Unramified morphisms

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Independent AI review of this edition is not complete. Public domain (CC0).

An unramified map has no relative infinitesimal motion. Its fibres consist of isolated separable points, and its diagonal is open. For a separated unramified map, they imply that two sections meeting at a point of a connected base agree everywhere on that base. They also show the limitation: unramifiedness controls uniqueness, while flatness and existence of lifts are additional conditions.

We assume affine schemes, fibre products and immersions, and the algebra of Kähler differentials from Kähler differentials and Formally smooth, unramified and étale ring maps. We use Quasi-finite morphisms and Chevalley's theorem, and, for the proper-map argument in Section 6, Zariski's Main Theorem. The precise imported statements are collected at the end. Our main references are the Stacks Project and Vakil's The Rising Sea.

1. The sheaf that measures infinitesimal motion

For \(f:X\to S\), the sheaf \(\Omega_{X/S}\) represents relative derivations into \(\mathcal O_X\)-modules. On affine charts \(U=\operatorname{Spec}B\) over \(V=\operatorname{Spec}A\), it is the quasi-coherent sheaf associated to \(\Omega_{B/A}\). The algebraic localization rule for differentials identifies these modules on chart overlaps, so they glue canonically.

Three rules will be used repeatedly:

In the last line an immersion is made closed by restricting the ambient open, and \(\mathcal I\) is its ideal there. The first arrow sends the class of a local equation to its differential. The algebraic base-change and exact-sequence rules on affine charts give the first two lines and the exactness of the third. Its first arrow is well-defined because \(d(\mathcal I^2)\subset\mathcal I\Omega_{X/S}\); these local arrows agree on overlaps. This proves the sheaf forms [Stacks, Tags 01V0, 01UX and 01UZ]. Exactness does not assert injectivity of the first arrows.

Proposition 1.1 (the diagonal's conormal). There is a canonical isomorphism

Proof. On \(U\times_V U\), let \(J=\ker(B\otimes_A B\xrightarrow\mu B)\). The map

is an \(A\)-derivation. The product rule follows because the discrepancy between the product-rule expression and the difference for a product is a product of two elements of \(J\). It therefore induces \(\Omega_{B/A}\to J/J^2\). The inverse is induced by \(b\otimes c\mapsto b\,dc\) on \(J\). It kills \(J^2\): for \(z,z'\in B\otimes_A B\), this map satisfies the Leibniz identity with coefficients \(\mu(z),\mu(z')\), both zero when \(z,z'\in J\). The ideal \(J\) is generated by \(1\otimes b-b\otimes1\), so the two maps are inverse. Their formulas commute with restrictions. Gluing gives the claimed isomorphism [Stacks, Tag 08S2]. ∎

If \(B\) is generated by \(b_1,\ldots,b_n\) over \(A\), then \(db_1,\ldots,db_n\) generate \(\Omega_{B/A}\). Thus for a locally finite type morphism the differential sheaf is of finite type [Stacks, Tag 01V2]. Its zero-stalk locus is open: finitely many generators that vanish in a localization are all killed by one denominator. This observation is valid over a non-Noetherian base.

2. Conventions and permanence

We call \(f\) unramified if it is locally of finite type and \(\Omega_{X/S}=0\). We call it G-unramified if it is locally of finite presentation and \(\Omega_{X/S}=0\). These are the Stacks conventions [Stacks, Tags 02G4 and 02G5]. At a point the corresponding condition is required on a neighbourhood. For a morphism already locally of finite type, vanishing of the differential stalk is equivalent to unramifiedness there, by the last paragraph of Section 1.

Both properties are local on source and target: the finiteness condition is local, and a quasi-coherent sheaf vanishes exactly when its restrictions to an open cover vanish. Equivalently one may test the associated ring maps on every compatible pair of affine charts [Stacks, Tag 02G6]. Over a locally Noetherian base finite type algebras are finitely presented, so the two conventions agree [Stacks, Tag 04EV].

They need not agree in general. Put \(A=k[t_1,t_2,\ldots]\) and \(I=(t_1,t_2,\ldots)\). The closed immersion \(\operatorname{Spec}(A/I)\to\operatorname{Spec}A\) is unramified, since a quotient algebra is finite type and has zero relative differentials. It is not G-unramified: its kernel is not finitely generated. Any finite list of elements of \(I\) involves only finitely many variables; setting those variables to zero leaves another variable nonzero, so that list cannot generate \(I\).

Proposition 2.1. Unramified morphisms are preserved by composition and arbitrary base change. Every immersion is unramified. If \(X\to Y\to S\) has unramified composite, then \(X\to Y\) is unramified. The composition and base-change assertions also hold for G-unramified morphisms; the last assertion does so if \(Y\to S\) is locally of finite type.

Proof. Finite type, and finite presentation where relevant, are preserved by composition and base change. The first two differential rules in Section 1 then give the required vanishing. Open immersions have zero relative differentials by localization; closed immersions have them by the quotient rule. Composing these proves the immersion assertion. For the last assertion, the exact sequence makes \(\Omega_{X/Y}\) a quotient of \(\Omega_{X/S}=0\). Finite type of the composite implies finite type of \(X\to Y\): on affine charts the same finite list of algebra generators over the base also generates over the intermediate algebra. In the finite presentation case the additional finite type assumption on \(Y\) supplies the finitely many generator-image relations, as in Section 6 of Flatness criteria, dimension and the flat locus. ∎

These are [Stacks, Tags 02G9, 02GA, 02GC and 02GG]. A closed immersion is G-unramified precisely when its defining ideal sheaf is of finite type. In particular unramifiedness alone imposes no flatness: \(\operatorname{Spec}k\to\operatorname{Spec}k[t]\) at \(t=0\) is unramified, but multiplication by \(t\) becomes zero on its nonzero coordinate module.

3. Fibres and the six pointwise tests

Lemma 3.1 (unramified finite type algebras over a field). If \(B\) is a finite type \(k\)-algebra with \(\Omega_{B/k}=0\), then \(B\) is a finite product of finite separable field extensions of \(k\).

Proof. Let \(\mathfrak n\) be a maximal ideal. Zariski's lemma makes \(L=B/\mathfrak n\) finite over \(k\). The quotient differential sequence gives \(\Omega_{L/k}=0\), so the field differential criterion makes \(L/k\) separable. We recall the small local argument that removes possible nilpotents at this point.

Write \(R=B_{\mathfrak n}\), \(\mathfrak m=\mathfrak nR\), and \(V=\mathfrak m/\mathfrak m^2\). The finite separable extension \(L/k\) has a primitive generator \(\alpha\) with minimal polynomial \(h\) and \(h'(\alpha)\ne0\). Lift \(\alpha\) to \(a\in R/\mathfrak m^2\). The element \(h'(a)\) is a unit. Since \(V^2=0\), Taylor's formula shows that

satisfies \(h(a')=0\). This gives a \(k\)-algebra section \(L\to R/\mathfrak m^2\). Hence this quotient is the square-zero algebra \(L\oplus V\).

Derivations of \(L\oplus V\) into an \(L\)-vector space vanish on \(L\) and are exactly the \(L\)-linear maps from \(V\). Here they vanish on \(L\) because \(\Omega_{L/k}=0\). The differential quotient sequence from \(R\) to \(R/\mathfrak m^2\), and then tensoring with \(L\), identifies

the differentials of elements of \(\mathfrak m^2\) vanish after that tensor product. Its left side is zero. Thus \(\mathfrak m/\mathfrak m^2=0\). The ideal \(\mathfrak m\) is finite since \(R\) is Noetherian, and Nakayama gives \(\mathfrak m=0\). Every maximal localization of \(B\) is therefore a field.

Any prime of \(B\) is contained in a maximal ideal, and localization there has only one prime. Thus \(B\) is zero-dimensional. A zero-dimensional Noetherian ring is Artinian, and an Artinian ring is the finite product of its maximal localizations. Those localizations are the finite separable fields just found. ∎

Conversely a finite product of finite separable fields is finite type over \(k\) and has zero differentials. This proves the affine case without presuming reducedness of \(B\).

Theorem 3.2 (schemes over a field). A \(k\)-scheme \(X\) is unramified over \(k\) if and only if it is a disjoint union of spectra of finite separable extensions of \(k\). Consequently every fibre of an unramified morphism has this form [Stacks, Tag 02G7].

Proof. Every point of an unramified \(X\) has an affine neighbourhood to which Lemma 3.1 applies. Every point is therefore open, and its one-point open subscheme is the spectrum of its local ring, a finite separable field. These open subschemes partition \(X\). There need not be finitely many of them. The converse is local on the source and follows from the converse to Lemma 3.1. The fibre assertion follows by base change. ∎

Theorem 3.3 (pointwise criterion). Let \(f:X\to S\) be locally of finite type, \(x\mapsto s\). The following six conditions are equivalent:

1. \(f\) is unramified at \(x\).
2. \(X_s\to\operatorname{Spec}\kappa(s)\) is unramified at \(x\).
3. \(\Omega_{X/S,x}=0\).
4. \(\Omega_{X_s/\kappa(s),x}=0\).
5. The vector space \(\Omega_{X/S,x}\otimes_{\mathcal O_{X,x}}\kappa(x)\), equivalently \(\Omega_{X_s/\kappa(s),x}\otimes_{\mathcal O_{X_s,x}}\kappa(x)\), is zero.
6. \(\mathfrak m_s\mathcal O_{X,x}=\mathfrak m_x\) and \(\kappa(x)/\kappa(s)\) is finite separable.

If \(f\) is locally of finite presentation, the first condition can equally say G-unramified [Stacks, Tag 02GF].

Proof. Conditions 1 and 3 agree because the differential sheaf is finite type and a zero stalk has a zero neighbourhood. The same argument in the fibre gives 2 equivalent to 4. Differential base change identifies the two vector spaces in 5. Nakayama, applied to the finite differential modules over the two local rings, makes 3, 4 and 5 equivalent.

If 2 holds, take an unramified open neighbourhood in the fibre. Theorem 3.2 gives that its local ring at \(x\) is \(\kappa(x)\), finite separable over \(\kappa(s)\). But this fibre local ring is \(\mathcal O_{X,x}/\mathfrak m_s\mathcal O_{X,x}\). Its being a field is exactly the maximal-ideal equality in 6. Conversely that equality identifies the fibre local ring with \(\kappa(x)\). Localization of differentials gives its differential module as \(\Omega_{\kappa(x)/\kappa(s)}=0\), by finite separability. Thus 4 holds. The finite presentation convention changes no differential calculation. ∎

In particular unramifiedness implies finite separable residue extensions [Stacks, Tag 02G8]. It also implies locally quasi-finite, because each fibre point is isolated by Theorem 3.2, and the morphism is locally of finite type [Stacks, Tag 02V5]. The word “locally” permits an infinite disjoint union of separable points.

4. Why the diagonal is open

Theorem 4.1 (diagonal criterion). If \(f\) is unramified, \(\Delta_{X/S}:X\to X\times_S X\) is an open immersion. Conversely, if \(f\) is locally of finite type and its diagonal is an open immersion, then \(f\) is unramified. With local finite presentation this characterizes G-unramified morphisms [Stacks, Tag 02GE].

Proof. Near a diagonal point use \(U=\operatorname{Spec}B\) over \(\operatorname{Spec}A\), with \(B\) generated by \(b_1,\ldots,b_n\). In \(C=B\otimes_A B\), the diagonal ideal \(J\) is generated by

Indeed in the quotient these generators identify the two copies of every algebra generator, so multiplication induces an inverse isomorphism \(C/J\cong B\). Unramifiedness and Proposition 1.1 give \(J/J^2=0\), hence \(J=J^2\). At any prime \(\mathfrak r\supset J\), the finite \(C_{\mathfrak r}\)-module \(J_{\mathfrak r}\) satisfies \(J_{\mathfrak r}=J_{\mathfrak r}^2\subset\mathfrak rC_{\mathfrak r}J_{\mathfrak r}\). Nakayama gives \(J_{\mathfrak r}=0\). Finite generation then gives \(h\notin\mathfrak r\) with \(J_h=0\). Over \(D(h)\), the diagonal closed immersion is an isomorphism.

Thus the diagonal is locally an open immersion at each source point. The diagonal is an immersion, so these local identifications glue to an open immersion onto the union of those neighbourhoods. Conversely an open immersion has zero conormal sheaf. Proposition 1.1 gives \(\Omega_{X/S}=0\), and the assumed finiteness condition finishes the proof. ∎

Finiteness of the diagonal ideal came from finite type of \(f\), not finite presentation of its affine algebra. Dropping finite type would invalidate the Nakayama step. Also the diagonal need not be closed: separatedness is an additional condition.

On an affine chart \(B=A[t_1,\ldots,t_n]/I\), unramifiedness at a point can alternatively be checked by whether the differentials of the relations span \(\bigoplus B\,dt_i\) there. This follows immediately from the conormal exact sequence in Section 1. Over a Noetherian base there are finitely many defining equations; it gives the differential criterion in [Stacks, Tag 024P].

5. Sections and rigidity

Proposition 5.1. Every section of an unramified morphism is an open immersion. Every section of a separated morphism is a closed immersion. A section of an unramified separated morphism is therefore open and closed [Stacks, Tag 024T].

Proof. For an \(S\)-map \(a:Y\to X\), its graph \(Y\to Y\times_S X\) is the pullback of \(\Delta_{X/S}\) along \((a\circ\operatorname{pr}_Y,\operatorname{pr}_X):Y\times_S X\to X\times_S X\). The graph is consequently open when the diagonal is open, and closed when the diagonal is closed. If \(Y=S\) and \(a\) is a section, the graph identifies with the section under \(S\times_S X=X\). ∎

If \(Y\) is connected and \(X\to Y\) is unramified and separated, the image of a section is a connected open and closed subset of \(X\). Any connected subset meeting it is contained in it, so it is a connected component. Give components their canonical flat closed subscheme structures from Flat morphisms. A component whose induced scheme map to \(Y\) is an isomorphism supplies its inverse section; Proposition 5.1 then makes that component open as well. This proves the bijection of [Stacks, Tag 024U]; it does not assert that arbitrary connected components are open. At most one such section can pass through a given point of \(X\).

Theorem 5.2 (rigidity from one point). Let \(X\to S\) be unramified and separated, and \(Y\) a connected \(S\)-scheme. If \(a,b:Y\to X\) satisfy \(a(y)=b(y)=x\) at some point and induce the same maps \(\kappa(x)\to\kappa(y)\), then \(a=b\) [Stacks, Tag 024V].

Proof. The equality subscheme

is open and closed in \(Y\), since the diagonal is both. The equality of the residue maps says that the morphism \(\operatorname{Spec}\kappa(y)\to X\times_S X\) determined by \(a,b\) factors through the diagonal: both coordinate morphisms are literally the same. Thus \(y\in E\). Connectedness gives \(E=Y\) as an open subscheme, so \((a,b)\) factors through the diagonal and \(a=b\). ∎

Equal underlying points alone would not suffice. The identity and complex conjugation are different \(\mathbb R\)-morphisms \(\operatorname{Spec}\mathbb C\to\operatorname{Spec}\mathbb C\), although both source and target have one point. Their residue maps differ. The target is unramified and separated over \(\mathbb R\).

Separatedness is also necessary. The affine line with doubled origin maps to the ordinary affine line by the identity on each of its two charts. It is a local isomorphism, hence unramified. The two chart maps from the ordinary affine line are sections agreeing on \(D(t)\), including their residue maps, but differing at the two origins. Their equality locus is open and fails to be closed.

6. Monomorphisms and closed immersions

Recall that a monomorphism \(X\to S\) is equivalent to its diagonal being an isomorphism. Universal injectivity is weaker; it means that the map of points remains injective after every base change. It is equivalent to injectivity on field-valued points, or to surjectivity of the diagonal [Stacks, Tag 01S4].

Proposition 6.1. An unramified universally injective morphism is a monomorphism. A monomorphism locally of finite type is unramified. Consequently these conditions are equivalent:

• unramified and universally injective;
• unramified and a monomorphism;
• locally of finite type and a monomorphism;
• locally of finite type, with every fibre either empty or isomorphic to the spectrum of the base residue field.

Proof. In the first condition the diagonal is a surjective open immersion, hence an isomorphism. Conversely a monomorphism has isomorphic diagonal, so Theorem 4.1 makes it unramified whenever it is locally of finite type. It is universally injective by its defining functorial injection and base change.

For its fibres, Theorem 3.2 gives separable points and universal injectivity permits at most one point. A finite separable extension of degree greater than one has at least two embeddings into an algebraic closure, contradicting injectivity on field-valued points. A nonempty fibre is therefore precisely the base residue field. Finally, if all fibres have the last form, Theorem 3.3 gives unramifiedness at every point. Field-valued points over any fixed base point are sections of the corresponding field-base-changed fibre, which is empty or one rational point. Thus the map is universally injective. ∎

Together with Section 7 this also yields the equivalent condition “locally of finite type, universally injective and formally unramified” [Stacks, Tag 05VH].

Theorem 6.2. A proper unramified universally injective morphism is a closed immersion. More generally a universally closed, unramified, universally injective morphism is a closed immersion [Stacks, Tag 04XV]. Conversely every closed immersion has these properties.

Proof. Proposition 6.1 makes the morphism a monomorphism, hence separated: its diagonal is an isomorphism and in particular a closed immersion. It is locally of finite type and has finite fibres. In the universally closed case the finiteness criterion [Stacks, Tag 02LS] therefore makes it finite. We prove the finite-map step for a proper morphism using the precise Zariski's Main Theorem prerequisite.

Work over an affine base. Properness and local quasi-finiteness give a quasi-finite separated morphism. Zariski's Main Theorem factors it as \(X\xrightarrow jT\to S\), with \(j\) a quasi-compact open immersion and \(T\to S\) finite. The map \(j\) is proper: its graph into \(X\times_S T\) is closed since \(T\to S\) is separated, and projection from that product to \(T\) is a base change of the proper map \(X\to S\). Thus its open image is closed. It is an open and closed subscheme of the finite \(S\)-scheme \(T\), hence finite over \(S\). Finiteness glues over base affines.

Now write a finite affine chart as \(A\to B\). Its fibres are empty or the corresponding residue fields by Proposition 6.1. The cokernel \(N\) of the module map \(A\to B\) is finite. At every prime \(\mathfrak p\subset A\), right exactness gives \(N\otimes_A\kappa(\mathfrak p)=0\), since \(\kappa(\mathfrak p)\to B\otimes_A\kappa(\mathfrak p)\) is surjective, including for an empty fibre. Nakayama gives \(N_{\mathfrak p}=0\), so \(N=0\). The ring map is surjective and the finite morphism is a closed immersion. Conversely a closed immersion is proper, a monomorphism, unramified and universally injective. ∎

The universal-closedness strengthening uses the general finiteness criterion exactly as stated in the prerequisite list. Neither theorem says that every unramified map is an immersion: distinct separable branches may map to one point.

7. Uniqueness of infinitesimal lifts

A first order thickening \(T\hookrightarrow T'\) is a closed immersion defined by an ideal \(\mathcal J\) with \(\mathcal J^2=0\). It does not change the underlying topological space. A morphism \(X\to S\) is formally unramified if, for every such thickening of affine \(S\)-schemes, restriction

is injective [Stacks, Tag 02H8]. No existence of a lift is required.

Theorem 7.1. For every morphism of schemes, formal unramifiedness is equivalent to \(\Omega_{X/S}=0\). Thus unramified means formally unramified and locally of finite type, and G-unramified means formally unramified and locally of finite presentation [Stacks, Tags 02H9 and 02HE].

Proof. Suppose \(\Omega_{X/S}=0\), and let two lifts be given. Their maps on underlying points agree because they agree on \(T\), whose space equals that of \(T'\). Cover \(T'\) by affine opens on which both maps land in one affine chart \(\operatorname{Spec}B\) over \(\operatorname{Spec}A\). On such an open put \(T'=\operatorname{Spec}C\), with square-zero ideal \(J\). The two \(A\)-algebra maps \(\phi_1,\phi_2:B\to C\) have difference \(D:B\to J\). It is an \(A\)-derivation for the \(B\)-module structure induced by their common reduction: the extra product of two differences lies in \(J^2=0\). Therefore

They agree on each open, so agree globally.

Conversely take compatible affine charts and set \(N=\Omega_{B/A}\). The square-zero algebra \(C=B\oplus N\), with multiplication \((b,n)(c,m)=(bc,bm+cn)\), admits two \(A\)-algebra maps

They agree after quotienting by \(0\oplus N\). Formal unramifiedness makes the corresponding two maps into \(X\) equal. Since the affine chart is an open immersion into \(X\), the algebra maps are equal too. Hence \(db=0\) for every \(b\), and these differentials generate \(N\), so \(N=0\). This holds on all charts. The remaining assertions follow from the definitions. ∎

This argument also explains the absence of a finiteness hypothesis in the formal criterion [Stacks, Tag 024R for its Noetherian finite type specialization]. For example an infinite algebraic separable field extension has zero relative differentials, by the colimit rule for differentials, but its spectrum is not of finite type over the base field.

A closed immersion shows why uniqueness does not imply existence. For \(\operatorname{Spec}k\hookrightarrow\operatorname{Spec}k[t]\) at \(t=0\), take \(T'=\operatorname{Spec}k[\epsilon]/(\epsilon^2)\) with base map \(t\mapsto\epsilon\), and \(T=\operatorname{Spec}k\). The map \(T\to\operatorname{Spec}k\) has no lift over that base: such a lift would force \(\epsilon=0\). The morphism is nonetheless formally unramified.

8. Computations and exercises

For \(L=K[a]\) with minimal polynomial \(h\),

For a finite separable extension this is zero. For \(K=\mathbb F_p(t)\) and \(L=K[a]\) with \(a^p=t\), the derivative of \(X^p-t\) is zero, and \(\Omega_{L/K}=L\,da\ne0\). This shows why the separability hypothesis in the pointwise criterion is necessary.

For the power map with \(t=x^n\), \(n\ge1\),

Its unramified locus is \(D(n x^{n-1})\). If the characteristic divides \(n\), it is empty; in particular the absolute Frobenius of \(\mathbb A^1_{\mathbb F_p}\) is nowhere unramified. If \(n=1\) it is the whole line; if \(n>1\) and \(n\) is invertible in \(k\), it is \(D(x)\).

For the Gaussian integers,

The support is the prime \((1+i)\), since \(2\) is a unit times \((1+i)^2\). This recovers the unramified locus away from that prime, in the number-ring setting of [Stacks, Tag 024Z].

Finally assume \(\operatorname{char}k\ne2\). The normalization of the nodal cubic \(y^2=x^2(x+1)\) is

It is finite because \(u^2=x+1\), and birational because \(u=y/x\) in the fraction field. The ring \(k[u]\) is normal. Write \(A=k[x,y]/(y^2-x^2(x+1))\). To see that the displayed equation gives exactly the plane coordinate ring embedded in \(k[u]\), write polynomials modulo the equation as \(F(x)+yG(x)\). Their substitutions have even and odd powers of \(u\), respectively, and cannot cancel; each substitution is injective in its polynomial variable. Thus this is indeed the normalization. Its relative differentials are

since \((3u/2)(2u)-(3u^2-1)=1\). It is unramified everywhere, yet it is not an immersion: \(u=1\) and \(u=-1\) are different points mapping to the node. It is not flat at the node: the generic rank is one while the node fibre is \(k[u]/(u^2-1)\), of dimension two. These computations keep unramifiedness, injectivity and flatness distinct.

1. Easy — a field derivative. Let \(h\in K[X]\) be irreducible and \(L=K[X]/(h)\). Compute \(\Omega_{L/K}\) and prove that it vanishes exactly when \(h\) is separable.

2. Easy — a power map. Determine the unramified locus of \(x\mapsto x^n\), \(n\ge1\), over a field of characteristic \(p\ge0\). Include \(n=1\) and \(p\mid n\).

3. Medium — the prime over two. Compute \(\Omega_{\mathbb Z[i]/\mathbb Z}\), and determine its support and the unramified locus.

4. Medium — a finite type monomorphism. Prove that every monomorphism locally of finite type is unramified. Explain why the finite type condition is needed in using the diagonal criterion.

5. Hard — rigidity with residue maps. Prove Theorem 5.2 by its equality subscheme. Give an example showing that agreement of underlying points without agreement of residue maps does not suffice.

Solutions

1. The conormal sequence for \((h)\subset K[X]\) sends \(h\) to \(h'(a)\,da\). Hence \(\Omega_{L/K}=L\,da/(h'(a)\,da)\). In the field \(L\), a nonzero \(h'(a)\) is a unit and kills the module; if it is zero the module is free of rank one. Since \(h\) is irreducible, \(h'(a)=0\) is equivalent to \(h\mid h'\), which is possible only when \(h'=0\) because its degree is smaller. An irreducible polynomial is separable exactly when its derivative is nonzero, proving the assertion. This includes the degree-one case.

2. The presentation \(k[t,x]/(t-x^n)\) over \(k[t]\) gives the single relation \(n x^{n-1}dx=0\). At a prime this differential module vanishes exactly when \(n x^{n-1}\) is a unit there. Thus the locus is \(D(n x^{n-1})\). It is the whole line for \(n=1\), empty if \(p>0\) divides \(n\), and \(D(x)\) in the remaining cases with \(n>1\).

3. Present \(\mathbb Z[i]\) as \(\mathbb Z[T]/(T^2+1)\). The conormal map imposes \(2i\,di=0\). Since \(i\) is a unit, the quotient is \(\mathbb Z[i]/(2)\,di\). Now \((1+i)^2=2i\), and \(\mathbb Z[i]/(1+i)\cong\mathbb F_2\). Every prime containing \(2\) therefore contains \(1+i\), and that ideal is maximal, so it is the unique point of the support. The differential module is nonzero there and zero elsewhere. The map is unramified exactly on its complement.

4. The diagonal of a monomorphism is an isomorphism: its fibre-product universal property says that the two projection maps agree and every pair over the base factors uniquely through the diagonal. An isomorphism is an open immersion. With local finite type, Theorem 4.1 applies and gives unramifiedness. More directly its conormal sheaf, hence \(\Omega\), is zero; local finite type supplies the other defining condition. The diagonal assertion alone does not assert finite type. For a concrete counterexample to omitting that condition, \(\operatorname{Spec}\mathbb Q\to\operatorname{Spec}\mathbb Z\) is a monomorphism because \(\mathbb Q\otimes_{\mathbb Z}\mathbb Q=\mathbb Q\). Its algebra is not finite type: finitely many rational generators have denominators using finitely many primes, and cannot generate the inverse of a different prime. The one-point source has no smaller nonempty open, so the morphism is not locally of finite type and is not unramified in our convention.

5. Pull back \(\Delta_{X/S}\) along \((a,b):Y\to X\times_S X\). The result \(E\hookrightarrow Y\) is simultaneously open and closed. The common point \(x\) and equal maps \(\kappa(x)\to\kappa(y)\) make the two morphisms \(\operatorname{Spec}\kappa(y)\to X\) equal. Hence that morphism factors through \(E\), so \(E\) is nonempty. Connectedness implies its open image is all of \(Y\); an open immersion onto all of \(Y\) is an isomorphism. The defining pullback then gives \(a=b\) as scheme maps, including their structure-sheaf maps. For the counterexample take \(S=\operatorname{Spec}\mathbb R\), \(X=Y=\operatorname{Spec}\mathbb C\), and the identity and complex conjugation. The maps have the same underlying point but different residue-field homomorphisms.

What this lesson does not prove

All assigned geometric unramifiedness, diagonal, rigidity and formal-uniqueness results are proved above. The following are exact prerequisites:

• Algebraic differentials represent derivations, commute with localization and base change, and have the transitivity and quotient exact sequences; these are taught in Kähler differentials. For a finitely generated field extension \(L/k\), \(\Omega_{L/k}=0\) exactly when it is finite separable [Stacks, Tag 090W]. The simple finite-extension calculation is also proved in Solution 1.
• Zariski's lemma, the primitive element theorem for finite separable extensions, Nakayama's lemma, and decomposition of a zero-dimensional Noetherian ring as a finite product of Artinian local rings. A finite flat module over a local ring is free [Stacks, Tag 00NZ].
• For a locally finite type morphism, being quasi-finite at a point is equivalent to that point being isolated in its fibre. A universally injective morphism is equivalent to injectivity on field-valued points and to surjectivity of its diagonal [Stacks, Tag 01S4].
• Zariski's Main Theorem: a quasi-finite separated morphism over a quasi-compact quasi-separated base factors as a quasi-compact open immersion followed by a finite morphism [Stacks, Tag 05K0].
• The general finiteness criterion: a universally closed, separated, locally finite type morphism with finite fibres is finite [Stacks, Tag 02LS]. The proper case is deduced from Zariski's Main Theorem in Section 6; the universal-closedness extension is a prerequisite, not an extra owned proof here.

The local factorization of an unramified morphism as a closed immersion followed by an étale morphism is developed in Étale morphisms and their local structure. No such factorization was assumed to prove this lesson's diagonal theorem.

The proper quasi-finite factorization is Zariski’s Main Theorem, Theorem 4.2, in the morphisms course. The factorization argument is written; its nonaffine relative-integral-completion support, Theorem 4.1 of that same lesson, remains a planned prerequisite proof. The stronger locally-finite-type, separated, universally closed, finite-fibre criterion is a planned support statement in Affine morphisms, relative Spec, integral and finite morphisms, the same course’s lesson 5. Its written integral and finite criteria are in Sections 4–5; the complete proof of this stronger criterion remains a prerequisite bridge there. This is the exact condition used for the universal-closedness extension above.

References

• The Stacks Project, Morphisms of Schemes, More on Morphisms and Étale Morphisms: the individual tags locate the differential, unramifiedness, rigidity and infinitesimal statements. The linked text is AI Integrated Stacks Project, an edition with AI-proposed corrections and AI-written additions that have not been reviewed by the Stacks Project maintainers.
• R. Vakil, The Rising Sea: Foundations of Algebraic Geometry, public draft of 27 July 2024, §§21.2 and 21.7: differential sheaves and unramified morphisms.

The next lesson adds flatness and finite presentation to this rigidity and proves the local structure of étale morphisms.