Kähler differentials

Written and self-checked by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. This edition incorporates an AI Integrated Stacks Project proof under GNU FDL 1.2; see the attribution and licence notice below.

A polynomial equation constrains both its points and its first-order variations. Differentiating the equation produces a linear relation among those variations. The module of Kähler differentials makes this calculation intrinsic: it represents every derivation, works over any base ring, and keeps track of the relations introduced by a quotient. Its fibre dimensions also explain why a singular curve can require two infinitesimal directions even though its generic point requires only one.

All rings are commutative with identity. No Noetherian hypothesis is implicit. Modules called finite are finitely generated. We use the localization lesson for the universal property of localization and tensor products with fractions, and the flatness lesson for right exactness of tensor product. One auxiliary result uses the proved exactness of Koszul complexes, Theorem 4.1 of the projective-dimension lesson. We write \(\overline s\) for the image of an element under a quotient map.

1. A module representing differentiation

For a ring map \(R\to S\) and an \(S\)-module \(M\), an \(R\)-derivation is an additive map \(D:S\to M\) which vanishes on the image of \(R\) and satisfies

It is \(R\)-linear. It is usually not \(S\)-linear. The identity with \(s=t=1\) gives \(D(1)=0\), and induction gives \(D(s^n)=ns^{n-1}D(s)\) for positive integers \(n\). Derivations form an \(S\)-module, denoted \(\operatorname{Der}_R(S,M)\). [Stacks, Tag 00RN.]

Theorem 1.1 (universal differential). There is an \(S\)-module \(\Omega_{S/R}\) and an \(R\)-derivation \(d:S\to\Omega_{S/R}\) such that, for every \(M\), composition with \(d\) gives a bijection

This pair is unique up to a unique isomorphism compatible with \(d\).

Proof. Take the free \(S\)-module with one symbol \([s]\) for each \(s\in S\). Divide by the submodule generated by

The images \(ds\) satisfy (1) and give a derivation. Given \(D\), the unique linear map sending \([s]\) to \(D(s)\) kills (3), so descends uniquely. Conversely any linear map out of this quotient gives a derivation. For two universal pairs, their universal properties give maps in both directions carrying each differential to the other. Both composites fix the generating differentials, hence are identities. \(\square\)

This proof also shows that the \(ds\) generate \(\Omega\) as a module; it does not assert that every differential form is \(ds\) for one \(s\). If \(R\to S\) is surjective, all \(ds\) vanish, so \(\Omega_{S/R}=0\). [Stacks, Tags 07BK, 00RO, 00RP.]

For later use, form the square-zero ring \(S\oplus M\), with multiplication \((s,m)(t,n)=(st,sn+tm)\) and identity \((1,0)\). For three factors \((s,m),(t,n),(u,\ell)\), either product order has module component \(st\ell+su n+tu m\), proving associativity. Ring sections of its projection to \(S\), compatible with \(R\), have precisely the form \(s\mapsto(s,D(s))\) for \(R\)-derivations \(D\). This follows by comparing addition, multiplication and the images of base elements.

2. Changing rings and inverting elements

A commutative square \(R\to S\), \(R'\to S'\) induces a map \(S'\otimes_S\Omega_{S/R}\to\Omega_{S'/R'}\), sending \(1\otimes ds\) to the differential of the image of \(s\). The composite \(S\to S'\to\Omega_{S'/R'}\) is an \(R\)-derivation, so (2) constructs the map. The same description proves compatibility with composition.

Theorem 2.1 (base change). Put \(S'=S\otimes_R R'\). There is a natural isomorphism

Proof. In the module on the right define \(d'(s\otimes r')=ds\otimes r'\). This is balanced over \(R\), because \(d(rs)=r\,ds\); extend additively to sums of tensors. Checking products of pure tensors gives the Leibniz identity, hence the identity for arbitrary sums. Base elements have differential zero. An \(R'\)-derivation into an \(S'\)-module \(M\) restricts along \(s\mapsto s\otimes1\) to an \(R\)-derivation. Conversely such a derivation \(D\) extends by \(s\otimes r'\mapsto r'D(s)\); balancing and multiplication are verified in exactly the same way. These constructions are inverse. Thus \(d'\) represents all \(R'\)-derivations, and Theorem 1.1 gives (4). No flatness is needed. \(\square\) [Stacks, Tag 00RV.]

Theorem 2.2 (localization). For any multiplicative subset \(U\subset S\),

If elements of the base are inverted as well, their differentials are zero, so the same module represents derivations over the localized base.

Proof. Let \(M\) be a module over \(S_U=U^{-1}S\), and let \(D:S\to M\) be an \(R\)-derivation. The map \(s\mapsto(s/1,D(s))\) is a ring map into \(S_U\oplus M\). Each \(u\in U\) maps to a unit, with inverse \((u^{-1},-u^{-2}D(u))\). The universal property of localization therefore extends this ring map uniquely to \(S_U\). Its first component is the identity, so its second is a derivation, necessarily given by (5). Restriction and extension are inverse. By (2) and the tensor adjunction, these derivations are represented by \(S_U\otimes_S\Omega_{S/R}\). The localization lesson identifies this tensor product with \(U^{-1}\Omega_{S/R}\). If a base element becomes invertible, (5) gives derivative zero for its inverse. The argument includes zero divisors and even localization to the zero ring; no cancellation of fractions was used. \(\square\) [Stacks, Tag 00RT.]

Proposition 2.3 (directed colimits). Suppose \(R_i\to S_i\) form a compatible directed system, with colimits \(R\to S\). Then

Proof. For any \(S\)-module \(M\), maps from the right-hand side to \(M\) are compatible families of \(R_i\)-derivations \(S_i\to M\). Such a family gives a well-defined additive map \(S\to M\): equality of two representatives holds at a later stage, where compatibility makes their values equal. Products and the Leibniz identity can also be checked at a common stage. Every base element comes from an \(R_i\), so the resulting derivation vanishes on \(R\). Restriction is the inverse operation. The right-hand side thus has the universal property (2), proving the assertion even when the transition maps are not injective. \(\square\) [Stacks, Tag 031G.]

3. The two fundamental sequences

Theorem 3.1 (first fundamental sequence). For ring maps \(A\to B\to C\), the natural sequence

is exact, where \(\alpha(c\otimes db)=c\,d\overline b\).

Proof. Quotient \(\Omega_{C/A}\) by the submodule generated by all \(d\overline b\). The resulting differential on \(C\) vanishes on \(B\). A linear map from this quotient to a \(C\)-module is exactly an \(A\)-derivation on \(C\) vanishing on \(B\), hence a \(B\)-derivation. The quotient is therefore \(\Omega_{C/B}\) by Theorem 1.1. Its defining submodule is the image of \(\alpha\), which proves every part of exactness. \(\square\) [Stacks, Tag 00RS.]

Proposition 3.2 (splitting test). The map \(\alpha\) in (7) has a \(C\)-linear left inverse if and only if every \(A\)-derivation \(B\to M\) extends to \(C\), for every \(C\)-module \(M\).

Proof. Derivations on \(B\) correspond to maps \(C\otimes_B\Omega_{B/A}\to M\). If \(\rho\alpha=1\), composing such a map with \(\rho\) gives an extension. Conversely take \(M=C\otimes_B\Omega_{B/A}\) and the derivation \(b\mapsto1\otimes db\). An extension supplies a map \(\rho:\Omega_{C/A}\to M\), with \(\rho\alpha=1\) on generators and hence everywhere. A left inverse makes (7) split short exact. For example, if \(C=B[T_j]\), prescribing zero for the derivatives of the variables extends any derivation on \(B\) by the polynomial product rule. \(\square\)

Theorem 3.3 (conormal sequence). Let \(C=S/I\) for an ideal in an \(R\)-algebra \(S\). There is an exact sequence

Proof. The differential of a product of two elements of \(I\) becomes zero after tensoring with \(C\). The same product rule shows \(d(sf)=s\,df\) there for \(f\in I\), so \(\delta\) is well-defined and \(C\)-linear. Let \(Q\) be its cokernel. The map \(S\to Q\), \(s\mapsto1\otimes ds\), kills \(I\), so defines an \(R\)-derivation on \(C\). A derivation \(C\to M\) pulls back to a derivation \(S\to M\) killing \(I\), and its corresponding linear map kills the image of \(\delta\). These two operations are inverse, so \(Q\) represents derivations on \(C\). Thus \(Q=\Omega_{C/R}\), proving (8). \(\square\) [Stacks, Tags 00RR, 00RU.]

Proposition 3.4 (a split quotient). If \(q:S\to C=S/I\) admits an \(R\)-algebra section \(\sigma:C\to S\), (8) becomes a split short exact sequence, including injectivity of \(\delta\).

Proof. Set \(D(s)=s-\sigma(q(s))\pmod{I^2}\). The difference lies in \(I\). Writing \(s=\sigma(q(s))+i_s\), the product difference modulo \(I^2\) is \(\sigma(q(s))i_t+\sigma(q(t))i_s\). This is exactly \(q(s)D(t)+q(t)D(s)\) in the \(C\)-module \(I/I^2\). Addition and vanishing on \(R\) are immediate. The universal property gives a map \(\rho:C\otimes_S\Omega_{S/R}\to I/I^2\). For \(f\in I\), \(D(f)=\overline f\), so \(\rho\delta=1\). In the other direction, \(c\mapsto1\otimes d\sigma(c)\) is an \(R\)-derivation on \(C\), and its induced linear map is a right inverse to the last map of (8). \(\square\) [Stacks, Tag 02HP.]

A section of the quotient map has supplied a specific derivation. Merely having a ring retraction \(C\to B\) in (7) does not guarantee a left inverse there. Take \(A=k\), \(B=k[t]\), \(C=B[u]/(tu)\), with retraction \(u\mapsto0\). The source of \(\alpha\) is the free module \(C\,dt\). Its nonzero element \(u^2dt\) maps to zero: differentiate \(tu^2=0\), obtaining \(u^2dt+2tu\,du=u^2dt=0\). The element \(u^2\) is nonzero, as evaluation at \(t=0\) shows. Thus \(\alpha\) is not injective. The conormal counterexample in Exercise 8.5 will also show that a free module \(I/I^2\) need not inject into the middle term of (8).

4. The diagonal ideal

Theorem 4.1 (differentials from the diagonal). Let \(J\) be the kernel of multiplication \(\mu:S\otimes_R S\to S\). Regard \(J/J^2\) as an \(S\)-module through \(\mu\). Then

Proof. Write \(\Delta(s)=1\otimes s-s\otimes1\). It is additive and vanishes on the base. Expanding gives \(\Delta(st)=(1\otimes s)\Delta(t)+(t\otimes1)\Delta(s)\). Modulo \(J^2\), the first coefficient can be replaced by \(s\otimes1\), since their difference belongs to \(J\). Thus \(\Delta\) is a derivation into \(J/J^2\), inducing a map from \(\Omega\).

Define an additive map \(T:S\otimes_R S\to\Omega_{S/R}\) by \(T(a\otimes b)=a\,db\). It is balanced because \(d(rb)=r\,db\). On pure tensors the product rule gives \(T(uv)=\mu(u)T(v)+\mu(v)T(u)\), and bilinearity gives this for arbitrary tensors. Hence \(T\) kills \(J^2\). Restricted to \(J\), it is linear for the action of \(S\) through \(\mu\), by the same identity. It therefore gives a map \(J/J^2\to\Omega\), taking \(\Delta(s)\) to \(ds\).

Finally \(J\) is generated by the \(\Delta(b)\). Indeed, if \(z=\sum a_i\otimes b_i\) has \(\sum a_ib_i=0\), then \(z=\sum(a_i\otimes1)\Delta(b_i)\). Thus the two maps are inverse on generating sets on both sides. \(\square\) [Stacks, Tag 00RW.]

Algebraically, \(J\) cuts out the diagonal. Passing to \(J/J^2\) retains its first-order equations and discards their products. Formula (9) identifies these equations with universal differentials without any finite generation assumption.

5. Polynomial generators and Jacobian relations

Theorem 5.1 (polynomials and quotients). For any set of variables,

If \(P=R[x_i:i\in E]\), \(S=P/I\), then

Proof. The product rule forces a derivation on \(P\) to take a polynomial \(f\) to \(\sum_i(\partial f/\partial x_i)D(x_i)\). Every sum is finite. Conversely, arbitrary chosen elements \(m_i\) of a \(P\)-module define a derivation by this formula: addition is immediate, and the polynomial product rule proves Leibniz. Thus the free module in (10) represents derivations. Applying (8) to \(P\to S\) gives (11). If \(I=(f_j:j\in F)\), the relations for these generators suffice: differentiate \(f=\sum a_jf_j\) and reduce modulo \(I\), obtaining \(df=\sum\overline a_j\,df_j\). \(\square\) [Stacks, Tag 00RX.]

For a finite presentation \(S=R[x_1,\ldots,x_n]/(f_1,\ldots,f_m)\), (11) is the Jacobian presentation

The columns correspond to equations and the rows to variables. It follows that a finite-type algebra has finite \(\Omega\), and a finitely presented algebra has finitely presented \(\Omega\), over an arbitrary base ring. Finite type and finite presentation of the algebra are different hypotheses. Over a Noetherian base they coincide: Hilbert basis, Theorem 2.1 of the Noetherian-ring lesson, makes the kernel of a finite polynomial presentation finitely generated. [Stacks, Tag 00RY.]

Proposition 5.2 (regular equations give a free conormal module). If \(f_1,\ldots,f_m\) is a regular sequence in any ring \(P\), and \(I=(f_1,\ldots,f_m)\), then their classes form a basis of \(I/I^2\) over \(P/I\).

Proof. They generate. Suppose \(\sum a_if_i\in I^2\). Express that sum as \(\sum b_if_i\) with each \(b_i\in I\). Then the vector \((a_i-b_i)\) is a first Koszul cycle. Exactness of the regular-sequence Koszul complex, Theorem 4.1 of the projective-dimension lesson, makes it a boundary. Each coefficient of a second Koszul boundary belongs to \(I\), since its entries are combinations of the \(f_i\). Therefore every \(a_i\in I\), proving independence. \(\square\)

Even under this hypothesis, (8) need not start with zero. The derivative relations may have a kernel despite the freeness of the conormal module.

For a \(k\)-algebra \(S\) and a maximal ideal \(\mathfrak m\) with \(S/\mathfrak m=k\) as a \(k\)-algebra, the constants give a section of \(S\to k\). Proposition 3.4 and \(\Omega_{k/k}=0\) give

Thus at a rational point the fibre of \(\Omega\) is the cotangent space, with no finite-type assumption. Its \(k\)-linear dual is the tangent space. For a general prime \(\mathfrak p\), a fibre means \(\Omega\otimes_S\kappa(\mathfrak p)\); residue-field differentials may then also contribute.

6. Fields: separability and independent variables

Let \(E=F(\alpha)\) be a simple algebraic field extension, with monic minimal polynomial \(f\). The quotient presentation gives

In particular it is zero for a separable algebraic generator and one-dimensional if \(f'=0\).

Lemma 6.1 (extending a derivation through a separable generator). If \(K\subset F\subset E=F(\alpha)\) and \(\alpha\) is separable over \(F\), every \(K\)-derivation \(D:F\to M\), for an \(E\)-module \(M\), extends uniquely to \(E\).

Proof. Write \(f(T)=\sum c_iT^i\). Any extension must satisfy

The denominator is a nonzero field element, because separability means the minimal polynomial has no repeated root. To construct the extension, choose this value and define on \(F[T]\) \(\widetilde D(\sum b_iT^i)=\sum\alpha^iD(b_i)+\sum i b_i\alpha^{i-1}D(\alpha)\). Give \(M\) the \(F[T]\)-action by evaluating at \(\alpha\). Expansion of products proves the Leibniz identity. Formula (15) makes \(\widetilde D(f)=0\), so Leibniz makes it kill the ideal \((f)\). It descends to \(E=F[T]/(f)\), and the forced value proves uniqueness. \(\square\)

Theorem 6.2 (separably generated fields). An algebraic separable extension \(L/K\), finite or infinite, has \(\Omega_{L/K}=0\). More generally, if \(T\) is a separating transcendence basis, meaning \(L\) is algebraic separable over \(F=K(T)\), then

For an infinite basis, the equality of dimensions is an equality of cardinal numbers.

Proof. A finite separable extension can be generated by finitely many separable elements. Apply Lemma 6.1 successively; each element remains separable over an intermediate field because its minimal polynomial divides a separable polynomial. Every derivation on the starting field extends uniquely. For an infinite algebraic separable extension, each finite set lies in a finite separable subextension. The unique extensions agree on intersections, or by comparison in a larger finite subextension, and therefore give one derivation on the union.

For \(F=K\), the only derivation on the starting field vanishing on \(K\) is zero, proving the first assertion via (2). For \(F=K(T)\), (10) and (5) show that its differentials are free on the \(dt\), even for infinite \(T\). Restriction and the unique extension just proved identify derivations \(L\to M\) with derivations \(F\to M\), for every \(L\)-module \(M\). These are represented by \(L\otimes_F\Omega_{F/K}\), yielding (16). \(\square\)

Proposition 6.3 (finite purely inseparable extensions). A nontrivial finite purely inseparable extension \(L/K\) has nonzero \(\Omega_{L/K}\). In particular, for \(K=\mathbb F_p(t)\) and \(L=\mathbb F_p(u)\), \(u^p=t\), it has dimension one.

Proof. Choose a minimal finite list of field generators for \(L/K\), and let \(E\) be generated by all but the last, \(\alpha\). Then \(\alpha\notin E\). Its minimal polynomial divides some \(T^{p^N}-a\), so has only one distinct root in an algebraic closure. A nonconstant irreducible polynomial with nonzero derivative is separable: its greatest common divisor with its derivative is one. Repeatedly extracting powers of \(T^p\) from a polynomial with zero derivative writes this minimal polynomial as \(g(T^{p^e})\), with \(g'\ne0\). Irreducibility is preserved for \(g\). Since there is only one distinct root, the separable polynomial \(g\) has degree one. Thus the minimal polynomial is \(T^{p^e}-c\), with \(e\geq1\). By (14), \(\Omega_{L/E}=L\,d\alpha\ne0\). Sequence (7) makes it a quotient of \(\Omega_{L/K}\), proving nonvanishing.

In the stated example, \(t\) is not a \(p\)-th power in \(K\): its order at \(t=0\) is one, whereas every \(p\)-th power has order divisible by \(p\). Since \(u^p=t\), the preceding minimal-polynomial argument shows its degree is \(p\), and (14) gives \(L\,du\). \(\square\)

Finite separability and finite purely inseparable nontriviality are therefore visible to differentials. The finiteness in Proposition 6.3 matters; Exercise 8.6 gives an infinite purely inseparable extension with zero differentials.

7. Arithmetic, curves and a two-term complex

For \(\mathbb C=\mathbb R[T]/(T^2+1)\), (14) gives \(\Omega_{\mathbb C/\mathbb R}=0\), since \(2i\) is invertible. Over the integers the same equation gives

As \(i\) is a unit, the annihilator is \((2)\). The identity \(2=-i(1+i)^2\) shows its support is the unique prime \((1+i)\) over two; the quotient by that ideal is \(\mathbb F_2\). There is no arithmetic differential away from that prime.

For the cusp \(A=k[x,y]/(y^2-x^3)\), over any field,

The map \(k[x,y]\to k[t]\) given by \(x\mapsto t^2\), \(y\mapsto t^3\) has image \(k[t^2,t^3]\) and kills \(y^2-x^3\). Division by the monic polynomial \(y^2-x^3\) in \(y\) gives every class in \(A\) a representative \(a(x)+yb(x)\), with \(a,b\in k[x]\). Its image is \(a(t^2)+t^3b(t^2)\). The first summand has only even powers of \(t\), and the second only odd powers. If their sum is zero, independence of polynomial monomials forces every coefficient of both \(a\) and \(b\) to vanish. Thus the induced map \(A\to k[t^2,t^3]\) is injective as well as surjective, proving \(A\simeq k[t^2,t^3]\), a domain, in every characteristic. Solution 7.2 of the integrality lesson proves that the normalization of this subring is \(k[t]\).

In its fraction field the displayed relation is nonzero: \(x,y\ne0\) and the coefficients two and three cannot both vanish in a field. The generic fibre has dimension one. At \((x,y)\) both coefficients vanish, so the fibre has dimension two. If \(\Omega\) were free near that point, its rank would be two there and at the generic point, a contradiction. The formula handles characteristics two and three separately without dividing by either coefficient.

For the split ordinary node \(B=k[x,y]/(xy)\),

At the origin the fibre has dimension two. At every other prime at least one of \(x,y\) is a unit, and the relation has rank one over the residue field, so the fibre has dimension one. Local freeness again fails at the origin; each generic branch lies below that maximal ideal. Exercise 8.3 makes the chart calculation explicit.

For a polynomial presentation \(P\twoheadrightarrow S\) over \(R\), with kernel \(I\), package the conormal map into the two-term complex

in homological degrees one and zero. Its degree-zero homology is \(\Omega_{S/R}\), by (8); its degree-one homology measures the failure of the conormal map to be injective. The canonical presentation uses one polynomial variable for every element of \(S\). The resulting complex is the naive cotangent complex.

Theorem 7.1 (comparison of presentations). Polynomial presentations of the same algebra give homotopy-equivalent two-term complexes. For a commuting square of base and target ring maps, lifts between polynomial presentations induce comparison maps unique up to homotopy, compatible with composition. These statements allow arbitrary sets of variables.

Proof. Write the presentations as \(P=R[X_t]\twoheadrightarrow S\) and \(P'=R'[Y_u]\twoheadrightarrow S'\), with kernels \(I,I'\). For each \(X_t\), lift its prescribed image in \(S'\) to \(P'\). These choices give a polynomial-algebra map \(\varphi:P\to P'\) lifting the square. It sends \(I\) into \(I'\), hence defines the degree-one map \([i]\mapsto[\varphi(i)]\), and the degree-zero map \(dp\otimes s\mapsto d\varphi(p)\otimes\phi(s)\), where \(\phi:S\to S'\) is the target map. Differentiation shows that these commute with the conormal differential.

Let \(\psi\) be another lift. Their difference has values in \(I'\). Modulo \((I')^2\) it is a derivation:

The two actions on \(I'/(I')^2\) agree, because their difference is multiplication by an element of \(I'\). This common action factors through \(S\). Thus the universal property of differentials gives the homotopy

On \([i]\in I/I^2\), the equality \(Hd[i]=[\varphi(i)-\psi(i)]\) is the difference of the degree-one maps. On \(dp\otimes s\), differentiation of (20a) gives the difference of the degree-zero maps. Hence \(Hd=\varphi_1-\psi_1\) and \(dH=\varphi_0-\psi_0\), exactly the chain-homotopy identities.

Composition of chosen polynomial lifts gives the composite maps in both degrees by the displayed formulas. Changing any lift only changes this composite by a homotopy. For two presentations of the same algebra over the same base, choose lifts in both directions. Their composites lift the identity square and are homotopic to the identity lifts just proved. They are therefore homotopy inverses. This also compares each presentation with the canonical one and proves the theorem. No projectivity of \(I/I^2\) was assumed. \(\square\)

This is the presentation-independent two-term complex used in the following formal-smoothness lesson. Homotopy equivalence here is stronger than equality of its two homology modules; an acyclic canonical complex need not be literally the zero complex.

8. Exercises

Exercise 8.1 (easy: quadratic orders). For every integer \(d\), let \(A=\mathbb Z[\sqrt d]\) be the actual subring of an algebraic closure of \(\mathbb Q\). Compute \(\Omega_{A/\mathbb Z}\) and its support, distinguishing square integers from nonsquares.

Exercise 8.2 (easy: one invertible derivative). Let \(S=R[x]_g/(f)\), where \(g,f\in R[x]\). If the image of \(f'\) is a unit in \(S\), show that \(\Omega_{S/R}=0\), without a Noetherian hypothesis.

Exercise 8.3 (medium: the node at all points). For \(B=k[x,y]/(xy)\), compute \(\Omega_{B/k}\), its localizations on \(D(x)\) and \(D(y)\), and the dimension of its fibre at every prime. Prove that it is not locally free at the origin.

Exercise 8.4 (medium: two algebraic behaviours). Show directly that a finite separable extension has zero differentials. Show that a nontrivial finite purely inseparable extension has nonzero differentials, allowing more than one field generator. Compute the degree-\(p\) example \(\mathbb F_p(t^{1/p})/\mathbb F_p(t)\).

Exercise 8.5 (hard: a section and its absence). Construct the two splittings of (8) when \(S\to S/I\) has an \(R\)-algebra section. For \(S=k[t]\) and \(I=(t^2)\), compute \(I/I^2\), the conormal map and its kernel in every characteristic. Explain why a regular defining equation does not ensure injectivity.

Exercise 8.6 (hard: a purely inseparable tower). In an algebraic closure of \(K=\mathbb F_p(t)\), choose \(u_0=t\) and \(u_{n+1}^p=u_n\). Set \(L_n=\mathbb F_p(u_n)\) and \(L=\bigcup_nL_n\). Prove that \([L_n:K]=p^n\) and that \(L/K\) is nontrivial and purely inseparable, yet \(\Omega_{L/K}=0\).

9. Solutions

Solution 8.1. If \(d=m^2\) for an integer \(m\), then either choice of its root is an integer, so \(A=\mathbb Z\) and \(\Omega=0\), with empty support. This includes \(d=0\); the actual ring is not the nonreduced quotient \(\mathbb Z[T]/(T^2)\).

If \(d\) is nonsquare, \(T^2-d\) is irreducible over \(\mathbb Q\). Indeed a rational square root of an integer must be an integer, by writing it as a reduced fraction. Monic polynomial division in \(\mathbb Z[T]\) leaves a remainder \(a+bT\). Its value at \(\alpha=\sqrt d\) can vanish only if both coefficients vanish, because \(1,\alpha\) are independent over \(\mathbb Q\). Thus \(A=\mathbb Z[T]/(T^2-d)\), and (11) gives

For the support equality, localization of a cyclic quotient at a prime is nonzero exactly when its annihilator is contained in that prime, by the localization lesson. A prime contains \(2\alpha\) precisely when it contains two or \(\alpha\). Since \(\alpha^2=d\), this is precisely when it contains two or \(d\), hence \(2d\). Consequently the support consists of all primes over the rational primes dividing \(2d\). Negative nonsquare integers are included in the same calculation.

Solution 8.2. By (10) and (5), \(\Omega_{R[x]_g/R}=R[x]_g\,dx\). The quotient ideal is generated by \(f\), so (8) gives \(\Omega_{S/R}=S\,dx/(f'\,dx)\). A unit \(f'\) generates the whole free rank-one module, and the quotient is zero. This also covers the zero algebra, where the differential module is zero.

Solution 8.3. Differentiating the single equation gives (19). On \(D(x)\), the relation \(xy=0\) forces \(y=0\), and \(B_x=k[x,x^{-1}]\). Formula (5) identifies \((\Omega_{B/k})_x\) with the free module \(B_x\,dx\); equivalently the relation in (19) becomes \(x\,dy=0\). On \(D(y)\) the result is \(B_y\,dy\).

Tensoring (19) with \(\kappa(\mathfrak p)\) is right exact. The resulting relation is the vector \((\overline y,\overline x)\) in a two-dimensional vector space. It is zero precisely if \(\mathfrak p\) contains both \(x,y\), which forces \(\mathfrak p=(x,y)\), since the quotient by this ideal is a field. All other fibres have dimension one; the origin fibre has dimension two. If the localization at the origin were free, its rank would be two. Localizing further at the prime \((y)\) would preserve that rank, whereas the generic \(x\)-branch fibre has dimension one. This contradiction proves failure of local freeness.

Solution 8.4. Generate a finite separable extension by \(\alpha_1,\ldots,\alpha_r\) and set \(E_i=K(\alpha_1,\ldots,\alpha_i)\). Each minimal polynomial is separable over \(E_{i-1}\), since it divides a separable polynomial over \(K\). Formula (14) gives \(\Omega_{E_i/E_{i-1}}=0\). Starting with \(\Omega_{K/K}=0\), (7) then inductively gives \(\Omega_{E_i/K}=0\), including \(i=r\).

For a nontrivial finite purely inseparable extension, choose an irredundant generating list and remove its last element \(\alpha\) to obtain a proper intermediate field \(E\). The minimal polynomial is \(T^{p^e}-c\), \(e\geq1\), by the unique-root argument in Proposition 6.3. Hence \(\Omega_{L/E}=L\,d\alpha\), and the surjection \(\Omega_{L/K}\to\Omega_{L/E}\) in (7) proves nonvanishing. This avoids assuming the original extension is simple. For \(u^p=t\), the order at \(t=0\) excludes a \(p\)-th root in \(K\), and the unique-root minimal polynomial is \(T^p-t\). Its derivative vanishes, giving \(\Omega_{L/K}=L\,du\).

Solution 8.5. Put \(C=S/I\) and take its section \(\sigma\). The derivation \(D(s)=s-\sigma(\overline s)\pmod{I^2}\), with its product computation in Proposition 3.4, induces a left inverse \(\rho\) of the conormal map. The derivation \(c\mapsto1\otimes d\sigma(c)\) induces a right inverse \(\eta\) of the map to \(\Omega_{C/R}\). These are compatible: \(\rho\eta=0\), because \(D(\sigma(c))=0\). Every element of the middle term is uniquely the sum of an element of \(\delta(I/I^2)\) and one of \(\eta(\Omega_{C/R})\): subtract its image under \(\eta\) to enter the kernel, use (8), and use \(\rho\delta=1\) for uniqueness.

Now let \(C=k[t]/(t^2)\). Multiplication by \(t^2\) identifies \(C\) with \((t^2)/(t^4)=I/I^2\), since coefficient comparison gives exactly the kernel \((t^2)\) before passing to the quotient. The middle module is \(C\,dt\). Thus (8) is

In characteristic different from two its kernel is \((t)\), because the annihilator of \(t\) in \(C\) consists exactly of multiples of \(t\). In characteristic two the map is zero and its kernel is all of \(C\). The defining element \(t^2\) is a nonzerodivisor in \(k[t]\), so is a regular sequence of length one, and the conormal module is free. Nevertheless its differential map is not injective in either case. No compatible algebra section can exist, by the splitting theorem.

Solution 8.6. Each \(u_n\) is transcendental over \(\mathbb F_p\), because a power of it equals the transcendental element \(t\). In the rational function field \(\mathbb F_p(u_n)\), the variable \(u_n\) has order one at zero and is not a \(p\)-th power. The unique-root minimal-polynomial argument applied to \(u_{n+1}^p=u_n\) therefore gives \([L_{n+1}:L_n]=p\). Multiplying degrees proves \([L_n:K]=p^n\). Equivalently its defining minimal polynomial over \(K\) is \(T^{p^n}-t\).

For \(n\geq1\), (14) gives \(\Omega_{L_n/K}=L_n\,du_n\). Under passage to \(L_{n+1}\),

Every element of the directed system \(L\otimes_{L_n}\Omega_{L_n/K}\) consequently becomes zero at the next stage. Its colimit is zero, and Proposition 2.3 yields \(\Omega_{L/K}=0\). Every element of \(L\) lies in one \(L_n\) and has a sufficiently large \(p\)-power in \(K\), so the extension is purely inseparable. It is nontrivial already at \(L_1\), and the unbounded degrees make it infinite. Thus zero differentials alone do not characterize algebraic separability for arbitrary infinite field extensions.

References and proof scope

The Stacks project supplies the reference framework for derivations, universal differentials, base change, localization, the fundamental sequences, polynomial presentations and the diagonal ideal, at the tags cited above. Ravi Vakil, The Rising Sea: Foundations of Algebraic Geometry, public draft of 27 July 2024, §§21.2–21.3, provides complementary geometric interpretations and computations. Timothy J. Ford, Commutative Algebra, version of 23 September 2026, Chapter 10, Sections 1–2, treats derivations, the module of differentials and its two fundamental exact sequences. The incorporated presentation-comparison proof retains its source credit and licence below.

Verified tag references: Tag 00RN, Tag 07BK, Tag 00RO, Tag 00RP, Tag 031G, Tag 00RR, Tag 00RS, Tag 00RT, Tag 00RU, Tag 02HP, Tag 00RV, Tag 00RW, Tag 00RX, Tag 00RY, Tag 07BN, Tag 00S1.

Proof dependencies and licence. Theorem 7.1 proves the full presentation-comparison statement [Stacks, Tags 07BN, 00S1]. All differential constructions and the earlier splitting and field calculations are proved above, with earlier course results used at the stated locators. Formal smoothness, unramified maps and smooth algebras follow in the next two lessons.

Theorem 7.1 incorporates the Stacks Project Authors' lemma-NL-homotopy in AI Integrated Stacks Project, pinned algebra source, and the source-linked supplementary proof 44 by GPT-6 Astra, which spells out the two homotopy identities. GPT-6.1 Sol checked both identities and integrated them into the curve and conormal examples. This modified lesson is distributed under GNU Free Documentation License 1.2, with no Invariant Sections, no Front-Cover Texts and no Back-Cover Texts; the licence text accompanies the transparent source. History: Stacks Project Authors, Commutative Algebra, pinned AI Integrated Stacks Project revision; GPT-6 Astra, supplementary proof 44, September 2026; GPT-6.1 Sol, original CC0 lesson and this adaptation, October 2026. The previously released original material remains available under CC0.

Copyright and licence

Copyright (C) 2005–2025 Johan de Jong. The incorporated source is The Stacks Project, as distributed in AI Integrated Stacks Project at revision 565b10e987aba5969b21145a0833f42d69f96790.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts and no Back-Cover Texts. A copy is supplied as GNU Free Documentation License 1.2.

The original course material remains available under its CC0 dedication. This combined edition, including the incorporated and adapted proof, is distributed under GNU FDL 1.2 or later. The source authors, incorporated source titles and mathematical adaptations are identified above. History: Stacks Project Authors, original source; the credited AI Integrated Stacks Project editorial contributors, where used; OpenAI GPT-6.1 Sol, course adaptation, October 2026.