Tensor product of quadratic forms

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In mathematics, 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 (that is, R has characteristic ${\displaystyle {\text{char}}(R)\neq 2}$), and if ${\displaystyle (V_{1},q_{1})}$ and ${\displaystyle (V_{2},q_{2})}$ are two quadratic spaces over R, then their tensor product ${\displaystyle (V_{1}\otimes V_{2},q_{1}\otimes q_{2})}$ is the quadratic space whose underlying R-module is the tensor product ${\displaystyle V_{1}\otimes V_{2}}$ of R-modules and whose quadratic form is the quadratic form associated to the tensor product of the bilinear forms associated to ${\displaystyle q_{1}}$ and ${\displaystyle q_{2}}$.

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

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

(which does uniquely characterize it however). It follows from this that if the quadratic forms are diagonalizable (which is always possible if 2 is invertible in R), i.e.,

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

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

then the tensor product has diagonalization

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

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