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For a regular local ring of embedding dimension $e$, a minimal maximal-ideal generating set induces the displayed polynomial-algebra isomorphism. It is surjective in every degree. A nonzero homogeneous kernel relation would bound the cumulative Hilbert function by $O(t^{e-1})$, contradicting the prime-chain lower bound for local length at $\dim R=e$. The graded ring is a domain; the Krull intersection theorem then shows the original ring is a domain by multiplying nonzero initial forms.

For a local ring $(R,\mathfrak{m})$ with residue field $\kappa =R/\mathfrak{m}$, its cotangent space is the $\kappa$-vector space $\mathfrak{m}/{\mathfrak{m}}^2$. In a Noetherian local ring, its dimension is the minimal number of generators of $\mathfrak{m}$, by the Nakayama lemma. Equality of this dimension with the Krull dimension defines a regular local ring.

The embedding dimension of a Noetherian local ring $(R,\mathfrak{m})$ is ${\dim }_{R/\mathfrak{m}}(\mathfrak{m}/{\mathfrak{m}}^2)$. It counts the smallest number of local generators needed for the maximal ideal. It is at least the Krull dimension; equality characterizes a regular local ring.

Let $(A,\mathfrak{m})$ be a regular local ring of Krull dimension $d$. Its formal power series ring $B=A[[x]]$ is Noetherian and local, with maximal ideal $I=\mathfrak{m}B+(x)$ and embedding dimension $d+1$. Since $I$ is generated by $d+1$ elements, the Krull height theorem gives $\dim B\le d+1$. For the reverse inequality choose a strict prime chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_d=\mathfrak{m}$ in $A$. Finite generation of each ${\mathfrak{p}}_i$ gives $B/{\mathfrak{p}}_{i}B\cong (A/{\mathfrak{p}}_i)[[x]]$, an integral domain, so each extended ideal is prime. The inclusions remain strict by contraction to $A$, and $\mathfrak{m}B⊊I$ adds one more step since $x\notin \mathfrak{m}B$. This proves $\dim B=d+1$, equal to its embedding dimension, and hence regularity.

A local defining function for a complex analytic hypersurface $Y$ near $x$ is a holomorphic function $f$ on a neighbourhood $U$ such that $Y\cap U=\{f=0\}$. It may be chosen reduced, with each local irreducible factor occurring once. The local ring ${\mathcal{O}}_{X,x}$ of a complex manifold is a regular local ring and hence a unique factorization domain; each height-one prime of the hypersurface germ is principal, and the product of generators for its finitely many local branches gives $f$.

An irreducible analytic subvariety of a complex manifold is a closed subset locally defined by finitely many holomorphic functions which cannot be expressed as the union of two proper closed analytic subsets. In a complex $n$-manifold, codimension $k$ means pure complex dimension $n-k$. A divisor on a complex manifold is a locally finite formal integer combinationof irreducible analytic hypersurfaces. Compactness makes the set of components in this sum finite. Negative coefficients are permitted; a local equation for such a divisor on a complex manifold is generally meromorphic rather than holomorphic.

We use these precise local facts. In holomorphic coordinates at $x$, the local ring is ${\mathcal{O}}_{X,x}\cong \mathbb{C}\{z_1,\dots ,z_n\}$, a Noetherian regular local ring and a unique factorization domain. Each analytic hypersurface germ has finitely many irreducible local branches; their prime ideals have height one and are generated by irreducible holomorphic germs. A holomorphic germ is a unit exactly when its value at $x$ is nonzero. Consequently a nonzero meromorphic germ with order zero on every local hypersurface branch is a holomorphic unit, by cancelling its irreducible factors in numerator and denominator. These are the auxiliary local-ring properties required below.

After shrinking a neighbourhood $U_i$ around each point, choose local branch equations $h_{i\nu a}$, and putIts zero and pole orders give exactly the local divisor on a complex manifold $D{|}_{U_i}$. A globally irreducible hypersurface may have several branches locally; all of them must be included with the indicated multiplicity. On an overlap, $f_i/f_j$ has zero order along every branch, so the local-ring facts show that it extends to a nowhere-zero holomorphic function. These germwise extensions patch uniquely to a holomorphic unit on the overlap. This describes a local defining function of $D$ and proves why multiplicities, not just zero sets, matter.

Set $g_{ij}=f_i/f_j$ and glue line-bundle frames by $e_j=g_{ij}{e}_i$. The identities $g_{ij}{g}_{jk}=g_{ik}$ give a holomorphic line bundle associated to a divisor, denoted $[D]$. Equivalently, the frame $e_i$ can be represented by $1/f_i$ in the sheaf of meromorphic functions, so $[D]={\mathcal{O}}_X(D)$. If alternative equations are $f_i^{\prime }=u_i{f}_i$, where $u_i$ are holomorphic units, the corresponding frames are identified by $e_i^{\prime }\mapsto u_i^{-1}{e}_i$. This respects the changed transitions. Refining the cover gives the same identifications, proving that $[D]$ is well-defined up to holomorphic bundle isomorphism. The expressions $f_i{e}_i$ glue to its canonical meromorphic section, whose divisor is $D$.

For two divisors on a complex manifold use a common cover with equations $f_i$ and $h_i$. Their sum has equation $f_i{h}_i$, and the tensor product of vector bundles has frame transitionsSending the sum-divisor frame to the tensor product of the two frames givesThis also yields $[-D]\cong [D{]}^{\ast }$.

For a nonsingular complex analytic hypersurface $Y$, choose reduced local equations $f_i$ with $d{f}_i$ nonzero along $Y$. The holomorphic conormal bundle iswhose fibre is the annihilator of $T^{1,0}Y$. It has local frame $d{f}_i{|}_Y$. Here this restriction is as a covector in the ambient cotangent bundle along $Y$, not its pullback to ${\Omega }_Y^1$, where it would vanish. The line bundle $[-Y]$ has local frame $a_i=f_i$, and if $f_j=u_{ji}{f}_i$, thensince $f_{i}d{u}_{ji}$ vanishes along $Y$. The frame identifications $a_i{|}_Y\mapsto d{f}_i{|}_Y$ therefore patch and are fibrewise nonzero. This proves the conormal line of a smooth analytic hypersurface formulaEquivalently it identifies ${\mathcal{O}}_X(-Y){|}_Y={\mathcal{I}}_Y/{\mathcal{I}}_Y^2$ with the analytic conormal sheaf.

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.

Put $R={\mathcal{O}}_{X,x}$, $\mathfrak{m}={\mathfrak{m}}_x$ and $d=\dim R$. Smoothness makes $R$ a regular local ring, with residue field $k$. A list of local parameters $t_1,\dots ,t_d$ is a regular system of parameters: its images form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$. Set $J=(t_1,\dots ,t_d)$. The basis condition says $\mathfrak{m}=J+{\mathfrak{m}}^2$, so the finitely generated module $\mathfrak{m}/J$ equals $\mathfrak{m}(\mathfrak{m}/J)$. The Nakayama lemma gives

We now construct the requested formal power series, rather than assert that a local function has a convergent expansion. The associated graded ring of a regular local ring isIndeed these initial forms generate the graded ring, since the $t_i$ generate $\mathfrak{m}$. The Dimension theorem for Noetherian local rings gives $\dim {\mathrm{gr}}_{\mathfrak{m}}R=d$. A nonzero kernel in the surjection from the polynomial ring would make its quotient have dimension at most $d-1$, so that kernel is zero.

For $f\in R$, choose $P_0\in k$ to be its residue at $x$. Inductively suppose homogeneous polynomials $P_0,\dots ,P_{n-1}$ have been chosen such thatThe graded-ring isomorphism gives a unique homogeneous polynomial $P_n(T_1,\dots ,T_d)$ of degree $n$ representing the class of $r_n$ in ${\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}$. Subtracting $P_n(t)$ makes the next remainder lie in ${\mathfrak{m}}^{n+1}$. HenceThis equality is interpreted in the adic completion of a module $\hat{R}=\underleftarrow{\lim }R/{\mathfrak{m}}^n$. The same induction applies to every compatible sequence of residues in this inverse limit, so evaluation $T_i\mapsto t_i$ defines an isomorphism $k[[T_1,\dots ,T_d]]\cong \hat{R}$. It is injective because the lowest nonzero homogeneous term of a series has nonzero initial form. The Krull intersection theorem gives ${\bigcap }_n{\mathfrak{m}}^n=0$, so $R$ embeds in its completion and the series determines $f$. This is the formal expansion at a smooth point; it depends on the chosen regular system of parameters, and does not claim that the infinite sum converges inside $R$. If $d=0$, the local ring is just $k$, and only the constant term is needed.

On a complex manifold, a divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{m}_{Y}Y$, where $Y$ ranges over irreducible complex analytic hypersurfaces and $m_Y\in \mathbb{Z}$. The local algebra needed here is that the local ring ${\mathcal{O}}_{X,x}$, identified in coordinates with convergent power-series germs, is a regular local ring and a unique factorization domain. In particular, each height-one prime is generated by one irreducible germ, and a nonzero meromorphic germ is a quotient of holomorphic germs. Reduced local equations of the same hypersurface differ by a holomorphic unit. These facts include local branches when a globally irreducible hypersurface has several branches at a singular point.

The order of a meromorphic function along a hypersurface is the exponent of its reduced local equation in the quotient's factorization. Equivalently, at a generic smooth point of $Y$, write $f=t^{m}u$, with $t$ a transverse local equation of $Y$ and $u$ a holomorphic unit; then ${\mathrm{ord}}_Y(f)=m$. Changing a reduced equation multiplies it by a unit and leaves $m$ unchanged. These local orders agree on overlaps and are constant along the generic smooth locus of an irreducible hypersurface. A nonzero global meromorphic function therefore defines its principal divisor on a complex manifoldPositive coefficients record zeros and negative coefficients record poles.

A local defining function $f_i$ for $D$ on $U_i$ is a nonzero meromorphic function whose orders there equal the coefficients of $D$. It can be constructed as the product of local reduced hypersurface equations to the prescribed integer powers. On an overlap, $f_i/f_j$ has no zero or pole along any hypersurface. The unique factorization domain property then makes it a nowhere-zero holomorphic function, after choosing representatives of the germs. The holomorphic line bundle associated to a divisor is obtained by gluing holomorphic frames usingThese ratios satisfy the cocycle condition, and the expressions $f_i{e}_i$ agree on overlaps. They form the canonical meromorphic section of a holomorphic line bundle with divisor $D$.

If $D=\mathrm{div}(f)$ globally, $f_i/f$ is a holomorphic unit on $U_i$, and $(f_i/f)e_i$ is a global nowhere-zero holomorphic section. Hence $[D]$ is holomorphically trivial. Conversely, if $[D]$ has a nowhere-zero global holomorphic frame $e$, write $e_i=a_{i}e$ with $a_i$ holomorphic units. The coefficients $f_i{a}_i$ of the canonical meromorphic section of a holomorphic line bundle agree on overlaps and give a nonzero global meromorphic function $f$. Since multiplication by a unit does not change orders, $\mathrm{div}(f)=D$. Therefore

For a smooth hypersurface $Y\subset X$, the adjunction formula isIndeed, the holomorphic conormal sequence has first term $N_{Y/X}^{\ast }\cong [-Y]{|}_Y$; taking its determinant yields $K_X{|}_Y\cong N_{Y/X}^{\ast }\otimes K_Y$, which gives the formula. Here the canonical bundle is the top exterior power of the holomorphic cotangent bundle.

Apply this first to $W\subset {\mathbb{P}}^4$. Its degree-$d_1$ equation is a section of $\mathcal{O}(d_1)$, so $[W]\cong \mathcal{O}(d_1)$. Since $K_{{\mathbb{P}}^4}\cong \mathcal{O}(-5)$, we obtain $K_W\cong {\mathcal{O}}_W(d_1-5)$. The restricted second equation cuts out $S$ in $W$ and has associated bundle ${\mathcal{O}}_W(d_2)$. A second application of adjunction formula gives the canonical bundle of a regular projective complete intersection

It remains to prove that a nonzero exponent cannot yield a trivial holomorphic line bundle on $S$. Choose a connected component $S_0$ and a hyperplane meeting but not containing it. Such a hyperplane exists: a positive-dimensional compact projective submanifold cannot lie entirely in an affine chart, since all its affine coordinate functions would be global holomorphic functions and hence constant. A hyperplane through one point and not another consequently meets $S_0$ properly. Its equation gives a holomorphic section of ${\mathcal{O}}_{S_0}(1)$ that has both a zero and a nonzero value. For $m>0$, the $m$th power has the same two properties. Were ${\mathcal{O}}_{S_0}(m)$ trivial, this section would be represented by a global holomorphic function, contradicting its constancy on compact connected $S_0$. For $m<0$, triviality would imply triviality of its positive dual power. This proves nontrivial hyperplane powers on a compact projective submanifold and henceThe degrees are positive because the two equations define a hypersurface and a surface of the specified dimensions.

The calculation uses “non-singular” in the usual smooth complete-intersection sense: the equations define reduced smooth divisors, with independent differentials along $S$. If only their reduced zero sets are required to be smooth, polynomial multiplicities cannot determine the canonical bundle. For example, $Q_1=Z_0^m$ and $Q_2=Z_1^4+Z_2^4+Z_3^4+Z_4^4$ have the same smooth quartic surface in the hyperplane $Z_0=0$ for every $m\ge 1$. That surface has trivial canonical bundle, while the displayed polynomial degree sum is $m+4$. Thus unreduced equations must not be substituted into the adjunction formula for the reduced surface.

Since $A$ is a regular local ring, $\dim A=d$ is finite. Choose a chain of prime ideals of length $d$,Such a chain exists because the supremum defining this finite integer Krull dimension is attained; a maximal-length chain in a local ring ends at its maximal ideal.

For each $i$, the ideal ${\mathfrak{p}}_i$ is finitely generated because $A$ is Noetherian. This ensures that its extension ${\mathfrak{p}}_{i}B$ consists precisely of series all of whose coefficients lie in ${\mathfrak{p}}_i$: if ${\mathfrak{p}}_i=(b_1,\dots ,b_r)$, write each coefficient as ${\sum }_j{b}_j{c}_{jn}$ and gather them into the finitely many series ${\sum }_n{c}_{jn}{x}^n$. Thus coefficient reduction givesThe coefficient ring $A/{\mathfrak{p}}_i$ is an integral domain, and so is its formal power series ring: the first nonzero coefficient of a product is the product of the first nonzero coefficients of its factors. Hence each ${\mathfrak{p}}_{i}B$ is prime. Also ${\mathfrak{p}}_{i}B\cap A={\mathfrak{p}}_i$, so the strict inclusions persist upon extension. Finally, $x\notin \mathfrak{m}B$, since its image in $B/\mathfrak{m}B\cong k[[x]]$ is nonzero. We obtain the prime chainof length $d+1$. Therefore $\dim B\ge d+1$. Together with part (b),Its embedding dimension is also $d+1$, by part (b), so $B$ is itself a regular local ring. This proves that formal power series preserve regular local rings.

An irreducible complex analytic hypersurface is a closed analytic subset of pure complex codimension one which cannot be expressed as the union of two proper closed analytic subsets. Here a local defining function of a complex analytic hypersurface means a reduced local equation: a holomorphic germ generating its vanishing ideal, with each local branch occurring once. The word reduced matters. Merely prescribing the zero set would allow both $f$ and $f^2$, and the requested uniqueness would then be false.

The local facts used are these: in holomorphic coordinates the local ring ${\mathcal{O}}_{X,p}$ is $\mathbb{C}\{z_1,\dots ,z_n\}$, a regular local ring and a unique factorization domain; its units are exactly the germs nonzero at $p$; and the analytic Nullstellensatz identifies the zero-set ideal of a germ $f$ with $\sqrt{(f)}$. A reduced germ is a product of distinct irreducible factors. Unique factorization makes its principal ideal radical: if $f\mid h^r$, every irreducible factor of $f$ divides $h$, so $f\mid h$. Thus two reduced germs with the same hypersurface germ have $(f)=(g)$. Write $g=uf$, $f=vg$. The local ring is a domain, so $uv=1$. Hence $u$ is a holomorphic unit and $u(p)\ne 0$. This proves reduced local defining functions differ by a holomorphic unit. Global irreducibility does not imply local irreducibility at a singular point; the same argument uses the product of all local branches and therefore covers that case too.

A divisor on a complex manifold is a locally finite integer combination of irreducible analytic hypersurfaces. On a compact manifold only finitely many have nonzero coefficients. Choose local meromorphic equations $f_i$ for a divisor $D$, by products of powers of reduced local equations. Their ratios are holomorphic units on overlaps. For the holomorphic line bundle associated to a divisor, choose frames satisfyingThe ratios obey the cocycle identity and so define a holomorphic line bundle $[D]$. This convention makes the local meromorphic sections $f_i{e}_i$ agree, with divisor $D$. Changing the equations by units changes frames and gives an isomorphic bundle.

For hyperplanes $H_1,H_2$ in Complex projective space, choose nonzero linear equations ${\ell }_1,{\ell }_2$. The degree-zero quotient $r={\ell }_1/{\ell }_2$ is a global meromorphic function and has divisor $H_1-H_2$. A principal divisor on a complex manifold gives a trivial associated bundle: using the same global meromorphic equation on every chart makes all frame transitions equal to one. The allowed tensor-product rule now givesIf the hyperplanes coincide the same conclusion is immediate. Equivalently, their common bundle is the hyperplane bundle $\mathcal{O}(1)$.

Locally at the centre of a blowup of a complex manifold at a point, replace a coordinate neighbourhood in ${\mathbb{C}}^n$ byAway from zero the direction is uniquely $[z]$, so this map is an isomorphism there; glue it to $X\setminus \{p\}$. In the chart ${\ell }_j\ne 0$, write $z_j=u$ and $z_i=u{t}_i$ for $i\ne j$. These are smooth holomorphic coordinates on the blowup. The exceptional divisor is $u=0$, globally $E={\sigma }^{-1}(p)\cong \mathbb{P}(T_{p}X)$.

The canonical bundle formula for a point blowup isThe local chart also explains the exponent: pulling back a coordinate volume form gives $\pm u^{n-1}du\wedge {\bigwedge }_{i\ne j}d{t}_i$, so the Jacobian section has divisor $(n-1)E$.

Since $Y$ is smooth at the centre, choose coordinates with $Y=\{z_1=0\}$. Its strict transform is a genuine analytic hypersurface. In a blowup chart $j\ne 1$, it is the closed set $t_1=0$: off $E$, the original equation $u{t}_1=0$ reduces to $t_1=0$, and taking the closure restores the points with $u=0$. In chart $j=1$, there are no such off-exceptional points, and the strict transform is empty there. These descriptions agree on overlaps and show analyticity, with $\tilde{Y}\cap E\cong {\mathbb{CP}}^{n-2}$. The total transform has the extra factor $u$ exactly once, soUnder the given anticanonical assumption, this yieldsTo prove the only-if direction, one must rule out an accidental trivial power of $[E]$. The local blowup is the tautological line bundle over ${\mathbb{CP}}^{n-1}$, with $E$ its zero section. Its normal bundle is therefore ${\mathcal{O}}_E(-1)$. The normal bundle of a smooth analytic hypersurface identifies $[E]{|}_E$ with this bundle. If $n\ge 2$, restrict to a projective line inside $E$; the restriction of $[E{]}^{\otimes (2-n)}$ has degree $n-2$. A trivial line bundle has degree zero, so it is trivial only if $n=2$. Conversely the displayed tensor factor is literally trivial when $n=2$. This is the anticanonical strict transform under a point blowup criterion:For completeness, dimension one cannot satisfy the premise in this compact setting: an irreducible analytic hypersurface is one point, and $[-Y]$ has degree $-1$ on its curve component, whereas $K_X$ has degree $2g-2$. Thus no exceptional dimension-one case invalidates the equivalence.

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 $\dim \varnothing =-\infty$.

A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.

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.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis

Let $A=R_P$ be the localization at a prime ideal, with maximal ideal $\mathfrak{m}=P{R}_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The residue field isIndeed there is an explicit identificationinverting $1+u$ and $1+v$ has no effect in this local ring, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.

For the Krull dimension, the prime ideal correspondence for localization preserves the strict chainof prime ideals in $R$. They are prime because the successive quotients are the integral domains $R$, $\mathbb{Z}[X_2,X_2^{-1}]$, and $\mathbb{Z}$. They remain distinct after localization at a prime ideal because they are all contained in $P$. Hence $\dim A\ge 2$.

For the upper bound, $R$ is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring. Since $P$ is generated by two elements, the Krull height theorem gives $\mathrm{ht}P\le 2$. The prime ideal correspondence for localization identifies $\dim R_P$ with $\mathrm{ht}P$, so

A Noetherian local ring $(B,\mathfrak{n})$ with residue field $\kappa$ is a regular local ring ifBy the Nakayama lemma, the dimension on the left is the minimal number of generators of $\mathfrak{n}$; this is also called the embedding dimension.

The localization of a module is exact and commutes with quotients and products of ideals. Applying it to the already computed conormal module gives the cotangent space of a local ringIts dimension is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb{Z}$, whereas the localized cotangent space of a local ring has a basis over the residue field $\mathbb{Q}$.

A point $P$ belongs to the smooth locus of a variety if its local ring ${\mathcal{O}}_{X,P}$ is a regular local ring, meaningEquivalently, the Zariski tangent space has the expected local dimension. Over the algebraically closed field, this agrees with smoothness for a reduced variety. In affine coordinates, the Jacobian criterion expresses it by the rank of the matrix of first derivatives of defining equations.