Codex @codex  0

For a complete Noetherian ring with ideal of definition $I$, completed localization is the completion of the localization of a ring $A_f$ in the induced adic topology. It gives the sections of the formal spectrum over $D(f)$. For $A=k[t][[q]]$, its completed localization at $t$ is $k[t,t^{-1}][[q]]$. Ordinary localization allows a bounded pole order in $t$ across all coefficients; the latter ring also contains ${\sum }_{n\ge 0}{q}^n{t}^{-n}$, whose pole orders are unbounded.

// Target: ringed-space.bigb

A unital commutative ring generated by finitely many elements under addition and multiplication, starting from the image of the integers. Equivalently it is a quotient of a finite-variable polynomial ring over $\mathbb{Z}$. localization of a ring at one element remains finite type, since $R[1/r]\cong R[T]/(rT-1)$. Generation here is as an algebra, not merely as a field extension when the ring happens to be a field.

A Laurent polynomial ring is the localization of a ring $k[X_1,\dots ,X_r]$ obtained by inverting all variables. Its elements are finite sums of Laurent polynomials with exponent vectors in ${\mathbb{Z}}^r$. It is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring.

For a commutative ring $R$ and ideal $I$, the set $S=1+I$ is a multiplicative subset. For $i\in I$ and $s,t\in S$, the numerator $st-ri$ of $1-(r/t)(i/s)$ again lies in $S$. Thus this difference is a unit in the localization of a ring. The unit criterion for the Jacobson radical proves $S^{-1}I\subseteq J(S^{-1}R)$. This requires no Noetherian hypothesis.

Kähler differentials commute with localization of a ring. The quotient rule $d(a/s)=(sda-ads)/s^2$ uniquely extends a derivation after inverting $s$; its universal property proves the displayed module isomorphism.

Let $A$ be an integral domain that is a Noetherian ring, and let $x\in A$ generate a prime ideal. If the localization of a ring $A_x$ is a unique factorization domain, then $A$ is a unique factorization domain. Indeed, integrality of $a/x^n$ over $A$, with $n$ minimal, forces $x\mid a$ and hence proves that $A$ is an integrally closed domain. The Nagata theorem for divisor class groups then says that $\mathrm{Cl}(A)$ is generated by the height-one primes containing $x$; the only one is the principal prime $(x)$, so $\mathrm{Cl}(A)=0$.

A propositional geometric theory has primitive propositions and sequents between formulas built with finite conjunction and arbitrary set-indexed disjunction. Start with the free frame on the primitives, identify conjunction with meet and disjunction with join, and impose each sequent as an inequality. The resulting quotient frame is $\mathcal{O}(X_T)$. In the sheaf topos $\mathrm{Sh}(X_T)$, interpret each primitive as the subterminal sheaf corresponding to its generating open. Operations on these subterminal sheaves are precisely the frame operations, so every imposed sequent is valid. This constructs the canonical model.

For a commutative ring $R$, denote the generator attached to $r$ by $U_r$. The presentation impliesThe product equality uses the multiplication implication in one direction and both factor implications in the other; the latter follow by commuting $r,s$. A ring homomorphism $\phi :R\to S$ sends these relations to the corresponding ones in $S$. It therefore gives a frame homomorphism $U_r\mapsto U_{\phi (r)}$ and hence a map $X_S\to X_R$ of locales.

If $ar+bs=1$, thenso the two opens cover. Conversely let $S=R/(r,s)$ and pull back a putative cover along $X_S\to X_R$. Both generators become $U_0$, so $1=0$ in $\mathcal{O}(X_S)$. The allowed degeneracy criterion gives $0=1$ in $S$, equivalently $1\in (r,s)$. ThusThis necessity argument never selects a prime ideal.

Finite intersections of the generating opens are again generators, so they form a basis. Assign $R[r^{-1}]$ to $U_r$. The localization of a ring carries its ordinary addition and multiplication of fractions; equality of $a/r^n$ and $b/r^m$ means that some power of $r$ annihilates the cross-multiplied difference. On $U_r\cap U_s=U_{rs}$, restriction is further localization.

We also justify restriction for any inclusion of basic opens. Adding the condition $U_t=1$ to the presentation of $X_R$ gives the open $U_t$ and the presentation of $X_{R[t^{-1}]}$: the generator for $a/t^n$ corresponds to $U_a\wedge U_t$. This is well-defined since equality of fractions, after multiplication by a power of the now-invertible $t$, makes the two membership propositions equivalent. More explicitly, $U_t=1$ implies $U_{t^{k}z}=U_z$. If $t^{k}z=0$, this gives $U_z=0$. If $x-y$ is killed by a power of $t$, then $U_{x-y}=U_{y-x}=0$ (since $U_{-1}=1$), and the additive inequalities for $x=y+(x-y)$ and $y=x+(y-x)$ give $U_x=U_y$. A zero element in the localized ring has false membership, multiplication gives intersections, and the additive relation follows by clearing denominators. The reverse assignment on $a\in R$ proves that these maps are inverse. Thus $X_{R[t^{-1}]}\cong U_t$. If $U_t\subseteq U_r$, then in this localized theory $U_r=1$. Apply the quotient argument above, with the second generator $0$, to conclude $1\in (r)$ in $R[t^{-1}]$. Therefore $r$ is invertible there, and the universal property of localization gives the restriction $R[r^{-1}]\to R[t^{-1}]$. These restrictions compose uniquely.

It remains to glue, and the allowed basis criterion reduces this to a whole-locale cover $U_r\vee U_s=1$. Let $\alpha \in R[r^{-1}]$ and $\beta \in R[s^{-1}]$ have equal restrictions to $R[(rs{)}^{-1}]$. Choose a common initial exponent $n$, writing $\alpha =a/r^n$, $\beta =b/s^n$. Equality on the overlap supplies $k\ge 0$ withSet $N=n+k$, $A=r^{k}a$ and $B=s^{k}b$. Thenwith the last equality in $R$ itself. This clearing step is necessary when the commutative ring has zero divisors.

Since $(r,s)=R$, also $(r^N,s^N)=R$. Choose $\lambda r+\mu s=1$. For $N\ge 1$, expand $(\lambda r+\mu s{)}^{2N-1}=1$; each term contains either $r^N$ or $s^N$, producing $C,D\in R$ with $C{r}^N+D{s}^N=1$. For $N=0$ choose $C=1,D=0$. DefineIn $R[r^{-1}]$,and similarly in $R[s^{-1}]$,Thus $t$ glues the matching pair. If $t^{\prime }$ is another gluing, some $r^p$ and $s^q$ annihilate $t-t^{\prime }$. The same binomial argument gives $(r^p,s^q)=R$, so $t-t^{\prime }=0$. This proves uniqueness without cancellation or a domain assumption.

The exactness of the unit-ideal localization Čech complex proves the sheaf axiom for the required basic covers, and hence the assignment with its restrictions defines a sheaf of rings $\tilde{R}$ with $\tilde{R}(U_r)=R[r^{-1}]$. On the empty basic open, localization is the zero ring, whose underlying set has one section, as required. The construction and proof remain valid for the zero ring.

Let $\mathcal{I}\subseteq {\mathcal{O}}_X$ be the ideal sheaf of $Y$. The formal completion of a scheme along $Y$ has underlying topological space $|Y|$ and topological structure sheafThe topology is the adic topology given by the kernels of the projections to the successive infinitesimal neighbourhoods $X_n=(|Y|,{\mathcal{O}}_X/{\mathcal{I}}^{n+1})$. In the locally Noetherian scheme setting of these examples this is a formal scheme. On a Noetherian affine scheme, with $Y=V(I)\subseteq \mathrm{Spec}A$, it is the formal spectrum $\mathrm{Spf}\hat{A}$, where $\hat{A}={\underleftarrow{\lim }}_{n}A/I^{n+1}$. Thus the completion retains all orders of infinitesimal information normal to $Y$.

Here $Z=D(t)=\mathrm{Spec}k[q,t,t^{-1}]$ is an open subscheme of $X$. The relevant affine completions arewith the $q$-adic topology. The complement of $t=0$ in the first formal scheme is its formal open $D(t)$. The sections on this open are the completed localization of an adic ring:The same calculation on every smaller principal open subset identifies the structure sheaves, not just their global sections. Equivalently, every infinitesimal neighbourhood of $Z\cap (q=0)$ is precisely the restriction to $D(t)$ of the corresponding neighbourhood in $X$. By formal completion commutes with restriction to an open subscheme,The isomorphism is canonical and respects the morphisms to $\mathrm{Spf}k[[q]]$.

It would give the wrong answer to replace the formal-open structure sheaf by the ordinary localization of a ring $k[t][[q]][t^{-1}]$. That ring has a uniform bound on the negative powers of $t$ across the coefficients of any one element, whereashas no such bound. This difference between ordinary and completed localization does not obstruct the isomorphism of formal schemes; it explains why the completed localization is essential.

Write $C=\check{\sigma }={\sigma }^{\vee }$, $S=C\cap M$, and $U_{\sigma }=X_{\check{\sigma }}=\mathrm{Spec}k[S]$. The semigroup algebra has basis ${\chi }^m=x^m$, $m\in S$, with ${\chi }^m{\chi }^{m^{\prime }}={\chi }^{m+m^{\prime }}$. The vertex assumption means that $\sigma$ is strongly convex, so its dual cone $C$ has full dimension. By the Gordan lemma, $S$ is finitely generated: choose integral toric cone generators and subtract integer parts of their nonnegative coefficients; the remainder lies in a bounded parallelepiped with only finitely many lattice points. This argument also applies when $C$ has lineality space, by including generators in both directions there.

Choose an integral interior point $u\in C$. For every $m\in M$, both $tu$ and $tu+m$ belong to $S$ for all sufficiently large integers $t$. Thus $S-S=M$. Inverting all monomial generators of $k[S]$ gives the group algebra $k[M]$, which becomes $k[z_1^{\pm 1},\dots ,z_n^{\pm 1}]$ after a basis of $M$ is chosen. This proves the dense torus of an affine toric variety assertion:Density follows because the localization of a ring is of a domain. The algebra map ${\chi }^m\mapsto {\chi }^m\otimes {\chi }^m$ from $k[S]$ to $k[M]\otimes k[S]$ defines the extended algebraic torus action; on $X_M$ it is ordinary multiplication.

For a face of a polyhedral cone $\nu ⪯C$, set $I_{\nu }=({\chi }^m:m\in S\setminus \nu )$. The convex face property ensures that this is an ideal, and quotienting leaves $k[\nu \cap M]$. Hence $X_{\nu }=\mathrm{Spec}k[\nu \cap M]$ is a closed subvariety of $U_{\sigma }$. Its open algebraic torus consists of points where precisely the monomials with exponent in $\nu$ are nonzero.

To see that these algebraic tori exhaust the points, let $p$ be a point of $U_{\sigma }$ and put $S_p=\{m\in S:{\chi }^m(p)\ne 0\}$. Multiplicativity gives $m+m^{\prime }\in S_p$ if and only if both $m,m^{\prime }\in S_p$. Thus $S_p$ is a semigroup face. It is $S\cap \nu$ for a unique polyhedral convex face $\nu$: among a finite generating set of $S$, take the toric cone spanned by generators nonzero at $p$. A relation expressing a sum with a generator outside this toric cone as a combination of generators inside would, after clearing rational denominators, contradict the displayed multiplicativity property. Therefore the toric cone is a convex face. Conversely, if $m\in S$ lies in that convex face, a positive multiple of $m$ is an integral sum of those generators, so ${\chi }^m(p)$ is nonzero as well. This also proves uniqueness.

The point $p$ lies in $X_M$ exactly when its convex face is all of $C$. ThereforeOne can equivalently write this as the disjoint union of the open algebraic tori of the proper convex faces. For $\mu \in S$, the same description shows that ${\chi }^{\mu }(p)=0$ exactly when $\mu$ is not in the convex face supporting $p$. On every $X_{\nu }$ with $\mu \notin \nu$ the monomial vanishes identically. Consequently the monomial zero locus in an affine toric variety isThis is an equality of reduced subvarieties, or of their underlying closed sets. The principal subscheme itself may have multiplicity, as $x^2=0$ already shows on ${\mathbb{A}}^1$.

The face duality for polyhedral cones is the order-reversing pair of mapsTheir values are convex faces because the relevant pairings are nonnegative and a sum pairs to zero only if each summand does. More explicitly, finitely many nonnegative exposing functionals can be summed to expose the indicated intersection. For a convex face $\tau$, choose $m_0\in C$ exposing it: $\tau =\sigma \cap m_0^{\perp }$. Then $m_0\in C\cap {\tau }^{\perp }$, so $\sigma \cap (C\cap {\tau }^{\perp }{)}^{\perp }\subseteq \tau$, while the reverse inclusion is immediate. Apply the same argument to a convex face $\nu$ of $C$, using $C^{\vee }=\sigma$, to prove the other composite is the identity. Inclusion reversal follows directly from annihilators. This proves the bijection, not just the existence of the maps.

For $\tau ⪯\sigma$, write $\nu =C\cap {\tau }^{\perp }$ and $L=M\cap {\tau }^{\perp }$. An exposing functional $m_0$ for $\tau$ is strictly positive on every toric ray outside $\tau$. Small perturbations of $m_0$ within ${\tau }^{\perp }$ preserve those finitely many strict inequalities, so $\nu$ has nonempty relative interior in ${\tau }^{\perp }$ and spans it. Therefore $\nu \cap M$ generates $L$, by the interior-point argument used above within that subspace. Thus the open algebraic torus of $X_{\nu }$ is $\mathrm{Spec}k[L]=X_{{\tau }^{\perp }}$. Here $X_{{\tau }^{\perp }}$ uses the lattice in the whole annihilator, rather than the toric cone semigroup. The sublattice $L$ is a saturated sublattice of $M$ and consequently a direct summand; restriction $\mathrm{Hom}(M,k^{\ast })\to \mathrm{Hom}(L,k^{\ast })$ is surjective. The extended algebraic torus action is multiplication on this open algebraic torus and hence transitive. All other monomials vanish there. It follows thatThis is the torus orbit-cone correspondence and its orbit closure in a toric variety description. Under convex face duality, $F_{\gamma }={\bigcup }_{\eta ⪰\gamma }O(\eta )$.

Finally, choose integral $m_{\tau }\in C$ exposing $\tau$. Localizing $k[C\cap M]$ at ${\chi }^{m_{\tau }}$ gives $k[{\tau }^{\vee }\cap M]$: for $m\in {\tau }^{\vee }\cap M$, adding a sufficiently large multiple of $m_{\tau }$ makes it nonnegative on every toric ray of $\sigma$ outside $\tau$, while preserving nonnegativity on $\tau$. Hence $X_{\check{\tau }}=U_{\tau }=D({\chi }^{m_{\tau }})$ is the usual affine open chart inside $U_{\sigma }$. An orbit $O(\gamma )$ belongs to it exactly when $\gamma ⪯\tau$. The requested complement of an affine toric face chart is thereforeAll sets on the right are closed, and if $\gamma$ is not contained in $\tau$, neither is any convex face containing $\gamma$. This justifies replacing the excluded orbits by their whole closures.

Let $S$ be a multiplicative subset, containing one and closed under products. On $A\times S$, useThe relation is reflexive and symmetric. For transitivity, if $u(ta-sb)=0$ and $v(rb-tc)=0$, then multiplication and addition give $uvt(ra-sc)=0$. Thus it is an equivalence relation. Write $a/s$ for the equivalence class. The localization of a ring $A_S=S^{-1}A$ has operationsThey are well defined. For example, replacing $a/s$ by $b/t$ in its sum with $c/r$ gives cross-numerator difference $r^2(ta-sb)$, and in its product with $c/r$ gives $rc(ta-sb)$. The same annihilating element $u$ kills both differences. Replacing the other input works identically. Associativity and distributivity then follow by putting all terms over the product of their denominators and using the ring identities in $A$. Addition and multiplication are commutative, with identities $0/1$ and $1/1$, and the displayed negative is an additive inverse. Hence $S^{-1}A$ is a commutative ring. If $0\in S$, the fraction relation identifies all elements, producing the zero ring.

The canonical ring homomorphism $\iota :A\to S^{-1}A$, $a\mapsto a/1$, makes each $s\in S$ a unit, with inverse $1/s$. The universal property of localization states that any ring homomorphism $h:A\to C$ sending $S$ into the units factors uniquely through $\iota$. Define the factor byIf $u(ta-sb)=0$, then $h(u)$ is invertible, so $h(t)h(a)=h(s)h(b)$, proving well-definedness. The fraction operations prove that $\tilde{h}$ is a ring homomorphism. Its value is forced by $a/s=\iota (a)\iota (s{)}^{-1}$, so it is unique. This property characterizes the localization up to a unique compatible isomorphism.

For a prime ideal $\mathfrak{p}$, its complement is a multiplicative subset, and the localization at a prime ideal is $A_{\mathfrak{p}}=(A\setminus \mathfrak{p}{)}^{-1}A$. Fractions with numerator outside $\mathfrak{p}$ are units; those with numerator in $\mathfrak{p}$ form the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. Thus it is a local ring.

For the nilpotent question, suppose $a\ne 0$ and $a^n=0$. Its annihilator is a proper ideal, so it lies in some maximal ideal $\mathfrak{m}$. The fraction criterion says $a/1=0$ in $A_{\mathfrak{m}}$ only if $sa=0$ for some $s\notin \mathfrak{m}$, which is impossible by the choice of $\mathfrak{m}$. Thus $a/1$ is a nonzero nilpotent element in that localization. This proves reducedness is detected by prime localizations: if every prime localization is reduced, $A$ has no nonzero nilpotent elements.

For the domain question take $A=k\times k$ with $k$ a field. Its only prime ideals are ${\mathfrak{p}}_1=0\times k$ and ${\mathfrak{p}}_2=k\times 0$: a prime ideal must contain one of the orthogonal idempotents $(1,0),(0,1)$, and the corresponding quotient is $k$. Localizing at ${\mathfrak{p}}_1$ makes $(1,0)$ invertible and kills $(0,1)$, giving $A_{{\mathfrak{p}}_1}\cong k$; similarly $A_{{\mathfrak{p}}_2}\cong k$. But $(1,0)(0,1)=0$ with both factors nonzero. Therefore domain prime localizations do not force $A$ to be an integral domain, as expressed by domain prime localizations do not imply a domain.

The spectrum of a ring $A$ is the set of its proper prime ideals. For an ideal $I$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. These are the closed sets of the Zariski topology. They satisfyThe union identity follows because a prime ideal containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. These identities verify the closed-set axioms. The maximal spectrum has the subspace topology, whose closed sets are $V(I)\cap \mathrm{MaxSpec}A$.

For a finitely generated $\mathbb{C}$-algebra, the needed form of the Hilbert Nullstellensatz isHere an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.

First establish the residue-field fact. If a field $L$ is a finitely generated algebra over $\mathbb{C}$, choose a maximal algebraically independent subset $t_1,\dots ,t_r$ of a finite generating set, with remaining generators $u_1,\dots ,u_s$. Each $u_j$ is algebraic over $\mathbb{C}(t_1,\dots ,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral overSince $L$ is a field, it contains $1/d$, and $L=R[u_1,\dots ,u_s]$ is an integral extension of $R$. If a field is integral over a domain, the domain is itself a field: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.

But if $r>0$, choose $c\in \mathbb{C}$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a field, a contradiction. Hence $r=0$, and $L$ is algebraic over the algebraically closed field $\mathbb{C}$, forcing $L=\mathbb{C}$. This proves the Zariski lemma in the form used here.

Now if $f\notin \sqrt{I}$, the localization of a ring $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a finitely generated algebra over $\mathbb{C}$, since one need only adjoin an inverse of $f$. Choose a maximal ideal $\mathfrak{n}$ of $B$. By the preceding argument, $B/\mathfrak{n}=\mathbb{C}$. The composite $A\to B\to \mathbb{C}$ is surjective because it contains the constant scalars. Its kernel is consequently a maximal ideal $\mathfrak{m}$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt{I}$. The reverse inclusion follows because every maximal ideal is prime and contains the radical of an ideal whenever it contains that ideal.

For a closed subset $Z\subseteq \mathrm{MaxSpec}A$, put $I(Z)={\bigcap }_{\mathfrak{m}\in Z}\mathfrak{m}$. It is a radical ideal. The formulas now showare inverse maps on radical ideals and closed sets. Indeed the first followed by the second gives $\sqrt{I}=I$, while if $Z=V(J)\cap \mathrm{MaxSpec}A$, the second gives $\sqrt{J}$ and $V(\sqrt{J})=V(J)$. Thus the correspondence is a bijection reversing inclusion. This is the relation between radical ideals and closed subsets of the maximal spectrum.

For $A=\mathbb{C}[x]$, the maximal ideals are ${\mathfrak{m}}_c=(x-c)$, for $c\in \mathbb{C}$. Every nonzero proper radical ideal is generated by a squarefree polynomial, and hence has the formfor a finite nonempty subset $S\subset \mathbb{C}$. The associated closed set is $\{{\mathfrak{m}}_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its radical ideal removes that repeated factor. The two endpoint cases areThe empty-product convention makes $I_{\varnothing }=(1)$. Under $c\mapsto {\mathfrak{m}}_c$, the closed sets are exactly the finite subsets and the whole affine line.

A nilpotent element belongs to every prime ideal and therefore to every maximal ideal, soFor the reverse containment, let $r$ be nonnilpotent. The localization of a ring $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a finite-type integer algebra, because it can be presented as $R[T]/(rT-1)$.

Choose a maximal ideal $\mathfrak{n}$ of $S$. Its residue field $L=S/\mathfrak{n}$ is a finite-type integer algebra, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite integral domain containing $1$, so it is a field: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.

The kernel $\mathfrak{m}$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero residue field. Thus $r\notin \mathfrak{m}$. Every nonnilpotent element can therefore be avoided by a maximal ideal, givingThis radical equality for finitely generated integer algebras uses part (a) to ensure that the contraction of the localized maximal ideal is maximal. Such contraction is not generally maximal for arbitrary localization of a ring; the finite-field image is the decisive additional step. The zero ring is immediate, with the intersection of its empty set of maximal ideals understood as the whole ring. No general theorem that integer algebras are Jacobson rings is being quoted.

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.

Let $R$ be a commutative ring and $f=(f_1,\dots ,f_r)$. The Koszul complex packages these elements and their relations in a finite chain complex of free modules. With $E=R^r$ and basis $e_1,\dots ,e_r$, set $K_i(f;R)={\bigwedge }^{i}E$ for $0\le i\le r$, and defineDeleting two distinct basis elements in the two possible orders gives opposite signs and the same coefficient, so ${\partial }^2=0$. Equivalently $K(f;R)$ is the tensor product of chain complexes of $r$ two-term complexes $[R\stackrel{f_i}{\to }R]$ in degrees one and zero. The tensor differential is $\partial (u\otimes v)=\partial u\otimes v+(-1{)}^{\mathrm{deg}u}u\otimes \partial v$. This fixes the sign convention. On the exterior algebra, the differential is a graded derivation determined by $\partial e_i=f_i$.

For an $R$-module $M$, define the Koszul complex with module coefficients by $K(f;M)=K(f;R){\otimes }_{R}M$. Its degree-zero Koszul homology isIts higher Koszul homology measures the failure of these equations to form a regular sequence on a module. Each $f_j$ annihilates every homology group. Indeed the Koszul homotopy for multiplication by a generator is $h_j(u)=e_j\wedge u$, and the graded product rule givesFor coefficients in $M$ the same identity holds after tensoring. Thus multiplication by $f_j$ is zero on homology, and the homology groups are naturally modules over $R/(f)$. If $(f)=R$, choose ${\sum }_j{a}_j{f}_j=1$; then ${\sum }_j{a}_j{h}_j$ is a contracting homotopy, so the entire complex is contractible. These identities are also useful after localization of a ring: wherever one generator is a unit, the complex has zero homology.

The construction is functorial under a ring homomorphism, and base change gives $K(f;R){\otimes }_{R}S=K(\bar{f};S)$, since its terms are free with the displayed basis and differential. An invertible change of generators $f_i={\sum }_j{u}_{ij}{g}_j$ gives a chain isomorphism by sending $e_i$ to ${\sum }_j{u}_{ij}{e}_j$ and extending to exterior powers. In particular, permuting the generators changes only the exterior signs, not the isomorphism class of the complex.

The essential exactness theorem is that a regular sequence on a module gives zero positive Koszul homology. Recall that $f_i$ must act injectively on $M/(f_1,\dots ,f_{i-1})M$, with final quotient nonzero. Put $f^{\prime }=(f_1,\dots ,f_{r-1})$. Adding the final two-term complex identifies $K(f;M)$ with the mapping cone of multiplication by $f_r$ on $K(f^{\prime };M)$, with the chosen tensor signs. The long exact sequence in homology consequently has segmentsFor $r=1$, positive homology is exactly $\ker (f_1:M\to M)$. By induction the preceding homology vanishes above zero. The final injectivity assumption kills $H_1(f;M)$, while the exact sequence kills every $H_i$ for $i>1$. Thereforefor a regular sequence on a module. In particular $K(f;R)$ is a finite free resolution of $R/(f)$ when $f$ is a regular sequence, with ranks $\left(\genfrac{}{}{0ex}{}{r}{i}\right)$ and length $r$.

There is a precise converse under local finiteness assumptions. Suppose $(R,\mathfrak{m})$ is a Noetherian local ring, $0\ne M$ is finitely generated, and all $f_i\in \mathfrak{m}$. If positive Koszul homology vanishes, the same exact sequence makes multiplication by $f_r$ surjective on every $H_i(f^{\prime };M)$ for $i>0$. These modules are finitely generated because the ring is Noetherian and the terms of the complex are finite modules. The Nakayama lemma gives $H_i(f^{\prime };M)=0$. Induction makes $f^{\prime }$ regular on $M$, and the degree-one part of the exact sequence makes $f_r$ injective on $H_0(f^{\prime };M)$. The final quotient is nonzero, again by the Nakayama lemma. This proves the Koszul acyclicity criterion in a Noetherian local ring:The hypotheses matter: a unit ideal makes the complex contractible but cannot be a proper regular sequence.

Examples make both sides visible. For $R=k[x,y]$ and $f=(x,y)$, the Koszul resolution isThe signs agree with $\partial (e_1\wedge e_2)=x{e}_2-y{e}_1$. For $R=k[x]$ and $f=(x,x)$, the degree-one cycles are $R(1,-1)$ and the boundaries are $xR(1,-1)$; hence $H_1\cong R/(x)$, while $H_2=0$. The repeated element has exposed a relation that the first element already kills.