Given a commutative ring $K$ and two $K$ -modules $M$ and $N$ then a map $q:M\rightarrow N$ is called quadratic if

1. $q(\alpha x)=\alpha^2 q(x)$ for all $x\in M$ and $\alpha\in K$ .
2. $b(x,y):=q(x+y)-q(x)-q(y)$ , for $x,y\in M$ , is a bilinear map.

The only difference between quadratic maps and quadratic forms is the insistence on the codomain $N$ instead of a $K$ . So in this way every quadratic form is a special case of a quadratic map. Most of the properties for quadratic forms apply to quadratic maps as well. For instance, if $K$ has no 2-torsion ($2x=0$ implies $x=0$ ) then $$ 2c(x,y)=q(x+y)-q(x)-q(y) $$ defines a symmetric $K$ -bilinear map $c:M\times M\to N$ with $c(x,x)=q(x)$ . In particular if $1/2\in K$ then $c(x,y)=\frac{1}{2}b(x,y)$ . This definition is one instance of a polarization (i.e.: substituting a single variable in a formula with $x+y$ and comparing the result with the formula over $x$ and $y$ separately.) Continuing without $2$ -torsion, if $b$ is a symmetric $K$ -bilinear map (perhaps not a form) then defining $q_b(x)=b(x,x)$ determines a quadratic map since $$ q_b(\alpha x)=b(\alpha x,\alpha x)=\alpha^2 b(x,x)=\alpha^2 q(x $$ and $$ q_b(x+y)-q_b(x)-q_b(y) =b(x+y,x+y)-b(x,x)-b(y,y)=b(x,y)+b(y,x)=2 b(x,y) $$ Have have no $2$ -torsion we can recover $b$ form $q_b$ . So in odd and 0 characteristic rings we find symmetric bilinear maps and quadratic maps are in 1-1 correspondence.

An alternative understanding of $b$ is to treat this as the obstruction to $q$ being an additive homomorphism. Thus a submodule $T$ of $M$ for which $b(T,T)=0$ is a submodule of $M$ on which $q|_T$ is an additive homomorphism. Of course because of the first condition, $q$ is semi-linear on $T$ only when $\alpha\mapsto \alpha^2$ is an automorphism of $K$ , in particular, if $K$ has characteristic 2. When the characteristic of $K$ is odd or 0 then $q(T)=0$ if and only if $b(T,T)=0$ simply because $q(x)=b(x,x)$ (or up to a $1/2$ multiple depending on conventions). However, in characteristic 2 it is possible for $b(T,T)=0$ yet $q(T)\neq 0$ . For instance, we can have $q(x)\neq 0$ yet $b(x,x)=q(2x)-q(x)-q(x)=0$ . This is summed up in the following definition:

A subspace $T$ of $M$ is called totally singular if $q(T)=0$ and totally isotropic if $b(T,T)=0$ . In odd or 0 characteristic, totally singular subspaces are precisely totally isotropic subspaces.