplying a row by t2 increases the degree of the determinant by 2. The dimension and the degree of the determinant change in lockstep.
The last part of the proof subtracts one vector from another. This does not change the determinant, and it must not change the lattice count either. The reasoning is essentially the same as that given in the real space proof, so I'm going to leave the details to you. Replace fractions with quotients of polynomials, or the reciprocals in laurent series, and you're pretty much there. If you follow this through, you may be surprised to find that only the shifted ceiling of the parallelatope gains and loses lattice points. At every level below the ceiling, the lattice points lie in both c1 and c2.
If F is a finite field, such as Zp, then dimension and cardinality go hand in hand. The traditional lattice count, the number of lattice points in the base cell, is p raised to the degree of the determinant, which is the index of the determinant in the base ring R.