Application of Hilberts Nullstellensatz (strong form) Let $\langle f_1,\ldots,f_r \rangle $ be an ideal in $\mathbb{C}[x_1,\ldots,x_n]$. Then an element $g \in \mathbb{C}[x_1,\ldots,x_n]$ belongs to $\sqrt{\langle f_1,\ldots,f_r \rangle}$ if and only if $1 \in b= \langle f_1,\ldots,f_r,1-yg \rangle$, where $b$ is an ideal in  $\mathbb{C}[x_1,\ldots,x_n,y]$ (where $y$ is a new variable).
The first implication is easy and follows directly from the classical proof of the Hilbert Nullstellensatz, which is due to S.Rabinowitch (1929). However, I dont see how the implication $$b= \langle 1 \rangle \implies g \in \sqrt{\langle f_1,\ldots,f_r \rangle}$$ follows. Is this also from the proof? Please help.

Not sure if I am allowed to use the Nullstellensatz here (i.e. if it is instructive in your current context) but if I can, then we just get that $V(b) = \emptyset \subseteq \mathbb C^{n+1}$ so in particular for any $p = (x,t)\in \mathbb C^{n+1}$ we get the following: if $p\in V(f_1, \dots, f_r)\subset \mathbb C^{n+1}$, i.e. $f_1(x) = 0, \dots, f_r(x) = 0$ then $1-t g(x) \neq 0$ since otherwise we'd have $p\in V(b) = \emptyset$. So $g(x) = 0$ since we can choose $t$ arbitrarily, in particular set $t=g(x)^{-1}$ if $g(x) \neq 0$. This shows that $p\in V(f_1, \dots, f_r) \implies p\in V(g)$. Hence $V(f_1, \dots, f_r) \subset V(g) \subset \mathbb C^n$ and $I(V(g)) \subset I(V(f_1, \dots, f_r))$. The Nullstellensatz then yields $g\in \sqrt{f_1, \dots, f_r}$. 
