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Every ring integral over a Jacobson ring is Jacobson. After quotienting by a prime, an integral equation for a nonzero element has nonzero constant term; choose a maximal ideal of the base avoiding that term and apply the Lying-over theorem.

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 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.

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.

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

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 $B$.

We use three theorems for an integral extension $S\subseteq R$, none requiring Noetherian hypotheses. The Lying-over theorem says that each prime ideal $\mathfrak{p}$ of $S$ is $\mathfrak{q}\cap S$ for some prime ideal $\mathfrak{q}$ of $R$. The going-up theorem says that if $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ in $S$ and $\mathfrak{q}\cap S=\mathfrak{p}$, there is ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap S={\mathfrak{p}}^{\prime }$. The incomparability theorem for integral extensions says that comparable prime ideals of $R$ with the same contraction to $S$ are equal.

Take any strict chainin $R$. The contraction of an ideal operation gives ${\mathfrak{p}}_j={\mathfrak{q}}_j\cap S$, which are prime ideals of $S$, by the proof in Question 1(b), and remain strictly increasing by the incomparability theorem for integral extensions. Thus every chain length in $R$ occurs in $S$, giving $\dim R\le \dim S$.

Conversely, take any strict chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n$ in $S$. Use the Lying-over theorem to choose ${\mathfrak{q}}_0$ over ${\mathfrak{p}}_0$, then repeatedly use the going-up theorem to obtain ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ with the prescribed contractions. Each inclusion must be strict, since its contractions are distinct. Hence every chain length in $S$ occurs in $R$, giving $\dim S\le \dim R$.

Taking suprema proves integral extensions preserve Krull dimension:The argument works equally well when the dimensions are infinite, since it compares all finite chain lengths.

An element $t\in T$ is an integral element over $R$ when it satisfies a monic equationThe extension $R\subseteq T$ is an integral extension when every $t\in T$ is integral over $R$.

Let $P\in \mathrm{Spec}R$ and localize both rings at $S=R\setminus P$. The extension $R_P\subseteq S^{-1}T$ remains integral. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}T$. The contraction of a maximal ideal under an integral extension is maximal, and the local ring $R_P$ has unique maximal ideal $P{R}_P$, so $\mathfrak{n}\cap R_P=P{R}_P$. Contracting $\mathfrak{n}$ back to $T$ produces $Q\in \mathrm{Spec}T$ with $Q\cap R=P$. This proves the Lying-over theorem and hence the surjectivity of

Suppose $Q_1\subseteq Q_2$ and both contract to $P$. Quotient by $Q_1$ and localize the resulting integral domain at the nonzero elements of $R/P$. The localized ring is an integral domain integral over the field $\mathrm{Frac}(R/P)$, and is therefore itself a field. The localization of $Q_2/Q_1$ must consequently be zero; since the localized ring is a domain, $Q_2/Q_1=0$. Thus the incomparability theorem for integral extensions gives

The Krull dimension of a ring is the supremum of the lengths of its strict chains of prime ideals. Incomparability makes the contraction of every strict prime chain in $T$ strict, so $\dim T\le \dim R$. Conversely, start over the bottom prime of any chain in $R$ using lying over and lift each subsequent inclusion using the going-up theorem. Hence

For the concrete surface, write $x,y,z$ for the residue classes of $X,Y,Z$ and putThenand characteristic zero allows division by three. ThereforeDivision by the monic quadratic shows that $T$ is free over $k[u,v]$ with basis $1,z$. In particular, $u,v$ are algebraically independent and the requested Noether normalization of the quadratic surface xy plus yz plus zx is

Form the finite-dimensional $k$-algebraBecause a finite extension is integral, the Lying-over theorem supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne 0$. Therefore $C$ has at least one maximal ideal.

As an Artinian ring, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak{n}$, the quotient $C/\mathfrak{n}$ is a finite field extension of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.

For a commutative ring $R$, the Jacobson radical isLet $A\subseteq B$ be integral. If $\mathfrak{n}$ is maximal in $B$, then its contraction $\mathfrak{n}\cap A$ is maximal in $A$. Therefore every $a\in J(A)$ belongs to every $\mathfrak{n}$, andConversely, the Lying-over theorem puts a maximal ideal $\mathfrak{n}$ of $B$ above every maximal ideal $\mathfrak{m}$ of $A$. Hence an element of $J(B)\cap A$ lies in every $\mathfrak{m}$, provingThis is the Jacobson radical under an integral extension formula.

Let $d=\mathrm{deg}f$ and let $f_d$ be the nonzero highest homogeneous part. Choose one coordinate, after a permutation, such thatis not the zero polynomial. A polynomial of degree at most $d$ in each variable cannot vanish on the entire grid $\{0,1,\dots ,d{\}}^{n-1}$, by induction on the number of variables. Hence there are $a_i\in \{0,1,\dots ,d\}$ such that

SetThis is given, up to the initial coordinate permutation, by an integer matrix $M$ with determinant $\pm 1$ andIn the inverse coordinates $t_i=y_i+a_i{y}_n$, the coefficient of $y_n^d$ in $f$ is the nonzero real number $f_d(a_1,\dots ,a_{n-1},1)$. Dividing by it makes the defining equation monic in $y_n$. Thus $A$ is integral over $\mathbb{R}[y_1,\dots ,y_{n-1}]$ by linear Noether normalization for a hypersurface. The Lying-over theorem now makessurjective.