Suppose *k* is a field and *V* a vector space over *k*. If *b* is an alternating nondegenerate bilinear form in *V*, it has a symplectic basis. A symplectic basis is a basis where the basis vectors come in pairs, with each pair making a hyperbolic plane and the hyperbolic planes orthogonal, so that the $2 \times 2$ matrix for the bilinear form on each hyperbolic plane looks like:

$$\begin{pmatrix}0 & 1 \\\\ -1 & 0 \\\\ \end{pmatrix}$$

More generally, if *b* is any alternating bilinear form, it has a basis comprising degenerate vectors and a symplectic basis for a complement to the subspace spanned by the degenerate vectors.

I want to generalize this to local rings.

Let *R* be a local ring, e.g., $R = \mathbb{Z}/p^k\mathbb{Z}$. Suppose *M* is a finitely generated *R*-module and *b* is an alternating (not necessarily nondegenerate) bilinear form on *M*. What is the right analogue for a symplectic basis for *b*?

What I seem to have got is that there is a generating set that includes some degenerate vectors, and other pairs of vectors such that the form looks as follows on the plane spanned by these:

$$\begin{pmatrix}0 & p^r \\\\ -p^r & 0 \\\\ \end{pmatrix}$$

where $0 \le r \le k - 1$. Moreover, it seems that the number of times each *r* occurs should be independent of the choice of basis.

I have the following questions:

- Is the result stated above correct? What is a precise and correct formulation of this result?
- Is there a standard reference or theorem that proves a result similar to what I've outlined above?
- Is the result valid, and how is it best interpreted, when
*M*is not a free*R*-module? In that case, the generating set in terms of which we are writing the matrix is not a freely generating set -- some of the elements may have torsion too.