Integral extension

Let $A\to B$ be a homomorphism of commutative rings . The map (or the extension) is called integral if every element $b\in B$ is integral over A ; equivalently, for each $b\in B$ there is a monic polynomial in $A[T]$ having $b$ as a root.

A frequently used sufficient condition is:

Finite algebra implies integral.
If $B$ is finitely generated as an $A$-module, then $A\to B$ is integral.

Integral extensions behave well with localization: localizing an integral extension at any multiplicative set preserves integrality (compare localization ). They also satisfy strong prime ideal behavior, formalized by results such as lying over and going up .

Examples

1. Adjoining a root of a monic polynomial.
   For any ring $A$ and monic $f(T)\in A[T]$, the quotient

   $$B := A[T]/(f)$$

   is integral over $A$: the class of $T$ in $B$ satisfies the monic equation $f(T)=0$.

2. Classical quadratic extensions.
   The inclusion $\mathbb{Z}\subseteq \mathbb{Z}[i]$ is integral since $i$ satisfies $T^2+1=0$ over $\mathbb{Z}$, and every element of $\mathbb{Z}[i]$ is a polynomial in $i$ with integer coefficients.

3. A cusp subring inside a polynomial ring.
   Let $k$ be a field, $A=k[x^2,x^3]\subseteq B=k[x]$. Then $B$ is integral over $A$ because $x\in B$ satisfies the monic equation $T^2-x^2=0$ with $x^2\in A$, and hence every polynomial in $x$ is integral over $A$.

4. Non-example: polynomial extensions are not integral.
   The inclusion $A\subseteq A[x]$ is typically not integral: the element $x$ does not satisfy any monic polynomial with coefficients in $A$.