Integral Extensions and Going Up

If $A=0$, then $1_A=0_A$, so every unital image of $A$ is the zero ring; hence $B=0$ and then $C=0$, making the composite integral trivially. For the rest of the proof assume $A\mathrm{\ne }0$.

Let $c\in C$. By [L1], the element $c$ is integral over $B$, so there is a monic equation $c^n+b_{n-1}{c}^{n-1}+\cdots +b_0=0$ with $b_i\in B$. Again by [L1], each coefficient $b_i$ is integral over $A$, so [L2] makes the $A$-subalgebra $D:=A[b_0,\dots ,b_{n-1}]$ module-finite over $A$.

The same equation for $c$ has coefficients in $D$, so $c$ is integral over $D$. Because $D$ is an $A$-subalgebra of $B$, it contains the image of $1_A=1_B$, so it is nonzero. Therefore [L4] makes $D[c]$ a finitely generated $D$-module, and then [L3] gives that $D[c]$ is a finitely generated $A$-module.

The ring $D[c]$ is a faithful module over the subring $A[c]$, and step 2.1 shows that this faithful $A[c]$-module is finitely generated over $A$. By [L4], $c$ is integral over $A$. Since $c\in C$ was arbitrary, the map $A\to C$ is integral.

Because $A$ is a domain, it is nonzero, so [L2] applies to the inclusion $A\subseteq K$. Therefore [L1] implies that $\bar{A}$ is a subring of the field $K$ containing $A$, and [A1] makes $\bar{A}$ a domain.

Let $x\in \mathrm{Frac}(\bar{A})$ be integral over $\bar{A}$. By [A1] we may regard $x$ as an element of $K$. Then [L3] shows that $x$ is integral over $A$.

Since $\bar{A}$ is, by [L1], exactly the set of elements of $K$ integral over $A$, step 1.2 gives $x\in \bar{A}$. Thus every element of $\mathrm{Frac}(\bar{A})$ integral over $\bar{A}$ already lies in $\bar{A}$, so $\bar{A}$ is integrally closed.

If $b$ satisfies $b^n+a_{n-1}{b}^{n-1}+\cdots +a_0=0$ with $a_i\in A$, then the same identity in $S^{-1}B$ reads $(b\mathrm{/}1{)}^n+(a_{n-1}\mathrm{/}1)(b\mathrm{/}1{)}^{n-1}+\cdots +a_0\mathrm{/}1=0$. By [L1], this makes $b\mathrm{/}1$ integral over $S^{-1}A$.

Conversely, assume $b\mathrm{/}1$ is integral over $S^{-1}A$. Choose a monic relation $(b\mathrm{/}1{)}^n+(a_{n-1}\mathrm{/}{s}_{n-1})(b\mathrm{/}1{)}^{n-1}+\cdots +a_0\mathrm{/}{s}_0=0$ with $a_i\in A$ and $s_i\in S$. Let $t:=s_0\cdots s_{n-1}$ and put $y:=tb$. Multiplying the relation by $t^n$ gives $(y\mathrm{/}1{)}^n+c_{n-1}(y\mathrm{/}1{)}^{n-1}+\cdots +c_0\mathrm{/}1=0$ with each $c_i\in A$. Hence [L4] gives some $u\in S$ with $u(y^n+c_{n-1}{y}^{n-1}+\cdots +c_0)=0$ in $B$. For $s:=ut$ and $z:=sb=uy$, multiplying that equation by $u^{n-1}$ yields $z^n+c_{n-1}u{z}^{n-1}+\cdots +c_0{u}^n=0$, a monic equation over $A$. So $sb$ is integral over $A$.

Now assume $A$ is a domain, $S\subseteq A\setminus \{0\}$, $K\mathrm{/}\mathrm{Frac}(A)$ is a field extension, and $\bar{A}$ is the integral closure of $A$ in $K$. If $x=a\mathrm{/}s$ with $a\in \bar{A}$ and $s\in S$, then $a$ is integral over $A$, so step 1.1 makes $x$ integral over $S^{-1}A$.

Conversely, let $x\in K$ be integral over $S^{-1}A$. Step 1.2 gives $s\in S$ with $sx$ integral over $A$, so [L2] gives $sx\in \bar{A}$. Therefore $x=(sx)\mathrm{/}s\in S^{-1}\bar{A}$. Combining this with step 2.1 proves that the integral closure of $S^{-1}A$ in $K$ is exactly $S^{-1}\bar{A}$.

Assume $A$ is a field, and let $0\mathrm{\ne }b\in B$. By [L1] there is a monic relation $b^n+a_{n-1}{b}^{n-1}+\cdots +a_0=0$ over $A$ of minimal degree. The constant term cannot vanish: if $a_0=0$, then $b(b^{n-1}+a_{n-1}{b}^{n-2}+\cdots +a_1)=0$, and [L2] with $b\mathrm{\ne }0$ would give a smaller monic relation, contradicting minimality. Since $a_0\mathrm{\ne }0$ and $A$ is a field, $a_0^{-1}\in A$, and rearranging gives $b^{-1}=-a_0^{-1}(b^{n-1}+a_{n-1}{b}^{n-2}+\cdots +a_1)\in B$. Thus every nonzero element of $B$ is invertible.

Assume $B$ is a field, and let $0\mathrm{\ne }a\in A$. Then $a^{-1}\in B$, and [L1] gives a monic equation $(a^{-1}{)}^n+c_{n-1}(a^{-1}{)}^{n-1}+\cdots +c_0=0$ with $c_i\in A$. Multiplying by $a^n$ yields $1+c_{n-1}a+\cdots +c_0{a}^n=0$, so $a(-c_{n-1}-c_{n-2}a-\cdots -c_0{a}^{n-1})=1$. Hence $a$ is invertible in $A$.

Step 1.1 proves that $A$ field implies $B$ field, and step 1.2 proves the converse. Therefore $A$ is a field if and only if $B$ is a field.

Because $\mathfrak{q}$ is prime, [L2] makes $B\mathrm{/}\mathfrak{q}$ a domain. The map $A\mathrm{/}\mathfrak{p}\to B\mathrm{/}\mathfrak{q}$ is injective by definition of $\mathfrak{p}$, so $A\mathrm{/}\mathfrak{p}$ is a subring of a domain and is therefore a domain. Thus [L2] also shows that $\mathfrak{p}$ is prime.

By [A1], $A\mathrm{/}\mathfrak{p}\subseteq B\mathrm{/}\mathfrak{q}$ is an integral extension of domains. Therefore [L1] says that $A\mathrm{/}\mathfrak{p}$ is a field if and only if $B\mathrm{/}\mathfrak{q}$ is a field. Using [L3] on both quotients, this is exactly the statement that $\mathfrak{p}$ is maximal if and only if $\mathfrak{q}$ is maximal.

Let $\bar{A}:=A\mathrm{/}\ker f$, let $\pi :A\to \bar{A}$ be the quotient map, and let $\bar{\mathfrak{p}}:=\mathfrak{p}\mathrm{/}\ker f$. By [L6], prime ideals of $\bar{A}$ correspond to prime ideals of $A$ containing $\ker f$, so $\bar{\mathfrak{p}}$ is prime. The map $f$ factors through an injective integral map $\bar{f}:\bar{A}\hookrightarrow B$, so it is enough to find a prime of $B$ contracting to $\bar{\mathfrak{p}}$ and then pull it back through [L6].

Replace $A$ by $\bar{A}$ and write again $\mathfrak{p}$ for the chosen prime. Set $S:=A\setminus \mathfrak{p}$. By [L5], $A\mathrm{/}\mathfrak{p}$ is a domain, so $S$ contains no zero element of $A$; since $A\hookrightarrow B$ is injective, $0\mathrm{\notin }S$ inside $B$ as well. Hence [L9] shows that $S^{-1}B$ is nonzero. By [L2], the map $S^{-1}A\to S^{-1}B$ is integral.

By [L7], prime ideals of $A_{\mathfrak{p}}=S^{-1}A$ correspond to prime ideals of $A$ contained in $\mathfrak{p}$. Therefore every prime of $A_{\mathfrak{p}}$ lies inside $\mathfrak{p}{A}_{\mathfrak{p}}$, so $\mathfrak{p}{A}_{\mathfrak{p}}$ is the unique maximal ideal of $A_{\mathfrak{p}}$. By [L8], the nonzero ring $S^{-1}B$ has a maximal ideal $\mathfrak{n}$. Then [L3] implies that $\mathfrak{n}\cap A_{\mathfrak{p}}=\mathfrak{p}{A}_{\mathfrak{p}}$.

By [L7], the prime ideal $\mathfrak{n}$ of $S^{-1}B$ is $S^{-1}\mathfrak{q}$ for a unique prime ideal $\mathfrak{q}$ of $B$ disjoint from $S$, and its contraction to $A$ is the contraction of $\mathfrak{p}{A}_{\mathfrak{p}}$, namely $\mathfrak{p}$. Therefore $\mathfrak{q}$ lies over $\mathfrak{p}$. Pulling back through step 1.1 gives a prime of $B$ lying over the original prime of $A$.

Let $S:=A\setminus \mathfrak{p}$. By [L3], the primes ${\mathfrak{q}}_1$ and ${\mathfrak{q}}_2$ correspond to primes $S^{-1}{\mathfrak{q}}_1\subseteq S^{-1}{\mathfrak{q}}_2$ of $S^{-1}B$, and by [L1] the localized map $A_{\mathfrak{p}}\to S^{-1}B$ remains integral.

By [L2], the contraction of each $S^{-1}{\mathfrak{q}}_i$ to $A_{\mathfrak{p}}$ is the maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. Therefore [L4] makes both $S^{-1}{\mathfrak{q}}_1$ and $S^{-1}{\mathfrak{q}}_2$ maximal ideals of $S^{-1}B$. Since one is contained in the other, they are equal.

Applying the inverse bijection of [L3] to the equality of step 2.1 gives ${\mathfrak{q}}_1={\mathfrak{q}}_2$.

The map $f$ induces a ring map $\bar{f}:A\mathrm{/}{\mathfrak{p}}_1\to B\mathrm{/}{\mathfrak{q}}_1$, and this induced map is integral because a monic equation for $b\in B$ over $A$ descends to the same monic equation for $b+{\mathfrak{q}}_1$ over $A\mathrm{/}{\mathfrak{p}}_1$. By [L3], the prime ${\mathfrak{p}}_2$ corresponds to the prime ${\mathfrak{p}}_2\mathrm{/}{\mathfrak{p}}_1$ of $A\mathrm{/}{\mathfrak{p}}_1$.

Apply [L2] to $\bar{f}$ and the prime ${\mathfrak{p}}_2\mathrm{/}{\mathfrak{p}}_1$. This yields a prime ${\bar{\mathfrak{q}}}_2$ of $B\mathrm{/}{\mathfrak{q}}_1$ with contraction ${\mathfrak{p}}_2\mathrm{/}{\mathfrak{p}}_1$.

By [L3], the prime ${\bar{\mathfrak{q}}}_2$ corresponds to a prime ideal ${\mathfrak{q}}_2$ of $B$ containing ${\mathfrak{q}}_1$. Its contraction to $A$ is exactly ${\mathfrak{p}}_2$. Therefore ${\mathfrak{q}}_2$ is the required prime above ${\mathfrak{p}}_2$.

First establish finite-module lying over. Fix any prime $\mathfrak{p}$ of $B$, put $S=B\setminus \mathfrak{p}$, $R=B_{\mathfrak{p}}$, $M=S^{-1}A$, and $\mathfrak{m}=\mathfrak{p}R$. The module $M$ is finite over $R$; it is a nonzero ring because $B\subseteq A$ are domains and no element of $S$ annihilates $1\in A$. If $\mathfrak{m}M=M$, [L1] gives $a\in \mathfrak{m}$ with $(1-a)M=0$. Write $a=b\mathrm{/}s$ with $b\in \mathfrak{p}$ and $s\mathrm{\notin }\mathfrak{p}$. Then $1-a=(s-b)\mathrm{/}s$ is an explicit unit of $R$, since $s-b\mathrm{\notin }\mathfrak{p}$. It would force $M=0$, a contradiction. Thus $M\mathrm{/}\mathfrak{m}M\mathrm{\ne }0$.

The quotient $C=M\mathrm{/}\mathfrak{m}M$ is a nonzero finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})=R\mathrm{/}\mathfrak{m}$. Among its proper ideals, choose one $J$ of maximal $\kappa (\mathfrak{p})$-vector-space dimension; this is possible because the zero ideal is proper and the dimensions form a nonempty subset of the finite set $\{0,\dots ,{\dim }_{\kappa (\mathfrak{p})}C-1\}$. A strictly larger proper ideal would have larger vector-space dimension, so $J$ is a maximal, hence prime, ideal of $C$. The field map $\kappa (\mathfrak{p})\to C$ is injective because $C\mathrm{\ne }0$, and therefore the preimage of $J$ in $M$ contracts to $\mathfrak{m}$ in $R$. Pulling it back along the localization $A\to M$ gives a prime of $A$ contracting to $\mathfrak{p}$ in $B$.

Apply steps 1.1–2.1 to $B\subseteq A$ at ${\mathfrak{p}}_0$ to obtain ${\mathfrak{q}}_0$ above ${\mathfrak{p}}_0$. Suppose ${\mathfrak{q}}_i$ has been selected above ${\mathfrak{p}}_i$. The induced inclusion $B\mathrm{/}{\mathfrak{p}}_i\subseteq A\mathrm{/}{\mathfrak{q}}_i$ is again an injective module-finite extension of domains. Apply the finite-module lying-over construction to the prime ${\mathfrak{p}}_{i+1}\mathrm{/}{\mathfrak{p}}_i$ in the lower quotient. By [L2], its prime above pulls back to a prime ${\mathfrak{q}}_{i+1}\supseteq {\mathfrak{q}}_i$ of $A$ contracting to ${\mathfrak{p}}_{i+1}$. The inclusion is strict because its contractions are strict.

Repeating step 3.1 for the supplied finite number $r$ of links produces ${\mathfrak{q}}_0⊊\cdots ⊊{\mathfrak{q}}_r$ with the required contractions. Only finite choices of witnesses occurred: the finite-dimensional ideal in step 2.1 is selected from a bounded set of integer dimensions, and the chain has a supplied finite length.

For $n=0$, the given prime ${\mathfrak{q}}_0$ already lifts the chain.

Fix $n\ge 0$ and assume the statement for chains of length $n$. Let ${\mathfrak{p}}_0\subseteq \cdots \subseteq {\mathfrak{p}}_n\subseteq {\mathfrak{p}}_{n+1}$ be a chain of length $n+1$. By the induction hypothesis, there are primes ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ over ${\mathfrak{p}}_0,\dots ,{\mathfrak{p}}_n$.

Apply [L1] to the inclusion ${\mathfrak{p}}_n\subseteq {\mathfrak{p}}_{n+1}$ and the prime ${\mathfrak{q}}_n$. This yields a prime ${\mathfrak{q}}_{n+1}\supseteq {\mathfrak{q}}_n$ with contraction ${\mathfrak{p}}_{n+1}$.

Step 1.1 is the base case, and steps 1.2 and 2.1 provide the induction step. Therefore every finite prime chain in $A$ lifts to one in $B$ once the first prime upstairs is fixed.

Contraction is inclusion-preserving, so $f^{-1}({\mathfrak{q}}_0)\subseteq \cdots \subseteq f^{-1}({\mathfrak{q}}_n)$.

Suppose two adjacent contractions were equal: $f^{-1}({\mathfrak{q}}_i)=f^{-1}({\mathfrak{q}}_{i+1})$ for some $i<n$. Then the comparable primes ${\mathfrak{q}}_i⊊{\mathfrak{q}}_{i+1}$ would have the same contraction, contradicting [L1]. Therefore every adjacent contraction is strict.

Since every adjacent inclusion is strict, the whole contracted chain is strict.

Both dimensions in the statement are defined by [L1] because $A$ and $B$ are nonzero.

Let ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n$ be any strict prime chain in $A$. By [L2], choose a prime ${\mathfrak{q}}_0$ of $B$ over ${\mathfrak{p}}_0$. Then [L3] gives primes ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ of $B$ over the whole chain. These inclusions are strict, because ${\mathfrak{q}}_i={\mathfrak{q}}_{i+1}$ would force their contractions ${\mathfrak{p}}_i$ and ${\mathfrak{p}}_{i+1}$ to agree. Hence $B$ has a strict prime chain of length $n$, so $\dim B\ge \dim A$.

Conversely, every strict prime chain in $B$ contracts to a strict prime chain of the same length in $A$ by [L4]. Therefore $\dim A\ge \dim B$.

The two inequalities of steps 1.2 and 1.3 imply $\dim A=\dim B$.

By [L3], the minimal polynomial $m_u(T)\in K[T]$ of $u$ is monic and irreducible. Since $u$ is integral over $A$, it satisfies some monic polynomial in $A[T]$, so [L3] implies that $m_u$ is nonzero. By [L4], choose a splitting field $E\mathrm{/}K$ for $m_u$ and write $m_u(T)={\prod }_{i=1}^n(T-{\alpha }_i)$ with ${\alpha }_1=u$.

Let $g(T)\in A[T]$ be a monic polynomial with $g(u)=0$. For each root ${\alpha }_i$ of $m_u$, the universal property [L5] gives a $K$-homomorphism $K[u]\to E$ sending $u$ to ${\alpha }_i$. Applying that homomorphism to the identity $g(u)=0$ shows $g({\alpha }_i)=0$. Thus every ${\alpha }_i$ is integral over $A$.

The coefficients of $m_u$ are, up to sign, the elementary symmetric polynomials in the integral elements ${\alpha }_1,\dots ,{\alpha }_n$. Since $A$ is a domain and therefore nonzero, [L2] shows that these symmetric polynomials are integral over $A$. But the coefficients also lie in $K$, so [L1] forces them to lie in $A$. This proves the first statement.

For the factor statement, choose a splitting field of $f$ over $K$ and write $f(T)=(T-a){\prod }_{i=2}^n(T-{\beta }_i)$. Each ${\beta }_i$ is integral over $A$ because it is a root of the monic polynomial $f\in A[T]$. Therefore the coefficients of $h(T)={\prod }_{i=2}^n(T-{\beta }_i)$ are integral over $A$ by the same symmetric-polynomial argument as in step 3.1, and they lie in $K$ because $h\in K[T]$. Hence [L1] forces all coefficients of $h$ to lie in $A$.

Assume $A$ is integrally closed, and let $\mathfrak{p}$ be a prime ideal. If $x\in \mathrm{Frac}(A)$ is integral over $A_{\mathfrak{p}}$, then [L2] gives some $s\mathrm{\notin }\mathfrak{p}$ with $sx$ integral over $A$. By [L1], $sx\in A$, so $x=(sx)\mathrm{/}s\in A_{\mathfrak{p}}$. Hence every prime localisation is integrally closed.

Assume every maximal localisation $A_{\mathfrak{m}}$ is integrally closed, and let $x\in \mathrm{Frac}(A)$ be integral over $A$. Because $A$ is a domain, it is nonzero, so [L3] applies and makes the $A$-algebra $M:=A[x]$ a finite $A$-module. For each maximal ideal $\mathfrak{m}$, the same element $x$ is integral over $A_{\mathfrak{m}}$ by [L2], so the hypothesis gives $x\in A_{\mathfrak{m}}$ and therefore $M_{\mathfrak{m}}=A_{\mathfrak{m}}$. Thus the localisation of the inclusion $i:A\hookrightarrow M$ at every maximal ideal is surjective.

Every maximal ideal is prime, so step 1.1 implies that if all prime localisations are integrally closed, then all maximal localisations are integrally closed.

By [L4], the map $i:A\hookrightarrow M$ is surjective. Hence $M=A$, so $x\in A$. By [L1], the domain $A$ is integrally closed.

Step 1.1 proves $(1)\Rightarrow (2)$, step 2.1 gives $(2)\Rightarrow (3)$, and step 2.2 proves $(3)\Rightarrow (1)$. Therefore the three conditions are equivalent.

Let $S:=B\setminus {\mathfrak{q}}_1$. By [L2], primes of $B_S=B_{{\mathfrak{q}}_1}$ correspond exactly to primes of $B$ contained in ${\mathfrak{q}}_1$. It is therefore enough to find a prime ideal of $B_{{\mathfrak{q}}_1}$ whose contraction to $A$ is ${\mathfrak{p}}_0$.

Let $b\in A\cap {\mathfrak{p}}_0{B}_{{\mathfrak{q}}_1}$. Write $b=y\mathrm{/}s$ with $y\in {\mathfrak{p}}_{0}B$ and $s\in B\setminus {\mathfrak{q}}_1$. Choose a representation $y={\sum }_{j=1}^m{a}_j{b}_j$ with $a_j\in {\mathfrak{p}}_0$ and $b_j\in B$. Because the extension is integral, [L1] says that each $b_j$ and $s$ is integral over $A$.

Let $M:=A[s,b_1,\dots ,b_m]$. By [L5], $M$ is module-finite over $A$. Since $y=\sum a_j{b}_j$ lies in ${\mathfrak{p}}_{0}B$, multiplication by $y$ sends $M$ into ${\mathfrak{p}}_{0}M$: for $m^{\mathrm{\prime }}\in M$, one has $y{m}^{\mathrm{\prime }}={\sum }_j{a}_j(b_j{m}^{\mathrm{\prime }})$ and each $b_j{m}^{\mathrm{\prime }}$ again lies in $M$. Choose $A$-generators $m_1,\dots ,m_r$ of $M$. Then there are coefficients $c_{ij}\in {\mathfrak{p}}_0$ with $y{m}_j={\sum }_i{c}_{ij}{m}_i$. Writing $C=(c_{ij})$, the adjugate identity [L6] applied to $y{I}_r-C$ yields a monic polynomial $P(T)=T^r+c_{r-1}{T}^{r-1}+\cdots +c_0$ with every $c_i\in {\mathfrak{p}}_0$ and $P(y)=0$.

Suppose $b\mathrm{\notin }{\mathfrak{p}}_0$. Let $K:=\mathrm{Frac}(A)$. Since $s$ is integral over $A$, [L4] gives a monic minimal polynomial $m_s(T)=T^d+d_{d-1}{T}^{d-1}+\cdots +d_0$ with coefficients in $A$. Because $y=bs$ and $P(y)=0$, dividing by $b^r$ in $K[T]$ gives a monic polynomial $Q(T)=T^r+\frac{c_{r-1}}{b}{T}^{r-1}+\cdots +\frac{c_0}{b^r}$ with coefficients in the localisation $A_{{\mathfrak{p}}_0}$ and satisfying $Q(s)=0$. Since $m_s$ is monic and divides every polynomial over $K$ that vanishes at $s$, it divides $Q$ in $K[T]$, hence also in $A_{{\mathfrak{p}}_0}[T]$.