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For an ideal $I\subseteq R$, its conormal module is $I/I^2$, naturally an $R/I$-module. It records generators and relations to first order along the quotient $R/I$. If $I$ is the maximal ideal of a local ring, it is the cotangent space of a local ring over the residue field. For the augmentation ideal $(X_1-1,\dots ,X_r-1)$ of $A[X_1^{\pm 1},\dots ,X_r^{\pm 1}]$, reduction modulo $I^2$ gives $I/I^2\cong A^r$, with basis the classes of $X_i-1$.

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.

Choose $a_1,\dots ,a_d\in \mathfrak{m}$ whose classes form a basis of the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$, and put $J=(a_1,\dots ,a_d)$. Spanning gives $\mathfrak{m}=J+{\mathfrak{m}}^2$. Thus the quotient module $N=\mathfrak{m}/J$ satisfies $N=\mathfrak{m}N$. It is a finitely generated module, since $A$ is a Noetherian ring.

Here is the needed form of the Nakayama lemma, with proof. Choose generators $n_1,\dots ,n_r$ of $N$ and write $n_i={\sum }_j{c}_{ij}{n}_j$ with $c_{ij}\in \mathfrak{m}$. For the column vector $n$, $(I-C)n=0$. Multiply by the adjugate to obtain $\det (I-C)n=0$. Since $\det (I-C)\equiv 1(\mathrm{mod}\mathfrak{m})$, this determinant is a unit in the local ring $A$, so every $n_i=0$. Therefore $N=0$, proving $\mathfrak{m}=J$. Conversely, any generators of $\mathfrak{m}$ must span its cotangent space, so at least $d$ are needed. HenceHere $\mu (M)$ denotes the minimal number of generators of a module $M$.

For the additional minimum, the maximal ideal $\mathfrak{m}$ is itself $\mathfrak{m}$-primary, so it supplies an eligible ideal generated by $d$ elements. ConsequentlyThe minimum here is over primary ideals. A bound $\mu (Q)\le d$ for every $\mathfrak{m}$-primary ideal would be false. Indeed, for $A=k[[u,v]]$ and $\mathfrak{m}=(u,v)$, $d=2$ but $Q={\mathfrak{m}}^2=(u^2,uv,v^2)$ is primary and needs three generators. To verify primaryness, if $ab\in Q$ and $a\notin Q$, then $b$ cannot be a unit, so $b\in \mathfrak{m}$ and $b^2\in Q$. The three displayed monomials form a basis of $Q/\mathfrak{m}Q={\mathfrak{m}}^2/{\mathfrak{m}}^3$, proving the lower bound of three, while they already generate $Q$. This illustrates that primary ideals can need more generators than the maximal ideal.

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.

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 $\mathbb{Q}$.