Q:

Hilbert's Nullstellensatz for Laurent Polynomials

I am picking up abstract algebra by myself and understood Hilbert's (weak) Nullstellensatz theorem for polynomials in $\mathbb{C}[\pmb{x}]=\mathbb{C}[x_1,\dots,x_n]$. I am now trying to understand the theorem for Laurent polynomials in $\mathbb{C}[\pmb{x},\pmb{x^{-1}}]=\mathbb{C}[x_1,\dots,x_n,x_1^{-1},\dots,x_n^{-1}]$. However, I came across two (seemingly) different manifestation of Nullstellensatz for Laurent polynomials from literature search and I want to know whether both stand as true or the two theorems are contradictory. Additionally, I would like to know how to derive these theorems from the usual polynomial Nullstellensatz. The two theorems I found are as follows.
An identity equation 
$$
f_1g_1 + \cdots + f_mg_m=1 \tag{1}
$$
is given, where, for each $k=1, \dots,m$, $f_k$ and $g_k$ belong to a ring $\mathcal{R}$.
Theorem 1. 
Let $\mathbb{F}$ be a subfield of $\mathbb{C}$ of complex numbers and $\mathcal{R}$ be the ring of Laurent polynomials $\mathcal{R}=\mathbb{F}[\pmb{x},\pmb{x^{-1}}]$. Furthermore, let $V$ be the set of common roots of the polynomials $f_1,\dots,f_m$, i.e.
$$
V:=\{\pmb{z}:\pmb{z} \in \mathbb{C}^n,f_1(\pmb{z})=\cdots=f_m(\pmb{z})=0\}.
$$
Then there exists a solution $g_1,\dots,g_m \in \mathcal{R}$ of equation $(1)$ if and only if 
$$
V \subset\{\pmb{z}:\pmb{z}=(z_1,\dots,z_n) \in \mathbb{C}^n, z_1 \cdots z_n = 0 \}. 
$$
In other words, at least one of $z_l$ is zero for $l=1,\dots, n$. 
I extracted this theorem from "Bézout Identities with Inequality Constraints" by WM Lawton and CA Micchelli.
Theorem 2.
Let $\mathit{K}$ be a field and $\bar{\mathit{K}}$ an algebraic closure of $\mathit{K}$, and $\mathcal{R}$ be the ring of Laurent polynomials $\mathcal{R}=\mathbb{\mathit{K}}[\pmb{x},\pmb{x^{-1}}]$. If $f_1,\dots,f_m$ do not have a common zero in $\bar{\mathit{K}}^{\ast}$, then there exists a solution $g_1,\dots,g_m \in \mathcal{R}$ of equation $(1)$. Here, I am assuming that I can set $\mathit{K} = \mathbb{C}=\bar{\mathbb{C}}$ and that $\mathbb{C}^{\ast} = \mathbb{C} \setminus \{0\}$. 
I have seen this theorem from "A Refinement of a Mixed Sparse Effective Nullstellensatz" by J Tuitman and "A Sparse Effective Nullstellensatz" by M Sombra.
Now back to my question. Are the two theorems contradictory? or can both stand true? How do you go about deriving the two from the usual Nullstellensatz for polynomials?
As I mentioned, I had limited formal training in abstract algebra, so any detailed comments with a pointer to a reference would be greatly appreciated.

A:

There is no contradiction. These are simply two different formulations, which are both true.
If we simplify things a little bit and take $\Bbb F = \Bbb C$ in Theorem 1 and also take $K = \Bbb C$ in Theorem 2, then they become the same, which is the following statement:

Let $\mathcal R = \Bbb C[\pmb{x},\pmb{x^{-1}}]$ and let $f_1, \dotsc, f_m$ be elements of $\mathcal R$. If they do not have a common zero in $(\Bbb C^*)^n$, then there exists a solution $g_1, \dotsc, g_m \in \mathcal R$ of equation $(1)$.

To deduce it from the usual Nullstellensatz, we first look at the definition of $\mathcal R$. You should convince yourself that $\mathcal R$ is defined as the quotient ring of the polynomial ring $\mathcal S = \Bbb C[\pmb x,\pmb y]$ by the ideal generated by the elements $x_iy_i - 1$.
We now view the polynomials $f_i(\pmb x)$ as elements of $\mathcal S$. These $m$ polynomials, together with the $n$ polynomials $x_iy_i - 1 \in \mathcal S$, define a system of polynomials in $\mathcal S$, which by our assumption do not have a common zero in $\Bbb C^{2n}$ (otherwise we have $(a_1, \dotsc, a_n, b_1, \dotsc, b_n)\in \Bbb C^{2n}$ such that $f_i(a_i) = 0$ and $a_i b_i = 1$, which implies that $(a_1, \dotsc, a_n)$ is a common zero of the $f_i$'s in $(\Bbb C^*)^n$).
Therefore by the usual Nullstellsatz, we know that there are polynomials $g_1, \dotsc, g_m$ and $h_1, \dotsc, h_n$ in $\mathcal S$ such that $$f_1 g_1 + \dotsc + f_mg_m + h_1(x_1y_1 - 1) + \dotsc + h_n(x_ny_n - 1) = 1$$ holds in $\mathcal S$.
Modulo the ideal $I$ gives us a solution in $\mathcal R$.

