= Solution
The key is a <Laurent normalization by a monomial change of variables>. We prove the slightly stronger assertion that every nonzero quotient of a <Laurent polynomial ring> in $r$ variables is finite as a <module> over an embedded <Laurent polynomial ring> in some number $m\leq r$ of variables.
Induct on $r$. If the defining <ideal> is zero, the ring itself works. Otherwise choose a nonzero Laurent polynomial
$$
f=\sum_{\mathbf a\in F}c_{\mathbf a}X^{\mathbf a}
$$
in that <ideal>, with nonzero coefficients. For a sufficiently large positive integer $N$, the weights
$$
N a_1+N^2a_2+\cdots+N^{r-1}a_{r-1}+a_r
$$
are distinct for all $\mathbf a\in F$. Indeed, each difference gives a nonzero polynomial in $N$, which has only finitely many integer roots. Make the invertible <monomial> substitution
$$
X_j=Y_jY_r^{N^j}\ (j<r),\qquad X_r=Y_r.
$$
Its inverse is $Y_j=X_jX_r^{-N^j}$, so it is an <automorphism> of the <Laurent polynomial ring>. When $r=1$ there is no substitution and distinct exponents already give distinct weights.
As a Laurent polynomial in $Y_r$, the transformed relation has unique lowest and highest exponents. Their coefficients are nonzero <scalar> multiples of <monomials> in $Y_1,\ldots,Y_{r-1}$, hence units in
$$
B_0=k[Y_1^{\pm1},\ldots,Y_{r-1}^{\pm1}].
$$
Multiplying by a power of $Y_r$ and the inverse leading coefficient gives
$$
Y_r^d+b_{d-1}Y_r^{d-1}+\cdots+b_0=0,\qquad b_j\in B_0,\quad b_0\in B_0^\times.
$$
Here $d>0$: a single-term Laurent polynomial would be a unit and the quotient would be zero. Reversing the equation and dividing by $b_0Y_r^d$ gives a monic equation for $Y_r^{-1}$. Thus both $Y_r$ and $Y_r^{-1}$ are integral over the image $B$ of $B_0$ in $T$. In fact $b_0$ being a unit also expresses $Y_r^{-1}$ as a polynomial in $Y_r$, so $T$ is generated as a $B$-<module> by $1,Y_r,\ldots,Y_r^{d-1}$.
The ring $B$ is itself a nonzero quotient of an $(r-1)$-variable <Laurent polynomial ring>. By induction, it is finite over an embedded <Laurent polynomial ring> in $m\leq r-1$ variables. Finiteness of <modules> is transitive, so the same is true of $T$; a finite ring extension is an <integral extension>, by the <determinant trick> applied to multiplication by an element.
Applying this to $r=3$ yields \b[an embedded subring]
$$
\boxed{k[Y_1^{\pm1},\ldots,Y_m^{\pm1}]\subseteq T,\qquad T\text{ integral over it}.}
$$
The base case $m=0$ means the subring $k$; finiteness over it would make $T$ a finite-dimensional <vector space>. The hypothesis excludes this, so \b[$1\leq m\leq3$], as required. No infinitude assumption on $k$ is used.
Back to article page
