How do you construct a symplectic basis on a lattice? Is this possible to do constructively? The only sources that I have for the possibility of this construction is an exercise in Lang's Algebra (on p. 598, I believe) which states that one can be constructed, and then a construction given in Milnor and Hussemoller's book which only applies in the case that the elementary divisors are all 1.
In my case, I have a rank 4 lattice of with a symplectic form of type (1, n), and I have some elements which span an index n sublattice. I would like to somehow relate them to a symplectic basis on the whole lattice, but it seems to me that I would need to have a constructive method for creating such a basis for this to be of any use whatsoever.
 A: Here is an algorithm, it may not be a good one. I will only explain how to find a basis $e_i$, $f_i$ such that $\langle e_i, e_j \rangle = \langle f_i, f_j \rangle =0$ and $\langle e_i, f_j \rangle = c_i \delta_{ij}$ for some constants $c_i$. I will punt on explaining how to make sure that $c_1$ divides $c_2$ which divides $c_3$ and so forth. This presentation is closely based on the algorithm in Wikipedia for computing Smith normal form. 
Find $e$ and $f$ so that $\langle e,f \rangle \neq 0$ (EDIT) and such that the lattice spanned by $e$ and $f$ is the intersection of the whole lattice with the vector space spanned by $e$ and $f$. 
Set $d := \langle e,f \rangle$. Complete $(e,f)$ to a basis $(e,f,g_1, g_2, \ldots, g_{2n})$ of our lattice.
Case 1: We have $\langle e,g_i \rangle = \langle f, g_i \rangle =0$ for all $i$. Take $e_1=e$, $f_1=f$ and apply our algorithm recursively to the sublattice spanned by the $g_i$.
Case 2: For all $i$, the integer $d$ divides  $\langle e,g_i \rangle$ and $\langle f, g_i \rangle =0$. Replace $g_i$ by 
$$g_i - \frac{1}{d}\langle g_i, f \rangle e- \frac{1}{d}\langle e, g_i \rangle f.$$
We have now reduced to Case 1.
Case 3: There is some $g_i$ so that $k:=\langle e, g_i \rangle$ is not divisible by $d$. Then, for some $q$, we have $0 < k-qd < d$. Set $f'=g_i-kf$ and $g'_i=f$. Then $(e,f',g_1,g_2,\ldots, g'_i, \ldots, g_n)$ is a basis, and $\langle e, f' \rangle$ is less than $d$. Return to the beginning with this new basis.
Case 4: There is some $g_i$ so that $k:=\langle f, g_i \rangle$ is not divisible by $d$. Just like Case 3, with the roles of $e$ and $f$ switched.
Since $d$ decreases at every step, eventually we will hit Case 1 and be able to reduce the dimension.
