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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.

Use the simple-root form of Hensel lemma: if $F\in {\mathbb{Z}}_p[T]$ and $t_0\in {\mathbb{Z}}_p$ satisfy $F(t_0)\equiv 0(\mathrm{mod}p)$ and $F^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, there is a unique $t\in {\mathbb{Z}}_p$ with $F(t)=0$ and $t\equiv t_0(\mathrm{mod}p)$. More generally the same assertion holds over a complete discrete valuation ring, with its maximal ideal in place of $(p)$.

Every nonzero 3-adic number has a unique expressionIndeed its normalized P-adic valuation fixes $k$, and its unit residue modulo three fixes $ϵ$. Cubing is an automorphism on $\{1,-1\}$, so it remains to compute the cube map on the principal units.

For $t\in {\mathbb{Z}}_3$,Thus $U_1^3\subseteq U_2:=1+9{\mathbb{Z}}_3$. Conversely, given $1+9a\in U_2$, apply Hensel lemma toModulo three it is $T-a$, and $F^{\prime }(T)=1+6T+9T^2$ is always a unit. The residue root $a\mathrm{mod}3$ therefore lifts, giving $t$ with $(1+3t{)}^3=1+9a$. HenceThis is the cube map on 3-adic principal units. Introducing $T$ is essential: applying simple-root lifting directly to $X^3-u$ would fail because its derivative is divisible by three at every unit.

The map $1+3t\mapsto t\mathrm{mod}3$ induces an isomorphism $U_1/U_2\cong (\mathbb{Z}/3\mathbb{Z},+)$, since multiplication adds the first principal-unit digits modulo three. The valuation factor contributes $\mathbb{Z}/3\mathbb{Z}$ independently, so the cube-class group of the 3-adic numbers isExplicit generators are the classes of $3$ and $4=1+3$. Representatives are $3^a{4}^b$ with $0\le a,b<3$. The class of $3$ is detected by valuation modulo three, and that of $4$ by its nonzero first principal-unit digit, so the two generators are independent.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Work over the complex scalar field. If $\lambda \in {\sigma }_A(x)$, the ideal generated by $x-\lambda 1$ is proper: otherwise commutativity would make $x-\lambda 1$ invertible. By Zorn lemma extend it to a maximal ideal $M$. This ideal is closed. Indeed, if its closure were all of $A$, it would contain an element of $M$ sufficiently close to $1$ to be invertible, since the invertible group is open; a proper ideal cannot contain such an element. Its closure is therefore proper, and maximality gives $\overline{M}=M$.

The quotient $A/M$ is a complex unital Banach algebra and a division algebra. The Gelfand-Mazur theorem identifies it with $\mathbb{C}$, giving a character of an algebra $\phi$ with $\phi (x)=\lambda$. Conversely, the argument in (a) shows that every $\phi (x)$ is in the spectrum. ThusHere, as usual, a unital Banach algebra has a nonzero identity; the zero algebra is excluded.

The holomorphic functional calculus assigns to every function holomorphic on an open neighbourhood $U$ of ${\sigma }_A(x)$ an element $f(x)\in A$. It is a unital algebra homomorphism, sends the coordinate function to $x$, agrees with evaluation of rational functions having no poles on the spectrum, depends only on the germ near the spectrum, and is continuous for uniform convergence on a fixed surrounding contour. Its formula isHere $\Gamma$ is a finite oriented contour cycle in $U\setminus {\sigma }_A(x)$, with winding number one on the spectrum and zero off $U$. A finite union of suitable polygonal boundary cycles suffices; no connectedness of $U$ is required. Such cycles exist by compactness of the spectrum, and the calculus is independent of the choice. The inverse is the resolvent of an element.

For the exponential, use a circle $|z|=R>\Vert x\Vert$. The uniformly convergent Neumann expansion $(z1-x{)}^{-1}={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$ may be integrated term by term. The scalar Cauchy integral formula gives $\frac{1}{2\pi i}\int e^z{z}^{-n-1}dz=1/n!$, provingThis series converges absolutely in the Banach algebra norm.

A character of an algebra is bounded and takes the resolvent to $(z-\phi (x){)}^{-1}$. Applying it inside the contour integral and using the scalar Cauchy formula gives $\phi (f(x))=f(\phi (x))$. The character of an algebra description of the spectrum, now applied to $f(x)$, proves the spectral mapping theorem:In particular this spectrum is contained in $V$ when $V\supset f(U)$.

For the composition rule for holomorphic functional calculus, choose a relatively compact open neighbourhood $D$ of $f({\sigma }_A(x))$ whose closure is contained in $V$, and a surrounding contour cycle $\Delta$ in $V$ with winding number one on $D$. Choose a sufficiently small neighbourhood $\Omega$ of ${\sigma }_A(x)$ with $\overline{\Omega }\subset U$ and $f(\overline{\Omega })\subset D$, and a contour $\Gamma$ in $\Omega$ surrounding the spectrum. The neighbourhoods and cycles may be finite unions, obtained from finite covers of the relevant compact sets. For $\zeta \in \Delta$ the function $h_{\zeta }(z)=(\zeta -f(z){)}^{-1}$ is holomorphic near $\overline{\Omega }$. Locality and multiplicativity of the calculus giveInsert this in the formula for $g(f(x))$ and interchange the two vector-valued integrals. The scalar inner integral is $g(f(z))$, soThis proves equality as algebra elements. Equality of their character of an algebra values alone would be insufficient in a Banach algebra with nonzero radical.

Finally, $\Vert x\Vert <1$ places the spectrum in the open unit disk. On that disk $f(z)=\log (1-z)=-{\sum }_{n\ge 1}{z}^n/n$ is a holomorphic branch. Its calculus value is the norm-convergent series $y=-{\sum }_{n\ge 1}{x}^n/n$: one may integrate the uniformly convergent scalar series on a circle with $\Vert x\Vert <R<1$. Since $\exp (f(z))=1-z$, the composition rule gives the logarithm of an element near the identity formula

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

Fix a non-Archimedean local field $K$, its valuation ring $R={\mathcal{O}}_K$, a uniformizer $\pi$, and a residue field with cardinality $q$. The construction below applies both in mixed and equal characteristic. A one-dimensional commutative formal group law is $F(X,Y)\in R[[X,Y]]$ with identity, associativity and commutativity, and $F(X,Y)=X+Y+$ higher-degree terms. It has a unique formal inverse. Its morphisms are series $h(T)\in TR[[T]]$ with $h(F(X,Y))=G(h(X),h(Y))$; a unit linear coefficient makes such a morphism invertible by recursive coefficient comparison. Evaluating its integral series on the maximal ideal of a finite extension of $K$ is legitimate because the series converge there.

Choose a Lubin–Tate series $f(T)$ satisfyingA useful choice is $f(T)=\pi T+T^q$. The Lubin–Tate functional equation lemma constructs a unique series with any prescribed linear form $\ell$ and satisfyingHere is its coefficient argument. If terms below degree $d$ have been found, the unknown homogeneous degree-$d$ term appears with multiplier $\pi -{\pi }^d$. The already known error has every coefficient divisible by $\pi$: after reduction, the equation becomes $H(X{)}^q=H(X_1^q,\dots ,X_r^q)$, valid because the residue coefficients lie in ${\mathbb{F}}_q$. Since $\pi -{\pi }^d$ is $\pi$ times a unit for $d\ge 2$, the next term is uniquely determined and integral. Recursion proves existence and uniqueness.

Use $\ell =X+Y$ to construct $F_f$, and $\ell =aT$ to construct $[a{]}_f$ for every $a\in R$. Uniqueness gives $F_f(X,0)=X$ and $F_f(0,Y)=Y$. Comparing $[a{]}_f(F_f(X,Y))$ with $F_f([a{]}_f(X),[a{]}_f(Y))$, which satisfy the same functional equation and have the same linear form, proves that $[a{]}_f$ is an endomorphism of the law. The same uniqueness applied to three variables proves associativity: $F_f(F_f(X,Y),Z)$ and $F_f(X,F_f(Y,Z))$ have the same linear form and functional equation. It also proves commutativity and the identitiesThus this Lubin–Tate formal group is equipped with an action of the whole ring $R$, not only its integers. For a unit $a$, the inverse of $[a{]}_f$ is $[a^{-1}{]}_f$. The same recursion, with two series for the fixed $\pi$, gives a unique strict Lubin–Tate change of series. Its integral series and inverse converge on torsion points, so it identifies their torsion fields. We may compute with $f(T)=\pi T+T^q$ without losing the general construction for that uniformizer.

Write $f_0(T)=T$ and $f_n=f^{\circ n}$. The Lubin–Tate torsion of level $n$ isFor $n\ge 1$, its primitive points are the roots ofThis is a monic polynomial of degree $D_n=(q-1)q^{n-1}$. Modulo $\pi$ it is $T^{D_n}$, and its constant coefficient is exactly $\pi$. Thus the Eisenstein layers of Lubin–Tate torsion are Eisenstein polynomials. A root ${\lambda }_n$ generates a totally ramified extension of degree $D_n$, and has $v_K({\lambda }_n)=1/D_n$ in the uniquely extended valuation. Also $f_{n-1}({\lambda }_n{)}^{q-1}=-\pi \ne 0$, so it has exact level $n$.

The mapis well-defined and injective. If $a={\pi }^{r}u$ with $r<n$ and $u$ a unit, then $[a]{\lambda }_n=[u]f_r({\lambda }_n)\ne 0$, because $[u]$ is invertible and ${\lambda }_n$ is primitive. Conversely every multiple of ${\pi }^n$ kills it. The image therefore contains $q^n$ distinct roots of the degree-$q^n$ polynomial $f_n$, so it is all of ${\Lambda }_n$. This proves, without merely assuming the torsion cardinality,Its primitive generators are exactly $[a]{\lambda }_n$ with $a\in (R/{\pi }^{n}R{)}^{\times }$.

Define the Lubin–Tate extension $K_{\pi ,n}=K({\Lambda }_n)$. Every integral endomorphism series evaluated at ${\lambda }_n$ converges in the complete field $K({\lambda }_n)$, so all torsion points already lie there. HenceIt is the splitting field of $f_n$, which has $q^n$ distinct roots by the module calculation. Thus it is Galois. Every K-automorphism preserves the uniquely extended absolute value on a field, hence is continuous and commutes with limits of the integral endomorphism series. Its automorphisms therefore commute with the endomorphism action and send a primitive generator to a primitive generator. There is a unique unit class $a_{\sigma }$ such that $\sigma ({\lambda }_n)=[a_{\sigma }]{\lambda }_n$. The Lubin–Tate Galois action on primitive torsion gives an injective homomorphismBoth sides have order $D_n=(q-1)q^{n-1}$, so it is an isomorphism. In particular every finite layer is abelian and totally ramified.

For an example, take $K={\mathbb{Q}}_p$, $\pi =p$, and the equally admissible series $f(T)=(1+T{)}^p-1$. Its formal group law is the formal multiplicative group $F(X,Y)=X+Y+XY$, with $[a](T)=(1+T{)}^a-1$ for $a\in {\mathbb{Z}}_p$. The binomial polynomials make its coefficients integral. Its level-$n$ torsion points are $\zeta -1$ with ${\zeta }^{p^n}=1$. Thus the Lubin–Tate cyclotomic example gives $K_{p,n}={\mathbb{Q}}_p({\mu }_{p^n})$ and Galois action $\zeta \mapsto {\zeta }^a$. The order formula includes $p=2,n=1$, when the first field and its Galois group are trivial over ${\mathbb{Q}}_2$.

Finally choose primitive points compatibly, $[\pi ]{\lambda }_{n+1}={\lambda }_n$. They exist by taking a root of $f(T)-{\lambda }_n$ in the algebraic closure; its roots have positive valuation and exact level $n+1$. Then $K_{\pi ,n}\subseteq K_{\pi ,n+1}$, and restriction of Galois automorphisms corresponds to reduction of unit classes. The Lubin–Tate tower $K_{\pi }={\bigcup }_n{K}_{\pi ,n}$ consequently has the topological Galois groupThe final isomorphism uses completeness of the discrete valuation ring $R$. This computes the finite-layer and infinite-tower Galois groups directly from the formal action and the primitive Eisenstein polynomials.

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.

Let $T=R\setminus P$. Since $P$ is a prime ideal, $T$ is a multiplicative subset containing $1$ and excluding $0$. The localization at a prime ideal isConcretely, its elements are equivalence classes of pairs $(r,s)\in R\times T$, written $r/s$, withThis extra multiplier is essential when $R$ has zero divisors. Addition and multiplication areThe equivalence criterion is compatible with these operations: after multiplying by the multipliers witnessing changes of representatives, the corresponding numerators agree. The commutative ring axioms follow from those of $R$; the identity is $1/1$.

The canonical ring homomorphism $\iota :R\to R_P$, $r\mapsto r/1$, supplies the algebra over a commutative ring structure, with scalar action $a(r/s)=ar/s$. It also has the universal property of localization: whenever a ring homomorphism $\psi :R\to A$ sends $T$ to units, the unique extension is $r/s\mapsto \psi (r)\psi (s{)}^{-1}$.

For later use, $R_P$ is a local ring with maximal ideal $P{R}_P$. Reduction of numerators and denominators gives a surjective ring homomorphism $R_P\to \mathrm{Frac}(R/P)$ with kernel $P{R}_P$, so this ideal is proper and maximal. Every fraction outside it has numerator outside $P$ and is a unit, with inverse $s/r$. Thus it is the unique maximal ideal.

No: $R$ cannot contain a nonzero nilpotent element. We prove the needed principle, localization detects zero elements, rather than assume it.

Let $a\ne 0$ in $R$. Its annihilator $\mathrm{Ann}(a)=\{r:ra=0\}$ is a proper ideal, because $1$ does not annihilate $a$. Every proper ideal is contained in a maximal ideal, by the Zorn lemma; choose such a maximal ideal $\mathfrak{m}$ containing $\mathrm{Ann}(a)$. If $a/1=0$ in $R_{\mathfrak{m}}$, then some $s\notin \mathfrak{m}$ satisfies $sa=0$. This would put $s$ in $\mathrm{Ann}(a)\subseteq \mathfrak{m}$, a contradiction. Thus every nonzero element survives in at least one localization at a prime ideal.

Now suppose $a^n=0$. For every prime ideal $P$, $(a/1{)}^n=0$ in $R_P$. The assumed absence of nonzero nilpotent elements forces $a/1=0$ in every $R_P$. The preceding argument then forces $a=0$. In the terminology of reduced rings, we have proved

Let $e_j$ be the sequence with $1$ in coordinate $j$ and zero elsewhere. In the direct product of rings $R=k^{\mathbb{N}}$, the idealsform a strict ascending chain: every element of $(e_1,\dots ,e_n)$ vanishes after coordinate $n$, whereas $e_{n+1}$ does not. Hence $R$ is not a Noetherian ring.

Nevertheless, each $a=(a_j)\in R$ has a coordinatewise generalized inverse $b=(b_j)$, defined by $b_j=a_j^{-1}$ when $a_j\ne 0$ and $b_j=0$ otherwise. It satisfiesThis is the defining property of a Von Neumann regular ring. Fix any prime ideal $P$, and take $a\in P$. Since $ab\in P$, $1-ab\notin P$; butConsequently $a/1=0$ in the localization at a prime ideal $R_P$. All elements of $P{R}_P$ therefore vanish. By the local ring description in part (a), $P{R}_P$ is the unique maximal ideal, so $R_P$ is a nonzero field. A field has only the ideals zero and itself, and thus is Noetherian.

We have proved the stronger conclusionThe argument applies to every prime ideal, without assuming that it comes from a coordinate projection.

False. In the polynomial ring $R=k[x,y]$ over a field, takeThe ideal $P$ is a maximal ideal, with quotient ring $R/P\cong k$. AlsoIn this last ring, an element with nonzero constant term is a unit, and an element with zero constant term is a multiple of $y$ and has square zero. Thus every zero divisor is a nilpotent element. This is equivalent to $I$ being a primary ideal: a zero product in the quotient ring with one nonzero factor forces the other factor to be nilpotent. Its radical of an ideal is $P$, so $I$ is $P$-primary.

However, $y\notin I$, so $I\ne P$. Moreover $x\in I$ while $x\notin P^2$, since every polynomial in $P^2$ has no term of total degree one. For every $n\ge 2$, $P^n\subseteq P^2$, so $I\ne P^n$. Hence

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.

Let $D$ be a Dedekind domain. The zero ideal is generated by $0$, and $D$ is generated by $1$, so consider a nonzero proper ideal $I$. Choose $0\ne a\in I$. We use these consequences of the unique factorization of ideals in a Dedekind domain: every nonzero ideal has finitely many nonzero prime-ideal valuations in a Dedekind domain; containment reverses the valuations as in part (b); andThe last formula follows from the containment rule: $J+K$ is the smallest ideal containing both $J$ and $K$, and the product with the indicated minimum exponents is exactly that ideal.

Let $T$ be the finite set of prime ideals dividing $(a)$. For each $P\in T$, set $r_P=v_P(I)$. Since $(a)\subseteq I$, we have $r_P\le v_P((a))$. The containment rule also gives $I⊈P^{r_P+1}$, so chooseDistinct nonzero prime ideals of $D$ are distinct maximal ideals and hence comaximal ideals; their positive powers are also comaximal. Indeed, if $u+v=1$ with $u\in P$, $v\in Q$, for positive integers $m,n$ expanding $(u+v{)}^{m+n-1}$ shows $1\in P^m+Q^n$, since each term contains either at least $m$ factors of $u$ or at least $n$ factors of $v$. We use the Chinese remainder theorem in its ideal form: for pairwise comaximal ideals $J_1,\dots ,J_s$, the map $D\to {\prod }_{j}D/J_j$ is surjective. Apply it to the powers $P^{r_P+1}$ to choose $e_P\in D$ withThen $b={\sum }_{P\in T}{e}_P{b}_P$ belongs to $I$ and satisfies $b\equiv b_P(\mathrm{mod}{P}^{r_P+1})$ for every $P\in T$. Thus $b\notin P^{r_P+1}$, whereas $b\in I\subseteq P^{r_P}$, proving $v_P((b))=r_P$.

For $P\in T$, therefore,For $P\notin T$, $v_P((a))=0$, and $(a)\subseteq I$ forces $v_P(I)=0$ as well. Hence $(a,b)$ and $I$ have the same valuation at every nonzero prime ideal. Uniqueness of their factorizations givesThis proves two generators for an ideal in a Dedekind domain; in fact the first nonzero generator $a\in I$ can be chosen in advance.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

For a discrete valuation ring $A$,Let $\pi$ be a uniformizer, so its unique nonzero maximal ideal is $\mathfrak{m}=(\pi )$. We use the defining description of a discrete valuation ring: it is a non-field local ring that is a principal ideal domain, and every nonzero element is $u{\pi }^n$ for a unit $u$ and an integer $n\ge 0$.

Every nonzero proper ideal is therefore $({\pi }^n)$ for some $n\ge 1$. Such an ideal is a prime ideal only when $n=1$: for $n\ge 2$, $\pi {\pi }^{n-1}\in ({\pi }^n)$ while neither factor belongs to $({\pi }^n)$. The ideal $(\pi )$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an integral domain. Thusand the longest strict chain has length one.