Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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:

  1. Is the result stated above correct? What is a precise and correct formulation of this result?
  2. Is there a standard reference or theorem that proves a result similar to what I've outlined above?
  3. 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.
share|improve this question
2  
Lang's "Algebra" discusses the "structure theorem" for alternating bilinear forms on finite free modules over a PID. Asking to "describe" the possibilities when the module isn't free sounds akin to (certainly no easier than!) asking to "describe" linear maps between finitely generated modules which aren't free: unclear what would really constitute a good answer. –  BCnrd Aug 26 '10 at 13:04
1  
To complement BCnrd's answer, the most general common setting for considering bilinear forms is f.g. projective modules - see e.g. Milnor and Husemoller's book, Ch.1, Cor 3.5, which proves the existence of a symplectic basis in the inner product case for $R$ local or Dedekind ring. –  Victor Protsak Aug 26 '10 at 16:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.