I'm stuck on the following exercise from Basic Quadratic Forms by Larry Gerstein.
In a radical splitting $V = \mbox{rad} V \perp V_1$, show that $V_1$ is regular.
I want to let $v \in \mbox{rad} V_1$ and deduce that $v = 0$. Here is how I've started. I'm looking for a hint or suggestion as how to proceed or if I'm on the right track.
Proof: Let $V = \mbox{rad} V \perp V_1$ and $w \in \mbox{rad} V_1$. There exists unique vectors $v \in V$ and $u \in \mbox{rad} V$ so that $v = u + w$. Let $v_1 \in V_1$ and consider $$0 = B(w, v_1) = B(v - u, v_1) = B(v, v_1) - B(u, v_1).$$ So $B(v, v_1) = B(u, v_1) = 0$ since $u \in \mbox{rad} V$.
From here I would like to conclude $v = u$ which would imply $w = 0$, but this doesn't seem correct to me. Any hints or comments will be appreciated (please don't just give me the answer).
Suppose $x\in V_1$ and $B(x,V_1)=\{0\}$.
Because $V_1$ is orthogonal to $rad(V)$, you also have $B(x,rad(V))=\{0\}$.
Thus $B(x,V)=\{0\}$.
Do you see the solution now?