5 Integral Dependence and Valuation

Definition

Let $A\subseteq B$ be a subring. We say $x\in B$ is integral over $A$, if there exists a monic polynomial $f(T)\in A[T]$ such that $f(x)=0$. Remark that if $A$ is a field, then integral iff algebraic.

Examples.

• Suppose $x=r/s\in \mathbb{Q}$ is integral over $\mathbb{Z}$ with $(r,s)=1$ then

  $$(r/s{)}^n+a_{n-1}(r/s{)}^{n-1}+\cdots +a_{1}r/s+a_0=0$$

  with $a_i\in \mathbb{Z}$. Multiply by $s^n$, then $s\mid r^n$ and $s=1$. Thus $x\in \mathbb{Z}$.

• $\mathbb{Z}\subseteq \mathbb{Q}[i]$, one can check $x\in \mathbb{Q}[i]$ integral over $\mathbb{Z}$ iff $x\in \mathbb{Z}[i]$.

Remark. A finite extension $E/\mathbb{Q}$ is called a number field. The subset ${\mathcal{O}}_E\subseteq E$ with elements integral over $\mathbb{Z}$, is called the ring of integers.

Proposition

Let $A\subseteq B$ be a subring. Let $x\in B$. TFAE:

• $x$ is integral over $A$;
• $A[x]$ is a finitely generated $A$-module;
• there exists subring $C$ such that $A[x]\subseteq C\subseteq B$, and $C$ is a finitely generated $A$-module;
• there exists a faithful $A[x]$-module $M$, such that $M$ is a finitely generated $A$-module.

\begin{proof} i)→ii) Since $x$ is integral over $A$, we have $x^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0=0$ and so $A[x]={⟨1,x,\cdots ,x^{n-1}⟩}_A$ is finitely generated.

ii)→iii) Let $C=A[x]$.

iii)→iv) Let $M=C$. For $y\in \mathrm{Ann}(C)$, $y\cdot 1_C=0$ and $y=0$, that is, $M$ is faithful.

iv)→i) Recall ^41223f, for a finitely generated $A$-module $M$, an ideal $\alpha \subseteq A$ and $\varphi :M\to M$ such that $\varphi (M)\subseteq \alpha M$, $\varphi$ satisfies

$${\varphi }^n+a_1{\varphi }^{n-1}+\cdots +a_n=0.$$

In our situation, $M$ is a finitely generated $A$-module, and let

$$\varphi :M\to M,m\mapsto xm$$

be a homomorphism of $A$-modules with $\alpha =A$. Then there exist $a_1,\cdots ,a_n\in A$ such that

$${\varphi }^n+a_1{\varphi }^{n-1}+\cdots +a_n$$

kills $M$. Then $x^n+a_1{x}^{n-1}+\cdots +a_n\in \mathrm{Ann}(M)$. Since $M$ is an $A[x]$-module, we know $x$ is integral over $A$. \end{proof}

Corollary

For a subring $A\subseteq B$, where $x_1,\cdots ,x_n\in B$ is integral over $A$, the ring $A[x_1,\cdots ,x_n]$ is a finitely generated $A$-module.

\begin{proof} By ^jy6o7s and induction. \end{proof}

Corollary

For subring $A\subseteq B$, let $C=\{x\in B:x\text{ is integral over }A\}$. Then $C$ is a subring.

\begin{proof} Let $x,y\in C$, then $A[x,y]$ is a finitely generated $A$-module by ^ig1gwo. By ^jy6o7s iii), since $A[x\pm y]\subseteq A[x,y]\subseteq B$ and $A[xy]\subseteq A[x,y]\subseteq B$, we know $x\pm y$ and $xy$ are integral. Now we finish the proof. \end{proof}

Definition

$C$ in ^4b091b is called the integral closure of $A$ in $B$.

• If $C$ above is equal to $A$, then we say $A$ is integrally closed in $B$.
• If $C=B$, then we say $B$ is integral over $A$.

Remark. Let $f:A\to B$ be a ring homomorphism. We say $f$ is an integral homomorphism if $B$ is integral over $f(A)$. In this case, we say $B$ is an integral $A$-algebra.

Note. If $f$ is of finite type and $f$ is an integral homomorphism, then $f$ is finite.

Corollary

If $A{\subseteq }_{\text{int}}B{\subseteq }_{\text{int}}C$, then $A$ is also integral over $C$.

\begin{proof} For any given $x\in C$, since $C$ is integral over $B$, there exists

$$x^n+b_1{x}^{n-1}+\cdots +b_n=0$$

with $b_i\in B$. Consider $B^{\prime }=A[b_1,\cdots ,b_n]\subseteq B$, then $x$ is also integral over $B^{\prime }$. Hence, $B^{\prime }[x]$ is a finitely generated $B^{\prime }$-module. Since each $b_i$ is integral over $A$, by ^ig1gwo $B^{\prime }$ is a finitely generated $A$-module and so for $B^{\prime }[x]$. By ^jy6o7s iii), $x$ is integral over $A$. \end{proof}

Corollary

For a subring $A\subseteq B$, let $C$ be the integral closure of $A$ in $B$. Then $C\subseteq B$ is integral closed.

\begin{proof} If $x\in B$ is integral over $C$, then by ^342aup, $x$ is integral over $A$ and so $x\in C$. \end{proof}

Proposition

Let $B$ is integral over $A$. Then:

• For an ideal $\beta \subseteq B$, let $\alpha =\beta \cap A$ then $B/\beta$ is integral over $A/\alpha$.
• If $S\subseteq A$ is multiplicative closed, then $S^{-1}B$ is integral over $S^{-1}A$.

\begin{proof} i) For any $x\in B$, $x^n+a_{n-1}{x}^{n-1}+\cdots +a_0=0$ for some $a_i\in A$. Then ${\overline{x}}^n+{\overline{a}}_{n-1}{\overline{x}}^{n-1}+\cdots +{\overline{a}}_0=0$ and we finish the proof.

ii) For any $x/s\in S^{-1}B$, there is

$$(x/s{)}^n+(a_1/s)(x/s{)}^{n-1}+\cdots +(a_0/s^n)=0$$

and so $x/s$ is integral over $S^{-1}A$. \end{proof}

Going-up Theorem

Proposition

Let $A,B$ be integral domain with $B/A$ integral. Then $B$ is a field iff $A$ is a field.

\begin{proof} "←" For any non-zero $y\in B$, there is

$$y^n+a_{n-1}{y}^{n-1}+\cdots +a_{1}y+a_0=0.$$

Suppose $n$ is the minimal degree, which yields $a_0\ne 0$ by $A$ integral domain. Then

$$y(y^{n-1}+a_{n-1}{y}^{n-2}+\cdots +a_1)=-a_0\ne 0$$

and so $y$ is a unit. Therefore, $A$ is a field.

"→" Let $x\in A$ be non-zero. Since $B$ is a field, $x^{-1}\in B$. Since $B/A$ is integral, we have

$$(x^{-1}{)}^n+a_{n-1}(x^{-1}{)}^{n-1}+\cdots +a_0=0$$

and so $x^{-1}=-(a_{n-1}+a_{n-2}x+\cdots +a_0{x}^{n-1})\in A$. \end{proof}

Corollary

Assume $B/A$ is integral. If $q\subseteq B$ is a prime ideal of $B$, then $q$ is maximal iff $q^c$ is maximal in $A$.

\begin{proof} By ^f6d0xm, $B/q$ is integral over $A/q^c$. Then apply ^mun8kw. \end{proof}

Corollary

Suppose $B/A$ is integral. Let $q\subseteq q^{\prime }\subseteq B$ be a prime ideal. If $q^c=q^{\prime c}=p\subseteq A$, then $q=q^{\prime }$.

\begin{proof} Consider the following diagram.

Note $f^{-1}(n)\supseteq p{A}_p=m$. Since $m$ is the unique maximal ideal of $A_p$, we know $f^{-1}(n)=m$. Similarly, $f^{-1}(n^{\prime })=m$.

By ^f6d0xm $B_p$ is integral over $A_p$.

By ^8zjrn5 $n$ and $n^{\prime }$ are maximal in $B_p$, which deduces $n=n^{\prime }$.

Then $q=q^{\prime }$ by ^961e0a. \end{proof}

Corollary

Assume $A{\subseteq }_{\text{int}}B$. Then $\dim A⩾\dim B$.

\begin{proof} Let $q_0⊊q_1⊊\cdots ⊊q_n$ be the largest chain of primes in $B$. Then by ^8f19d4

$$q_0^c⊊q_1^c⊊\cdots ⊊q_n^c$$

is also a chain in $A$. Then $\dim A⩾n$. \end{proof}

Theorem

Let $B/A$ be integral. For any prime ideal $p\subseteq A$, there exists prime $q\subseteq B$ such that $q^c=p$.

\begin{proof} Consider the following diagram

Let $n$ be any maximal ideal of $B_p$. By ^8zjrn5, $f^{-1}(n)$ is maximal and so $f^{-1}(n)=m$. Let $q={\beta }^{-1}(n)$, which is a prime. Further

$$q\cap A={\alpha }^{-1}(f^{-1}(n))={\alpha }^{-1}(m)=p$$

and so we finish the proof. \end{proof}

going-up theorem

Assume $A{\subseteq }_{\text{int}}B$. Let $p_1\subseteq \cdots \subseteq p_n$ be a chain of prime ideals in $A$, and let $q_1\subseteq \cdots \subseteq q_m$ be a chain of prime ideals in $B$ with $0⩽m<n$ and $q_i^c=p_i$. Then one can extend the second chain to

$$q_1\subseteq \cdots \subseteq q_m\subseteq q_{m+1}\subseteq \cdots \subseteq q_n$$

such that $q_i^c=p_i$ for all $i$.

\begin{proof} First, if $m=0$, then by ^k2927t there exists $q_1$. So by induction, it suffices to show the case $n=2$, $m=1$. Namely, we now have $p_1\subseteq p_2\subseteq A$ and $q_1\subseteq \text{?}\subseteq B$. Consider $A/p_1\hookrightarrow B/q_1$, which is integral by ^f6d0xm. Since ${\overline{p}}_2$ is prime in LHS, by ^k2927t there exists $q_2/q_1$ such that $(q_2/q_1{)}^c=p_2/p_1$. Therefore, $q_2^c=p_2$. \end{proof}

Corollary

If $A\subseteq B$ is integral, then $\dim A=\dim B$.

\begin{proof} By ^whjkcj, we have $\dim A⩽\dim B$. By ^2uikai we finish the proof. \end{proof}

Integrally Closed Integral Domain

Proposition

Let $A\subseteq B$, and let $C$ be an integral closure of $A$ in $B$. For a multiplicative closed set $S\subseteq A$, $S^{-1}C$ is a integral closure of $S^{-1}A$ in $S^{-1}B$.

\begin{proof} By ^f6d0xm, $S^{-1}C$ is integral over $S^{-1}A$. Now suppose $b/s\in S^{-1}B$ is integral over $S^{-1}A$, we aim to show $b/s\in S^{-1}C$. There is

$$(b/s{)}^n+(a_1/s_1)(b/s{)}^{n-1}+\cdots +(a_n/s_n)=0$$

with $a_i/s_i\in S^{-1}A$. Multiply by $s^n{s}_1^n\cdots s_n^n$ and we get

$$(b{s}_1\cdots s_n{)}^n+\cdots =0$$

and so $b{s}_1\cdots s_n$ is integral over $A$. Thus $b{s}_1\cdots s_n\in C$, leading to $b/s=(b{s}_1\cdots s_n)/(s{s}_1\cdots s_n)\in S^{-1}C$. \end{proof}

Definition

For any integral domain $A$, we say it is integrally closed, if $A$ is integral closed in $\mathrm{Frac}A$.

Example. $\mathbb{Z}$, UFD (see here).

integrally closed is a local property

For an integral domain $A$, TFAE:

• $A$ is integrally closed;
• $A_p$ is integrally closed for any prime ideal $p$;
• $A_m$ is integrally closed for any maximal ideal $m$.

\begin{proof} Let $C$ be the integral closure of $A$ in $\mathrm{Frac}A$. By ^9iz2fs, $C_p$ is the integral closed of $A_p$ in $\mathrm{Frac}(A_p)=\mathrm{Frac}A$. Since $A$ is integrally closed, we have $C=A$ and so $A_p=C_p$ for any prime ideal $p$. Then by ^766550, $A_p=C_p$ for all $p$ iff $A_m=C_m$ for all $m$ iff $A=C$. \end{proof}

Definition

Let $A\subseteq B$, and let $\alpha \subseteq A$ be an ideal. Say $x\in B$ is integral over $\alpha$ if $x^n+a_1{x}^{n-1}+\cdots +a_n=0$ with $a_i\in \alpha$. The integral closure of $\alpha$ in $B$ is the set of all such elements.

Lemma

Let $A\subseteq B$, and let $C$ be the integral closure of $A$ in $B$. For an ideal $\alpha \subseteq A$ with ${\alpha }^e\subseteq C$. then the integral closure of $\alpha$ in $B$ is $r({\alpha }^e)\subseteq C$, which is an ideal in $C$.

\begin{proof} "$\subseteq$" If $x\in B$ is integral over $\alpha$, then $x^n=-(a_1{x}^{n-1}+\cdots +a_n)\subseteq {\alpha }^e$ and so $x\in r({\alpha }^e)$.

"$\supseteq$" Consider $x\in r({\alpha }^e)$, then $x^n={\sum }_{i=1}^N{a}_i{x}_i$ with $a_i\in \alpha$ and $x_i\in C$. Consider $M=A[x_1,\cdots ,x_N]\subseteq C$. Since $x_i$ is integral over $A$, we know $M$ is a finitely generated $A$-module by ^jy6o7s. Consider $\varphi :M\to M,m\mapsto x^{n}m$, where $\mathrm{im}(\varphi )\subseteq \alpha M$. By ^41223f, there is

$${\varphi }^m+b_1{\varphi }^{m-1}+\cdots +b_m=0\text{with }{b}_i\in \alpha .$$

It deduces that $x^{mn}+b_1{x}^{n(m-1)}+\cdots +b_m=0$ by $\varphi (1)=x^n$ and so $x$ is integral over $\alpha$. \end{proof}

Proposition

Let $A\subseteq B$ be two integral domains. Suppose $A$ is integrally closed and $\alpha \subseteq A$ is an ideal. If $x\in B$ is integral over $\alpha$, then $x$ is algebraic over $k=\mathrm{Frac}A$. Furthermore, if $\mathrm{Irr}(x,K)=t^n+a_1{t}^{n-1}+\cdots +a_n$, then $a_i\in r(\alpha )$.

\begin{proof} Since $x$ is integral over $\alpha$, we have $g(x)=x^m+b_1{x}^{m-1}+\cdots +b_m=0$ with $b_i\in \alpha$. Then $b_i\in \alpha \subseteq A\subseteq \mathrm{Frac}A=k$ yields $x$ is algebraic over $k$.

However, note possibly $m>n=\mathrm{deg}(\mathrm{Irr}(x,K))$. Now let $x_1,\cdots ,x_n$ be all the roots of $f(t):=\mathrm{Irr}(x,K)$, then $x_1,\cdots ,x_n$ are also roots of $g(x)$ and $x_i$ is integral over $\alpha$. By Vita’s theorem, $a_i$ are symmetric functions in $x_i$. By ^vwwek4, $a_i$ are integral over $\alpha$.

Apply ^vwwek4 in the setting $A\subseteq A\subseteq \mathrm{Frac}A$, and we know $a_i\in r({\alpha }^e)=r(\alpha )$. Now we finish the proof. \end{proof}

Going-down theorem

Let $A\subseteq B$ be integral domains. Suppose $A$ is integrally closed and $B$ is integral over $A$. Let $p_1\supseteq \cdots \supseteq p_n$ be a chain of prime ideals in $A$, and let $q_1\supseteq \cdots \supseteq q_m$ be a chain of prime ideals in $B$ with $q_i^c=p_i$ and $m<n$. Then there exists $q_1\supseteq \cdots \supseteq q_m\supseteq \cdots \supseteq q_n$ such that $q_i^c=p_i$.

\begin{proof} The case of $n=1$ is trivial by ^whjkcj. For $n⩾2$, similarly it suffices to consider the case of $n=2$ and $m=1$. So we have $p_1\supseteq p_2$ and $q_1\supseteq ?$. Recall ^7xys15, for a ring homomorphism $A\to B$, prime ideal $p\subseteq A$ is a contraction iff $p^{ec}=p$.

Consider $A\to B\to B_{q_1}$. Note prime ideals of $B_{q_1}$ are precisely $q{B}_{q_1}$ with $q\subseteq q_1\subseteq B$. By ^7xys15, it suffices to prove $p_2{B}_{q_1}\cap A=p_2$. Notice that "$\supseteq$" is obvious. Now we prove "$\subseteq$".

For any $x\in p_2{B}_{q_1}$, $x$ can be written as $x=y/s$ with $y\in p_{2}B$ and $s\in B-q_1$. By ^vwwek4, integral closure of $p_2$ in $B$ is $r(p_{2}B)\supseteq p_{2}B$. Thus $y\in p_{2}B$ is integral over $p_2$. Since $A$ is integrally closed, by ^fbc39e,

$$\mathrm{Irr}(y,\mathrm{Frac}A)=t^r+u_1{t}^{r-1}+\cdots +u_r\text{\hspace{1em}(*)}$$

with $u_i\in r(p_2)=p_2$.

Now suppose $x\in p_2{B}_{q_1}\cap A$, $x=y/s$. Then $s=y{x}^{-1}$. Plug $y=xs$ to $(\ast )$, then divide $x^r$, get

$$s^r+(u_1/x)s^{r-1}+\cdots +u_r/x^r=0\text{\hspace{1em}(**)}$$

with coefficients in $\mathrm{Frac}A$. Thus $s$ is algebraic over $\mathrm{Frac}A$. Since $(\ast )$ is irreducible, we know $(\ast \ast )$ is irreducible and so $\mathrm{Irr}(s,\mathrm{Frac}A)=(\ast \ast )$.

But $s\in B-q_1\subseteq B$ yields $s$ is integral over $A$. Apply ^fbc39e with $\alpha =A$, then we get coefficients of $(\ast \ast )$ are in $r(A)=A$. Define $v_i=u_i/x^r\in A$.

Recall that we aim to show $x\in p_2$. Suppose otherwise, since $v_i{x}^r=u_i\in p_2$ and $x\notin p_2$, there is $v_i\in p_2$ and so $s^r\in p_{2}B\subseteq p_{1}B\subseteq q_1$, contradicting with $s\notin q_1$. \end{proof}

高辉说：

“这一章要考的话，主要考integral的判别法则。”

“going up going down 不考，主要是欣赏一下。”

Check

上升定理说的是，从下往上匹配素理想链总是可能的（只要是整扩张）。 下降定理说的是，在更好的条件下（A 整闭等），从上往下匹配素理想链也是可能的。