Integral element
Let $R$ be a commutative ring.
Integral elements
Definition. Let $A$ be a subring of $R$. An element $x\in R$ is said to be integral over $A$ if it is the root of a monic polynomial with coefficients in $A$. In other words, if there are $n\geq 1$ and $a_{ij}\in A$ such that $$ x^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0=0. $$
Integrally closed
Definition. If there are no elements outside of $A$ then we say $A$ is integrally closed in $R$.
Definition. An integral domain $A$ is called integrally closed if it is integrally closed in its field of fractions.
Properties
- Any element in $A$ is integral over $A$
- The following are equivalent:
- $x\in B$ is integral over $A$
- $A $ is a finitely generated $A$-module
- There is a finitely generated $A$-module $C$ such that $A \subset C\subset B$
Proofs 2 (1\Rightarrow 2) Since $x$ is integral over $A$, it is the root of a monic polynomial $P_x$ which has coefficients in $A$. Suppose $P_x$ has degree $n$. Then since it is monic, we have an expression of $x^n$ as an $A$-linear combination of the elements $1,x,\dots,x^{n-1}$. Then $x^{n+1}=x^nx$ can be expressed by multiplying the linear combination for $x^n$ above by $x$, then substituting in that expression if an $x^n$ term appears after multiplying by $x$. Thus we can express $x^{n+k}$ for $k\geq 0$ in terms of $1,x,\dots,x^{n-1}$. A natural basis for $A[x]$ is $1,x,x^2,\dots$, but we have just shown that we can take the finite basis $1,x,\dots,x^{n-1}$. Hence $A[x]$ is a finitely generated $A$-module.
(2\Rightarrow 3) Take $C=A[x]$, which is finitely generated by assumption. It only remains to show that $C\subset B$. But $A\subset B$ and $x\in B$. Thus the basis of $A[x]$, which is $1,x,\dots,x^{n-1}$, is a subset of $B$. So $A[x]\subset B$.
(3\Rightarrow 1) The inclusions are $A\subset A[x]\subset C\subset B$. A natural basis for $A[x]$ is ${1,x,x^2,\dots}$, and since $A[x]\subset C$ and $C$ is finitely generated as an $A$-module, it must be the case that not all of these are $A$-linearly independent. Hence there exists some $n> 0$ such that $x^n$ can be expressed as a finite $A$-linear combination of the $x^{n-k}$ for $0<k\leq n$. Subtracting this linear combination from $x^n$ and substituting $x$ for a variable $X$ we obtain a monic polynomial $P(X)$ such that $P(x)=0$.