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Put $S=R^G$, the ring of invariants. It is an integral domain because it is a subring of the domain $R$. Embed $\mathrm{Frac}S$ in $\mathrm{Frac}R$ in the evident way.

Suppose $q\in \mathrm{Frac}S$ is an integral element over $S$. Its monic equation has coefficients in $S\subseteq R$, so $q$ is integral over $R$. Since $R$ is a normal domain, $q\in R$.

Write $q=a/b$ with $a,b\in S$ and $b\ne 0$. Each ring automorphism in $G$ extends to the fraction field by acting on numerator and denominator, and fixes this fraction. Hence $q\in R^G=S$. We have proved that $S$ is integrally closed in its own fraction field:This normality of a ring of invariants argument actually works for any group; finiteness is not needed for this conclusion. For finite $G$, there is additionally an integral extension $R^G\subseteq R$, since for each $r\in R$ the orbit polynomial under a finite automorphism groupis monic, has coefficients fixed by $G$ and vanishes at $r$. Neither argument divides by $|G|$, so it is valid when the characteristic divides the group order. Here normality means integral closedness of a domain.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

Let $k\to A$ be a map of commutative rings. The Kähler differentials ${\Omega }_{A/k}$ are generated as an $A$-module by symbols $da$, with relationsThese relations make $d:A\to {\Omega }_{A/k}$ the universal $k$-derivation. Every $k$-derivation $D:A\to M$ into an $A$-module factors uniquely through the map $da\mapsto D(a)$. Thus the universal property of Kähler differentials isThis constructs the module and proves its universal property, rather than merely listing a formal derivative rule.

For the polynomial ring $P=k[X_i:i\in T]$, the Kähler differentials of a polynomial algebra form the free module ${⨁}_{i\in T}Pd{X}_i$. The usual formal partial derivatives prove that assigning arbitrary images to $d{X}_i$ defines a derivation, and every polynomial involves only finitely many variables. For $A=P/I$, the Conormal exact sequence for Kähler differentials givesThe map is well-defined because $d(I^2)$ vanishes after reduction modulo $I$. Its cokernel has exactly the universal property of derivations on $P$ that kill $I$, so is ${\Omega }_{A/k}$. In a finite polynomial presentation this givesThere need not be injectivity at $I/I^2$: in characteristic $p$, the relation $X^p$ has derivative zero.

Localization of Kähler differentials commutes with localization of a ring:The quotient rule follows by differentiating $s\cdot s^{-1}=1$. It extends every derivation uniquely and proves the isomorphism by the universal property. Likewise base change for Kähler differentials gives ${\Omega }_{(A{\otimes }_k{k}^{\prime })/k^{\prime }}\cong {\Omega }_{A/k}{\otimes }_k{k}^{\prime }$: a $k^{\prime }$-linear derivation is determined by its values on $A\otimes 1$, and the product rule extends those values to the tensor product.

For a tower $k\to F\to L$, the Transitivity exact sequence for Kähler differentials isQuotienting ${\Omega }_{L/k}$ by the submodule generated by differentials of elements of $F$ represents precisely the $F$-derivations, which proves exactness. The first map need not be injective in general. To relate this to a transcendence basis, we need the stronger property supplied by a separable field extension.

If $L/F$ is finite separable and $u$ has minimal polynomial $g(T)=T^r+c_{r-1}{T}^{r-1}+\cdots +c_0$, differentiating its equation forcesThe denominator is nonzero by separability. Conversely this formula extends an arbitrary $k$-derivation $F\to M$, for an $L$-module $M$, to $F[u]$; it kills the relation $g$ and hence descends to $L$. A tower of simple separable extensions proves unique extension for all finite separable $L/F$. Therefore Kähler differentials under a separable field extension satisfyThis proves injectivity in this case, which would not follow from right exactness alone.

A transcendence basis $t_1,\dots ,t_d$ for a finitely generated field extension $L/k$ has algebraic independence over $k$ and makes $L/k(t_1,\dots ,t_d)$ algebraic, hence finite. A separating transcendence basis additionally makes that finite extension separable. Apply the polynomial computation and the quotient rule to the rational function field $F=k(t_1,\dots ,t_d)$, and then the separable-extension isomorphism. We obtain the central link:when the basis is separating. The dual statement says that any prescribed values of the $D(t_i)$ extend uniquely to a $k$-derivation of $L$ with values in $L$. In characteristic zero every transcendence basis of a finitely generated field extension is separating, so Kähler differentials measure transcendence degree exactly.

There is also a characteristic-zero test for algebraic independence. If $t_1,\dots ,t_r$ are algebraically dependent, choose a nonzero polynomial relation of minimum total degree. Some formal partial derivative is nonzero in characteristic zero, and it cannot also vanish on the tuple, since it has smaller degree. Differentiation therefore gives a nontrivial linear relation among the $d{t}_i$. Conversely, an algebraically independent tuple extends to a transcendence basis, whose differentials form a basis as just proved. Thus differentials detect algebraic independence in characteristic zero: a finite tuple is algebraically independent exactly when its differentials are linearly independent. Its differentials form a basis of ${\Omega }_{L/k}$ exactly when the tuple is a transcendence basis.

The separability qualification is essential in positive characteristic. For $k={\mathbb{F}}_p(s)$ and $L=k(u)$ with $u^p=s$, the extension is purely inseparable of degree $p$, with transcendence degree zero. The polynomial presentation yields ${\Omega }_{L/k}=Ldu\ne 0$, because the defining relation has derivative zero. Also, in ${\mathbb{F}}_p(t)/{\mathbb{F}}_p$, the tuple $\{t^p\}$ is a transcendence basis but $d(t^p)=0$, while $dt$ is a basis of the one-dimensional differential module. A chosen arbitrary transcendence basis therefore need not give differential coordinates in characteristic $p$.

Finally, the presentation makes the relation useful geometrically. For $A=k[x,y]/(y^2-x^3)$ in characteristic zero,After passage to the fraction field, it has dimension one, the transcendence degree of the curve. At the origin, tensoring with its residue field leaves both $dx,dy$ independent, so the differential fibre has dimension two. For a $k$-rational point with maximal ideal $\mathfrak{m}$, that fibre is the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$: write elements as their constant value plus an element of $\mathfrak{m}$, and note that derivations into $k$ kill ${\mathfrak{m}}^2$. This explains how Kähler differentials record both generic transcendental parameters and the extra tangent direction at a singular point.

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.

On an affine chart $\mathrm{Spec}A\subseteq V$, the Module of Kähler differentials ${\Omega }_{A/k}$ is generated by symbols $da$ subject to $k$-linearity and $d(ab)=adb+bda$. It represents $k$-derivations. These modules commute with localization, so their associated quasi-coherent sheaves glue to the Kähler differential sheaf ${\Omega }_{V/k}^1$. It is coherent: if $A=k[x_1,\dots ,x_N]/I$ and $I=(f_1,\dots ,f_s)$, then it has the finite presentation of a moduleFor a closed point $P$, put $R={\mathcal{O}}_{V,P}$ with maximal ideal $\mathfrak{m}$ and residue field $k$. The Zariski tangent space is ${\mathrm{Hom}}_k(\mathfrak{m}/{\mathfrak{m}}^2,k)$, equivalently ${\mathrm{Der}}_k(R,k)$. A derivation kills constants and ${\mathfrak{m}}^2$, so it factors through $a\mapsto a-a(P)$; conversely every linear functional on $\mathfrak{m}/{\mathfrak{m}}^2$ defines such a derivation by the product rule. The universal property of Kähler differentials therefore identifies this space withIn particular, ${\Omega }_{V,P}^1/{\mathfrak{m}}_P{\Omega }_{V,P}^1\cong {\mathfrak{m}}_P/{\mathfrak{m}}_P^2$, the algebraic cotangent space.

A point is a smooth point of a variety when its local ring is a regular local ring; over this algebraically closed field this says ${\dim }_k{T}_{V,P}=n=\dim V$. Tensor the finite presentation of a module above with $k(P)$. The tangent space is the kernel of the Jacobian matrix $J(P)$, hence has dimension $N-\mathrm{rank}J(P)$. This proves the Jacobian criterionFor a smooth irreducible variety $V$, choose at each point an invertible $(N-n)$-minor and shrink the affine chart so that it remains invertible. Its relations eliminate $N-n$ of the generators of ${\Omega }_{A/k}$, yielding a surjection $A^n\to {\Omega }_{A/k}$. At the generic point, its target has dimension $n$: over a perfect ground field, a separating transcendence basis of the function field has differentials forming a basis. Equivalently this follows from the assumed density of the smooth locus. The kernel therefore becomes zero over the fraction field of the integral domain $A$. As a submodule of $A^n$, it is a torsion-free module, so it is already zero. Thus these maps give local isomorphisms with ${\mathcal{O}}_V^n$, proving local freeness of differentials on a smooth variety with rank $n$.

For an affine chart $\mathrm{Spec}A\subseteq V$, write $W\cap \mathrm{Spec}A=\mathrm{Spec}(A/I)$ and $\mathcal{M}=\tilde{M}$. The restriction of a module sheaf to a closed subvariety is $\widetilde{M{\otimes }_{A}A/I}=\widetilde{M/IM}$. The Conormal exact sequence for Kähler differentials isIt follows from the generators and relations: passing to $A/I$ imposes precisely the additional relations $df=0$ for $f\in I$. The first map is well defined because $d(I^2)$ lies in $I{\Omega }_{A/k}$. Glue these exact module sequences, using exactness of localization, to obtainIn the final assertion, interpret locally principal subvariety as a proper local hypersurface. Its ideal on each chart is $(f)$ with $f\ne 0$. Since $V$ is an irreducible variety and reduced, $f$ is a non-zero-divisor, and $A/(f)\to (f)/(f^2)$, $\bar{a}\mapsto af$, is an isomorphism. These local rank-one descriptions make the conormal sheaf ${\mathcal{I}}_W/{\mathcal{I}}_W^2$ invertible.

Because $W$ is not contained in the singular locus of $V$, there is a dense open subset of $W$ where both $V$ and $W$ are smooth varieties. At a closed point there, $\dim W=\dim V-1$ by the Krull principal ideal theorem. The tangent description $T_{W,P}=\ker (df:T_{V,P}\to k)$ then forces $df\ne 0$. Hence the conormal map is injective at the generic point of $W$. Its kernel is a subsheaf of a line bundle on the integral variety $W$, so it is a torsion-free sheaf; a torsion-free sheaf with zero generic fibre is zero. This proves conormal injectivity for a generically smooth Cartier divisor. If zero equations were allowed in the phrase locally principal, $W=V$ would be a counterexample to invertibility; the proper-hypersurface convention is essential.

The fraction field of an integral domain $R$ is a field $F$ containing an identified copy of $R$ in which every element has the form $a/b$ with $a,b\in R$ and $b\ne 0$. Equality means $a/b=c/d$ exactly when $ad=bc$; the operations are the usual addition and multiplication of fractions. We use unital ring homomorphisms.

Since $\phi$ is injective, $\phi (b)\ne 0$ whenever $b\ne 0$. DefineIf $ad=bc$, applying the ring homomorphism and dividing by $\phi (b)\phi (d)$ proves the value is independent of the representative. The fraction addition and multiplication rules show that $\Phi$ preserves both operations and one, and $\Phi (a/1)=\phi (a)$. If $\Phi (a/b)=0$, then $\phi (a)=0$, so $a=0$ because $\phi$ is an injective function; therefore its kernel is zero and $\Phi$ is injective. It is also the unique extension, because any ring homomorphism extending $\phi$ must send $b^{-1}$ to $\phi (b{)}^{-1}$.

For a counterexample, take $R=\mathbb{Z}$, $F=\mathbb{Q}$, and reduction modulo two $\psi :\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$. An extension would give $1=\Psi (2)\Psi (1/2)=0$, a contradiction. Nonzero denominators must become invertible in the target; an arbitrary ring homomorphism need not have that property.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the Going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The Going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the Going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

A unique factorization domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements, uniquely up to reordering and multiplication of factors by a unit in a ring. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Gauss lemma for polynomials says that over a unique factorization domain $D$, the product of two primitive polynomials is primitive, and the polynomial content satisfies $c(fg)=c(f)c(g)$ up to a unit. In particular a primitive polynomial of positive degree is irreducible over $D$ exactly when it is irreducible over the fraction field of $D$. Consequently a polynomial ring over a unique factorization domain is itself a unique factorization domain.

The Eisenstein criterion says that if $f(X)=a_n{X}^n+\cdots +a_0\in D[X]$ has $n\ge 1$ and there is a prime element $p$ with $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$, then $f$ is irreducible over the fraction field. If $f$ is primitive, it is also irreducible in $D[X]$. Over $\mathbb{Z}$ one may take $p$ to be an ordinary prime number. The primitivity qualification prevents a nonunit constant content from being overlooked.

Use $q\in k^{\times }$ and $YX=qXY$ for the quantum plane. Its monomials $X^i{Y}^j$, $i,j\ge 0$, form a basis, with multiplicationOne way to verify the basis assertion without assuming it is to define this multiplication on the vector space with the displayed formal basis. This defines an associative algebra: for a third monomial $X^w{Y}^z$, the two products of three monomials have the same exponent $ju+jw+vw$ of $q$, and its generators satisfy the required relation. Conversely, the relation puts every word into this form, establishing the presentation. Order exponent pairs by the lexicographic order. The largest monomials of two nonzero finite sums give the uniquely largest monomial of their product, with coefficient $cd{q}^{ju}\ne 0$. Therefore the quantum plane is a domain. If one allows $q=0$, the assertion fails because $YX=0$ with both factors nonzero.

A uniform module is a nonzero module in which any two nonzero submodules have nonzero intersection. Suppose the right regular module of a right Noetherian domain $A$ were not a uniform module. Choose nonzero $a,b$ from two right ideals with zero intersection. Then $aA\cap bA=0$. The right idealsform a direct sum. For if ${\sum }_{i=0}^m{b}^{i}a{c}_i=0$, then $a{c}_0\in aA\cap bA$ is zero, so $c_0=0$ by the noncommutative domain property. Cancel the nonzero factor $b$ on the left and repeat to obtain every $c_i=0$. Each summand is nonzero, so their finite partial sums form a strictly ascending chain of right ideals. This contradicts the ascending chain condition of a right Noetherian ring. Hence $A$ is a uniform right module.

It follows that $aA\cap sA\ne 0$ whenever $a,s\ne 0$: there are nonzero $u,v$ with $au=sv$. This is the right Ore condition for the multiplicative set $S=A\setminus \{0\}$; zero numerators cause no difficulty. The Ore localization theorem therefore constructs the ring of right fractionsThe map is injective because an element mapping to zero is annihilated on the right by some nonzero denominator, impossible in a noncommutative domain. To make the denominator convention concrete, if $su=tv\ne 0$ thenFor multiplication, choose $bu=sc$ with $u\ne 0$; thenCommon right multiples make these operations independent of the chosen representatives. Every nonzero $a{s}^{-1}$ has inverse $s{a}^{-1}$, so $Q$ is a division ring. The original field $k$ is central in $A$ and therefore in the inverses as well, making $Q$ a division algebra over $k$. No commutative fraction field construction is being assumed.

Let $A=A(X)=k[x_1,\dots ,x_m]/I(X)$. Polynomial restrictions are regular functions, and a restriction is zero precisely when its class in the coordinate ring is zero. We prove surjectivity.

Using the usual irreducible-variety convention, view a regular function $f$ as an element of the fraction field of $A$. On a neighbourhood of every point it has the form $a/b$, where $a,b\in A$ and $b$ does not vanish at that point. Equality on a nonempty open subset of the irreducible variety is equality of rational functions. DefineThis is an ideal and contains a denominator nonzero at each point. Thus its zero locus is empty. The Hilbert Nullstellensatz implies $J=A$, so $1\in J$ and $f\in A$. ThereforeThe result also holds under a convention allowing reduced reducible affine varieties. One may cover by finitely many principal opens $D(b_i)$, represent $f$ in each localization, and increase powers until the local representatives glue after clearing denominators. Since these opens cover, the ideal generated by those powers is the unit ideal by the Nullstellensatz; a unit linear combination glues their numerators to one element of $A$. Irreducibility is used only for the shorter fraction-field proof, not for the assertion itself.

The semicubical parabola has coordinate ringunder $x\mapsto t^3$, $y\mapsto t^2$. The isomorphism is valid in every characteristic: after division by the monic polynomial in $x$, a remainder $P(y)+xQ(y)$ maps to a sum of distinct even powers and odd powers at least three, and can vanish only when both polynomials are zero.

Its fraction field contains $t=t^3/t^2$, which is an integral element over $A$ because it satisfies $z^2-y=0$. However $t\notin A$, since a polynomial in $t^2,t^3$ has no term of degree one. Thus $A$ is not an integrally closed domain. The ring $k[t]$ of the affine line is a principal ideal domain and therefore an integrally closed domain. This property is preserved by ring isomorphism. By coordinate-ring functoriality,The parametrization is a bijection on points over an algebraically closed field, but that does not make its inverse a morphism at the cusp.

A Dedekind domain is a Noetherian ring $D$ which is an integral domain, is an integrally closed domain, and has Krull dimension one. Explicitly, every ideal is finitely generated; every element of the fraction field that is an integral element over $D$ already belongs to $D$; and every nonzero prime ideal is a maximal ideal, with $D$ itself not a field.

With this convention, an equivalent local characterization is that $D$ is a Noetherian integral domain that is not a field and whose localization at a prime ideal $D_P$ is a discrete valuation ring for every nonzero prime ideal $P$. The integrally closed domain condition rules out singularities such as those represented by the subring $k[t^2,t^3]\subset k[t]$, where $t$ is integral but missing.

Some conventions also include fields by allowing Krull dimension at most one; the conclusions about ideals below remain valid under that convention.

Write $f:U={\mathbb{A}}_{\mathbb{C}}^1\setminus \{1\}\to {\mathbb{A}}_{\mathbb{C}}^1$ for $f(t)=t^2$. For any closed point $a\in \mathbb{C}$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb{C}$, and every finite morphism is proper. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.

The morphism itself fails the valuative criterion for properness. Take the discrete valuation ring $R=\mathbb{C}[[s]]$ with fraction field $K=\mathbb{C}((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined byAny extension $\mathrm{Spec}R\to {\mathbb{A}}^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that proper closed-point fibers do not imply properness.

The Going-down theorem states: let $R\subseteq A$ be an integral extension of integral domains, with $R$ integrally closed in its fraction field. If ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ are prime ideals of $R$ and ${\mathfrak{q}}_1$ is a prime ideal of $A$ lying over ${\mathfrak{p}}_1$, then there is a prime ideal ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ lying over ${\mathfrak{p}}_0$.

Let $R$ be a valuation ring with fraction field $K$. For every commutative squarethe valuative criterion for separatedness says that a finite type morphism $f$ between Noetherian schemes is separated exactly when there is at most one dotted lift $\mathrm{Spec}R\to X$ completing the diagram.

Under the same finiteness hypotheses, the valuative criterion for properness says that $f$ is proper exactly when every such square has a unique lift. Thus separatedness supplies uniqueness, while properness supplies existence as well.

Let $R$ be a discrete valuation ring with uniformizer $\pi$, fraction field $K$, and normalized discrete valuation $\nu :K^{\times }\to \mathbb{Z}$. A map $\mathrm{Spec}K\to {\mathbb{P}}_{\mathbb{Z}}^n$ is a projective pointwith at least one nonzero coordinate. Put $m={\min }_{a_i\ne 0}\nu (a_i)$ and set $b_i={\pi }^{-m}{a}_i$. Then every $b_i$ lies in $R$, and at least one $b_j$ is a unit.

On the standard affine chart $U_j$ of projective space, the ratios $b_i/b_j$ all lie in $R$. They therefore define a map $\mathrm{Spec}R\to U_j\subseteq {\mathbb{P}}_{\mathbb{Z}}^n$ whose generic restriction is the original point. This proves existence in the DVR case. The assumed separatedness of ${\mathbb{P}}_{\mathbb{Z}}^n\to \mathrm{Spec}\mathbb{Z}$, through the valuative criterion for separatedness, gives uniqueness. Thus the morphism satisfies the requested DVR form of the valuative criterion for properness.

Write $A=k[x,y]$ and $\mathfrak{m}=(x,y)$. Since $f(0,0)\ne 0$, the principal open subscheme $D(f)=\mathrm{Spec}{A}_f$ contains $\mathfrak{m}$, andCover $V$ by the two affine opens $D(xf)$ and $D(yf)$, whose intersection is $D(xyf)$. The degree-zero part of the resulting Čech cohomology complex givesinside the fraction field of $A$. The last equality follows because $A_f$ is a unique factorization domain and a rational function regular after localizing at both $x$ and $y$ has no possible prime factor left in its denominator.

The same affine cover is acyclic, so its degree-one Čech group computes sheaf cohomology and givesBefore localizing at $f$, the quotienthas the $k$-basisWriting $f=c+h$ with $c=f(0,0)\in k^{\times }$ and $h\in (x,y)$, multiplication by $h$ is locally nilpotent on $Q$: for each negative monomial, a sufficiently high power of $(x,y)$ moves every term into $A_x+A_y$. Hence $c+h$ acts invertibly on $Q$ by a finite geometric series on each element. Localizing at $f$ therefore leaves $Q$ unchanged, andThe displayed infinite basis proves that this vector space is infinite-dimensional.