---- > [!definition] Definition. ([[integral element of an algebra]]) > Let $A$ be a [[commutative ring|(commutative)]] $R$-[[algebra]]. An element $x \in A$ is said to be **integral over $R$**, or **$R$-integral**, if there is a [[monic polynomial|monic]] [[polynomial 4|polynomial]] $f \in R[T]$ for which it is a [[root of a polynomial|root]]: $f(x)=0$.[^1] > > The set $\mathcal{O}$ of $R$-integral elements of $A$ forms an $R$-[[subalgebra]] of $A$. ^definition > [!generalization] > The notion of [[integrality over an ideal]] is more general in the sense that it concerns integrality over any [[ideal]] $\mathfrak{a} \subset R$ rather than the specific ideal $\langle 1 \rangle=R$. ^generalization > [!basicexample] > - If $k$ is a [[field]], and $A$ is a $k$-[[algebra]], then $x \in A$ is $k$-integral if and only if $x$ is $k$-[[algebraic element|algebraic]], because we can always divide out by the leading coefficient to 'turn a solution monic'. > - The integral elements of $\mathbb{Q}$ over $\mathbb{Z}$ are $\mathbb{Z}$ (inspiring in part the term 'integral') ^basic-example > [!equivalence] Equivalence (a corollary of [[Cayley-Hamilton Theorem|Cayley-Hamilton]]) > Let $R \subset A$ be [[ring|rings]] and $x \in A$. Considering the [[ring adjunction|adjunction]] $R[x] \subset A$, we see that $A$ is also an $R[x]$-[[module]]. Then $x$ is $R$-integral if and only if there is $M \subset A$ such that both of the following conditions hold: > > 1. $M$ is a [[faithful module|faithful]] $R[x]$-[[submodule]] of $A$, i.e.: > 1. $M$ is an $R$-[[submodule]] of $A$; > 2. $xM \subset M$; > 3. For all $0 \neq p \in R[x]$ there is $m \in M$ such that $pm \neq 0$. > 2. $M$ is [[submodule generated by a subset|finitely generated]] as an $R$-[[module]]. > [!proof] Proof of Equivalence. > First, assume $(1)$ and $(2)$ hold. Since $xM \subset M$, have an $R$-[[linear map|linear map]] $f:M \to M$, $f(m)=xm$. Since $M$ is [[submodule generated by a subset|finitely generated]] over $R$, we can apply [[Cayley-Hamilton Theorem|Cayley-Hamilton]] to get $f^{n}+r_{1}f^{n-1}+\dots+r_{n}f^{0}=0\text{ in }\text{End}_{R}M$ > for $r_{1},\dots,r_{n} \in R$, $n \geq 1$. This means that, for all $m \in M$, $x^{n}m+r_{1} x^{n-1}m+\dots+r_{n}m=(x^{n}+r_{1}x^{n-1} + \dots + r_{n})m=0.$ > Since $M$ is [[faithful module|faithful]] as an $R[x]$-[[module]], it must be the case that $x^{n}+r_{1}x^{n-1}+\dots+r_{n}=0$, witnessing that $x$ is $R$-integral. > > Conversely suppose $x \in A$ is $R$-integral, write $x^{n}+r_{1}x^{n-1}+\dots+r_{n}=0$. Define $M:=\text{span}_{R}\{ x^{0},\dots,x^{n-1} \}.$ > Note that $xM \subset M$ because $x^{n}=-(r_{1}x^{n-1}+\dots+r_{n})$. [^1] Faithfulness follows because $1=x^{0} \in M$. Finally, $M$ is finitely generated by construction. > [^1]: $f(x)=0$ is sometimes called the **integrality equation**. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```