# Proving the Nullstellensatz: Problem with Zariski's Lemma

Having problem understanding an important step in Zariski's lemma. I'm following Zariski's original paper, and the step that I'm not getting is at the beginning of page 364 (or, well, page 3 of this particular pdf document):

Each $\xi_i$, $i=2, 3, \dots, n$ is a root of a polynomial $f_i(X)$ taking its coefficients in $K[\xi_1]$. Let $b_i$ be the leading coefficient of $f_i (X)$, $b_i \neq 0$. If $\omega$ is any element of $R_n$, there will exist an integer $\rho$ such that the product $\omega \cdot (b_2 \dots b_n)^{\rho}$ can be expressed as a linear combination, with coefficients in $K[\xi_1]$, of the $m_2 m_3 \dots m_n$ power products $\xi_2^{j_2} \xi_3^{j_3} \dots \xi_n^{j_n}$, $0 \leq j_i \leq m_i - 1$, where $m_i$ is the degree of $f_i (X)$.

($R_n$ for the record, is how Zariski denotes $K[\xi_1 , \dots , \xi_n]$ to save space, since he uses it a lot.)

Now, I can understand the first part, why each $\xi_i$, $i=2, 3, \dots, n$ is a root of a polynomial $f_i(X)$ taking its coefficients in $K[\xi_1]$. What I cannot understand is how it from this follows that there exists a $\rho$ such that $\omega \cdot (b_2 \dots b_n)^{\rho}$ is a linear combination of the power products taking all their coefficients in $K[\xi_1]$.

Any and all help would be deeply appreciated.

• What do you denote $R_n$? – Bernard Aug 25 '17 at 19:51
• $R_n$ is short for $K [\xi_1 , \dots , \xi_n] = K(\xi_1) [\xi_2 , \dots , \xi_n]$. – StormyTeacup Aug 25 '17 at 20:10
• Maybe if one rephrased the question as follows it would be easier: – StormyTeacup Aug 26 '17 at 11:26
• Let $A$ be a field, $K$ its field of fractions, and $K[a_1 , \dots , a_n]$ an algebraic extension of $K$. Since each $a_i$ is algebraic over $K$, there exists a polynomial $g_i (X) \in K[X]$ such that $g_i (a_i) = 0$. Letting $v_i$ be the product of all the denominators in $g_i$, we have a polynomial $f_i (X) = v_i g_i (X) \in A[X]$ such that $f_i (a_i) = 0$. – StormyTeacup Aug 26 '17 at 11:32
• Let $b_i$ be the leading coefficient of $f_i$. Then, for every element $\omega \in K[a_1 , \dots , a_n]$, there exists an integer $\rho$ such that $\omega \cdot (b_1 \dots b_n)^{\rho}$ can be expressed as a sum of power products $a_1^{j_1} a_2^{j_2} \dots a_n^{j_n}$, with all coefficients being in $A$. – StormyTeacup Aug 26 '17 at 11:32

Let $\omega$ be any element of $R_n$, which we now regard as $K[\xi_1] [\xi_2 , \dots , \xi_n]$. It is clear that for any given other element element $c \in K[\xi_1]$, for sufficiently large $N$, it holds true that $\omega c^N \in K[\xi_1] [\xi_2 , \dots , \xi_n]$. So let $c = (b_2 \dots b_n)$ where $b_i$ is defined as above. Then clearly, there exist some integer $\rho$ such that $\omega (b_2 \dots b_n)^{\rho} \in K[\xi_1] [(b_2 \dots b_n)\xi_2 , \dots , (b_2 \dots b_n) \xi_n]$. For notational economism, let's write $\beta = (b_2 \dots b_n)$ from hereon.
Now, $\omega \beta^{\rho}$ is of the form \begin{equation*} \omega \beta^{\rho} = \sum_{i_2 , \dots , i_n} k_{i_2 , \dots , i_n} (\beta \xi_2)^{i_2} \dots (\beta \xi_n)^{i_n} , \end{equation*} where all $k_{i_2 , \dots , i_n} \in K[\xi_1]$. We can further always "simplify it down" so that the terms of the power products $\xi_2^{i_2} \dots \xi_n^{i_n}$ don't go on increasingly further and further. Because we know that $f_i (\xi_i) = b_i \xi_i^{\deg (f_i)} + \mathcal{O} (\xi_i^{\deg (f_i)-1}) = 0$, we have $b_i \xi_i^{\deg (f_i)} = \mathcal{O} (\xi_i^{\deg (f_i)-1})$, and so since $b_i$ is a factor of $\beta$, when a term with a factor $\xi_i^{\deg (f_i)}$ appears, we can always couple it with the $b_i$ in the $\beta$ and exchange this for the $\mathcal{O} (\xi_i^{\deg (f_i)-1})$. Consequently, the only power products that we end up actually looking at are $\xi_2^{j_2} \xi_3^{j_3} \dots \xi_n^{j_n}$, where $0 \leq j_i \leq \deg(f_i) - 1$.