= Solution
The <Laurent polynomial ring> is a <localization of a ring>:
$$
R=k[X_1,X_2]_{X_1X_2}.
$$
The <Hilbert basis theorem> makes $k[X_1,X_2]$ a <Noetherian ring>, and <Localization of a Noetherian ring> preserves this property. Thus $\boxed{R\text{ is Noetherian}}$.
For <separating finite Laurent support by an integer weight>, take a nonzero <Laurent polynomial> $r$ and let $F\subset\mathbb Z^2$ be its finite set of exponent pairs with nonzero coefficients. Choose an integer $N$ such that
$$
N>\max_{m,m'\in F}|m_1-m'_1|,
$$
where the maximum is zero if $F$ has only one member. Set $(a_1,a_2)=(1,N)$. If $m_1+Nm_2=m'_1+Nm'_2$, then
$$
N|m_2-m'_2|=|m_1-m'_1|<N,
$$
so $m_2=m'_2$ and then $m_1=m'_1$. Consequently all powers of $T$ in $\theta_{(1,N)}(r)$ are distinct. The <linear independence> of the Laurent <monomials> prevents cancellation, giving
$$
\boxed{\theta_{(1,N)}(r)\ne0.}
$$
The choice depends on the finite support of $r$; a single fixed substitution need not detect every nonzero element.
For <Noetherianity of subalgebras of a one-variable Laurent polynomial ring>, let $A$ be a unital $k$-<subalgebra> of $k[T,T^{-1}]$. If $A=k$, it is already a <Noetherian ring>. Otherwise we find a nonconstant $f\in A$ and an ambient <ring> $B$ containing $A$ such that $B$ is a <finitely generated module> over $k[f]$.
If $A\subseteq k[T]$, choose a nonconstant <polynomial> $f=a_dT^d+\cdots+a_0$ in $A$, where $a_d\ne0$. Then $T$ is an <integral element> over $k[f]$, since it satisfies the <monic polynomial> equation
$$
T^d+\frac{a_{d-1}}{a_d}T^{d-1}+\cdots+\frac{a_1}{a_d}T+\frac{a_0-f}{a_d}=0.
$$
Thus $B=k[T]$ is a <finitely generated module> over $k[f]$, generated by $1,T,\ldots,T^{d-1}$. The case $A\subseteq k[T^{-1}]$ is identical after replacing $T$ by $T^{-1}$.
In the remaining case $A$ is contained in neither of these two <polynomial rings>. There is an element with a positive exponent and an element with a negative exponent. If either already has both, use it as $f$. Otherwise their sum has both, since the positive powers of the first and the negative powers of the second cannot cancel. We obtain
$$
f=\sum_{j=-p}^{q}a_jT^j,\qquad p,q>0,\quad a_{-p}a_q\ne0.
$$
Multiplying this identity by $T^p$ gives a <polynomial equation> for $T$ with leading coefficient $a_q$:
$$
\sum_{j=-p}^{q}a_jT^{j+p}-fT^p=0.
$$
Dividing by $a_q$ makes it monic. Similarly, multiplication by $T^{-q}$ gives a monic equation for $T^{-1}$ after division by $a_{-p}$. Both generators are therefore <integral elements> over $k[f]$, so $B=k[T,T^{-1}]$ is a <finitely generated module> over $k[f]$.
In all three cases $k[f]$ is a <Noetherian ring>, as a quotient of a one-variable <polynomial ring>. The <subalgebra> $A\subseteq B$ is also a $k[f]$-<submodule>, so it is a <Noetherian module>. Every <ideal> of $A$ is a $k[f]$-<submodule> of $A$ and is therefore finitely generated over $k[f]$, hence over $A$. This proves
$$
\boxed{\text{Every }k\text{-subalgebra of }k[T,T^{-1}]\text{ is Noetherian}.}
$$
For the final request, take the following <subalgebra> of the two-variable <Laurent polynomial ring>:
$$
\boxed{A=k+X_1k[X_1,X_2]=k[X_1,X_1X_2,X_1X_2^2,\ldots].}
$$
Let $J=X_1k[X_1,X_2]$, an <ideal> of $A$. We have
$$
A/J\cong k,\qquad J^2=X_1^2k[X_1,X_2],\qquad J/J^2\cong X_1k[X_2]
$$
as $k$-<vector spaces>. In particular, the classes of $X_1X_2^j$, for $j\ge0$, are <linearly independent>. If $J$ had finitely many generators as an $A$-<module>, their classes would span $J/J^2$ over $A/J=k$, contradicting its infinite <dimension>. Hence $J$ is not finitely generated and $A$ is not a <Noetherian ring>. This illustrates why the preceding result is special to one Laurent variable.
Back to article page
