Say A is a torsion-free Abelian group, with basis {v_1,..,v_n} ; AFAIK , a torsion-free modules over a PID are free (after $9.99 S&H )

and we have a symmetric quadratic bilinear form

Q(a,b) : AxA-->Z , with a matrix representation

Q_ij = Q(v_i,v_j); i,j=1,..,n.

Now, we tensor A with IR over Z , i.e., A(X)IR , tensor over Z -integers, where IR

is the reals (with IR as an Abelian group , aka, Z-module, I assume;

not clear in the paper, but I assume it is tensoring of Z-modules, which would seem obvious)

Question:

How does the matrix rep. of Q(a,b) change after

the tensoring above (given that the map on A(x)IR is

now (Z-)linear and not bilinear anymore.)?

Do we just use the basis rep. of A(x)R for the

matrix, using basis {v_i(x)1; i=1,..,n} ?.

Is the matrix rep. still diagonalizable? ( the

original matrix is symmetric, so diagonalizable)

Thanks.