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

1. 1.

$q(\alpha x)=\alpha^{2}q(x)$ for all $x\in M$ and $\alpha\in K$.

2. 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

 $q_{b}(x+y)-q_{b}(x)-q_{b}(y)=b(x+y,x+y)-b(x,x)-b(y,y)=b(x,y)+b(y,x)=2b(x,y).$

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.