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If $f\in k[Y][X]$ is monic of positive degree in $X$, monic division makes $k[Y][X]/(f)$ free of rank ${\mathrm{deg}}_{X}f$ over $k[Y]$. The inclusion of $k[Y]$ is injective and the quotient is an integral extension, so its Krull dimension is one. This argument applies even when the polynomial factors or the field has positive characteristic.

Going up and going down concern lifting inclusions of prime ideals through a ring extension. Going-up theorem lifts upward from a chosen lower prime for an integral extension. Going-down theorem lifts downward from a chosen upper prime under additional hypotheses, including an integrally closed base domain for integral extensions of domains.

If $A\subseteq B$ is an integral extension, then $\dim A=\dim B$. The incomparability theorem for integral extensions contracts every strict prime chain in $B$ to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, the Lying-over theorem starts over the bottom of any prime chain in $A$, and repeated use of the going-up theorem lifts the whole chain to $B$.

For an integral extension $A\subseteq B$ of commutative rings,Contraction sends maximal ideals of $B$ to maximal ideals of $A$, while the Lying-over theorem puts a maximal ideal of $B$ over every maximal ideal of $A$.

For any algebraic field extension $K/\mathbb{Q}$, including an infinite one, its ring ${\mathcal{O}}_K$ of elements integral over $\mathbb{Z}$ is an integral extension of $\mathbb{Z}$. The Lying-over theorem, going-up theorem and incomparability theorem for integral extensions give $\dim {\mathcal{O}}_K=\dim \mathbb{Z}=1$. The argument does not require that ${\mathcal{O}}_K$ be Noetherian or finitely generated over $\mathbb{Z}$.

An integral extension $A\subset B$ need not give an integral extension $A_{P\cap A}\subset B_P$. For example, take $\mathbb{Z}\subset \mathbb{Z}[i]$ and $P=(2+i)$, whose contraction is $(5)$. The element $1/(2-i)$ belongs to $\mathbb{Z}[i{]}_P$ but is not integral over ${\mathbb{Z}}_{(5)}$: multiplying a hypothetical monic equation by a power of $2-i$ and reducing modulo the other Gaussian prime ideal $(2-i)$ gives $1=0$ in ${\mathbb{F}}_5$. Localizing both rings by the same multiplicative subset does preserve integrality.

If $A\subseteq B$ is an integral extension, then every prime ideal $\mathfrak{p}$ of $A$ is the contraction of a prime ideal $\mathfrak{q}$ of $B$. Thus the contraction map $\mathrm{Spec}B\to \mathrm{Spec}A$ is surjective.

Write $X=\mathrm{Spec}B$, $S=\mathrm{Spec}A$ with $A,B$ integral domains, let $\alpha :A\to B$ be the ring homomorphism, and put $K=\mathrm{Frac}(B)$. We do not assume that $\alpha$ is injective. Write $A^{\prime }=\alpha (A)\subseteq B\subseteq K$.

Let $D\subseteq K$ be any valuation ring containing $A^{\prime }$, with fraction field $K$. The canonical morphism of schemes $\mathrm{Spec}K\to \mathrm{Spec}B$ and the ring homomorphism $A\to A^{\prime }\hookrightarrow D$ form a valuative problem for $f$. Since $f$ is a proper morphism, there is an extension. On coordinate rings it is a ring homomorphism $\beta :B\to D$ such thatis the original inclusion of $B$ into $K$. Therefore every element of $B$ lies in $D$. Since $D$ was arbitrary, the stated description of integral closure as an intersection of valuation rings givesThus every element of $B$ is an integral element over $A^{\prime }$, and hence over $A$: lift the coefficients of its monic polynomial along the surjection $A\to A^{\prime }$. The ring homomorphism $\alpha$ is therefore an integral extension in the sense of the question.

A proper morphism is of finite type, so in this affine situation $B=A[b_1,\dots ,b_s]$. If $b_i$ satisfies a monic polynomial of degree $n_i$, repeatedly reducing its powers shows that the finitely many monomialsgenerate $B$ as an $A$-module. Equivalently, apply the allowed equivalence between finite and integral plus finite type. Hence

The spectrum of a ring $A$ is the set of its proper prime ideals. For an ideal $I$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. These are the closed sets of the Zariski topology. They satisfyThe union identity follows because a prime ideal containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. These identities verify the closed-set axioms. The maximal spectrum has the subspace topology, whose closed sets are $V(I)\cap \mathrm{MaxSpec}A$.

For a finitely generated $\mathbb{C}$-algebra, the needed form of the Hilbert Nullstellensatz isHere an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.

First establish the residue-field fact. If a field $L$ is a finitely generated algebra over $\mathbb{C}$, choose a maximal algebraically independent subset $t_1,\dots ,t_r$ of a finite generating set, with remaining generators $u_1,\dots ,u_s$. Each $u_j$ is algebraic over $\mathbb{C}(t_1,\dots ,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral overSince $L$ is a field, it contains $1/d$, and $L=R[u_1,\dots ,u_s]$ is an integral extension of $R$. If a field is integral over a domain, the domain is itself a field: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.

But if $r>0$, choose $c\in \mathbb{C}$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a field, a contradiction. Hence $r=0$, and $L$ is algebraic over the algebraically closed field $\mathbb{C}$, forcing $L=\mathbb{C}$. This proves the Zariski lemma in the form used here.

Now if $f\notin \sqrt{I}$, the localization of a ring $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a finitely generated algebra over $\mathbb{C}$, since one need only adjoin an inverse of $f$. Choose a maximal ideal $\mathfrak{n}$ of $B$. By the preceding argument, $B/\mathfrak{n}=\mathbb{C}$. The composite $A\to B\to \mathbb{C}$ is surjective because it contains the constant scalars. Its kernel is consequently a maximal ideal $\mathfrak{m}$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt{I}$. The reverse inclusion follows because every maximal ideal is prime and contains the radical of an ideal whenever it contains that ideal.

For a closed subset $Z\subseteq \mathrm{MaxSpec}A$, put $I(Z)={\bigcap }_{\mathfrak{m}\in Z}\mathfrak{m}$. It is a radical ideal. The formulas now showare inverse maps on radical ideals and closed sets. Indeed the first followed by the second gives $\sqrt{I}=I$, while if $Z=V(J)\cap \mathrm{MaxSpec}A$, the second gives $\sqrt{J}$ and $V(\sqrt{J})=V(J)$. Thus the correspondence is a bijection reversing inclusion. This is the relation between radical ideals and closed subsets of the maximal spectrum.

For $A=\mathbb{C}[x]$, the maximal ideals are ${\mathfrak{m}}_c=(x-c)$, for $c\in \mathbb{C}$. Every nonzero proper radical ideal is generated by a squarefree polynomial, and hence has the formfor a finite nonempty subset $S\subset \mathbb{C}$. The associated closed set is $\{{\mathfrak{m}}_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its radical ideal removes that repeated factor. The two endpoint cases areThe empty-product convention makes $I_{\varnothing }=(1)$. Under $c\mapsto {\mathfrak{m}}_c$, the closed sets are exactly the finite subsets and the whole affine line.

An extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$, meaning thatfor some monic polynomial. The Krull dimension of a ring is the supremum of the lengths $r$ of strict chains ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_r$ of prime ideals; the number counted is the number of inclusions.

We prove the prime-chain facts needed for integral extensions preserve Krull dimension. Taking quotients preserves an integral equation. Localizing both rings by the same multiplicative subset of $A$ also preserves integrality: for $b/s$, divide the equation for $b$ by $s^n$, so its new coefficients are $a_i/s^{n-i}$. Localizing just $B$ at all elements outside one of its primes would not be the same operation.

Two elementary field observations are useful. If a field $C$ is integral over a subdomain $R$, then $R$ is a field, by the inverse argument given in Question 2. Conversely, if an integral domain $C$ is integral over a field $R$, every nonzero $c\in C$ has a monic equation with nonzero constant term: cancel a factor of $c$ from an equation if necessary. Dividing by $c$ and by that constant term expresses $c^{-1}$ as a polynomial in $c$. Thus $C$ is a field. In particular, the contraction of a maximal ideal under an integral extension is maximal.

For the Lying-over theorem, fix $\mathfrak{p}\in \mathrm{Spec}A$ and localize by $S=A\setminus \mathfrak{p}$. The nonzero domain $S^{-1}B$ is integral over the local ring $A_{\mathfrak{p}}$. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}B$. Its contraction is maximal in $A_{\mathfrak{p}}$, so it is the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. The prime ideal correspondence for localization now pulls $\mathfrak{n}$ back to a prime ideal $\mathfrak{q}$ of $B$ with $\mathfrak{q}\cap A=\mathfrak{p}$.

For the going-up theorem, suppose $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ and $\mathfrak{q}\cap A=\mathfrak{p}$. The injection $A/\mathfrak{p}\hookrightarrow B/\mathfrak{q}$ is still integral. Apply the just-proved Lying-over theorem to ${\mathfrak{p}}^{\prime }/\mathfrak{p}$. Pulling the resulting prime back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap A={\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ have the same contraction $\mathfrak{p}$. Write $R=A/\mathfrak{p}$ and $C=B/\mathfrak{q}$. The latter is a domain integral over $R$. Localizing both by $R\setminus \{0\}$ makes $S^{-1}C$ integral over the fraction field of $R$, so $S^{-1}C$ is a field. The prime ${\mathfrak{q}}^{\prime }/\mathfrak{q}$ is disjoint from $S$ and therefore localizes to a proper prime, which must be zero in this field. Since $C$ is a domain and all denominators are nonzero, an element of $C$ whose localization is zero was already zero. Hence ${\mathfrak{q}}^{\prime }/\mathfrak{q}=0$, proving ${\mathfrak{q}}^{\prime }=\mathfrak{q}$.

A strict prime chain in $B$ therefore contracts to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, start over the bottom of any finite prime chain in $A$ using Lying-over theorem, and successively use going-up theorem to lift the entire chain. Distinct contractions ensure that the lifted inclusions are strict, giving the reverse inequality. Taking suprema, including when arbitrarily long chains exist, proves

For an algebraic extension $K/\mathbb{Q}$, its ring $\mathcal{O}$ of algebraic integers is an integral extension of $\mathbb{Z}$. The primes of $\mathbb{Z}$ are $(0)$ and the maximal ideals $(p)$, so the longest strict chains have length one; for example $(0)⊊(2)$. The preceding argument yieldsThis Krull dimension of rings of algebraic integers argument also covers infinite algebraic extensions. It uses integrality, not an assumption that $K$ is a number field or that $\mathcal{O}$ is Noetherian.

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

Put $S=R^G$, the ring of invariants. It is an integral domain because it is a subring of the domain $R$. Embed $\mathrm{Frac}S$ in $\mathrm{Frac}R$ in the evident way.

Suppose $q\in \mathrm{Frac}S$ is an integral element over $S$. Its monic equation has coefficients in $S\subseteq R$, so $q$ is integral over $R$. Since $R$ is a normal domain, $q\in R$.

Write $q=a/b$ with $a,b\in S$ and $b\ne 0$. Each ring automorphism in $G$ extends to the fraction field by acting on numerator and denominator, and fixes this fraction. Hence $q\in R^G=S$. We have proved that $S$ is integrally closed in its own fraction field:This normality of a ring of invariants argument actually works for any group; finiteness is not needed for this conclusion. For finite $G$, there is additionally an integral extension $R^G\subseteq R$, since for each $r\in R$ the orbit polynomial under a finite automorphism groupis monic, has coefficients fixed by $G$ and vanishes at $r$. Neither argument divides by $|G|$, so it is valid when the characteristic divides the group order. Here normality means integral closedness of a domain.

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.

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 key is a Laurent normalization by a monomial change of variables. We prove the slightly stronger assertion that every nonzero quotient of a Laurent polynomial ring in $r$ variables is finite as a module over an embedded Laurent polynomial ring in some number $m\le r$ of variables.

Induct on $r$. If the defining ideal is zero, the ring itself works. Otherwise choose a nonzero Laurent polynomialin that ideal, with nonzero coefficients. For a sufficiently large positive integer $N$, the weightsare distinct for all $\mathbf{a}\in F$. Indeed, each difference gives a nonzero polynomial in $N$, which has only finitely many integer roots. Make the invertible monomial substitutionIts inverse is $Y_j=X_j{X}_r^{-N^j}$, so it is an automorphism of the Laurent polynomial ring. When $r=1$ there is no substitution and distinct exponents already give distinct weights.

As a Laurent polynomial in $Y_r$, the transformed relation has unique lowest and highest exponents. Their coefficients are nonzero scalar multiples of monomials in $Y_1,\dots ,Y_{r-1}$, hence units inMultiplying by a power of $Y_r$ and the inverse leading coefficient givesHere $d>0$: a single-term Laurent polynomial would be a unit and the quotient would be zero. Reversing the equation and dividing by $b_0{Y}_r^d$ gives a monic equation for $Y_r^{-1}$. Thus both $Y_r$ and $Y_r^{-1}$ are integral over the image $B$ of $B_0$ in $T$. In fact $b_0$ being a unit also expresses $Y_r^{-1}$ as a polynomial in $Y_r$, so $T$ is generated as a $B$-module by $1,Y_r,\dots ,Y_r^{d-1}$.

The ring $B$ is itself a nonzero quotient of an $(r-1)$-variable Laurent polynomial ring. By induction, it is finite over an embedded Laurent polynomial ring in $m\le r-1$ variables. Finiteness of modules is transitive, so the same is true of $T$; a finite ring extension is an integral extension, by the determinant trick applied to multiplication by an element.

Applying this to $r=3$ yields an embedded subringThe base case $m=0$ means the subring $k$; finiteness over it would make $T$ a finite-dimensional vector space. The hypothesis excludes this, so $1\le m\le 3$, as required. No infinitude assumption on $k$ is used.

An integral extension $R\subseteq T$ means every $t\in T$ satisfies a monic polynomialTo prove the Lying-over theorem, localize at $S=R\setminus P$. The injection $R_P\hookrightarrow S^{-1}T$ remains injective, and $S^{-1}T$ is integral over $R_P$: a monic relation for $t$ divided by suitable powers of $s$ gives one for $t/s$. This is a nonzero ring, so choose a maximal ideal $M$ of it.

The contraction of $M$ to $R_P$ is maximal. To see this directly, a field integral over a subring forces that subring to be a field. For nonzero $a$ in the subring, $a^{-1}$ is integral; multiply its monic equation of degree $d$ by $a^{d-1}$ to express $a^{-1}$ as a polynomial in $a$ with coefficients in the subring. Apply this to the integral extensionSince $R_P$ is a local ring with unique maximal ideal $P{R}_P$, we obtain $M\cap R_P=P{R}_P$.

Contract $M$ to $T$, obtaining a prime ideal $Q$. Its contraction to $R$ is the contraction of $P{R}_P$, namely $P$. Thus

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.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

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.