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The map $\phi :M\to N$, $x\mapsto rx$, is a module homomorphism, and it is injective because $r\ne 0$ and $M$ is a torsion-free module. The linear independence in a module of $T$ makes the map $R^{|T|}\to N$ sending the standard basis to $T$ an isomorphism. Thus $N$ is a finite free module, and $M\cong \phi (M)\subseteq N\cong R^{|T|}$. This proves embedding a finitely generated torsion-free module in a finite free module. No claim that the submodule itself is free over an arbitrary integral domain is needed.

For each $s\in S\setminus T$, maximality implies a nontrivial relationHere $a_s\ne 0$, since otherwise this would contradict the linear independence in a module of $T$. Thus $a_{s}s\in N$. Take $r={\prod }_{s\in S\setminus T}{a}_s$, with empty product equal to one. Because $R$ is an integral domain, $r\ne 0$. Every $s\in S\setminus T$ satisfies $rs\in N$, and this is also true for $s\in T$. Expressing any element of the finitely generated module as an $R$-linear combination of $S$ now gives $rM\subseteq N$. This is clearing denominators relative to an independent module subset; it does not require $R$ to be a field.

An irreducible scheme is a nonempty scheme whose underlying topological space cannot be expressed as the union of two proper closed subsets. Equivalently, any two nonempty open subsets meet. A reduced scheme is one whose local rings have no nonzero nilpotent elements. Equivalently, every affine open subscheme has a reduced ring of regular functions. We may define an integral scheme as a nonempty scheme for which the coordinate ring of every nonempty affine open subscheme is an integral domain. We shall show that this is equivalent to being reduced and irreducible. The nonempty convention matters: the empty scheme is reduced but is not irreducible or integral.

For a commutative ring $A$, the spectrum of a ring is reduced exactly when $A$ is a reduced ring. One direction follows since localization preserves reducedness. Conversely, if $a$ is a nonzero nilpotent element, choose a prime ideal containing its proper annihilator. Then $a/1$ cannot vanish at that localization, contradicting reducedness of its local ring.

The spectrum of a ring is irreducible exactly when its nilradical $N=\sqrt{(0)}$ is a prime ideal. To see the essential implication directly, if $ab\in N$, then $D(a)\cap D(b)=D(ab)$ is empty. Irreducibility forces $D(a)$ or $D(b)$ to be empty, hence $a\in N$ or $b\in N$. Also $N$ is proper because the spectrum is nonempty. Conversely, if $N$ is a prime ideal, the point $N$ has closure $V(N)=\mathrm{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives

Now suppose $X$ is reduced and irreducible. Every nonempty affine open subscheme is also reduced and irreducible: irreducibility passes to nonempty open subsets, because their nonempty open subsets are nonempty opens of $X$. The affine criterion makes each coordinate ring an integral domain, so $X$ is integral. Conversely, suppose all its nonempty affine coordinate rings are integral domains. Their localizations show that $X$ is reduced. If $X$ were reducible, there would be disjoint nonempty open subsets; choose nonempty affine open subschemes $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ inside them. Their disjoint union is itself an affine open subscheme $\mathrm{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore

The generic point of an integral scheme $X$ is the unique point $\eta$ with $\overline{\{\eta \}}=X$. For existence, take a nonempty affine open subscheme $V=\mathrm{Spec}A$. Its point $(0)$ has closure containing $V$, which is dense in $X$, so its closure in $X$ is all of $X$. For uniqueness, both proposed generic points lie in every nonempty open subset, hence in $V$, where the only dense point is $(0)$. The function field isHere the stalk description shows that the field of fractions is independent of the choice of nonempty affine open subscheme.

Every nonempty open subscheme $U$ contains $\eta$, so taking a germ there defines $\Gamma (U,{\mathcal{O}}_X)\to K(X)$. If a section has zero germ, restrict it to any nonempty affine open subscheme $V\subseteq U$. Its image in $\mathrm{Frac}\Gamma (V,{\mathcal{O}}_X)$ is zero. Since this coordinate ring is an integral domain, the section is zero on $V$. Such affines cover $U$, and the sheaf gluing axiom makes the section zero on $U$. Thus the generic-point embedding of regular functions is

For a nonreduced reducible fibre between integral schemes, take both source and target to be the affine line over $\mathbb{C}$, and use the ring homomorphismBoth coordinate rings are integral domains, so both schemes are integral. At the target closed point $t=0$, the scheme-theoretic fibre has ringby the Chinese remainder theorem. Its underlying space has two distinct closed points, so it is reducible. The class of $x(x-1)$ is nonzero but has square zero, so it is not reduced. Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.

For the Sheaf of relative Kähler differentials on $X={\mathbb{P}}_{\mathbb{C}}^3$, the cotangent form of the Euler sequence isThe last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed Sheaf of relative Kähler differentials. The cohomology of twisting sheaves on projective space gives $H^q(X,{\mathcal{O}}_X(-1))=0$ for every $q$, while $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$ and its positive-degree groups vanish. The long exact sequence in sheaf cohomology therefore gives the cotangent-sheaf cohomology of projective space in this dimension:The isomorphism for $q=1$ is the connecting map from the constant global sections of ${\mathcal{O}}_X$. The Euler characteristic of a coherent sheaf is the alternating sum of dimensions, explaining the minus sign.

For the curve, put $S=\mathbb{C}[t_0,t_1,t_2,t_3]$. It is a unique factorization domain, so irreducibility of $F$ makes $(F)$ a prime ideal. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the integral domain $S/(F)$. It follows that $(F,G)$ is a regular sequence. Its Koszul complex is an exact sequence, and graded sheafification giveswhere $i:Z\hookrightarrow X$ is the closed immersion. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.

Let ${\mathcal{I}}_Z$ be the ideal sheaf of a closed subscheme. Split this Koszul resolution intoFor all integers $d$, $H^1(X,{\mathcal{O}}_X(d))=H^2(X,{\mathcal{O}}_X(d))=0$. Also $H^1(X,{\mathcal{O}}_X)=H^2(X,{\mathcal{O}}_X)=0$. The two long exact sequences in sheaf cohomology, together with sheaf cohomology under a closed inclusion, consequently identifyBy Serre duality, $H^3(X,{\mathcal{O}}_X(-m))\cong S_{m-4}^{\ast }$ for $m\ge 4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Thus the source has dimension $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$, and the two target spaces have dimensions $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The rank-nullity theorem proves the requested bound:

In fact equality holds. Under the Serre duality pairings, the dual of this map isIf $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. Thus the lower bound is attained for every pair allowed in the question. No smoothness assumption is needed. The regular sequence cuts out a projective complete intersection of dimension one. Also $H^0(X,{\mathcal{I}}_Z)=H^1(X,{\mathcal{I}}_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,{\mathcal{O}}_Z)=\mathbb{C}$. This is the genus of a complete-intersection space curve: the first cohomology has dimension $141$, equal to its arithmetic genus.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

Let $D$ be a finite integral domain. For any $a\ne 0$, multiplication $m_a:D\to D$ is injective: $ax=ay$ implies $a(x-y)=0$, hence $x=y$ because there are no zero divisors. An injective map of a finite set to itself is surjective, so $ab=1$ for some $b$. Every nonzero element is invertible; therefore a finite integral domain is a field.

For an irreducible polynomial $f\in F[X]$, consider a nonzero class $[g]$ in the quotient ring $F[X]/(f)$. Since $f$ does not divide $g$, irreducibility gives $\gcd (f,g)=1$. The Euclidean algorithm in the polynomial ring gives polynomials $u,v$ with $uf+vg=1$. Modulo $(f)$, $[v][g]=1$. Thus the quotient by an irreducible polynomial is a field.

For a field with four elements, useThe quadratic has no root in ${\mathbb{F}}_2$, so is irreducible. Write $\alpha =[X]$. Polynomial division shows the elements are $0,1,\alpha ,\alpha +1$, with ${\alpha }^2=\alpha +1$. Its multiplication table is

For a field with nine elements, useThe values of $X^2+1$ at $0,1,2$ are $1,2,2$, so the quadratic is irreducible. Every class is uniquely $a+b\beta$, $a,b\in {\mathbb{F}}_3$, with ${\beta }^2=-1=2$. These nine classes form a finite field; multiplication is $(a+b\beta )(c+d\beta )=(ac-bd)+(ad+bc)\beta$, with coefficients modulo three.

With the usual commutative, unital conventions, the chain isA field is an integral domain, and every nonzero ideal contains an invertible element and is consequently the whole ring. Its only ideals are $(0)$ and $(1)$, so it is a principal ideal domain.

To prove the middle implication, first show that a principal ideal domain has the ascending chain condition on ideals. The union of an ascending chain is an ideal, hence is $(a)$; its generator belongs to one member of the chain, and all later members equal $(a)$. Now suppose a nonzero nonunit admitted no factorization into irreducible elements. It would factor into two nonunits, at least one of which again has no such factorization. Repeating would produce a strictly increasing chain of principal ideals: if $a=bc$ with $c$ a nonunit, $(a)⊊(b)$, since equality and cancellation would make $c$ invertible. This contradicts the ascending chain condition. Existence of factorization follows. By the permitted fact that every irreducible element is a prime element in a principal ideal domain, any irreducible factor in one factorization divides a factor in the other; cancellation then proves uniqueness up to units and order. This is a unique factorization domain. The final implication is part of the definition of a unique factorization domain.

None of the adjacent implications reverses. The principal ideal domain $\mathbb{Z}$ is not a field, since $2$ is not invertible. The polynomial ring $\mathbb{Q}[x,y]$ is a unique factorization domain, by applying the polynomial extension theorem for unique factorization domains twice, but its ideal $(x,y)$ is not principal: a generator would divide both $x$ and $y$, so would be a unit, whereas every member of this ideal vanishes at the origin. Finally, $\mathbb{Z}[\sqrt{-5}]$ is an integral domain as a subring of $\mathbb{C}$, but is not a unique factorization domain. Its multiplicative norm is $N(a+b\sqrt{-5})=a^2+5b^2$. There is no element of norm $2$ or $3$. Hence $2,3,1+\sqrt{-5},1-\sqrt{-5}$, of norms $4,9,6,6$, are all irreducible: a nontrivial factorization would require one of those missing norms. The only units have norm one and are $\pm 1$. The two factorizationsare therefore not related by associates and reordering.

The set is the ideal quotient $K=(I:J)$. It contains zero. If $r,s\in K$, then $(r-s)J\subseteq I$, since $I$ is an subgroup. If $a\in R$ and $r\in K$, then $(ar)J=a(rJ)\subseteq I$ because $I$ is an ideal. Thus $K$ is an ideal.

Write $J=(a)$ and $K=(b)$. Because $I\subseteq J$, each $x\in I$ has the form $ar$. Then $rJ=r(aR)=xR\subseteq I$, so $r\in K$, and $x\in aK$. Conversely $aK\subseteq I$ by the definition of $K$. Thereforeso $I$ is a principal ideal. This argument uses only the commutative ring convention, not cancellation or an integral domain hypothesis.

A ring is Artinian when its ideals satisfy the descending chain condition: every chain $I_1\supseteq I_2\supseteq \cdots$ eventually becomes constant. It is Noetherian when its ideals satisfy the ascending chain condition, equivalently when every ideal is finitely generated. We first prove finite length of a commutative Artinian ring; this gives the stronger structural reason for its Noetherian property.

If $\mathfrak{p}$ is a prime ideal of an Artinian ring, the quotient $R/\mathfrak{p}$ is an Artinian integral domain. For a nonzero element $x$ of that domain, the chain $(x)\supseteq (x^2)\supseteq \cdots$ stabilizes. Thus $x^n=b{x}^{n+1}$ for some $b$, and cancellation gives $bx=1$. The quotient is a field, so every prime ideal is a maximal ideal.

There are only finitely many maximal ideals. Otherwise, choose distinct ones ${\mathfrak{m}}_1,{\mathfrak{m}}_2,\dots$. The intersections ${\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ form a strictly descending chain. To see strictness, for each $i\le n$ choose $a_i\in {\mathfrak{m}}_i\setminus {\mathfrak{m}}_{n+1}$; their product belongs to the first $n$ ideals but not to the next one, since ${\mathfrak{m}}_{n+1}$ is prime. This contradicts the descending chain condition.

Put $J={\bigcap }_i{\mathfrak{m}}_i$. This Jacobson radical is also the nilradical, since all primes are maximal. We need the stronger conclusion that $J$ is nilpotent, without assuming Noetherianity. Its powers stabilize, say $K=J^n=J^{n+1}=JK$. Suppose $K\ne 0$. By the descending chain condition, choose an ideal $L$ minimal subject to $KL\ne 0$. Some $x\in L$ has $Kx\ne 0$, so minimality gives $L=Rx$. Moreover $K(JL)=KL\ne 0$, and minimality gives $JL=L$. Hence $x=jx$ for some $j\in J$. But $1-j$ is a unit: it cannot lie in any maximal ideal, because $j$ lies in all of them. Thus $x=0$, a contradiction. Therefore $J^n=0$.

The Chinese remainder theorem gives $R/J\cong {\prod }_{i}R/{\mathfrak{m}}_i$, a finite product of fields. Each quotient $J^i/J^{i+1}$ is an Artinian module over this product, and each field component must be a finite-dimensional vector space; an infinite-dimensional vector space admits a strictly descending chain of subspaces. Consequently every layer has finite composition length. The finite filtrationshows that $R$ itself has finite composition length. A strict inclusion of submodules strictly increases length, so an ascending chain cannot continue indefinitely. Every commutative Artinian ring is therefore Noetherian. This is the Artinian rings are Noetherian result.

For the formal power series ring, let $B=R[[X]]$ with $R$ Noetherian, and let $H$ be any ideal of $B$. For $n\ge 0$ define a coefficient idealThese coefficient ideals of a formal power series ideal satisfy $A_n\subseteq A_{n+1}$, by multiplication by $X$. The ascending chain condition gives $A_n=A_N$ for all $n\ge N$. For each $0\le n\le N$, choose finitely many series $f_{n,j}\in H\cap X^{n}B$ whose coefficients at $X^n$ generate $A_n$.

We claim that these finitely many series generate $H$ as an ordinary ideal. Given $f\in H$, cancel its coefficient at $X^k$ successively. After coefficients below $k$ have vanished, its coefficient at $X^k$ lies in $A_k$. If $k\le N$, use an $R$-linear combination of the $f_{k,j}$. If $k>N$, use a combination of $X^{k-N}{f}_{N,j}$, since $A_k=A_N$. The remainder then belongs to $H\cap X^{k+1}B$.

Collect all the cancellations against each fixed generator. For $n<N$ its multiplier is a polynomial, while the multipliers of the $f_{N,j}$ are well-defined formal power series: at any fixed degree, only finitely many cancellation steps contribute. The remainder has every coefficient zero. ThusThis is a finite sum of ideal generators, rather than merely a topological closure assertion. Since $H$ was arbitrary,This coefficient-cancellation argument proves Noetherianity of a formal power series ring.

The corresponding Artinian assertion is false. For any nonzero ring $R$, the idealsin $R[[X]]$ are strictly decreasing, since $X^n$ has a nonzero coefficient in degree $n$ and no multiple of $X^{n+1}$ does. In particular, a field $k$ is Artinian, but $k[[X]]$ is not. The zero ring is the harmless exception.

An integral extension $R\subseteq T$ means that every element $t\in T$ satisfies a monic polynomial with coefficients in $R$:The Krull dimension is the supremum of the lengths of strict chains of primes:There need not be a finite bound on these lengths.

Here are the prime-ideal facts behind dimension preservation, with their relevant proofs. If a domain $B$ is integral over a subdomain $A$ and $B$ is a field, then $A$ is a field: for $0\ne a\in A$, a monic equation for $a^{-1}\in B$, multiplied by a suitable power of $a$, expresses $a^{-1}$ as an element of $A$. Conversely, an integral domain integral over a field is itself a field: a nonzero element has a polynomial equation with nonzero constant term after removing any factor of the indeterminate, and that equation expresses its inverse. Applied to quotients, these observations show that a prime in an integral extension is maximal if and only if its contraction is maximal.

For the Lying-over theorem, localize $R\subseteq T$ at $S=R\setminus \mathfrak{p}$. The inclusion remains injective and integral, and $S^{-1}T$ is nonzero. Any maximal ideal of $S^{-1}T$ contracts to the unique maximal ideal of $R_{\mathfrak{p}}$, by the field criterion just proved. The prime ideal correspondence for localization then gives a prime of $T$ contracting to $\mathfrak{p}$.

For the Going-up theorem, suppose $\mathfrak{q}$ lies over $\mathfrak{p}$ and $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$. The quotient inclusion $R/\mathfrak{p}\subseteq T/\mathfrak{q}$ is integral. Apply Lying-over theorem to the prime ${\mathfrak{p}}^{\prime }/\mathfrak{p}$; lifting back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ contracting to ${\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ contract to the same $\mathfrak{p}$. After localizing at $R\setminus \mathfrak{p}$, both are maximal ideals, because they lie over the maximal ideal of $R_{\mathfrak{p}}$. Their inclusion is therefore equality. The bijection between primes under localization gives $\mathfrak{q}={\mathfrak{q}}^{\prime }$.

Now contract a strict chain of primes in $T$. Incomparability theorem for integral extensions ensures that every contraction remains strict, so $\dim T\le \dim R$. Conversely, Lying-over theorem lifts the first member of any finite chain in $R$, and repeated Going-up theorem lifts the remaining members; different contractions ensure a strict chain in $T$. Taking suprema, including the possibility of infinity, givesThis proves that integral extensions preserve Krull dimension.

For the given quotient, put $B=k[Y][X]/(X^2+YX+Y^3)$. The relation is monic in $X$, so monic polynomial division gives a unique representative $a(Y)+b(Y)X$. Thus $k[Y]\hookrightarrow B$ is injective and $B$ is free of rank two as a $k[Y]$-module. In particular, $B$ is integral over $k[Y]$. The one-variable polynomial ring has dimension one: its zero prime is strictly below $(Y)$, and every nonzero prime is maximal because $k[Y]$ is a principal ideal domain. HenceThis is an instance of dimension of a monic plane hypersurface. No irreducibility or algebraic-closure assumption on $k$ is needed; monicity supplies the integral extension in every characteristic.

The integral closure of $R$ in $T$ isIt is a subring: finitely many integral elements generate a finite $R$-module algebra, and the determinant trick shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.

A valuation ring is an integral domain $V$ such that, for every nonzero $z$ in its fraction field, either $z\in V$ or $z^{-1}\in V$. Equivalently, its ideals are totally ordered by inclusion. For principal ideals, comparability is precisely the condition on $a/b$; if two arbitrary ideals were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a ring is local. Its nonunits form an ideal: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique maximal ideal consists of those nonunits.

First, valuation rings are integrally closed. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its maximal ideal $\mathfrak{m}$. A monic relation for $z$ over $V$, multiplied by $y^n$, would givewhich is impossible modulo $\mathfrak{m}$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.

For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the valuation domination lemma: a local subring $(A,\mathfrak{n})$ of a field $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and ${\mathfrak{m}}_V\cap A=\mathfrak{n}$. Here is a proof, including the crucial maximality step.

Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the rings and of their maximal ideals. The union is a local ring: an element outside the union ideal is already a unit in a member of the chain, while an element in that ideal cannot become a unit in a later dominating member. The union still dominates $A$. Thus Zorn's lemma supplies a maximal pair $(V,\mathfrak{m})$.

For any $z\in K^{\times }$, at least one of $\mathfrak{m}V[z]$ and $\mathfrak{m}V[z^{-1}]$ is proper. Suppose otherwise. There would be relationswith $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\ge s$, the second relation givesRepeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak{m}$. This contradicts minimality of $n$ (or gives $1\in \mathfrak{m}$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.

Choose whichever extension has a proper extended ideal, then a maximal ideal containing it. Localizing that extension at the chosen maximal ideal produces a local ring dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its fraction field is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the valuation domination lemma.

Now let $x\in K$ be nonintegral over $R$, so $x\ne 0$, and set $y=x^{-1}$, $B=R[y]$. The ideal $yB$ is proper: otherwise $1=y(r_0+r_{1}y+\cdots +r_n{y}^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a maximal ideal $\mathfrak{n}$ of $B$ containing $y$, and apply the valuation domination lemma to $B_{\mathfrak{n}}\subset K$. Its dominating valuation ring $V$ contains $R$ and has $y\in {\mathfrak{m}}_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.

We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the integral closure as an intersection of valuation rings isNeither Noetherianity nor a discrete valuation is required for this separation argument.

Use the convention that a multiplicative subset contains $1$ and is closed under multiplication. Usually $0$ is excluded; allowing it gives the harmless zero-ring localization. A prime ideal disjoint from $S$ is mapped toin the localization of a ring. This ideal is proper, andis an integral domain, since the denominators are nonzero in the integral domain $R/P$. Thus $S^{-1}P$ is a prime ideal.

Conversely, for a prime ideal $Q$ in $S^{-1}R$, its contractionis a prime ideal disjoint from $S$: each $s/1$ is a unit and cannot belong to a proper ideal. Membership in $Q$ is equivalent to numerator membership in $P$, since $a/s=(a/1)(s/1{)}^{-1}$. Hence $Q=S^{-1}P$.

To check the other composite, $a/1\in S^{-1}P$ implies $sa\in P$ for some $s\in S$, by the equality criterion for fractions in a localization of a ring. Since $s\notin P$, primality gives $a\in P$. Therefore extension and contraction are mutually inverse:If $0\in S$, both sets are empty. This is the prime ideal correspondence for localization.

A prime ideal $P$ is a proper ideal such that $ab\in P$ implies $a\in P$ or $b\in P$. Equivalently $R/P$ is an integral domain.

For each $i$, and each $j\ne i$, choose $a_{ji}\in P_j\setminus P_i$, using incomparability. PutThen $b_i$ belongs to the other two prime ideals and does not belong to $P_i$, by primality. In particular each $b_i$ lies in the union. But $b_1+b_2+b_3\notin P_i$ for every $i$: modulo $P_i$, the sum equals $b_i\ne 0$. The union is not closed under addition and is not an ideal. This is a concrete instance of prime avoidance.

The union of three incomparable nonprime ideals example can retain pairwise incomparability. In the ringtakeThese distinct two-element ideals are pairwise incomparable, yetwhich is an ideal. None is prime: its quotient ring still contains a nonzero element of square zero. The three one-dimensional vector subspaces cover the two-dimensional vector space over ${\mathbb{F}}_2$.

A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.