Integral element

An element $a$ of an $R$-algebra is integral over $R$ when it satisfies a monic polynomial with coefficients in $R$.

An element $x$ of an $A$-algebra is an integral element precisely when $A[x]$ is a finite $A$-module. More generally, a finite $A$-submodule containing one and stable under multiplication by $x$ proves integrality by the determinant trick. Conversely a monic polynomial equation reduces all powers of $x$ to finitely many generators.

An extension $R\subseteq A$ is integral when every element of $A$ is integral over $R$.

If a field $L$ is integral over a subdomain $A$, every nonzero $a\in A$ has an inverse in $L$ satisfying a monic equation over $A$. Multiplying that equation by $a^{r-1}$ expresses $a^{-1}$ as a polynomial in $a$ with coefficients in $A$. Hence $a^{-1}\in A$, and $A$ is itself a field. The embedding and integrality assumptions are essential; a fraction field is not generally integral over its source domain.

An integral extension $A\subset B$ need not give an integral extension $A_{P\cap A}\subset B_P$. For example, take $\mathbb{Z}\subset \mathbb{Z}[i]$ and $P=(2+i)$, whose contraction is $(5)$. The element $1/(2-i)$ belongs to $\mathbb{Z}[i{]}_P$ but is not integral over ${\mathbb{Z}}_{(5)}$: multiplying a hypothetical monic equation by a power of $2-i$ and reducing modulo the other Gaussian prime ideal $(2-i)$ gives $1=0$ in ${\mathbb{F}}_5$. Localizing both rings by the same multiplicative subset does preserve integrality.

If $A\subseteq B$ is an integral extension, then $\dim A=\dim B$. The incomparability theorem for integral extensions contracts every strict prime chain in $B$ to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, the Lying-over theorem starts over the bottom of any prime chain in $A$, and repeated use of the going-up theorem lifts the whole chain to $B$.

For any algebraic field extension $K/\mathbb{Q}$, including an infinite one, its ring ${\mathcal{O}}_K$ of elements integral over $\mathbb{Z}$ is an integral extension of $\mathbb{Z}$. The Lying-over theorem, going-up theorem and incomparability theorem for integral extensions give $\dim {\mathcal{O}}_K=\dim \mathbb{Z}=1$. The argument does not require that ${\mathcal{O}}_K$ be Noetherian or finitely generated over $\mathbb{Z}$.

If $f\in k[Y][X]$ is monic of positive degree in $X$, monic division makes $k[Y][X]/(f)$ free of rank ${\mathrm{deg}}_{X}f$ over $k[Y]$. The inclusion of $k[Y]$ is injective and the quotient is an integral extension, so its Krull dimension is one. This argument applies even when the polynomial factors or the field has positive characteristic.

If $A\subseteq B$ is an integral extension, then every prime ideal $\mathfrak{p}$ of $A$ is the contraction of a prime ideal $\mathfrak{q}$ of $B$. Thus the contraction map $\mathrm{Spec}B\to \mathrm{Spec}A$ is surjective.

Going up and going down concern lifting inclusions of prime ideals through a ring extension. Going-up theorem lifts upward from a chosen lower prime for an integral extension. Going-down theorem lifts downward from a chosen upper prime under additional hypotheses, including an integrally closed base domain for integral extensions of domains.

For an integral extension $A\subseteq B$, primes ${\mathfrak{p}}_1\subseteq {\mathfrak{p}}_2$ of $A$, and a prime ${\mathfrak{q}}_1$ of $B$ over ${\mathfrak{p}}_1$, there is a prime ${\mathfrak{q}}_2\supseteq {\mathfrak{q}}_1$ over ${\mathfrak{p}}_2$.

Let $R\subseteq A$ be an integral extension of domains with $R$ integrally closed. Given primes ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ of $R$ and a prime ${\mathfrak{q}}_1$ of $A$ over ${\mathfrak{p}}_1$, there is a prime ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ over ${\mathfrak{p}}_0$.

If $A\subseteq B$ is integral and ${\mathfrak{q}}_1\subseteq {\mathfrak{q}}_2$ are primes of $B$ with equal contractions to $A$, then ${\mathfrak{q}}_1={\mathfrak{q}}_2$. Consequently strict prime chains remain strict under contraction.

Let $A\subseteq B$ be integral and let $I\subseteq J$ be ideals of $B$. If $I$ is prime and $I\cap A=J\cap A$, then $I=J$. After quotienting by $I$, a nonzero element of $J/I$ has an integral equation of least degree whose nonzero constant term lies in $(J/I)\cap (A/(I\cap A))$, a contradiction.

If $A\subseteq B$ is integral and $\mathfrak{n}$ is a maximal ideal of $B$, then $\mathfrak{n}\cap A$ is maximal in $A$. Indeed, $B/\mathfrak{n}$ is an integral domain integral over $A/(\mathfrak{n}\cap A)$, and a subring over which a field is integral is itself a field.

For an integral extension $A\subseteq B$ and $\mathfrak{p}\in \mathrm{Spec}A$, localization gives a bijectionIt combines the prime ideal correspondence for localization with the contraction of a maximal ideal under an integral extension.

If $a_1,\dots ,a_n$ are integral over $R$, then $R[a_1,\dots ,a_n]$ is a finitely generated $R$-module. In particular, every element of this subalgebra is integral over $R$.

An integral domain is integrally closed when every element of its fraction field that is integral over the domain already belongs to the domain.

The integral closure of $R$ in an $R$-algebra $A$ is the subring of elements of $A$ integral over $R$.

For a normal Noetherian ring $A$ that is an integral domain, the integral closure in a finite separable field extension of its fraction field is finite as an $A$-module. An integral field basis by denominator clearing gives a finite lattice. Integral traces place the closure in its finite trace-dual lattice; Noetherianity makes that submodule finitely generated.

If $A$ is an integral domain with fraction field $K$ and $L/K$ is finite, a $K$-basis of $L$ can be rescaled to consist of integral elements over $A$. For a monic algebraic equation, choose $a\in A$ clearing the coefficient denominators; replacing $u$ by $au$ multiplies the coefficient of degree $r-j$ by $a^j$, putting it in $A$.

For a domain $R$ with fraction field $K$, its integral closure is the intersection of all valuation rings of $K$ containing $R$. Integral elements belong to every such ring because valuation rings are integrally closed. For a nonintegral $x$, the ideal $x^{-1}R[x^{-1}]$ is proper, since its containing $1$ would give a monic equation for $x$. Localize at a maximal ideal containing $x^{-1}$ and apply the valuation domination lemma. The resulting ring contains $R$ but not $x$, proving the reverse inclusion.

If a graded ring $A$ is a graded extension of a graded subring $R$, then the integral closure of $R$ in $A$ is graded: every homogeneous component of an integral element is integral. This follows by separating the extremal degrees in a monic integral equation and inducting on the number of nonzero homogeneous components.