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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 ideal $I$ is idempotent when $I^2=I$. Then $I^j=I$ for every $j\ge 1$, so a nonzero idempotent ideal gives a nonzero intersection of all its powers. A semigroup algebra with arbitrarily divisible positive exponents supplies examples in a non-Noetherian integral domain. If $I$ is finitely generated, the determinant trick applied to $I=I^2$ produces $r\in I$ such that $(1+r)I=0$.

Let $R$ be a one-dimensional normal Noetherian local domain. Choose $0\ne x$ in its maximal ideal $\mathfrak{m}$ and a minimal $n$ with ${\mathfrak{m}}^n\subseteq (x)$. If $n>1$, choose $y\in {\mathfrak{m}}^{n-1}\setminus (x)$ and put $z=y/x$. Then $z\notin R$ but $z\mathfrak{m}\subseteq R$. If $z\mathfrak{m}\subseteq \mathfrak{m}$, the determinant trick makes $z$ integral, contrary to normality. Thus $zu$ is a unit for some $u\in \mathfrak{m}$, and $v=u(zv)/(zu)$ for every $v\in \mathfrak{m}$ proves $\mathfrak{m}=(u)$. The case $n=1$ is already principal. Such a local domain is a discrete valuation ring.

Let $S=R[\varphi ]$ be the commutative subring of endomorphisms generated by scalar multiplication and $\varphi$. Commutativity follows because the scalar maps commute with the $R$-linear map $\varphi$, and every element of $S$ is a polynomial in $\varphi$ with scalar coefficients. Thus determinants and the adjugate identity apply to matrices over $S$.

With the convention for the coefficients in the question, the column of generators satisfiesMultiply by $\mathrm{adj}(C)$ over $S$. Since $\mathrm{adj}(C)C=(\det C)I_n$, the endomorphism $\det C$ kills every generator $m_i$. It therefore kills every element $\sum r_i{m}_i$ of the module, by $R$-linearity. Hence it is the zero endomorphism, andThe argument does not require the generating set to be a basis; relations among the generators cause no difficulty. This is the determinant trick.

A finite basis is a generating set, so the preceding determinant trick applies with $R=k$ and $M=V$. If $p(t)=\det (A-tI)$, the map $k[t]\to k[\varphi ]$ sending $t$ to $\varphi$ is a ring homomorphism. Since a determinant is a polynomial expression in its entries, it sends $p(t)$ to $\det (A-\varphi I)=0$. ThusWith the more usual monic characteristic polynomial ${\chi }_A(t)=\det (tI-A)=(-1{)}^{\dim V}p(t)$, this reads ${\chi }_A(\varphi )=0$, the Cayley-Hamilton theorem. If the matrix convention uses column coordinates instead of the row convention above, transpose $A$; its characteristic polynomial is unchanged.

We first prove the local algebra fact behind regularity in codimension one. Let $(A,\mathfrak{m})$ be a one-dimensional normal Noetherian local domain. Choose $0\ne a\in \mathfrak{m}$. Its radical is $\mathfrak{m}$, so some ${\mathfrak{m}}^n$ lies in $(a)$; take minimal $n$. If $n=1$, then $\mathfrak{m}=(a)$. Otherwise choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$ and put $z=b/a$ in the fraction field. We have $z\notin A$ but $z\mathfrak{m}\subseteq A$.

If $z\mathfrak{m}\subseteq \mathfrak{m}$, express multiplication by $z$ on a finite generating set of $\mathfrak{m}$. The determinant trick gives a monic polynomial in $z$ annihilating $\mathfrak{m}$; as this ideal is nonzero in a domain, the polynomial itself vanishes. Normality would then give $z\in A$, a contradiction. Hence there is $u\in \mathfrak{m}$ with $zu$ a unit of $A$. Every $v\in \mathfrak{m}$ satisfies $v=u(zv)/(zu)$, proving $\mathfrak{m}=(u)$. The maximal ideal therefore has one generator in dimension one, so $A$ is a regular local ring, indeed a discrete valuation ring. This proves one-dimensional normal local rings are discrete valuation rings directly.

Now a codimension-one point of the normal variety has such a local ring, and a codimension-zero local ring is a field. Both are regular. Since the algebraically closed ground field is perfect, regularity and smoothness agree for this finite-type variety, and the Jacobian criterion makes the singular locus closed. It contains no codimension-zero or codimension-one point. Therefore every irreducible component of the singular locus has codimension at least two, givingFor $d\le 1$ this means the singular locus is empty. In particular the singular set itself is the required closed subset, not merely a set contained in some larger one.

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving

Work on affine open subsets. Their coordinate rings are Noetherian rings and integrally closed domains. At a prime ideal of height of a prime ideal one, the local ring is a one-dimensional Noetherian local ring which is an integrally closed domain, hence a discrete valuation ring and therefore a regular local ring. At the generic point the local ring is a field, hence a regular local ring. For a brief justification of the one-dimensional local ring fact, take $0\ne a\in \mathfrak{m}$ in a one-dimensional Noetherian local ring $R$ which is an integrally closed domain. Some ${\mathfrak{m}}^n\subset (a)$; choose the least such $n$, and choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$. Then $q=b/a\notin R$, but $q\mathfrak{m}\subset R$. If $q\mathfrak{m}\subset \mathfrak{m}$, the determinant trick applied to the finitely generated faithful module $\mathfrak{m}$ would make $q$ integral over $R$, a contradiction. Hence the ideal $q\mathfrak{m}$ contains a unit and equals $R$. Thus $\mathfrak{m}=q^{-1}R$ is principal. A one-dimensional Noetherian local ring with principal maximal ideal is a discrete valuation ring. Therefore a normal variety is regular in codimension one.

The ground field is an algebraically closed field, hence a perfect field, so having a regular local ring is equivalent to being a smooth point of a variety here. Moreover, the smooth locus of a variety is open, making the singular locus closed. If an irreducible component of the singular locus had algebraic codimension zero or one, its generic point would have a regular local ring by the preceding argument, a contradiction. Consequently every such component has algebraic codimension at least two.

One can read the numerical bound directly from chains of prime ideals, without identifying algebraic dimension with transcendence degree. For a prime ideal $P$ of height of a prime ideal at least two in an affine chart, append a chain $(0)⊊P_1⊊P$ below any chain in the quotient by $P$. Its length increases by two. Thus $\dim A/P\le \dim A-2$. Since a nonempty affine open subset of the irreducible variety $X$ has algebraic dimension $d$,For $d=0$ or $1$, this means the singular locus is empty; take $\dim \varnothing =-\infty$.

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.

The integral closure of $R$ in $T$ isIt is a subring: finitely many integral elements generate a finite $R$-module algebra, and the determinant trick shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.

A valuation ring is an integral domain $V$ such that, for every nonzero $z$ in its fraction field, either $z\in V$ or $z^{-1}\in V$. Equivalently, its ideals are totally ordered by inclusion. For principal ideals, comparability is precisely the condition on $a/b$; if two arbitrary ideals were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a ring is local. Its nonunits form an ideal: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique maximal ideal consists of those nonunits.

First, valuation rings are integrally closed. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its maximal ideal $\mathfrak{m}$. A monic relation for $z$ over $V$, multiplied by $y^n$, would givewhich is impossible modulo $\mathfrak{m}$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.

For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the valuation domination lemma: a local subring $(A,\mathfrak{n})$ of a field $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and ${\mathfrak{m}}_V\cap A=\mathfrak{n}$. Here is a proof, including the crucial maximality step.

Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the rings and of their maximal ideals. The union is a local ring: an element outside the union ideal is already a unit in a member of the chain, while an element in that ideal cannot become a unit in a later dominating member. The union still dominates $A$. Thus Zorn's lemma supplies a maximal pair $(V,\mathfrak{m})$.

For any $z\in K^{\times }$, at least one of $\mathfrak{m}V[z]$ and $\mathfrak{m}V[z^{-1}]$ is proper. Suppose otherwise. There would be relationswith $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\ge s$, the second relation givesRepeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak{m}$. This contradicts minimality of $n$ (or gives $1\in \mathfrak{m}$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.

Choose whichever extension has a proper extended ideal, then a maximal ideal containing it. Localizing that extension at the chosen maximal ideal produces a local ring dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its fraction field is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the valuation domination lemma.

Now let $x\in K$ be nonintegral over $R$, so $x\ne 0$, and set $y=x^{-1}$, $B=R[y]$. The ideal $yB$ is proper: otherwise $1=y(r_0+r_{1}y+\cdots +r_n{y}^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a maximal ideal $\mathfrak{n}$ of $B$ containing $y$, and apply the valuation domination lemma to $B_{\mathfrak{n}}\subset K$. Its dominating valuation ring $V$ contains $R$ and has $y\in {\mathfrak{m}}_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.

We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the integral closure as an intersection of valuation rings isNeither Noetherianity nor a discrete valuation is required for this separation argument.

The key is a Laurent normalization by a monomial change of variables. We prove the slightly stronger assertion that every nonzero quotient of a Laurent polynomial ring in $r$ variables is finite as a module over an embedded Laurent polynomial ring in some number $m\le r$ of variables.

Induct on $r$. If the defining ideal is zero, the ring itself works. Otherwise choose a nonzero Laurent polynomialin that ideal, with nonzero coefficients. For a sufficiently large positive integer $N$, the weightsare distinct for all $\mathbf{a}\in F$. Indeed, each difference gives a nonzero polynomial in $N$, which has only finitely many integer roots. Make the invertible monomial substitutionIts inverse is $Y_j=X_j{X}_r^{-N^j}$, so it is an automorphism of the Laurent polynomial ring. When $r=1$ there is no substitution and distinct exponents already give distinct weights.

As a Laurent polynomial in $Y_r$, the transformed relation has unique lowest and highest exponents. Their coefficients are nonzero scalar multiples of monomials in $Y_1,\dots ,Y_{r-1}$, hence units inMultiplying by a power of $Y_r$ and the inverse leading coefficient givesHere $d>0$: a single-term Laurent polynomial would be a unit and the quotient would be zero. Reversing the equation and dividing by $b_0{Y}_r^d$ gives a monic equation for $Y_r^{-1}$. Thus both $Y_r$ and $Y_r^{-1}$ are integral over the image $B$ of $B_0$ in $T$. In fact $b_0$ being a unit also expresses $Y_r^{-1}$ as a polynomial in $Y_r$, so $T$ is generated as a $B$-module by $1,Y_r,\dots ,Y_r^{d-1}$.

The ring $B$ is itself a nonzero quotient of an $(r-1)$-variable Laurent polynomial ring. By induction, it is finite over an embedded Laurent polynomial ring in $m\le r-1$ variables. Finiteness of modules is transitive, so the same is true of $T$; a finite ring extension is an integral extension, by the determinant trick applied to multiplication by an element.

Applying this to $r=3$ yields an embedded subringThe base case $m=0$ means the subring $k$; finiteness over it would make $T$ a finite-dimensional vector space. The hypothesis excludes this, so $1\le m\le 3$, as required. No infinitude assumption on $k$ is used.

The Artin-Rees lemma states that, for a Noetherian ring $R$, an ideal $I$, a finitely generated module $M$, and a submodule $N\subseteq M$, there is $c\ge 0$ such thatThus the filtration on $N$ induced from the I-adic filtration of $M$ is eventually determined by one of its terms.

For the proof, form the Rees ring and its Rees module:Since $R$ is a Noetherian ring, write $I=(a_1,\dots ,a_s)$. Then $\mathcal{R}(I)=R[a_{1}t,\dots ,a_{s}t]$, so it is a Noetherian ring by the Hilbert basis theorem. Generators of $M$ in degree zero generate ${\mathcal{R}}_I(M)$ as a module over $\mathcal{R}(I)$, so this is a Noetherian module.

The graded submoduleis therefore finitely generated. Choose homogeneous generators $z_{\alpha }{t}^{d_{\alpha }}$ with $d_{\alpha }\le c$ for some $c$. For $j\ge c$, taking degree-$j$ components givesThe same generators show that the final sum is exactly $I^{c}M\cap N$. This proves the Artin-Rees lemma. If $L=0$, one may simply take $c=0$.

Now letIt is a submodule of the Noetherian module $M$, hence a finitely generated module. Apply the Artin-Rees lemma with $N=K$ and $j=c+1$. Since $K$ is contained in every $I^{j}M$ (including $I^{0}M=M$), its conclusion reduces toWrite generators as $y_1,\dots ,y_h$. There is a matrix $A=(a_{ij})$ with entries in $I$ such that $y_i={\sum }_j{a}_{ij}{y}_j$. Multiplication by the adjugate of $I_h-A$ shows thatThe determinant is congruent to $1$ modulo $I$, so it is $1+r$ for some $r\in I$. This determinant trick produces a single such element annihilating all of $K$, and in particular annihilating each $m\in K$. If $K=0$, use $r=0$.

Conversely, if $(1+r)m=0$ with $r\in I$, then $m=-rm$. Iterating gives $m=(-r{)}^{j}m\in I^{j}M$ for every $j\ge 1$. Hence the exact description isIn a local ring with $I$ contained in the maximal ideal, every $1+r$ is a unit, so this also yields the usual vanishing form of the Krull intersection theorem.

For a failure without the Noetherian ring hypothesis, let $D=\mathbb{Z}[1/2{]}_{\ge 0}$ and use the semigroup algebraIts elements are finite sums of formal monomials $t^q$, with multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. Each ring in the union is a polynomial ring and the inclusions are injective, so $R$ is an integral domain. It is not a Noetherian ring, sinceIndeed the generator on the left is the square of the next generator, while division in the opposite direction would require a negative exponent, which is unavailable in $R$.

Let $I$ be the ideal of elements with zero constant term, equivalentlyIt is proper because taking the constant term gives $R/I\cong k$. Every generator $t^q$ is $(t^{q/2}{)}^2$, with $t^{q/2}\in I$, so $I^2=I$. Thus this is a nonzero idempotent ideal, with $I^j=I$ for every $j\ge 1$. TakeFor every $r\in I$, the constant term of $1+r$ is $1$, so $1+r\ne 0$. Since $R$ is an integral domain and $t\ne 0$, we have $(1+r)t\ne 0$. This gives all three requested objects and proves that no element of the required form annihilates $m$.

Choose an integral basis $u_1,\dots ,u_d$ of the nonzero ideal $\mathfrak{a}$. Its rational span is $L$, since it contains a nonzero integer multiple of ${\mathcal{O}}_L$. The condition $x\mathfrak{a}\subseteq \mathfrak{a}$ gives integers $m_{ij}$ with $x{u}_j={\sum }_i{m}_{ij}{u}_i$. Thus multiplication by $x$ on $L$ has an integer matrix $M$ in this basis.

Its characteristic polynomial $P(t)=\det (tI-M)$ is monic with integer coefficients. By the Cayley-Hamilton theorem, $P(M)=0$. On the field $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an algebraic integer, andThis is the determinant trick for an ideal preserved by multiplication.

A ring extension $A\subseteq B$ is finite when $B$ is a finitely generated module over $A$. Every finite extension is integral by the determinant trick.

Choose $0\ne a\in \mathfrak{m}$. Since the one-dimensional local domain has no nonzero prime ideal other than $\mathfrak{m}$, one has $\sqrt{(a)}=\mathfrak{m}$. Finite generation of $\mathfrak{m}$ therefore gives some $n\ge 1$ with ${\mathfrak{m}}^n\subseteq (a)$. Choose $n$ minimal andIn the fraction field of $A$, put $y=x/a$. Then $y\notin A$ but

If $y\mathfrak{m}\subseteq \mathfrak{m}$, multiplication by $y$ would preserve the nonzero finitely generated faithful $A$-module $\mathfrak{m}$. The determinant trick would make $y$ integral over $A$, contradicting that $A$ is integrally closed and $y\notin A$. Hence some $t\in \mathfrak{m}$ satisfies $yt\notin \mathfrak{m}$. Since $yt\in A$ and $A$ is local, $yt$ is a unit.

For any $z\in \mathfrak{m}$, one has $yz\in A$, and thereforeThus $\mathfrak{m}\subseteq (t)$, while the reverse inclusion follows from $t\in \mathfrak{m}$. ConsequentlyThis proves the principal maximal ideal in a one-dimensional normal local domain result.