Integrality and Integral Closures

Integrality

Let $S$ be an commutative integral domain with $1$.

Definition

Let $R$ be a subring of $S$ containing $1_S$. We say $\alpha$ is an integral element over $R$, if there exists a monic polynomial $f(x)\in R[x]$ such that $f(\alpha )=0$.

Proposition

Let $R\subseteq S$ be rings with $1_S\in R$ and $\alpha \in S$. TFAE:

• $\alpha$ is an integral domain over $R$
• $R[\alpha ]$ is a finitely generated $R$-module.
• There exists a subring $T$ of $S$ with $R\subseteq T$ such that $T$ is a finitely generated $R$-module and $\alpha \in T$.

\begin{proof} i)→ii) There is $f(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$ with $f(\alpha )=0$. Then ${\alpha }^n=-{\sum }_{i=0}^{n-1}{a}_i{\alpha }^i$ is a linear combination of $1,\alpha ,\cdots ,{\alpha }^{n-1}$. It deduces that $R[\alpha ]$ is generated by $1,\alpha ,\cdots ,{\alpha }^{n-1}$.

ii)→iii) Take $T=R[\alpha ]$.

iii)→i) Let ${ϵ}_1,\cdots ,{ϵ}_n$ be a generator set of $T$ as an $R$-module. Write $\alpha {ϵ}_i=a_{i1}{ϵ}_1+\cdots +a_{in}{ϵ}_n$ with $a_{ij}\in R$. It deduces that $-a_{i1}{ϵ}_1+\cdots +(\alpha -a_{ii}){ϵ}_i+\cdots +(-a_{in}{ϵ}_n)=0$ and so

$$\left(\begin{array}{ccc}{ϵ}_1& \cdots & {ϵ}_n\end{array}\right)\left(\begin{array}{cccc}\alpha -a_{11}& -a_{21}& \cdots & -a_{n1}\\ -a_{12}& \alpha -a_{22}& \cdots & -a_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& -a_{2n}& \cdots & \alpha -a_{nn}\end{array}\right)=\left(\begin{array}{ccc}0& \cdots & 0\end{array}\right)$$

where $A=\left(\begin{array}{cccc}\alpha -a_{11}& -a_{21}& \cdots & -a_{n1}\\ -a_{12}& \alpha -a_{22}& \cdots & -a_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& -a_{2n}& \cdots & \alpha -a_{nn}\end{array}\right)$. Multiplying $A^{\ast }$, we have $({ϵ}_1,\cdots ,{ϵ}_n)\mathrm{diag}(\det A,\cdots ,\det A)=(0,\cdots ,0)$. Hence ${ϵ}_i\det A=0$ and so $\det A=0$. Therefore, $\alpha$ is integral over $R$. \end{proof}

Lemma

Define $S,R$ as above. Define subrings $M_1,M_2$ of $S$ such that $R\subseteq M_i\subseteq S$. If $M_1,M_2$ are finitely generated $R$-modules, then the ring $R[M_1,M_2]$ is finitely generated as an $R$-module.

\begin{proof} Since $M_1,M_2$ are finitely generated $R$-modules, there exists $\{m_1,\cdots ,m_s\}$ and $\{m_1^{\prime },\cdots ,m_r^{\prime }\}$ such that $M_1={⟨m_1,\cdots ,m_s⟩}_R$ and $M_2={⟨m_1^{\prime },\cdots ,m_r^{\prime }⟩}_R$. Note that each element of $R[M_1,M_2]$ is in the form of

$$\sum _{\alpha ,\beta }{r}_{\alpha ,\beta }{m}_1^{{\alpha }_1}\cdots m_s^{{\alpha }_s}{m}_1^{\prime {\beta }_1}\cdots m_r^{\prime {\beta }_r}$$

where $m_1^{{\alpha }_1}\cdots m_s^{{\alpha }_s}\in M_1$ can be written as $m_1^{{\alpha }_1}\cdots m_s^{{\alpha }_s}=\sum r_j{m}_j$ and similarly for $m_1^{\prime {\beta }_1}\cdots m_r^{\prime {\beta }_r}$. Therefore, $R[M_1,M_2]$ is generated by $\{m_i{m}_j^{\prime }:1⩽i⩽s,1⩽j⩽r\}$ as an $R$-module. Now we finish the proof. \end{proof}

Theorem

Let $R$ be a subring of $S$ containing $1_S$. The integral elements of $S$ over $R$ form a subring of $S$ containing $R$.

\begin{proof} Since $\alpha ,\beta$ are integral, by ^274bb5 we know $R[\alpha ],R[\beta ]$ are finitely generated. Then by ^0f6d0e, $R[\alpha ,\beta ]$ is finitely generated. It deduces that $\alpha -\beta$, $\alpha \beta$ are integral and so the integral elements of $S$ form a subring. \end{proof}

Definition

Let $R$ be a subring of $S$ containing $1_S$.

• The subring of $S$ containing of integral of $S$ over $R$ is called the integral closure of $R$ in $S$.
• $R$ is called integral closed in $S$ if every integral elements of $S$ over $R$ is containing in $R$.
• We say an integral domain $R$ is integral closed if it is integral closed in its fractional field.
• $\mathbb{Z}\subseteq \mathbb{C}$, if $z\in \mathbb{C}$ is integral over $\mathbb{Z}$, we call $z$ an algebraic integer. All algebraic integers form a ring.

Theorem

A UFD is integral closed.

\begin{proof} Let $A$ be a UFD, and let $K=\mathrm{Frac}(A)$ be its fraction field. To show $A$ is UFD, it suffices to show for any $r\in K$ which is integral over $A$, $r\in A$. Assume that $r=r_1/r_2$ with $r_1,r_2\in A$ and $\gcd (r_1,r_2)=1$. If $r$ is integral over $A$, there exist $a_0,\cdots ,a_{n-1}\in A$ such that

$$\frac{r_1^n}{r_2^n}+a_{n-1}\frac{r_1^{n-1}}{r_2^{n-1}}+\cdots +a_0=0⟹-a_0{r}_2^n=r_1(r_1^{n-1}+a_{n-1}{r}_2{r}_1^{n-2}+\cdots +a_1{r}_2^{n-1})\in A.$$

Since $A$ is a UFD, we know $r_2\mid r_1(r_1^{n-1}+a_{n-1}{r}_2{r}_1^{n-2}+\cdots +a_1{r}_2^{n-1})$ and so $r_2\mid r_1^n$. It deduces that $r_2=1$ by $\gcd (r_1,r_2)=1$. Then $r=r_1\in R$ and we have done. \end{proof}

Corollary

A DVR is integral closed.

\begin{proof} By ^ve57xd, a DVR is PID and then is UFD. By ^437xgo, a DVR is integral closed. \end{proof}