# 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.

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Do you mean to distinguish $K$ and $k$? If so, what is $K$? – Thomas Andrews Oct 12 '12 at 16:09
Does it help if you replace the phrase "substitute $Y \mapsto f^{-1}$" with "evaluate at $Y = f^{-1}$"? – Hurkyl Oct 12 '12 at 16:16
@ThomasAndrews: I fixed that. – Manos Oct 12 '12 at 16:22
@Hurkyl: It does not help because, if i had started in the beginning with $f^{-1}$, then $1-ff^{-1}=0$ and i have done nothing new. – Manos Oct 12 '12 at 16:23
@Manos: But you didn't start in the beginning with $f^{-1}$: you started with the indeterminate variable $Y$ and derived a polynomial equation. – Hurkyl Oct 12 '12 at 16:29

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 $f^{-1}\in S$.

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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.

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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.

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It was something else, thanks anyway. – Manos Oct 12 '12 at 19:38