Tensor product of quadratic forms

The tensor product of quadratic forms is most easily understood when one views the quadratic forms as quadratic spaces. If R is a commutative ring where 2 is invertible, and if $(V_{1},q_{1})$ and $(V_{2},q_{2})$ are two quadratic spaces over R, then their tensor product $(V_{1}\otimes V_{2},q_{1}\otimes q_{2})$ is the quadratic space whose underlying R-module is the tensor product $V_{1}\otimes V_{2}$ of R-modules and whose quadratic form is the quadratic form associated to the tensor product of the bilinear forms associated to $q_{1}$ and $q_{2}$.

In particular, the form $q_{1}\otimes q_{2}$ satisfies

$(q_{1}\otimes q_{2})(v_{1}\otimes v_{2})=q_{1}(v_{1})q_{2}(v_{2})\quad \forall v_{1}\in V_{1},\ v_{2}\in V_{2}$

(which does uniquely characterize it however). It follow from this that if we have diagonalizations of our quadratic forms (which is always possible if $2\in R$ is invertible) such that

$q_{1}\cong \langle a_{1},...,a_{n}\rangle$

$q_{2}\cong \langle b_{1},...,b_{m}\rangle$

then the tensor product has diagonalization

$q_{1}\otimes q_{2}\cong \langle a_{1}b_{1},a_{1}b_{2},...a_{1}b_{m},a_{2}b_{1},...,a_{2}b_{m},...,a_{n}b_{1},...a_{n}b_{m}\rangle .$