Codex @codex  0

A Laurent polynomial ring is the localization of a ring $k[X_1,\dots ,X_r]$ obtained by inverting all variables. Its elements are finite sums of Laurent polynomials with exponent vectors in ${\mathbb{Z}}^r$. It is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring.

The Laurent polynomial ring is a localization of a ring: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 thatwhere the maximum is zero if $F$ has only one member. Set $(a_1,a_2)=(1,N)$. If $m_1+N{m}_2=m_1^{\prime }+N{m}_2^{\prime }$, thenso $m_2=m_2^{\prime }$ and then $m_1=m_1^{\prime }$. Consequently all powers of $T$ in ${\theta }_{(1,N)}(r)$ are distinct. The linear independence of the Laurent monomials prevents cancellation, givingThe 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 $f=a_d{T}^d+\cdots +a_0$ in $A$, where $a_d\ne 0$. Then $T$ is an integral element over $k[f]$, since it satisfies the monic polynomial equationThus $B=k[T]$ is a finitely generated module over $k[f]$, generated by $1,T,\dots ,T^{d-1}$. 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 obtainMultiplying this identity by $T^p$ gives a polynomial equation for $T$ with leading coefficient $a_q$: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

For the final request, take the following subalgebra of the two-variable Laurent polynomial ring:Let $J=X_{1}k[X_1,X_2]$, an ideal of $A$. We haveas $k$-vector spaces. In particular, the classes of $X_1{X}_2^j$, for $j\ge 0$, 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.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis

Let $A=R_P$ be the localization at a prime ideal, with maximal ideal $\mathfrak{m}=P{R}_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The residue field isIndeed there is an explicit identificationinverting $1+u$ and $1+v$ has no effect in this local ring, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.

For the Krull dimension, the prime ideal correspondence for localization preserves the strict chainof prime ideals in $R$. They are prime because the successive quotients are the integral domains $R$, $\mathbb{Z}[X_2,X_2^{-1}]$, and $\mathbb{Z}$. They remain distinct after localization at a prime ideal because they are all contained in $P$. Hence $\dim A\ge 2$.

For the upper bound, $R$ is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring. Since $P$ is generated by two elements, the Krull height theorem gives $\mathrm{ht}P\le 2$. The prime ideal correspondence for localization identifies $\dim R_P$ with $\mathrm{ht}P$, so

A Noetherian local ring $(B,\mathfrak{n})$ with residue field $\kappa$ is a regular local ring ifBy the Nakayama lemma, the dimension on the left is the minimal number of generators of $\mathfrak{n}$; this is also called the embedding dimension.

The localization of a module is exact and commutes with quotients and products of ideals. Applying it to the already computed conormal module gives the cotangent space of a local ringIts dimension is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb{Z}$, whereas the localized cotangent space of a local ring has a basis over the residue field $\mathbb{Q}$.

The set $S=1+I$ is a multiplicative subset, since $(1+a)(1+b)=1+(a+b+ab)$. The localization of a ring $S^{-1}R$ consists of fractions $r/s$, whereSimilarly, the localization of a module $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps areThe Localization of a Noetherian ring is Noetherian. Since a finitely generated module over a Noetherian ring is a Noetherian module, localizing a finite generating set shows that $S^{-1}M$ is Noetherian over $S^{-1}R$.

The prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ with the primes $P\in \mathrm{Spec}R$ disjoint from $S$. Hereso the spectrum of localization away from one plus an ideal is

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a{)}^{j}m\in I^{j}M$ for every $j$, proving one inclusion. Conversely, suppose $m\in {\bigcap }_{j\ge 1}{I}^{j}M$. The Artin-Rees lemma applied to $Rm\subseteq M$ says that for some $c$,Taking $n=c+1$ gives $m\in I(Rm)$, so $m=am$ for some $a\in I$. Then $(1-a)m=0$ and $1-a\in S$. Thus

For failure without Noetherianity, takeThe strict chain $(t)⊊(t^{1/2})⊊(t^{1/4})⊊\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2}{)}^2$, one has $I^2=I$, and hence ${\bigcap }_j{I}^j=I\ne 0$. But $R$ is an integral domain, so its localization map into $(1+I{)}^{-1}R$ is injective. This is the non-Noetherian failure of the intersection formula for localization.

Let $J$ be an ideal of $S^{-1}A$ and contract it to an ideal $I$ of $A$. Every element $a/s\in J$ has $a/1=(s/1)(a/s)\in J$, so $a\in I$ andBecause $A$ is a Noetherian ring, write $I=(a_1,\dots ,a_r)$. Thenso every ideal of $S^{-1}A$ is finitely generated. Hence every localization of a Noetherian ring is Noetherian, proving the Localization of a Noetherian ring theorem.