Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/\mathrm{Rad}(b) \to \mathbb{Q/Z}$, s.t. $\pi^\*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:


Is there some definition/construction of the kernel of quadratic forms on abelian groups?



Is there a good notion of kernels of quadratic forms on abelian groups?
As requested, I post my comment (with mild changes) as answer:


In the theory of quadratic forms over a field of characteristic $2$, the radical of a quadratic form $q$ is sometimes defined as $R(q)=\{x\in \mathrm{rad}(b):q(x)=0\}$, where $b$ is the polar form of $q$. This should work in your situation as well: $R(q)$ is a subgroup of $G$ and $q$ factors via a $\mathbb{Q}/\mathbb{Z}$-valued quadratic map on $G/R(q)$.


The subgroup $R(q)$ should coincide with the kernel suggested in @YCor's comment.


