Here we strengthen the results of the text to include information about the sheaf of differentials at a not necessarily closed point of a scheme \(X\).
Generalize (8.7) as follows. Let \(B\) be a local ring containing a field \(k\), and assume that the residue field \(k(B) = B/\mfm\) of \(B\) is a separably generated extension of \(k\). Then the exact sequence of (8.4A), \[ 0 \to \mfm/\mfm^2 \mapsvia{\delta} \Omega_{B/k} \tensor k(B) \to \Omega_{k(B)/k} \to 0 \] is exact on the left also.
Generalize (8.8) as follows. With \(B, k\) as above, assume furthermore that \(k\) is perfect and that \(B\) is a localization of an algebra of finite type over \(k\). Show that \(B\) is a regular local ring if and only if \(\Omega_{B/k}\) is free of rank \(\dim B + \trdeg k(B)/k\).
Strengthen (8.15) as follows. Let \(X\) be an irreducible scheme of finite type over a perfect field \(k\), and let \(\dim X = n\). For any point \(x \in X\), not necessarily closed, show that the local ring \(\OO_{x, X}\) is a regular local ring if and only if the stalk \((\Omega_{X/k})_x\) is free of rank \(n\).
Strengthen (8.16) as follows. If \(X\) is a variety over an algebraically closed field \(k\), then \(U = \ts{x \in X \st \OO_x \text{ is a regular local ring}}\) is an open dense subset of \(X\).
Hint
For (a), in copying the proof of Proposition II.8.7, first pass to \(B/\mfm^2\), which is a complete local ring, and then use Theorem II.8.25A to choose a field of representatives containing \(k\).
Solution
For the local statements put \(\mathfrak m=\mfm\) and \(K=B/\mathfrak m\). Tensor products with \(K\) are over \(B\) unless indicated otherwise. Use the universal derivation \(d:B\to\Omega_{B/k}\).
1. The quotient map \(C=B/\mathfrak m^2\to K\) has a \(k\)-algebra section \(j:K\to C\).
Put \(N=\mathfrak m/\mathfrak m^2\), the maximal ideal of \(C\); then \(N^2=0\). Choose a separating transcendence basis \((t_i)\) for \(K/k\), so that \(K\) is separable algebraic over \(F=k(t_i)\). Lift the \(t_i\) to elements of \(C\). Every nonzero polynomial in the lifts has nonzero residue, by algebraic independence, and is therefore a unit of the local ring \(C\). The lifts consequently define a copy of the field \(F\) in \(C\), with the required residue map.
Consider embeddings into \(C\) of intermediate fields \(F\subseteq E\subseteq K\) that extend this copy of \(F\) and are sections of the residue map on \(E\). An increasing chain has its union as an upper bound, so Zorn’s lemma gives a maximal such embedded field \(E\). If \(\alpha\in K\setminus E\), its minimal polynomial \(p(T)\in E[T]\) is separable. Choose a lift \(a\in C\) of \(\alpha\) and view the coefficients of \(p\) in \(C\). Then \(p(a)\in N\) and \(p'(a)\) is a unit because its residue is \(p'(\alpha)\ne0\). The element \[ a'=a-\frac{p(a)}{p'(a)} \] satisfies \(p(a')=0\): expanding the polynomial, the linear correction cancels \(p(a)\) and all higher correction terms lie in \(N^2=0\). Thus \(E[T]/(p)\) embeds in \(C\), since a unital map from this field is injective, and its residue map identifies it with \(E(\alpha)\subseteq K\). This contradicts maximality. Hence \(E=K\), giving \(j\). This proves the coefficient-field assertion needed from the hint without a noetherian assumption on \(B\).
2. The map \(\delta:N\to\Omega_{B/k}\otimes_B K\) is injective, proving (a).
1.Define\[ D:C\longrightarrow N,\qquad c\longmapsto c-j(\rho(c)). \]It is additive and annihilates \(k\).Writing \(c=j(\bar c)+a\) and \(c'=j(\bar c')+a'\) with \(a,a'\in N\) gives\[ D(cc')=\bar c\,D(c')+\bar c'\,D(c), \]since \(aa'=0\).Thus \(D\) is a \(k\)-derivation into the \(C\)-module \(N\), on which \(C\) acts through \(K\).Its composite with \(B\to C\) is also a \(k\)-derivation and induces a \(K\)-linear map\[ r:\Omega_{B/k}\otimes_B K\longrightarrow N. \]For \(b\in\mathfrak m\), one has \(r(db\otimes1)=b\bmod\mathfrak m^2\).Therefore \(r\circ\delta=\id_N\), proving injectivity.The middle and right exactness are the quotient differential sequence [@Har10a, Proposition II.8.4A], whose first map is \(b\bmod\mathfrak m^2\mapsto db\otimes1\).This proves the entire short exact sequence; the chosen splitting is not asserted to be canonical.
3. For the ring in (b), put \(t=\operatorname{trdeg}_kK\) and \(e=\dim_K(\mathfrak m/\mathfrak m^2)\).Then\[ \dim_K(\Omega_{B/k}\otimes_B K)=e+t. \]
The ring \(B\) is noetherian because it is a localization of a finite-type \(k\)-algebra. Its differential module is generated by the differentials of finitely many algebra generators before localization: differentiation of a polynomial uses these generators, and \(d(s^{-1})=-s^{-2}ds\) adds none after localization. Thus \(\Omega_{B/k}\) is a finite \(B\)-module. The residue field is finitely generated over \(k\). Since \(k\) is perfect, every finitely generated extension of \(k\) is separably generated [@Har10a, Theorem I.4.8A]. The field-differential formula gives \(\dim_K\Omega_{K/k}=t\) [@Har10a, Theorem II.8.6A]. Taking dimensions in the short exact sequence of step 2 proves the equality.
4. The local ring \(B\) is regular if and only if \(\Omega_{B/k}\) is free of rank \(\dim B+t\).
If \(\Omega_{B/k}\) is free of that rank, step 3 gives \(e+t=\dim B+t\), so \(e=\dim B\). This is the definition of regularity for the noetherian local ring \(B\).
Conversely, suppose \(B\) is regular and put \(r=\dim B+t\). Step 3 says its differential module has an \(r\)-dimensional residue space. Lift a basis and apply Nakayama’s lemma to obtain a surjection \[ B^{\oplus r}\twoheadrightarrow\Omega_{B/k}. \] A regular local ring is a domain [@Har10a, Remark II.6.11.1A]; write \(L=\operatorname{Frac}B\). There is a finite-type \(k\)-domain \(A\) and a prime \(\mathfrak p\) with \(B=A_{\mathfrak p}\): start with a finite-type presentation for \(B\) and quotient by the kernel of its map to this domain. The dimension formula for finite-type algebras, together with \(\dim A=\operatorname{trdeg}_k L\), gives \[ \operatorname{trdeg}_k L=\dim B+\operatorname{trdeg}_kK=r. \] Again \(L/k\) is separably generated, so \(\dim_L\Omega_{L/k}=r\). Localizing differentials identifies \(\Omega_{B/k}\otimes_B L\) with \(\Omega_{L/k}\). Our surjection consequently becomes an isomorphism after tensoring with \(L\). Its kernel is therefore a torsion submodule of the free module \(B^{\oplus r}\) and must be zero, since \(B\) is a domain. Thus the original surjection is an isomorphism, proving (b).
5. For the irreducible finite-type scheme in (c), the rank in step 4 is \(n\) at every point.
Choose an affine neighborhood \(\Spec A\) of \(x\) and let \(\mathfrak p\) correspond to \(x\). The reduced ring \(A_{\mathrm{red}}\) is a domain because \(X\) is irreducible. It has dimension \(n\): its fraction field is the function field of \(X_{\mathrm{red}}\), and the dimension of every nonempty affine open in this integral finite-type scheme is its transcendence degree over \(k\). Reduction does not change dimensions or residue fields. The same dimension formula used in step 4 therefore gives \[ \dim\OO_{X,x}+\operatorname{trdeg}_k\kappa(x)=n. \] Localization in the construction of the differential sheaf gives \[ (\Omega_{X/k})_x\cong\Omega_{\OO_{X,x}/k}. \] Applying step 4 proves (c), without assuming that \(x\) is closed or that \(X\) is reduced.
6. The locus in (d) is open and dense.
The sheaf \(\Omega_{X/k}\) is coherent, since \(X\) is of finite type over the field \(k\). The locus where its stalk is free of rank \(n=\dim X\) is open. Indeed, a basis at a stalk extends to a morphism \(\OO^{\oplus n}\to\Omega_{X/k}\) on a neighborhood; its coherent kernel and cokernel vanish at that point, and hence vanish after shrinking the neighborhood. Step 5 identifies this open locus with the regular locus.
The variety \(X\) is integral, so the local ring at its generic point is a field and is regular. Thus the regular locus contains the generic point and is dense. This proves (d).
7. Q.E.D.
Steps <1>1–<1>2 prove (a), steps <1>3–<1>4 prove (b), step 5 proves (c), and step 6 proves (d).