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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 Artin-Rees lemma states that, for a Noetherian ring $R$, an ideal $I$, a finitely generated module $M$, and a submodule $N\subseteq M$, there is $c\ge 0$ such thatThus the filtration on $N$ induced from the I-adic filtration of $M$ is eventually determined by one of its terms.

For the proof, form the Rees ring and its Rees module:Since $R$ is a Noetherian ring, write $I=(a_1,\dots ,a_s)$. Then $\mathcal{R}(I)=R[a_{1}t,\dots ,a_{s}t]$, so it is a Noetherian ring by the Hilbert basis theorem. Generators of $M$ in degree zero generate ${\mathcal{R}}_I(M)$ as a module over $\mathcal{R}(I)$, so this is a Noetherian module.

The graded submoduleis therefore finitely generated. Choose homogeneous generators $z_{\alpha }{t}^{d_{\alpha }}$ with $d_{\alpha }\le c$ for some $c$. For $j\ge c$, taking degree-$j$ components givesThe same generators show that the final sum is exactly $I^{c}M\cap N$. This proves the Artin-Rees lemma. If $L=0$, one may simply take $c=0$.

Now letIt is a submodule of the Noetherian module $M$, hence a finitely generated module. Apply the Artin-Rees lemma with $N=K$ and $j=c+1$. Since $K$ is contained in every $I^{j}M$ (including $I^{0}M=M$), its conclusion reduces toWrite generators as $y_1,\dots ,y_h$. There is a matrix $A=(a_{ij})$ with entries in $I$ such that $y_i={\sum }_j{a}_{ij}{y}_j$. Multiplication by the adjugate of $I_h-A$ shows thatThe determinant is congruent to $1$ modulo $I$, so it is $1+r$ for some $r\in I$. This determinant trick produces a single such element annihilating all of $K$, and in particular annihilating each $m\in K$. If $K=0$, use $r=0$.

Conversely, if $(1+r)m=0$ with $r\in I$, then $m=-rm$. Iterating gives $m=(-r{)}^{j}m\in I^{j}M$ for every $j\ge 1$. Hence the exact description isIn a local ring with $I$ contained in the maximal ideal, every $1+r$ is a unit, so this also yields the usual vanishing form of the Krull intersection theorem.

For a failure without the Noetherian ring hypothesis, let $D=\mathbb{Z}[1/2{]}_{\ge 0}$ and use the semigroup algebraIts elements are finite sums of formal monomials $t^q$, with multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. Each ring in the union is a polynomial ring and the inclusions are injective, so $R$ is an integral domain. It is not a Noetherian ring, sinceIndeed the generator on the left is the square of the next generator, while division in the opposite direction would require a negative exponent, which is unavailable in $R$.

Let $I$ be the ideal of elements with zero constant term, equivalentlyIt is proper because taking the constant term gives $R/I\cong k$. Every generator $t^q$ is $(t^{q/2}{)}^2$, with $t^{q/2}\in I$, so $I^2=I$. Thus this is a nonzero idempotent ideal, with $I^j=I$ for every $j\ge 1$. TakeFor every $r\in I$, the constant term of $1+r$ is $1$, so $1+r\ne 0$. Since $R$ is an integral domain and $t\ne 0$, we have $(1+r)t\ne 0$. This gives all three requested objects and proves that no element of the required form annihilates $m$.

For a finitely generated module $V={⨁}_d{V}_d$ over the graded ring $S=k[X_1,\dots ,X_n]$, with each $X_i$ of degree one, its Poincare series of a graded module isThis is also its Hilbert series. One usually takes $d\ge 0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\dots ,a_h$ give a surjection ${⨁}_{j}S(-a_j)\to V$, where the graded shift is defined by $V(-a{)}_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional vector space and $V_d=0$ for all sufficiently negative $d$.

The Hilbert-Serre theorem in the standard grading saysFor a nonnegatively graded $V$, the numerator is in $\mathbb{Z}[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.

We prove the Hilbert-Serre theorem by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional graded module; thus $P_V(t)$ is a Laurent polynomial. Suppose $n>0$ and putThe Hilbert basis theorem makes $S$ a Noetherian ring, so $K$ is a finitely generated module, as is $C$. Both are graded modules killed by $X_n$, hence finitely generated over $S^{\prime }=k[X_1,\dots ,X_{n-1}]$.

The graded exact sequenceidentifies the kernel and cokernel of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yieldssoBy induction the right side has denominator $(1-t{)}^{n-1}$, which proves the required denominator $(1-t{)}^n$. Since $(1-t{)}^n{P}_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent polynomial numerator is an ordinary polynomial in that case.

For $n\ge 1$, this also implies the usual Hilbert polynomial consequence. Writing $Q(t)={\sum }_j{q}_j{t}^j$ and usinggives, for all sufficiently large $d$,a polynomial in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.

For the final request, a free resolution of $V$ is an exact sequencewith each $F_i$ a free module. In the graded setting we take finite direct sums of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The syzygy modules are the successive kernels which record the relations among generators, then relations among those relations, and so on.

Put $\mathfrak{m}=(X_1,\dots ,X_n)$, the homogeneous maximal ideal with $S/\mathfrak{m}=k$. Choose a homogeneous $k$-basis of $V/\mathfrak{m}V$ and lift it to $V$. These lifts generate $V$ by the graded Nakayama lemma: a bounded-below graded module $W$ satisfying $W=\mathfrak{m}W$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the submodule generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an isomorphism modulo $\mathfrak{m}$.

Its kernel is finitely generated because $S$ is a Noetherian ring. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a minimal graded free resolution, meaningIndeed, at each stage the free cover induces an isomorphism modulo $\mathfrak{m}$, so its kernel lies in $\mathfrak{m}$ times its source. The sequence is exact by construction, though it may at first appear infinite.

The Koszul complex on $X_1,\dots ,X_n$ iswith basis vectors $e_j$ given degree one, and differentialThe hat means omission of that factor. Every pair of terms in ${\partial }^2$ cancels with opposite signs, so this is a chain complex. The sequence $X_1,\dots ,X_n$ is a regular sequence: after quotienting by the first $j-1$ variables, the next variable is a non-zero-divisor in the remaining polynomial ring.

The Koszul complex is consequently exact in positive degrees and has $H_0=k$, so it is a Koszul resolution of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\stackrel{X_1}{\to }S\to S/(X_1)\to 0$. Appending $X_j$ constructs the mapping cone of multiplication by $X_j$ on the previous chain complex. Its homology is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\dots ,X_{j-1})$; its degree-zero homology is the further quotient. Here a mapping cone combines a chain complex with a shifted copy and adds the given multiplication map to the differential, thereby measuring its kernel and cokernel on homology.

The Tor functor ${\mathrm{Tor}}_i^S(V,k)$ is defined as the degree-$i$ homology of $F_{\bullet }{\otimes }_{S}k$ for a free resolution $F_{\bullet }$ of $V$. It can equally be computed as the homology of $V{\otimes }_S{K}_{\bullet }$, using the Koszul resolution of $k$. To see the equality, form the double complex $F_{\bullet }{\otimes }_S{K}_{\bullet }$. Taking homology first along $F_{\bullet }$ leaves $V{\otimes }_S{K}_{\bullet }$, since each $K_i$ is a free module; taking homology first along $K_{\bullet }$ leaves $F_{\bullet }{\otimes }_{S}k$, since each $F_i$ is free. The finite sums along each total degree identify both with the homology of the total complex. This also explains the symmetry used in computing the Tor functor.

Since $K_i=0$ for $i>n$, we have ${\mathrm{Tor}}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the minimal graded free resolution become zero after tensoring with $k=S/\mathfrak{m}$, soThus $F_i/\mathfrak{m}{F}_i=0$ for $i>n$, and the graded Nakayama lemma implies $F_i=0$ for those $i$. We have proved the required Hilbert syzygy theorem:where some initial $F_i$ may be zero. The argument supplies finitely generated graded free modules, which is stronger than merely giving an ungraded free resolution.

Consider all ideals containing $I$ and disjoint from the nonempty multiplicative subset $S$. This collection is nonempty because it contains $I$. The ascending chain condition for the Noetherian ring gives a maximal member $P$: otherwise repeatedly choosing a strictly larger member would give an infinite strictly ascending chain of ideals. Since $S$ is nonempty and $P$ misses it, $P\ne R$.

Suppose $ab\in P$ but $a,b\notin P$. Maximality implies that $P+(a)$ and $P+(b)$ both meet $S$, so write $s=p+ra\in S$ and $t=q+ub\in S$ with $p,q\in P$. Thenwhile closure under products gives $st\in S$, a contradiction. HenceThis proves the prime ideal avoiding a multiplicative subset result without needing $1\in S$. The assumptions themselves rule out $0\in S$, since every ideal contains zero.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

False: existence holds, but uniqueness fails. First, the Lasker–Noether theorem supplies existence; here is an argument from the preceding part. If some proper ideal in a Noetherian ring were not a finite intersection of irreducible ideals, the ascending chain condition would let us choose such an ideal $I$ maximal under inclusion. It is not irreducible, so $I=J\cap K$ with both $J$ and $K$ strictly larger than $I$. By maximality, both are finite intersections of irreducible ideals, and therefore so is $I$, a contradiction. Part (d) makes each factor a primary ideal. The whole ring is represented by the empty intersection if that convention is needed.

For a counterexample to uniqueness, work in $R=k[x,y]$ over a field $k$ and put $I=(x^2,xy)$. The Hilbert basis theorem makes $R$ a Noetherian ring. There are two distinct minimal primary decompositions:The ideal $(x)$ is a prime ideal, hence a primary ideal. Each of the other factors has quotient ring isomorphic to $k[x]/(x^2)$, by substituting respectively $y=0$ and $y=x$. As in part (b), each is $(x,y)$-primary.

To check both intersections, let $Q_c=(x^2,y-cx)$, for $c=0,1$. Substitution identifies $R/Q_c$ with $k[x]/(x^2)$. The class of $xg(x,y)$ is zero exactly when $g(0,0)=0$, equivalently when $g\in (x,y)$. Therefore $(Q_c:x)=(x,y)$ andThe two $Q_c$ are different, since $y$ belongs to $Q_0$ but has nonzero class $x$ modulo $Q_1$. Both decompositions are irredundant: $x\notin Q_c$, while $y-cx\notin (x)$. Applying the radical of an ideal operation to the two components in either decomposition gives $(x)$ and $(x,y)$, respectively.

The second uniqueness theorem for primary decomposition explains the distinction: the component at an isolated prime of a primary decomposition is unique, but embedded primary components need not be. The counterexample varies an embedded component.

A ring is a Noetherian ring if it satisfies the ascending chain condition on ideals: every chaineventually becomes constant. Equivalently, every ideal has a finite generating set of an ideal. To see this equivalence, if an ideal were not finitely generated, one could repeatedly choose an element outside the ideal generated by the preceding choices, producing a strictly ascending chain. Conversely, the union of an ascending chain is an ideal; if it is generated by finitely many elements, all of them lie in one term of the chain, and every later term equals that term.

A ring is an Artinian ring if it satisfies the descending chain condition on ideals: every chaineventually becomes constant. These are stabilization conditions on all ideals, not merely on principal ideals.

Yes. Taking the constant coefficient defines a surjective ring homomorphismIts kernel of a ring homomorphism is $(x)$, so the first isomorphism theorem for rings gives $R[[x]]/(x)\cong R$.

We prove the needed preservation result. If $A$ is a Noetherian ring and $J\subseteq A$ is an ideal, any ascending chain of ideals in $A/J$ lifts under the quotient ring homomorphism to an ascending chain of ideals in $A$. The lifted chain stabilizes; surjectivity makes the original chain stabilize as well. Hence every quotient ring of a Noetherian ring is Noetherian.

Applying this with $A=R[[x]]$ and $J=(x)$ gives

A Dedekind domain is a Noetherian ring $D$ which is an integral domain, is an integrally closed domain, and has Krull dimension one. Explicitly, every ideal is finitely generated; every element of the fraction field that is an integral element over $D$ already belongs to $D$; and every nonzero prime ideal is a maximal ideal, with $D$ itself not a field.

With this convention, an equivalent local characterization is that $D$ is a Noetherian integral domain that is not a field and whose localization at a prime ideal $D_P$ is a discrete valuation ring for every nonzero prime ideal $P$. The integrally closed domain condition rules out singularities such as those represented by the subring $k[t^2,t^3]\subset k[t]$, where $t$ is integral but missing.

Some conventions also include fields by allowing Krull dimension at most one; the conclusions about ideals below remain valid under that convention.

The height of a prime ideal $P$ is

The Krull principal ideal theorem states that if $R$ is Noetherian and $P$ is minimal among the primes containing a proper principal ideal $(a)$, thenSuppose otherwise that $P_0⊊Q⊊P$. Quotient by $P_0$ and localize at $P$; it is enough to consider a Noetherian local domain whose maximal ideal $P$ is the only prime containing $(a)$ and which has $0⊊Q⊊P$.

For $n\ge 1$, putThis is a $Q$-primary ideal. Since $R/(a)$ is a zero-dimensional Noetherian ring, it is Artinian, and the descending chain $(Q^{(n)}+(a))/(a)$ eventually stabilizes. Thus, for all sufficiently large $n$, every $q_n\in Q^{(n)}$ can be writtenNow $ra\in Q^{(n)}$, while $a\notin Q$; $Q$-primaryness gives $r\in Q^{(n)}$. HenceThe Nakayama lemma applied to $Q^{(n)}/Q^{(n+1)}$ gives $Q^{(n)}=Q^{(n+1)}$. Localizing at $Q$ makes all sufficiently large powers of the nonzero maximal ideal $Q{R}_Q$ equal. A nonzero element of the stable power then belongs to ${\bigcap }_n{Q}^n{R}_Q$, contradicting the Krull intersection theorem. This proves the theorem.

Now let $R$ be a Noetherian integral domain. If $R$ is a unique factorization domain and $P$ has height one, choose $0\ne x\in P$ and an irreducible factor $p$ of $x$. In a UFD, $p$ is prime, soHeight one forces $P=(p)$.

Conversely, suppose every height-one prime is principal. Noetherianity makes $R$ an atomic domain. Given an irreducible element $p$, choose a prime $P$ minimal over $(p)$. The principal ideal theorem gives $\mathrm{ht}P=1$, so $P=(q)$ by hypothesis. Since $p=qr$ and $p$ is irreducible, $r$ is a unit; hence $(p)=P$ is prime. Thus every irreducible is a prime element, and an atomic domain with this property is a UFD. Therefore

Suppose first that every ideal of $R$ is finitely generated. For an ascending chainthe union $I={\bigcup }_n{I}_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\ge N$. Thus $R$ satisfies the ascending chain condition and is a Noetherian ring.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\dots ,a_n$, chooseThis creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\phi :R\to S$ is surjective and $J⊲S$, then ${\phi }^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\dots ,a_m$, then $J$ is generated by $\phi (a_1),\dots ,\phi (a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The Hilbert basis theorem states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I⊲R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\dots ,f_s\in I$, and let $N={\max }_i\mathrm{deg}{f}_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha \in \mathbb{C}$, the ring $\mathbb{Z}[\alpha ]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb{Z}[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb{Z}[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb{Z}[X]$, and quotients preserve Noetherianity.

For a unique factorization domain that is not Noetherian, takeEvery polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. Butis a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb{R},\mathbb{R})$ is not Noetherian. For $n\ge 1$, letThese are ideals and $I_n⊊I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.

Suppose first that $M$ is finitely generated. Its quotient $M/N$ is generated by the images of any finite generating set of $M$. A principal ideal domain is Noetherian, so every submodule of the finitely generated $R$-module $M$ is finitely generated; in particular, $N$ is finitely generated.

Conversely, supposeand $M/N$ is generated by the cosets ${\overline{m}}_1,\dots ,{\overline{m}}_s$. For any $m\in M$, its coset is an $R$-linear combination of the ${\overline{m}}_j$, soExpressing this remainder in terms of the $n_i$ shows thatThus $M$ is finitely generated if and only if both the submodule $N$ and the quotient module $M/N$ are finitely generated.

The exact sequenceand part ii show that a direct sum of two modules is Noetherian exactly when both summands are. Induction proves the assertion for every finite direct sum.

Each $R/I_j$ is Noetherian as an $R$-module because its $R$-submodules are precisely its ideals as a Noetherian ring. The diagonal maphas kernel ${\bigcap }_j{I}_j=0$. Hence $R$ is an $R$-submodule of a finite direct sum of Noetherian modules, so $R$ is a Noetherian $R$-module; equivalently, $R$ is a Noetherian ring.