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If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

A Noetherian ring that is an integrally closed domain is a unique factorization domain exactly when its divisor class group vanishes. Indeed, vanishing says every prime Weil divisor, equivalently every height-one prime ideal, is principal.

Let $A$ be an integral domain that is a Noetherian ring, and let $x\in A$ generate a prime ideal. If the localization of a ring $A_x$ is a unique factorization domain, then $A$ is a unique factorization domain. Indeed, integrality of $a/x^n$ over $A$, with $n$ minimal, forces $x\mid a$ and hence proves that $A$ is an integrally closed domain. The Nagata theorem for divisor class groups then says that $\mathrm{Cl}(A)$ is generated by the height-one primes containing $x$; the only one is the principal prime $(x)$, so $\mathrm{Cl}(A)=0$.

An integral locally Noetherian scheme is normal when every local ring is an integrally closed domain.

More generally a scheme is normal when every local ring is an integrally closed domain; the scheme need not be globally integral, for example a disjoint union of normal integral components.

An algebraic variety whose local rings are integrally closed domains. In particular an irreducible variety that is a normal variety has normal affine coordinate rings, enabling codimension-two extension of regular functions on a normal variety.

Let $A$ be an integrally closed domain with fraction field $K$, and let $L/K$ be finite. If $b\in L$ is an integral element over $A$, its monic minimal polynomial lies in $A[T]$: its roots are integral, so its coefficients are integral elements of $K$, hence belong to $A$. If that polynomial has degree $r$ and constant coefficient $a_0$, then $N_{L/K}(b)=((-1{)}^r{a}_0{)}^{[L:K(b)]}\in A$. This remains true for inseparable extensions. An infinite algebraic extension has no usual determinant field norm.

A Dedekind domain is a Noetherian integrally closed domain in which every nonzero prime ideal is maximal. The equivalent local characterization is that its localizations at nonzero prime ideals are discrete valuation rings. A field is harmless as the dimension-zero case.

Write $R$ for the domain, $F$ for its fraction field and $I$ for the nonzero fractional ideal. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $r{I}^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a fractional ideal; the Noetherian hypothesis also makes it finitely generated.

For a nonzero maximal prime ideal $\mathfrak{p}$, the discrete valuation ring $R_{\mathfrak{p}}$ makes $I_{\mathfrak{p}}$ principal, say $I_{\mathfrak{p}}=a_{\mathfrak{p}}{R}_{\mathfrak{p}}$. We need to justify that inversion commutes with this localization. One inclusion is immediate. Conversely, if $x{I}_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$, finite generation of $I$ supplies a common $s\notin \mathfrak{p}$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, provingIt follows that $(I{I}^{-1}{)}_{\mathfrak{p}}=R_{\mathfrak{p}}$ at every such prime ideal. Globally $I{I}^{-1}\subseteq R$. If this were a proper ideal, it would be contained in a maximal ideal and would remain proper after localizing there, a contradiction. In the field case every nonzero fractional ideal is the whole field. Hence in all casesThis proves that every nonzero fractional ideal of a Dedekind domain is an invertible fractional ideal, rather than presupposing its inverse exists.

The usual field norm $N_{L/K}$ is defined for a finite field extension. The printed hypothesis “algebraic” alone allows an infinite extension, for which this determinant norm is not defined. We therefore assume $[L:K]<\infty$. For an arbitrary algebraic extension the meaningful elementwise statement is instead $N_{K(b)/K}(b)\in A$, and we prove this as well.

First a unique factorization domain is an integrally closed domain. Indeed, write an integral fraction as $u/v$ with relatively prime numerator and denominator. A monic integral equation, after multiplication by $v^r$, shows that $v$ divides $u^r$. Unique factorization forces $v$ to be a unit.

Let $m_b(T)=T^r+a_{r-1}{T}^{r-1}+\cdots +a_0\in K[T]$ be the monic minimal polynomial of $b$. Because $b$ is an integral element over $A$, it satisfies a monic polynomial $f\in A[T]$, and $m_b$ divides $f$ in $K[T]$. Every root of $m_b$ in an algebraic closure is therefore integral over $A$. This assertion includes repeated roots when the extension is inseparable. The coefficients of $m_b$ are elementary symmetric expressions in these roots. The integral elements form a ring: finitely many integral elements generate a finite $A$-module algebra, and the determinant trick shows every element of that algebra is integral. Thus all $a_i$ are integral over $A$. Since $a_i\in K$ and $A$ is integrally closed, $a_i\in A$.

Multiplication by $b$ on $K(b)$, in the basis $1,b,\dots ,b^{r-1}$, has the companion matrix of $m_b$, whose determinant is $(-1{)}^r{a}_0$. If $e=[L:K(b)]$, take a $K(b)$-basis of $L$. Multiplication by $b$ preserves each summand in the resulting direct sum of $e$ copies of $K(b)$, so its determinant over $K$ isIn particular $N_{K(b)/K}(b)=(-1{)}^r{a}_0\in A$. The argument requires no separability; in fact it holds for every integrally closed domain, not just a unique factorization domain.

For the connection with dimension, let $B$ be the coordinate ring of an irreducible affine variety. By Noether normalization there is a finite integral inclusion $A=k[t_1,\dots ,t_d]\subset B$, with $d=\dim B$. Their fraction fields form a finite extension. For any $0\ne b\in B$, its minimal-polynomial constant term $a_0$ is nonzero and lies in $A$. Its equation givesThus a nonzero function on the normalized variety produces a nonzero function on the base that vanishes wherever $b$ vanishes. Up to sign, the field norm is a power of this $a_0$, so it too belongs to $bB\cap A$.

Under the finite morphism to ${\mathbb{A}}^d$, the image of $V_B(b)$ is therefore contained in the proper Zariski-closed set $V_A(a_0)$ of dimension at most $d-1$. Finite morphisms preserve the dimension of a closed subvariety and its image. Applying this to a nonzero element of the ideal of any proper closed subvariety proves proper closed subvarieties have strictly smaller dimension. This is the useful role of the norm argument: nonzero equations upstairs give nonzero equations downstairs under a finite normalization.

Assume $K\subseteq L^p$. Every $a\in A$ has a unique $p$th root $r(a)$ in $L$: existence is the field inclusion and uniqueness follows from injectivity of the Frobenius endomorphism in a field. Since $r(a{)}^p=a$, this root satisfies the monic polynomial $T^p-a$ over $B$. Because $B$ is an integrally closed domain, this gives $r(a)\in B$. The identitiesand uniqueness show thatis a ring homomorphism.

On $D(a)$ the rule sends a fraction to the fraction of its unique roots. Since $r(a{)}^p=a$, inverting $a$ is equivalent to inverting $r(a)$, so the rule localizes. Also, for a prime $\mathfrak{q}\subset B$, $r(a)\in \mathfrak{q}$ exactly when $a\in \mathfrak{q}$; hence the induced point map agrees with that of $f$. The localized rules consequently glue to a morphism $h:X\to Y$. Its pullback followed by the Frobenius endomorphism sends $a$ to $r(a{)}^p=a$, which provesIf another morphism satisfies this equation, it has the same point map because the Absolute Frobenius morphism fixes points, and its pullback of $a$ must be a $p$th root of $a$ in the normal domain $B$. That root is unique. Thus $h$ is unique.

Assume $L^p\subseteq K$. For $b\in B$, its power $b^p$ belongs to $K$ and is integral over $A$, because $B$ is finite over $A$. Because $A$ is an integrally closed domain, $b^p\in A$. Thusis a ring homomorphism, defining a morphism $g_V:V\to f^{-1}V$. Its composition with $f:f^{-1}V\to V$ is the Absolute Frobenius morphism of $f^{-1}V$, because the composite ring map sends $b$ to $b^p$ in $B$.

These local morphisms glue. Explicitly, over a distinguished open $D(a)\subseteq V$, the ring rule isso it is exactly the localization of ${\gamma }_V$. For the underlying topology, each fibre of $f$ consists of a single point: if primes ${\mathfrak{q}}_1,{\mathfrak{q}}_2\subset B$ contract to the same $\mathfrak{p}$, thenso ${\mathfrak{q}}_1={\mathfrak{q}}_2$. The map $g_V$ sends that point of $V$ back to its unique point above it. Thus localization compatibility on a basis of a topology also gives compatibility on overlaps of arbitrary affine charts. We obtain

For uniqueness, $f$ is surjective on points, so the equation determines the point map of $g$. On each affine chart it then requires $f^{\ast }(g^{\ast }(b))=b^p$. The inclusion $A\hookrightarrow B$ is injective, so this determines $g^{\ast }(b)$ uniquely as $b^p$. Hence the sheaf map and the morphism are unique.

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 $0$.

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

Work on affine open subsets. Their coordinate rings are Noetherian rings and integrally closed domains. At a prime ideal of height of a prime ideal one, the local ring is a one-dimensional Noetherian local ring which is an integrally closed domain, hence a discrete valuation ring and therefore a regular local ring. At the generic point the local ring is a field, hence a regular local ring. For a brief justification of the one-dimensional local ring fact, take $0\ne a\in \mathfrak{m}$ in a one-dimensional Noetherian local ring $R$ which is an integrally closed domain. Some ${\mathfrak{m}}^n\subset (a)$; choose the least such $n$, and choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$. Then $q=b/a\notin R$, but $q\mathfrak{m}\subset R$. If $q\mathfrak{m}\subset \mathfrak{m}$, the determinant trick applied to the finitely generated faithful module $\mathfrak{m}$ would make $q$ integral over $R$, a contradiction. Hence the ideal $q\mathfrak{m}$ contains a unit and equals $R$. Thus $\mathfrak{m}=q^{-1}R$ is principal. A one-dimensional Noetherian local ring with principal maximal ideal is a discrete valuation ring. Therefore a normal variety is regular in codimension one.

The ground field is an algebraically closed field, hence a perfect field, so having a regular local ring is equivalent to being a smooth point of a variety here. Moreover, the smooth locus of a variety is open, making the singular locus closed. If an irreducible component of the singular locus had algebraic codimension zero or one, its generic point would have a regular local ring by the preceding argument, a contradiction. Consequently every such component has algebraic codimension at least two.

One can read the numerical bound directly from chains of prime ideals, without identifying algebraic dimension with transcendence degree. For a prime ideal $P$ of height of a prime ideal at least two in an affine chart, append a chain $(0)⊊P_1⊊P$ below any chain in the quotient by $P$. Its length increases by two. Thus $\dim A/P\le \dim A-2$. Since a nonempty affine open subset of the irreducible variety $X$ has algebraic dimension $d$,For $d=0$ or $1$, this means the singular locus is empty; take $\dim \varnothing =-\infty$.

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 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 $B$.