Prove that the quotient space $V/V^\perp$ is regular

Question

Given a quadratic vector space (V,b), where b is a non-regular quadratic form, how can I prove that the quotient space $V/V^\perp$ is regular?

Answer 1

If $x \in V$, I will denote by $[x] \in V^\perp/V$. The reduced form $b_0$ on $V^\perp/V$ is given by $$ b_0([x],[y]) = b(x,y),$$ which is well-defined (it does not depend of the choice of representatives $x$ and $y$).

What is an element of the kernel $\ker b_0$? It's a class $[z] \in V^\perp/V$ such that $$\forall \xi \in V^\perp/V, b_0([z],\xi) = 0.$$ By definition of $b_0$, this means that $$\forall x \in V, b(z,x) = 0.$$ So $x \in V^\perp$ and $[x] = 0$. This shows $(V^\perp/V,b_0)$ is regular.