Integral closure

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.