A subtle point in proving Hilbert's Nullstellensatz using Rabinowitsch's trick Consider the proof of the Hilbert's Nullstellensatz using the Rabinowitsch trick, as found e.g. in Lang's Algebra, Theorem 1.5, p. 380, which i reproduce here convenience.
In particular, let $\mathfrak{a}$ be an ideal of $k[x_1,\cdots,x_n]$ and $f$ such that every element in the zero set of $\mathfrak{a}$ is a zero of $f$, where the zero set is a subset of $\bar{k}^n$, where $\bar{k}$ is the algebraic closure of $k$. 
Then the Rabinowitsch trick is to introduce a new variable $Y$ and consider the ideal $\mathfrak{a}'$ generated by $\mathfrak{a}$ and the element $1-Yf$ in $k[x_1,\cdots,x_n,Y]$. If $\xi \in \bar{k}^{n+1}$ is a zero of $\mathfrak{a}'$, then the first $n$ coordinates of $\xi$ must form a zero of $\mathfrak{a}$, and so of $f$, contradiction, since $1-Yf$ evaluated at $\xi$ equals $1$. Hence $\mathfrak{a}'=k[x_1,\cdots,x_n,Y]$ and so $1=g_0(1-Yf)+g_1 h_1 + \cdots g_n h_n$, where $g_i \in k[x_1,\cdots,x_n,Y], \, h_i \in \mathfrak{a}$.
Then we substitute $Y \mapsto f^{-1}$ and multiplying by a suitable power of $f$ we obtain that $f$ is inside the radical of $\mathfrak{a}$.
Here is my concern: What values can we substitute for the variable $Y$? Observe that the value $Y = f^{-1}$, "cancels" our original construction of introducing the element $1-Yf$, since we effectively introduce $0$. So, it seems to me that the crucial equation
$1=g_0(1-Yf)+g_1 h_1 + \cdots g_n h_n$ is not valid for $Y=f^{-1}$. 
Any insights?
PS: By no means i am not implying anything about Rabinowitsch's trick, i just want to see what i am missing and how this "gap" can be filled.
 A: I'm not sure I completely understand your question but here, in theorem 3 of  "The Nullstellensatz:, the author basically looks at the quotient
$$K[x_1,...,x_n,Y]/\langle\,1-fY\,\rangle$$ 
Another way (pretty similar after all) is to "go up" to the fraction field of the ring with the new variable, make the substitution there and then "go down" to the original domain.
A: In general, if $R,S$ are rings, and $i:R\rightarrow S$ is an inclusion, then an element $g\in R[Y]$ can be evaluated in $g(s)$ at $s\in S$. (Actually, $i$ need not be an inclusion, even, just a ring homomorphism.)
In particular, if $R=k[x_1,...,x_n]$ and $S=k(x_1,...,x_n)$ is the field of rational functions, you can do this. In that case, $f\in R$ and $f\neq 0$ means $f^{-1}\in S$.
A: It is valid.
Observe that the $g_i$ live in $k[x_1,...,x_n,Y]$, hence substituting $f^{-1}$ for $Y$ makes the $g_i$ live in $k[x_1,...,x_n,f^{-1}]$. Then the equation $1=g_0(1-Yf)+g_1h_1+...+g_nh_n$ with $g_i \in k[x_1,...,x_n,Y]$ and $h_i \in \mathfrak{a}$ becomes
$$
1=g_1h_1+...+g_nh_n
$$
and now the $g_i$ live in $k[x_1,...,x_n,f^{-1}]$ (so not necessarily in $k[x_1,...,x_n]$)
Does that help? Or is your problem something else? I'm not quite sure.
A: We can substitute any polynomial for $Y$. As others have said, one way to think of this is to work in the field of rational functions, that is, the fraction field of $k[x_1,\dotsc,x_n]$.
We don't have to do that. We can stay entirely in the ring of polynomials $k[x_1,\dotsc,x_n]$ in the following way: instead of introducing fractions and then clearing denominators to get back to polynomials, we can "pre-clear the denominators", which is to say, multiply by a power of $f$ first, before we substitute $f^{-1}$ for $Y$. Then in fact we don't even need to mention $f^{-1}$; instead of substituting $f^{-1}$ for $Y$, we can substitute $1$ for $fY$.
In a little more detail: Suppose we have a relation
$$
  1 = g_0(x_1,\dotsc,x_n,Y) (1-Yf) + \sum_{i=1}^s g_i(x_1,\dotsc,x_n,Y) h_i,
$$
each $h_i \in k[x_1,\dotsc,x_n]$. (By the way, we shouldn't assume that the number of $h$'s is equal to the number $n$ of variables. So I'm using $h_1,\dotsc,h_s$ instead of $h_n$.) Multiply both sides of this equation by $f^m$, where $m$ is a large positive integer. Specifically, $m$ should be at least as large as the highest power of $Y$ that appears in any $g_0,g_1,\dotsc,g_s$. Now wherever there's a $Y$ to some power, say $Y^b$, we can group it with $f^b$ (and presumably have an extra factor $f^{m-b}$, whenever $b<m$). We can write
$$
  f^m = Q(x_1,\dotsc,x_n,Yf)(1-Yf) + \sum_{i=1}^s g'_i(x_1,\dotsc,x_n,Yf) h_i
\tag{1}
$$
for some $Q$ and $g'_1,\dotsc,g'_s$. At this point setting $Y=f^{-1}$ introduces no fractions and simply yields
$$
  f^m = \sum_{i=1}^s g'_i(x_1,\dotsc,x_n,1) h_i.
$$
If one wishes to avoid even this, then we can introduce another variable $Y'$ and get
$$
  f^m = Q(x_1,\dotsc,x_n,Y')(1-Y') + \sum_{i=1}^s g'_i(x_1,\dotsc,x_n,Y') h_i
\tag{2}
$$
and then finally substitute $Y'=1$. Here "introduce another variable" is basically just a simple way of saying to consider a ring homomorphism $k[x_1,\dotsc,x_n,Y'] \to k[x_1,\dotsc,x_n,Y]$ given by $Y' \mapsto Yf$ (and each $x_i \mapsto x_i$) and checking that it is injective. Since the left and right sides of (2) clearly map to the left and right sides of (1), by injectivity we get that the left and right sides of (2) are equal to each other. So (2) is a valid equation in $k[x_1,\dotsc,x_n,Y']$. Then setting $Y'=1$ gets us to the original destination, $f^m \in (h_1,\dotsc,h_s)$.
No fractions, no localization: all "honest" polynomials. But maybe it becomes evidence in favor of accepting fractions and localizations to make many arguments easier!
