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.
 A: Suppose $1\in b$.  This means we can write $1=\sum c_i f_i+ d(1-yg)$, for some $c_1,\dots,c_r,d\in \mathbb{C}[x_1,\dots,x_n,y]$.  Write $c_i=\sum c_{ij} y^j$ and $d=\sum d_j y^j$, for $c_{ij},d_j\in \mathbb{C}[x_1,\dots,x_n]$.  Consider the right-hand side of this equation as a polynomial in $y$ with coefficients in $\mathbb{C}[x_1,\dots,x_n]$, and consider the coefficient of $y^j$ in this polynomial.  Since $y$ does not appear in $f_i$ or $g$, we get that $$1=\sum c_{i0}f_i+d_0$$ and $$0=\sum c_{ij} f_i+d_j-d_{j-1}g$$ for every $j>0$.  The first equation tells us $d_0-1\in \langle f_1,\dots,f_r\rangle$, and the second tells us $d_j-d_{j-1}g\in \langle f_1,\dots f_r\rangle$ for all $j>0$.  It follows that for any $j$, $$d_j-g^j=(d_j-d_{j-1}g)+g(d_{j-1}-d_{j-2}g)+\dots+g^{j-2}(d_2-d_1g)+g^{j-1}(d_1-d_0g)+g^j(d_0-1)$$ must also be in $\langle f_1,\dots,f_r\rangle$.  But $d$ has finite degree, so $d_j=0$ for $j$ sufficiently large.  We thus obtain that $g^j\in\langle f_1,\dots,f_r\rangle$ for $j$ sufficiently large.
You can get a more intuitive argument by identifying the quotient $\mathbb{C}[x_1,\dots,x_n,y]/(1-yg)$ with the ring $A=\mathbb{C}[x_1,\dots,x_n,g^{-1}]\subset\mathbb{C}(x_1,\dots x_n)$ consisting of those rational functions in $x_1,\dots,x_n$ whose denominator is a power of $g$ (actually, you do not even have to prove that these rings are isomorphic; you can just use the fact that you have a homomorphism $\mathbb{C}[x_1,\dots,x_n,y]/(1-yg)\to A$ fixing $\mathbb{C}[x_1,\dots,x_n]$ and sending $y$ to $1/g$).  We then wish to show that if $1$ is in the ideal generated by $f_1,\dots,f_r$ in $A$, then some power of $g$ is in the ideal generated by $f_1,\dots,f_r$ in $\mathbb{C}[x_1,\dots,x_n]$.  But this is easy: if you have $1=\sum c_i f_i$ where $c_i\in A$, just multiply both sides by a large enough power of $g$ to clear the denominators, and you get an equation $g^j=\sum c_i' f_i$, where the coefficients $c_i'$ are now in $\mathbb{C}[x_1,\dots,x_n]$.
A: 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 $g\in \sqrt{f_1, \dots, f_r}$. 
