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 Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

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.