SMOOTHING RING MAPS 07BW Contents 1. Introduction 1 2

SMOOTHING RING MAPS

07BW

Contents 1. Introduction 1 2. Singular ideals 2 3. Presentations of algebras 4 4. Intermezzo: Néron desingularization 10 5. The lifting problem 13 6. The lifting lemma 16 7. The desingularization lemma 18 8. Warmup: reduction to a base field 21 9. Local tricks 22 10. Separable residue fields 24 11. Inseparable residue fields 26 12. The main theorem 32 13. The approximation property for G-rings 32 14. Approximation for henselian pairs 35 15. Other chapters 35 References 37

1. Introduction 07BX The main result of this chapter is the following: A regular map of Noetherian rings is a filtered colimit of smooth ones. This theorem is due to Popescu, see [Pop90]. A readable exposition of Popescu’s proof was given by Richard Swan, see [Swa98] who used notes by André and a paper of Ogoma, see [Ogo94]. Our exposition follows Swan’s, but we first prove an intermediate result which lets us work in a slightly simpler situation. Here is an overview. We first solve the following “lifting problem”: A flat infinitesimal deformation of a filtered colimit of smooth algebras is a filtered colimit of smooth algebras. This result essentially says that it suffices to prove the main theorem for maps between reduced Noetherian rings. Next we prove two very clever lemmas called the “lifting lemma” and the “desingularization lemma”. We show that these lemmas combined reduce the main theorem to proving a Noetherian, geometrically regular algebra Λ over a field k is a filtered colimit of smooth k-algebras. Next, we discuss the necessary local tricks that go into the Popescu-Ogoma-Swan-André proof. Finally, in the last three sections we give the proof.

This is a chapter of the Stacks Project, version fac02ecd, compiled on Sep 14, 2021. 1 SMOOTHING RING MAPS 2

We end this introduction with some pointers to references. Let A be a henselian Noetherian local ring. We say A has the approximation property if for any f1, . . . , fm ∈ A[x1, . . . , xn] the system of equations f1 = 0, . . . , fm = 0 has a solution in the completion of A if and only if it has a solution in A. This definition is due to Artin. Artin first proved the approximation property for analytic systems of equations, see [Art68]. In [Art69] Artin proved the approximation property for local rings essentially of finite type over an excellent discrete valuation ring. Artin conjec- tured (page 26 of [Art69]) that every excellent henselian local ring should have the approximation property. At some point in time it became a conjecture that every regular homomorphism of Noetherian rings is a filtered colimit of smooth algebras (see for example [Ray72], [Pop81], [Art82], [AD83]). We’re not sure who this conjecture1 is due to. The relationship with the approximation property is that if A → A∧ is a colimit of smooth algebras with A as above, then the approximation property holds (insert future reference here). Moreover, the main theorem applies to the map A → A∧ if A is an excellent local ring, as one of the conditions of an excellent local ring is that the formal fibres are geometrically regular. Note that excellent local rings were defined by Grothendieck and their definition appeared in print in 1965. In [Art82] it was shown that R → R∧ is a filtered colimit of smooth algebras for any local ring R essentially of finite type over a field. In [AR88] it was shown that R → R∧ is a filtered colimit of smooth algebras for any local ring R essentially of finite type over an excellent discrete valuation ring. Finally, the main theorem was shown in [Pop85], [Pop86], [Pop90], [Ogo94], and [Swa98] as discussed above. Conversely, using some of the results above, in [Rot90] it was shown that any Noetherian local ring with the approximation property is excellent. The paper [Spi99] provides an alternative approach to the main theorem, but it seems hard to read (for example [Spi99, Lemma 5.2] appears to be an incorrectly reformulated version of [Elk73, Lemma 3]). There is also a Bourbaki lecture about this material, see [Tei95].

2. Singular ideals 07C4 Let R → A be a ring map. The singular ideal of A over R is the radical ideal in A cutting out the singular locus of the morphism Spec(A) → Spec(R). Here is a formal deﬁnition. 07C5Deﬁnition 2.1. Let R → A be a ring map. The singular ideal of A over R, denoted HA/R is the unique radical ideal HA/R ⊂ A with

V (HA/R) = {q ∈ Spec(A) | R → A not smooth at q} This makes sense because the set of primes where R → A is smooth is open, see Algebra, Definition 137.11. In order to find an explicit set of generators for the singular ideal we first prove the following lemma.

07C6 Lemma 2.2. Let R be a ring. Let A = R[x1, . . . , xn]/(f1, . . . , fm). Let q ⊂ A be a prime ideal. Assume R → A is smooth at q. Then there exists an a ∈ A, a 6∈ q, an

1The question/conjecture as formulated in [Art82], [AD83], and [Pop81] is stronger and was shown to be equivalent to the original version in [CP84]. SMOOTHING RING MAPS 3

integer c, 0 ≤ c ≤ min(n, m), subsets U ⊂ {1, . . . , n}, V ⊂ {1, . . . , m} of cardinality c such that 0 a = a det(∂fj/∂xi)j∈V,i∈U for some a0 ∈ A and 2 af` ∈ (fj, j ∈ V ) + (f1, . . . , fm) for all ` ∈ {1, . . . , m}.

Proof. Set I = (f1, . . . , fm) so that the naive cotangent complex of A over R 2 L is homotopy equivalent to I/I → Adxi, see Algebra, Lemma 134.2. We will use the formation of the naive cotangent complex commutes with localization, see Algebra, Section 134, especially Algebra, Lemma 134.13. By Algebra, Definitions 2 L 137.1 and 137.11 we see that (I/I )a → Aadxi is a split injection for some a ∈ A, a 6∈ q. After renumbering x1, . . . , xn and f1, . . . , fm we may assume that f1, . . . , fc 2 form a basis for the vector space I/I ⊗A κ(q) and that dxc+1,..., dxn map to a 0 0 0 basis of ΩA/R ⊗A κ(q). Hence after replacing a by aa for some a ∈ A, a 6∈ q 2 we may assume f1, . . . , fc form a basis for (I/I )a and that dxc+1,..., dxn map N to a basis of (ΩA/R)a. In this situation a for some large integer N satisfies the conditions of the lemma (with U = V = {1, . . . , c}). We will use the notion of a strictly standard element in A over R. Our notion is slightly weaker than the one in Swan’s paper [Swa98]. We also define an elementary standard element to be one of the type we found in the lemma above. We compare the different types of elements in Lemma 3.7. 07C7Definition 2.3. Let R → A be a ring map of finite presentation. We say an element a ∈ A is elementary standard in A over R if there exists a presentation A = R[x1, . . . , xn]/(f1, . . . , fm) and 0 ≤ c ≤ min(n, m) such that 0 (2.3.1)07C8 a = a det(∂fj/∂xi)i,j=1,...,c for some a0 ∈ A and 2 (2.3.2)07C9 afc+j ∈ (f1, . . . , fc) + (f1, . . . , fm) for j = 1, . . . , m − c. We say a ∈ A is strictly standard in A over R if there exists a presentation A = R[x1, . . . , xn]/(f1, . . . , fm) and 0 ≤ c ≤ min(n, m) such that X (2.3.3)07ER a = aI det(∂fj/∂xi)j=1,...,c, i∈I I⊂{1,...,n}, |I|=c

for some aI ∈ A and 2 (2.3.4)07ES afc+j ∈ (f1, . . . , fc) + (f1, . . . , fm) for j = 1, . . . , m − c. The following lemma is useful to ﬁnd implications of (2.3.3).

07ET Lemma 2.4. Let R be a ring. Let A = R[x1, . . . , xn]/(f1, . . . , fm) and write I = (f1, . . . , fm). Let a ∈ A. Then (2.3.3) implies there exists an A-linear map L ⊕c ψ : i=1,...,n Adxi → A such that the composition

⊕c (f1,...,fc) 2 f7→df M ψ ⊕c A −−−−−−→ I/I −−−−→ Adxi −→ A i=1,...,n is multiplication by a. Conversely, if such a ψ exists, then ac satisﬁes (2.3.3).

Proof. This is a special case of Algebra, Lemma 15.5. SMOOTHING RING MAPS 4

07CA Lemma 2.5 (Elkik). Let R → A be a ring map of ﬁnite presentation. The singular ideal HA/R is the radical of the ideal generated by strictly standard elements in A over R and also the radical of the ideal generated by elementary standard elements in A over R.

Proof. Assume a is strictly standard in A over R. We claim that Aa is smooth over R, which proves that a ∈ HA/R. Namely, let A = R[x1, . . . , xn]/(f1, . . . , fm), 0 c, and a ∈ A be as in Definition 2.3. Write I = (f1, . . . , fm) so that the naive 2 L cotangent complex of A over R is given by I/I → Adxi. Assumption (2.3.4) 2 implies that (I/I )a is generated by the classes of f1, . . . , fc. Assumption (2.3.3) 2 L implies that the differential (I/I )a → Aadxi has a left inverse, see Lemma 2.4. Hence R → Aa is smooth by definition and Algebra, Lemma 134.13.

Let He,Hs ⊂ A be the radical of the ideal generated by elementary, resp. strictly standard elements of A over R. By definition and what we just proved we have He ⊂ Hs ⊂ HA/R. The inclusion HA/R ⊂ He follows from Lemma 2.2. 07CBExample 2.6. The set of points where a finitely presented ring map is smooth needn’t be a quasi-compact open. For example, let R = k[x, y1, y2, y3,...]/(xyi) S and A = R/(x). Then the smooth locus of R → A is D(yi) which is not quasi- compact. 07CC Lemma 2.7. Let R → A be a ring map of finite presentation. Let R → R0 be a ring map. If a ∈ A is elementary, resp. strictly standard in A over R, then a ⊗ 1 0 0 is elementary, resp. strictly standard in A ⊗R R over R .

Proof. If A = R[x1, . . . , xn]/(f1, . . . , fm) is a presentation of A over R, then A ⊗R 0 0 0 0 0 0 0 R = R [x1, . . . , xn]/(f1, . . . , fm) is a presentation of A ⊗R R over R . Here fj is 0 the image of fj in R [x1, . . . , xn]. Hence the result follows from the definitions. 07EU Lemma 2.8. Let R → A → Λ be ring maps with A of finite presentation over R. Assume that HA/RΛ = Λ. Then there exists a factorization A → B → Λ with B smooth over R. P Proof. Choose f1, . . . , fr ∈ HA/R and λ1, . . . , λr ∈ Λ such that fiλi = 1 in Λ. Set B = A[x1, . . . , xr]/(f1x1 + ... + frxr − 1) and define B → Λ by mapping xi to

λi. To check that B is smooth over R use that Afi is smooth over R by deﬁnition

of HA/R and that Bfi is smooth over Afi . Details omitted. 3. Presentations of algebras 07CD Some of the results in this section are due to Elkik. Note that the algebra C in the following lemma is a symmetric algebra over A. Moreover, if R is Noetherian, then C is of finite presentation over R. 07CE Lemma 3.1. Let R be a ring and let A be a finitely presented R-algebra. There exists finite type R-algebra map A → C which has a retraction with the following two properties

(1) for each a ∈ A such that R → Aa is a local complete intersection (More on Algebra, Deﬁnition 33.2) the ring Ca is smooth over Aa and has a presen- 2 tation Ca = R[y1, . . . , ym]/J such that J/J is free over Ca, and

(2) for each a ∈ A such that Aa is smooth over R the module ΩCa/R is free over Ca. SMOOTHING RING MAPS 5

Proof. Choose a presentation A = R[x1, . . . , xn]/I and write I = (f1, . . . , fm). Deﬁne the A-module K by the short exact sequence 0 → K → A⊕m → I/I2 → 0

where the jth basis vector ej in the middle is mapped to the class of fj on the right. Set ∗ 2 C = SymA(I/I ). The retraction is just the projection onto the degree 0 part of C. We have a 2 surjection R[x1, . . . , xn, y1, . . . , ym] → C which maps yj to the class of fj in I/I . The kernel J of this map is generated by the elements f1, . . . , fm and by elements P P hjyj with hj ∈ R[x1, . . . , xn] such that hjej deﬁnes an element of K. By Algebra, Lemma 134.4 applied to R → A → C and the presentations above and More on Algebra, Lemma 9.12 there is a short exact sequence 2 2 (3.1.1)07EW I/I ⊗A C → J/J → K ⊗A C → 0

of C-modules. Let h ∈ R[x1, . . . , xn] be an element with image a ∈ A. We will use as presentations for the localized rings 0 0 Aa = R[x0, x1, . . . , xn]/I and Ca = R[x0, x1, . . . , xn, y1, . . . , ym]/J 0 0 0 0 2 2 where I = (hx0 − 1,I) and J = (hx0 − 1,J). Hence I /(I ) = Aa ⊕ (I/I )a as 0 0 2 2 Aa-modules and J /(J ) = Ca ⊕ (J/J )a as Ca-modules. Thus we obtain 2 2 (3.1.2)07EX Ca ⊕ I/I ⊗A Ca → Ca ⊕ (J/J )a → K ⊗A Ca → 0

as the sequence of Algebra, Lemma 134.4 corresponding to R → Aa → Ca and the presentations above.

Next, assume that a ∈ A is such that Aa is a local complete intersection over R. 2 Then (I/I )a is ﬁnite projective over Aa, see More on Algebra, Lemma 32.3. Hence 2 ∼ ⊕m we see Ka ⊕ (I/I )a = Aa is free. In particular Ka is ﬁnite projective too. By More on Algebra, Lemma 33.6 the sequence (3.1.2) is exact on the left. Hence 0 0 2 ∼ 2 ∼ ⊕m+1 J /(J ) = Ca ⊕ I/I ⊗A Ca ⊕ K ⊗A Ca = Ca

This proves (1). Finally, suppose that in addition Aa is smooth over R. Then the

same presentation shows that ΩCa/R is the cokernel of the map 0 0 2 M M J /(J ) −→ Cadxi ⊕ Cadyj i j 0 0 2 The summand Ca of J /(J ) in the decomposition above corresponds to hx0 − 1 2 and hence maps isomorphically to the summand Cadx0. The summand I/I ⊗A Ca 0 0 2 L of J /(J ) maps injectively to i=1,...,n Cadxi with quotient ΩAa/R ⊗Aa Ca. The L summand K ⊗A Ca maps injectively to j≥1 Cadyj with quotient isomorphic to 2 2 I/I ⊗A Ca. Thus the cokernel of the last displayed map is the module I/I ⊗A 2 Ca ⊕ ΩAa/R ⊗Aa Ca. Since (I/I )a ⊕ ΩAa/R is free (from the deﬁnition of smooth ring maps) we see that (2) holds. The following proposition was proved for smooth ring maps over henselian pairs by Elkik in [Elk73]. For smooth ring maps it can be found in [Ara01], where it is also proven that ring maps between smooth algebras can be lifted.

07M8 Proposition 3.2. Let R → R0 be a surjective ring map with kernel I.

(1) If R0 → A0 is a syntomic ring map, then there exists a syntomic ring map ∼ R → A such that A/IA = A0. SMOOTHING RING MAPS 6

(2) If R0 → A0 is a smooth ring map, then there exists a smooth ring map ∼ R → A such that A/IA = A0.

Proof. Assume R0 → A0 syntomic, in particular a local complete intersection (More on Algebra, Lemma 33.5). Choose a presentation A0 = R0[x1, . . . , xn]/J0. Set C = Sym∗ (J /J 2). Note that J /J 2 is a ﬁnite projective A -module (Al- 0 A0 0 0 0 0 0 gebra, Lemma 136.16). By Lemma 3.1 the ring map A0 → C0 is smooth and 2 we can ﬁnd a presentation C0 = R0[y1, . . . , ym]/K0 with K0/K0 free over C0. By Algebra, Lemma 136.6 we can assume C0 = R0[y1, . . . , ym]/(f 1,..., f c) where 2 f 1,..., f c maps to a basis of K0/K0 over C0. Choose f1, . . . , fc ∈ R[y1, . . . , yc] lifting f 1,..., f c and set

C = R[y1, . . . , ym]/(f1, . . . , fc)

By construction C0 = C/IC. By Algebra, Lemma 136.11 we can after replacing C by Cg assume that C is a relative global complete intersection over R. We conclude that there exists a finite projective A -module P such that C = Sym∗ (P ) is 0 0 0 A0 0 isomorphic to C/IC for some syntomic R-algebra C. ⊕n Choose an integer n and a direct sum decomposition A0 = P0 ⊕ Q0. By More on Algebra, Lemma 9.11 we can find an étale ring map C → C0 which induces an isomorphism C/IC → C0/IC0 and a finite projective C0-module Q such that Q/IQ ∗ 0 is isomorphic to Q0 ⊗A0 C/IC. Then D = SymC0 (Q) is a smooth C -algebra (see More on Algebra, Lemma 9.13). Picture R / C / C0 / D

∼ = 0 0 R/I / A0 / C/IC / C /IC / D/ID Observe that our choice of Q gives ∗ D/ID = SymC/IC (Q0 ⊗A0 C/IC) = Sym∗ (Q ) ⊗ C/IC A0 0 A0 = Sym∗ (Q ) ⊗ Sym∗ (P ) A0 0 A0 A0 0 = Sym∗ (Q ⊕ P ) A0 0 0 = Sym∗ (A⊕n) A0 0

= A0[x1, . . . , xn]

Choose f1, . . . , fn ∈ D which map to x1, . . . , xn in D/ID = A0[x1, . . . , xn]. Set A = D/(f1, . . . , fn). Note that A0 = A/IA. We claim that R → A is syntomic in a neighbourhood of V (IA). If the claim is true, then we can find a f ∈ A mapping to 1 ∈ A0 such that Af is syntomic over R and the proof of (1) is finished. Proof of the claim. Observe that R → D is syntomic as a composition of the syntomic ring map R → C, the étale ring map C → C0 and the smooth ring map C0 → D (Algebra, Lemmas 136.17 and 137.10). The question is local on Spec(D), hence we may assume that D is a relative global complete intersection (Algebra, 0 0 Lemma 136.15). Say D = R[y1, . . . , ym]/(g1, . . . , gs). Let f1, . . . , fn ∈ R[y1, . . . , ym] be lifts of f1, . . . , fn. Then we can apply Algebra, Lemma 136.11 to get the claim. Proof of (2). Since a smooth ring map is syntomic, we can find a syntomic ring map R → A such that A0 = A/IA. By assumption the fibres of R → A are smooth SMOOTHING RING MAPS 7

over primes in V (I) hence R → A is smooth in an open neighbourhood of V (IA) (Algebra, Lemma 137.17). Thus we can replace A by a localization to obtain the result we want. We know that any syntomic ring map R → A is locally a relative global complete intersection, see Algebra, Lemma 136.15. The next lemma says that a vector bundle over Spec(A) is a relative global complete intersection. 07CG Lemma 3.3. Let R → A be a syntomic ring map. Then there exists a smooth R-algebra map A → C with a retraction such that C is a global relative complete intersection over R, i.e., ∼ C = R[x1, . . . , xn]/(f1, . . . , fc) flat over R and all fibres of dimension n − c. Proof. Apply Lemma 3.1 to get A → C. By Algebra, Lemma 136.6 we can write 2 C = R[x1, . . . , xn]/(f1, . . . , fc) with fi mapping to a basis of J/J . The ring map R → C is syntomic (hence flat) as it is a composition of a syntomic and a smooth ring map. The dimension of the fibres is n − c by Algebra, Lemma 135.4 (the fibres are local complete intersections, so the lemma applies). 07CH Lemma 3.4. Let R → A be a smooth ring map. Then there exists a smooth R-algebra map A → B with a retraction such that B is standard smooth over R, i.e., ∼ B = R[x1, . . . , xn]/(f1, . . . , fc)

and det(∂fj/∂xi)i,j=1,...,c is invertible in B. Proof. Apply Lemma 3.3 to get a smooth R-algebra map A → C with a retraction such that C = R[x1, . . . , xn]/(f1, . . . , fc) is a relative global complete intersection over R. As C is smooth over R we have a short exact sequence M M 0 → Cfj → Cdxi → ΩC/R → 0 j=1,...,c i=1,...,n

Since ΩC/R is a projective C-module this sequence is split. Choose a left inverse t P P ∂fj to the ﬁrst map. Say t(dxi) = cijfj so that ci` = δj` (Kronecker delta). i ∂xi Let 0 B = C[y1, . . . , yc] = R[x1, . . . , xn, y1, . . . , yc]/(f1, . . . , fc) 0 The R-algebra map C → B has a retraction given by mapping yj to zero. We claim that the map 0 X R[z1, . . . , zn] −→ B , zi 7−→ xi − cijyj j 0 0 is étale at every point in the image of Spec(C) → Spec(B ). In ΩB /R[z1,...,zn] we have X ∂fj X ∂fj 0 0 = dfj − dzi ≡ ci`dy` ≡ dyj mod (y1, . . . , yc)ΩB /R[z1,...,zn] i ∂xi i,` ∂xi 0 P 0 Since 0 = dzi = dxi modulo B dyj + (y1, . . . , yc)ΩB /R[z1,...,zn] we conclude that

0 0 ΩB /R[z1,...,zn]/(y1, . . . , yc)ΩB /R[z1,...,zn] = 0. 0 0 As ΩB /R[z1,...,zn] is a finite B -module by Nakayama’s lemma there exists a g ∈ 0 0 1 + (y1, . . . , yc) that (ΩB /R[z1,...,zn])g = 0. This proves that R[z1, . . . , zn] → Bg is unramified, see Algebra, Definition 151.1. For any ring map R → k where k is a SMOOTHING RING MAPS 8

0 field we obtain an unramified ring map k[z1, . . . , zn] → (Bg)⊗R k between smooth k- 0 algebras of dimension n. It follows that k[z1, . . . , zn] → (Bg)⊗R k is flat by Algebra, Lemmas 128.1 and 140.2. By the critère de platitude par fibre (Algebra, Lemma 0 128.8) we conclude that R[z1, . . . , zn] → Bg is flat. Finally, Algebra, Lemma 143.7 0 0 implies that R[z1, . . . , zn] → Bg is étale. Set B = Bg. Note that C → B is smooth and has a retraction, so also A → B is smooth and has a retraction. Moreover, R[z1, . . . , zn] → B is étale. By Algebra, Lemma 143.2 we can write

B = R[z1, . . . , zn, w1, . . . , wc]/(g1, . . . , gc)

with det(∂gj/∂wi) invertible in B. This proves the lemma. 07CI Lemma 3.5. Let R → Λ be a ring map. If Λ is a filtered colimit of smooth R-algebras, then Λ is a filtered colimit of standard smooth R-algebras. Proof. Let A → Λ be an R-algebra map with A of finite presentation over R. According to Algebra, Lemma 127.4 we have to factor this map through a standard smooth algebra, and we know we can factor it as A → B → Λ with B smooth over R. Choose an R-algebra map B → C with a retraction C → B such that C is standard smooth over R, see Lemma 3.4. Then the desired factorization is A → B → C → B → Λ. 07EY Lemma 3.6. Let R → A be a standard smooth ring map. Let E ⊂ A be a finite subset of order |E| = n. Then there exists a presentation A = R[x1, . . . , xn+m]/(f1, . . . , fc) with c ≥ n, with det(∂fj/∂xi)i,j=1,...,c invertible in A, and such that E is the set of congruence classes of x1, . . . , xn.

Proof. Choose a presentation A = R[y1, . . . , ym]/(g1, . . . , gd) such that the image of det(∂gj/∂yi)i,j=1,...,d is invertible in A. Choose an enumerations E = {a1, . . . , an} and choose hi ∈ R[y1, . . . , ym] whose image in A is ai. Consider the presentation

A = R[x1, . . . , xn, y1, . . . , ym]/(x1 − h1, . . . , xn − hn, g1, . . . , gd)

and set c = n + d. 07EZ Lemma 3.7. Let R → A be a ring map of ﬁnite presentation. Let a ∈ A. Consider the following conditions on a:

(1) Aa is smooth over R,

(2) Aa is smooth over R and ΩAa/R is stably free,

(3) Aa is smooth over R and ΩAa/R is free, (4) Aa is standard smooth over R, (5) a is strictly standard in A over R, (6) a is elementary standard in A over R. Then we have (a) (4) ⇒ (3) ⇒ (2) ⇒ (1), (b) (6) ⇒ (5), (c) (6) ⇒ (4), (d) (5) ⇒ (2), e (e) (2) ⇒ the elements a , e ≥ e0 are strictly standard in A over R, e (f) (4) ⇒ the elements a , e ≥ e0 are elementary standard in A over R. SMOOTHING RING MAPS 9

Proof. Part (a) is clear from the deﬁnitions and Algebra, Lemma 137.7. Part (b) is clear from Deﬁnition 2.3.

Proof of (c). Choose a presentation A = R[x1, . . . , xn]/(f1, . . . , fm) such that (2.3.1) and (2.3.2) hold. Choose h ∈ R[x1, . . . , xn] mapping to a. Then

Aa = R[x0, x1, . . . , xn]/(x0h − 1, f1, . . . , fm). 2 Write J = (x0h − 1, f1, . . . , fm). By (2.3.2) we see that the Aa-module J/J is generated by x0h − 1, f1, . . . , fc over Aa. Hence, as in the proof of Algebra, Lemma 136.6, we can choose a g ∈ 1 + J such that

Aa = R[x0, . . . , xn, xn+1]/(x0h − 1, f1, . . . , fm, gxn+1 − 1).

At this point (2.3.1) implies that R → Aa is standard smooth (use the coordinates x0, x1, . . . , xc, xn+1 to take derivatives).

Proof of (d). Choose a presentation A = R[x1, . . . , xn]/(f1, . . . , fm) such that (2.3.3) and (2.3.4) hold. Write I = (f1, . . . , fm). We already know that Aa 2 is smooth over R, see Lemma 2.5. By Lemma 2.4 we see that (I/I )a is free L on f1, . . . , fc and maps isomorphically to a direct summand of Aadxi. Since 2 L ΩAa/R = (ΩA/R)a is the cokernel of the map (I/I )a → Aadxi we conclude that it is stably free.

Proof of (e). Choose a presentation A = R[x1, . . . , xn]/I with I finitely generated. By assumption we have a short exact sequence 2 M 0 → (I/I )a → Aadxi → ΩA /R → 0 i=1,...,n a 2 which is split exact. Hence we see that (I/I )a ⊕ ΩAa/R is a free Aa-module. 2 Since ΩAa/R is stably free we see that (I/I )a is stably free as well. Thus replacing the presentation chosen above by A = R[x1, . . . , xn, xn+1, . . . , xn+r]/J with 2 J = (I, xn+1, . . . , xn+r) for some r we get that (J/J )a is (finite) free. Choose 2 f1, . . . , fc ∈ J which map to a basis of (J/J )a. Extend this to a list of generators f1, . . . , fm ∈ J. Consider the presentation A = R[x1, . . . , xn+r]/(f1, . . . , fm). Then (2.3.4) holds for ae for all sufficiently large e by construction. Moreover, 2 L since (J/J )a → i=1,...,n+r Aadxi is a split injection we can find an Aa-linear left inverse. Writing this left inverse in terms of the basis f1, . . . , fc and clearing ⊕n+r ⊕c denominators we find a linear map ψ0 : A → A such that

⊕c (f1,...,fc) 2 f7→df M ψ0 ⊕c A −−−−−−→ J/J −−−−→ Adxi −−→ A i=1,...,n+r

e0 is multiplication by a for some e0 ≥ 1. By Lemma 2.4 we see (2.3.3) holds for all ce0 e a and hence for a for all e with e ≥ ce0.

Proof of (f). Choose a presentation Aa = R[x1, . . . , xn]/(f1, . . . , fc) such that det(∂fj/∂xi)i,j=1,...,c is invertible in Aa. We may assume that for some m < n the classes of the elements x1, . . . , xm correspond ai/1 where a1, . . . , am ∈ A are N generators of A over R, see Lemma 3.6. After replacing xi by a xi for m < i ≤ n we may assume the class of xi is ai/1 ∈ Aa for some ai ∈ A. Consider the ring map

Ψ: R[x1, . . . , xn] −→ A, xi 7−→ ai. N This is a surjective ring map. By replacing fj by a fj we may assume that fj ∈ R[x1, . . . , xn] and that Ψ(fj) = 0 (since after all fj(a1/1, . . . , an/1) = 0 in Aa). Let J = Ker(Ψ). Then A = R[x1, . . . , xn]/J is a presentation and f1, . . . , fc ∈ J SMOOTHING RING MAPS 10

2 are elements such that (J/J )a is freely generated by f1, . . . , fc and such that det(∂fj/∂xi)i,j=1,...,c maps to an invertible element of Aa. It follows that (2.3.1) e and (2.3.2) hold for a and all large enough e as desired. 4. Intermezzo: Néron desingularization 0BJ1 We interrupt the attack on the general case of Popescu’s theorem to an easier but already very interesting case, namely, when R → Λ is a homomorphism of discrete valuation rings. This is discussed in [Art69, Section 4]. 0BJ2Situation 4.1. Here R ⊂ Λ is an extension of discrete valuation rings with ramification index 1 (More on Algebra, Definition 109.1). We assume given a factorization ϕ R → A −→ Λ −1 with R → A flat and of finite type. Let q = Ker(ϕ) and p = ϕ (mΛ). In Situation 4.1 let π ∈ R be a uniformizer. Recall that flatness of A over R signifies that π is a nonzerodivisor on A (More on Algebra, Lemma 22.10). By our assumption on R ⊂ Λ we see that π maps to a uniformizer of Λ. Since π ∈ p we can consider Néron’s affine blowup algebra (see Algebra, Section 70) 0 0 p ϕ : A = A[ π ] −→ Λ which comes endowed with an induced map to Λ sending a/πn, a ∈ pn to π−nϕ(a) in Λ. We will denote q0 ⊂ p0 ⊂ A0 the corresponding prime ideals of A0. Observe that the isomorphism class of A0 does not depend on our choice of uniformizer. Repeating the construction we obtain a sequence A → A0 → A00 → ... → Λ 0BJ3 Lemma 4.2. In Situation 4.1 Néron’s blowup is functorial in the following sense 0 (1) if a ∈ A, a 6∈ p, then Néron’s blowup of Aa is Aa, and (2) if B → A is a surjection of flat finite type R-algebras with kernel I, then A0 is the quotient of B0/IB0 by its π-power torsion. Proof. Both (1) and (2) are special cases of Algebra, Lemma 70.3. In fact, when- 0 ever we have A1 → A2 → Λ such that p1A2 = p2, we have that A2 is the quotient 0 of A1 ⊗A1 A2 by its π-power torsion. 0BJ4 Lemma 4.3. In Situation 4.1 assume that R → A is smooth at p and that R/πR ⊂ Λ/πΛ is a separable field extension. Then R → A0 is smooth at p0 and there is a short exact sequence 0 0 0 ⊕c 0 → ΩA/R ⊗A Ap0 → ΩA0/R,p0 → (A /πA )p0 → 0

where c = dim((A/πA)p). Proof. By Lemma 4.2 we may replace A by a localization at an element not in p; we will use this without further mention. Write κ = R/πR. Since smoothness is stable under base change (Algebra, Lemma 137.4) we see that A/πA is smooth over κ at p. Hence (A/πA)p is a regular local ring (Algebra, Lemma 140.3). Choose g1, . . . , gc ∈ p which map to a regular system of parameters in (A/πA)p. Then we see that p = (π, g1, . . . , gc) after possibly replacing A by a localization. Note that π, g1, . . . , gc is a regular sequence in Ap (ﬁrst π is a nonzerodivisor and then Algebra, Lemma 106.3 for the rest of the sequence). After replacing A by a localization we SMOOTHING RING MAPS 11

may assume that π, g1, . . . , gc is a regular sequence in A (Algebra, Lemma 68.6). It follows that 0 A = A[y1, . . . , yc]/(πy1 − g1, . . . , πyc − gc) = A[y1, . . . , yc]/I by More on Algebra, Lemma 31.2. In the following we will use the deﬁnition of smoothness using the naive cotangent complex (Algebra, Deﬁnition 137.1) and the criterion of Algebra, Lemma 137.12 without further mention. The exact sequence 0 of Algebra, Lemma 134.4 for R → A[y1, . . . , yc] → A looks like this

2 0 M 0 0 → H1(NLA0/R) → I/I → ΩA/R ⊗A A ⊕ A dyi → ΩA0/R → 0 i=1,...,c 2 where the class of πyi −gi in I/I is mapped to −dgi +πdyi in the next term. Here 0 we have used Algebra, Lemma 134.6 to compute NLA /A[y1,...,yc] and we have used

that R → A[y1, . . . , yc] is smooth, so H1(NLA[y1,...,yc]/R) = 0 and ΩA[y1,...,yc]/R is a finite projective (a fortiori flat) A[y1, . . . , yc]-module which is in fact the direct sum of ΩA/R ⊗A A[y1, . . . , yc] and a free module with basis dyi. To finish the proof it suffices to show that dg1,..., dgc forms part of a basis for the finite free 2 module ΩA/R,p. Namely, this will show (I/I )p is free on πyi − gi, the localization at p of the middle map in the sequence is injective, so H1(NLA0/R)p = 0, and that the cokernel ΩA0/R,p is finite free. To do this it suffices to show that the images of dgi are κ(p)-linearly independent in ΩA/R,p/π = Ω(A/πA)/κ,p (equality by Algebra, Lemma 131.12). Since κ ⊂ κ(p) ⊂ Λ/πΛ we see that κ(p) is separable over κ (Algebra, Definition 42.1). The desired linear independence now follows from Algebra, Lemma 140.4. 0BJ5 Lemma 4.4. In Situation 4.1 assume that R → A is smooth at q and that we have a surjection of R-algebras B → A with kernel I. Assume R → B smooth at −1 pB = (B → A) p. If the cokernel of 2 I/I ⊗A Λ → ΩB/R ⊗B Λ is a free Λ-module, then R → A is smooth at p.

2 Proof. The cokernel of the map I/I → ΩB/R ⊗B A is ΩA/R, see Algebra, Lemma 131.9. Let d = dimq(A/R) be the relative dimension of R → A at q, i.e., the dimension of Spec(A[1/π]) at q. See Algebra, Definition 125.1. Then ΩA/R,q is free over Aq of rank d (Algebra, Lemma 140.3). Thus if the hypothesis of the lemma holds, then ΩA/R ⊗A Λ is free of rank d. It follows that ΩA/R ⊗A κ(p) has dimension d (as it is true upon tensoring with Λ/πΛ). Since R → A is flat and since p is a specialization of q, we see that dimp(A/R) ≥ d by Algebra, Lemma 125.6. Then it follows that R → A is smooth at p by Algebra, Lemmas 137.17 and 140.3. 0BJ6 Lemma 4.5. In Situation 4.1 assume that R → A is smooth at q and that R/πR ⊂ Λ/πΛ is a separable extension of fields. Then after a finite number of affine Néron blowups the algebra A becomes smooth over R at p.

Proof. We choose an R-algebra B and a surjection B → A. Set pB = (B → −1 A) (p) and denote r the relative dimension of R → B at pB. We choose B such that R → B is smooth at pB. For example we can take B to be a polynomial algebra in r variables over R. Consider the complex 2 I/I ⊗A Λ −→ ΩB/R ⊗B Λ SMOOTHING RING MAPS 12 of Lemma 4.4. By the structure of ﬁnite modules over Λ (More on Algebra, Lemma 121.9) we see that the cokernel looks like M Λ⊕d ⊕ Λ/πei Λ i=1,...,n for some d ≥ 0, n ≥ 0, and ei ≥ 1. Observe that d is the relative dimension of A/R P at q (Algebra, Lemma 140.3). If the defect e = i=1,...,n ei is zero, then we are done by Lemma 4.4. Next, we consider what happens when we perform the Néron blowup. Recall that A0 is the quotient of B0/IB0 by its π-power torsion (Lemma 4.2) and that R → B0 is smooth at pB0 (Lemma 4.3). Thus after blowup we have exactly the same setup. Picture 0 / I0 / B0 / A0 / 0 O O O

0 / I / B / A / 0 0 0 Since I ⊂ pB, we see that I → I factors through πI . Looking at the induced map of complexes we get

0 0 2 0 I /(I ) ⊗A0 Λ / ΩB0/R ⊗B0 Λ M O O

2 I/I ⊗A Λ / ΩB/R ⊗B Λ M Then M ⊂ M 0 are ﬁnite free Λ-modules with quotient M 0/M annihilated by π, see Lemma 4.3. Let N ⊂ M and N 0 ⊂ M 0 be the images of the horizontal maps and denote Q = M/N and Q0 = M 0/N 0. We obtain a commutative diagram

0 / N 0 / M 0 / Q0 / 0 O O O

0 / N / M / Q / 0 Then N ⊂ N 0 are free Λ-modules of rank r − d. Since I maps into πI0 we see that N ⊂ πN 0.

Let K = Λπ be the fraction ﬁeld of Λ. We have a commutative diagram

0 0 0 0 0 / N / NK ∩ M / Qtor / 0 O O O

0 / N / NK ∩ M / Qtor / 0 whose rows are short exact sequences. This shows that the change in defect is given by 0 0 0 0 0 e − e = length(Qtor) − length(Qtor) = length(N /N) − length(NK ∩ M /NK ∩ M) 0 0 0 Since M /M is annihilated by π, so is NK ∩ M /NK ∩ M, and its length is at most 0 0 dimK (NK ). Since N ⊂ πN we get length(N /N) ≥ dimK (NK ), with equality if and only if N = πN 0. SMOOTHING RING MAPS 13

To finish the proof we have to show that N is strictly smaller than πN 0 when A is not smooth at p; this is the key computation one has to do in Néron’s argument. To do this, we consider the exact sequence 2 I/I ⊗B κ(pB) → ΩB/R ⊗B κ(pB) → ΩA/R ⊗A κ(p) → 0 (follows from Algebra, Lemma 131.9). Since R → A is not smooth at p we see that the dimension s of ΩA/R ⊗A κ(p) is bigger than d. On the other hand the first arrow factors through the injective map 2 pBp/p Bp → ΩB/R ⊗B κ(pB) of Algebra, Lemma 140.4; note that κ(p) is separable over k by our assumption on R/πR ⊂ Λ/πΛ. Hence we conclude that we can find generators g1, . . . , gt ∈ I such 2 0 2 0 that gj ∈ p for j > r − s. Then the images of gj in A are in π I for j > r − s. Since r−s < r−d we find that at least one of the minimal generators of N becomes 2 0 divisible by π in N . Thus we see that e decreases by at least 1 and we win. If R → Λ is an extension of discrete valuation rings, then R → Λ is regular if and only if (a) the ramification index is 1, (b) the extension of fraction fields is separable, and (c) R/mR ⊂ Λ/mΛ is separable. Thus the following result is a special case of general Néron desingularization in Theorem 12.1. 0BJ7 Lemma 4.6. Let R ⊂ Λ be an extension of discrete valuation rings which has ramification index 1 and induces a separable extension of residue fields and of fraction fields. Then Λ is a filtered colimit of smooth R-algebras. Proof. By Algebra, Lemma 127.4 it suffices to show that any R → A → Λ as in Situation 4.1 can be factored as A → B → Λ with B a smooth R-algebra. After replacing A by its image in Λ we may assume that A is a domain whose fraction field K is a subfield of the fraction field of Λ. In particular, A is separable over the fraction field of R by our assumptions. Then R → A is smooth at q = (0) by Algebra, Lemma 140.9. After a finite number of Néron blowups, we may assume R → A is smooth at p, see Lemma 4.5. Then, after replacing A by a localization at an element a ∈ A, a 6∈ p it becomes smooth over R and the lemma is proved. 5. The lifting problem 07CJ The goal in this section is to prove (Proposition 5.3) that the collection of algebras which are filtered colimits of smooth algebras is closed under infinitesimal flat de- formations. The proof is elementary and only uses the results on presentations of smooth algebras from Section 3. 07CK Lemma 5.1. Let R → Λ be a ring map. Let I ⊂ R be an ideal. Assume that (1) I2 = 0, and (2) Λ/IΛ is a filtered colimit of smooth R/I-algebras. Let ϕ : A → Λ be an R-algebra map with A of finite presentation over R. Then there exists a factorization A → B/J → Λ where B is a smooth R-algebra and J ⊂ IB is a finitely generated ideal. Proof. Choose a factorization A/IA → B¯ → Λ/IΛ SMOOTHING RING MAPS 14

with B¯ standard smooth over R/I; this is possible by assumption and Lemma 3.5. Write B¯ = A/IA[t1, . . . , tr]/(¯g1,..., g¯s)

and say B¯ → Λ/IΛ maps ti to the class of λi modulo IΛ. Choose g1, . . . , gs ∈ P A[t1, . . . , tr] lifting g¯1,..., g¯s. Write ϕ(gi)(λ1, . . . , λr) = ijµij for some ij ∈ I and µij ∈ Λ. Deﬁne 0 X A = A[t1, . . . , tr, δi,j]/(gi − ijδij) and consider the map 0 A −→ Λ, a 7−→ ϕ(a), ti 7−→ λi, δij 7−→ µij We have 0 0 ∼ A /IA = A/IA[t1, . . . , tr]/(¯g1,..., g¯s)[δij] = B¯[δij] This is a standard smooth algebra over R/I as B¯ is standard smooth. Choose a 0 0 ¯ ¯ ¯ presentation A /IA = R/I[x1, . . . , xn]/(f1,..., fc) with det(∂fj/∂xi)i,j=1,...,c in- 0 0 ¯ ¯ vertible in A /IA . Choose lifts f1, . . . , fc ∈ R[x1, . . . , xn] of f1,..., fc. Then

B = R[x1, . . . , xn, xn+1]/(f1, . . . , fc, xn+1 det(∂fj/∂xi)i,j=1,...,c − 1) is smooth over R. Since smooth ring maps are formally smooth (Algebra, Proposi- tion 138.13) there exists an R-algebra map B → A0 which is an isomorphism modulo I. Then B → A0 is surjective by Nakayama’s lemma (Algebra, Lemma 20.1). Thus 0 A = B/J with J ⊂ IB finitely generated (see Algebra, Lemma 6.3). 07CL Lemma 5.2. Let R → Λ be a ring map. Let I ⊂ R be an ideal. Assume that (1) I2 = 0, (2) Λ/IΛ is a filtered colimit of smooth R/I-algebras, and (3) R → Λ is flat. Let ϕ : B → Λ be an R-algebra map with B smooth over R. Let J ⊂ IB be a finitely generated ideal such that ϕ(J) = 0. Then there exists R-algebra maps β B −→α B0 −→ Λ such that B0 is smooth over R, such that α(J) = 0 and such that β ◦α = ϕ mod IΛ. Proof. If we can prove the lemma in case J = (h), then we can prove the lemma by induction on the number of generators of J. Namely, suppose that J can be generated by n elements h1, . . . , hn and the lemma holds for all cases where J is generated by n−1 elements. Then we apply the case n = 1 to produce B → B0 → Λ 0 0 where the first map kills of hn. Then we let J be the ideal of B generated by the 0 00 images of h1, . . . , hn−1 and we apply the case for n − 1 to produce B → B → Λ. It is easy to verify that B → B00 → Λ does the job. P Assume J = (h) and write h = ibi for some i ∈ I and bi ∈ B. Note that P 0 = ϕ(h) = iϕ(bi). As Λ is flat over R, the equational criterion for flatness (Algebra, Lemma 39.11) implies that we can find λj ∈ Λ, j = 1, . . . , m and aij ∈ R P P such that ϕ(bi) = j aijλj and i iaij = 0. Set X C = B[x1, . . . , xm]/(bi − aijxj)

with C → Λ given by ϕ and xj 7→ λj. Choose a factorization C → B0/J 0 → Λ SMOOTHING RING MAPS 15

as in Lemma 5.1. Since B is smooth over R we can lift the map B → C → B0/J 0 to a map ψ : B → B0. We claim that ψ(h) = 0. Namely, the fact that ψ agrees with B → C → B0/J 0 mod I implies that X ψ(bi) = aijξj + θi

0 0 for some ξi ∈ B and θi ∈ IB . Hence we see that X X X ψ(h) = ψ( ibi) = iaijξj + iθi = 0

2 because of the relations above and the fact that I = 0. 07CM Proposition 5.3. Let R → Λ be a ring map. Let I ⊂ R be an ideal. Assume that (1) I is nilpotent, (2) Λ/IΛ is a filtered colimit of smooth R/I-algebras, and (3) R → Λ is flat. Then Λ is a filtered colimit of smooth R-algebras. Proof. Since In = 0 for some n, it follows by induction on n that it suffices to consider the case where I2 = 0. Let ϕ : A → Λ be an R-algebra map with A of finite presentation over R. We have to find a factorization A → B → Λ with B smooth over R, see Algebra, Lemma 127.4. By Lemma 5.1 we may assume that A = B/J with B smooth over R and J ⊂ IB a finitely generated ideal. By Lemma 5.2 we can find a (possibly noncommutative) diagram

0 B α / B

ϕ β ~ Λ of R-algebras which commutes modulo I and such that α(J) = 0. The map D : B −→ IΛ, b 7−→ ϕ(b) − β(α(b))

is a derivation over R hence we can write it as D = ξ ◦ dB/R for some B-linear map ξ :ΩB/R → IΛ. Since ΩB/R is a finite projective B-module we can write P ξ = i=1,...,n iΞi for some i ∈ I and B-linear maps Ξi :ΩB/R → Λ. (Details omitted. Hint: write ΩB/R as a direct sum of a finite free module to reduce to the finite free case.) We define

00 ∗ M 0 B = SymB0 ΩB/R ⊗B,α B i=1,...,n and we deﬁne β0 : B00 → Λ by β on B0 and by 0 0 β |ith summand ΩB/R⊗B,αB = Ξi ⊗ β and α0 : B → B00 by

0 X α (b) = α(b) ⊕ idB/R(b) ⊗ 1 ⊕ 0 ⊕ ... At this point the diagram B / B00 α0

ϕ β0 ~ Λ SMOOTHING RING MAPS 16

does commute. Moreover, it is direct from the deﬁnitions that α0(J) = 0 as I2 = 0. Hence the desired factorization.

6. The lifting lemma 07CN Here is a ﬁendishly clever lemma. 07CP Lemma 6.1. Let R be a Noetherian ring. Let Λ be an R-algebra. Let π ∈ R and 2 2 assume that AnnR(π) = AnnR(π ) and AnnΛ(π) = AnnΛ(π ). Suppose we have R-algebra maps R/π2R → C¯ → Λ/π2Λ with C¯ of ﬁnite presentation. Then there exists an R-algebra homomorphism D → Λ and a commutative diagram

R/π2R / C¯ / Λ/π2Λ

R/πR / D/πD / Λ/πΛ with the following properties (a) D is of ﬁnite presentation, (b) R → D is smooth at any prime q with π 6∈ q, (c) R → D is smooth at any prime q with π ∈ q lying over a prime of C¯ where R/π2R → C¯ is smooth, and (d) C/π¯ C¯ → D/πD is smooth at any prime lying over a prime of C¯ where R/π2R → C¯ is smooth. Proof. We choose a presentation

C¯ = R[x1, . . . , xn]/(f1, . . . , fm)

2 We also denote I = (f1, . . . , fm) and I¯ the image of I in R/π R[x1, . . . , xn]. Since R is Noetherian, so is C¯. Hence the smooth locus of R/π2R → C¯ is quasi-compact, see Topology, Lemma 9.2. Applying Lemma 2.2 we may choose a ﬁnite list of elements a1, . . . , ar ∈ R[x1, . . . , xn] such that ¯ ¯ (1) the union of the open subspaces Spec(Cak ) ⊂ Spec(C) cover the smooth locus of R/π2R → C¯, and (2) for each k = 1, . . . , r there exists a ﬁnite subset Ek ⊂ {1, . . . , m} such that ¯ ¯2 (I/I )ak is freely generated by the classes of fj, j ∈ Ek. 2 Set Ik = (fj, j ∈ Ek) ⊂ I and denote I¯k the image of Ik in R/π R[x1, . . . , xn]. By ¯ ¯ 0 (2) and Nakayama’s lemma we see that (I/Ik)ak is annihilated by 1 + bk for some 0 ¯ 0 N bk ∈ Iak . Suppose bk is the image of bk/(ak) for some bk ∈ I and some integer N. After replacing ak by akbk we get ¯ ¯ (3) (Ik)ak = (I)ak .

Thus, after possibly replacing ak by a high power, we may write (4) a f = P hj f + π2g k ` j∈Ek k,` j k,` j for any ` ∈ {1, . . . , m} and some hi,`, gi,` ∈ R[x1, . . . , xn]. If ` ∈ Ek we choose j hk,` = akδ`,j (Kronecker delta) and gk,` = 0. Set

D = R[x1, . . . , xn, z1, . . . , zm]/(fj − πzj, pk,`). SMOOTHING RING MAPS 17

Here j ∈ {1, . . . , m}, k ∈ {1, . . . , r}, ` ∈ {1, . . . , m}, and

X j pk,` = akz` − hk,`zj − πgk,`. j∈Ek

Note that for ` ∈ Ek we have pk,` = 0 by our choices above. 2 The map R → D is the given one. Say C¯ → Λ/π Λ maps xi to the class of λi 2 modulo π . For an element f ∈ R[x1, . . . , xn] we denote f(λ) ∈ Λ the result of 2 substituting λi for xi. Then we know that fj(λ) = π µj for some µj ∈ Λ. Deﬁne D → Λ by the rules xi 7→ λi and zj 7→ πµj. This is well deﬁned because X j pk,` 7→ ak(λ)πµ` − hk,`(λ)πµj − πgk,`(λ) j∈Ek X j = π ak(λ)µ` − hk,`(λ)µj − gk,`(λ) j∈Ek

Substituting xi = λi in (4) above we see that the expression inside the brackets 2 is annihilated by π , hence it is annihilated by π as we have assumed AnnΛ(π) = 2 AnnΛ(π ). The map C¯ → D/πD is determined by xi 7→ xi (clearly well deﬁned). Thus we are done if we can prove (b), (c), and (d). Using (4) we obtain the following key equality

X j 2 πpk,` = πakz` − πhk,`zj − π gk,` j∈Ek X j X j 2 = −ak(f` − πz`) + akf` + hk,`(fj − πzj) − hk,`fj − π gk,` j∈Ek j∈Ek X j = −ak(f` − πz`) + hk,`(fj − πzj) j∈Ek

The end result is an element of the ideal generated by fj − πzj. In particular, we see that D[1/π] is isomorphic to R[1/π][x1, . . . , xn, z1, . . . , zm]/(fj − πzj) which is isomorphic to R[1/π][x1, . . . , xn] hence smooth over R. This proves (b). For ﬁxed k ∈ {1, . . . , r} consider the ring

Dk = R[x1, . . . , xn, z1, . . . , zm]/(fj − πzj, j ∈ Ek, pk,`)

The number of equations is m = |Ek| + (m − |Ek|) as pk,` is zero if ` ∈ Ek. Also, note that

(Dk/πDk)ak = R/πR[x1, . . . , xn, 1/ak, z1, . . . , zm]/(fj, j ∈ Ek, pk,`) ¯ ¯ X j = (C/πC)ak [z1, . . . , zm]/(akz` − hk,`zj) j∈Ek ∼ ¯ ¯ = (C/πC)ak [zj, j ∈ Ek] ¯ ¯ In particular (Dk/πDk)ak is smooth over (C/πC)ak . By our choice of ak we have ¯ ¯ that (C/πC)ak is smooth over R/πR of relative dimension n − |Ek|, see (2). Hence ¯ for a prime qk ⊂ Dk containing π and lying over Spec(Cak ) the ﬁbre ring of R → Dk is smooth at qk of dimension n. Thus R → Dk is syntomic at qk by our count of the number of equations above, see Algebra, Lemma 136.11. Hence R → Dk is smooth at qk, see Algebra, Lemma 137.17. To ﬁnish the proof, let q ⊂ D be a prime containing π lying over a prime where 2 R/π R → C¯ is smooth. Then ak 6∈ q for some k by (1). We will show that the surjection Dk → D induces an isomorphism on local rings at q. Since we know that the ring maps C/π¯ C¯ → Dk/πDk and R → Dk are smooth at the corresponding SMOOTHING RING MAPS 18

prime qk by the preceding paragraph this will prove (c) and (d) and thus ﬁnish the proof. First, note that for any ` the equation πp = −a (f −πz )+P hj (f −πz ) k,` k ` ` j∈Ek k,` j j

proved above shows that f`−πz` maps to zero in (Dk)ak and in particular in (Dk)qk . The relations (4) imply that a f = P hj f in I/I2. Since (I¯ /I¯2) is free k ` j∈Ek k,` j k k ak on fj, j ∈ Ek we see that j X j0 j a 0 h − h 0 h 0 k k,` 0 k ,` k,j j ∈Ek0 ¯ 0 is zero in Cak for every k, k , ` and j ∈ Ek. Hence we can ﬁnd a large integer N such that N j X j0 j a a 0 h − h 0 h 0 k k k,` 0 k ,` k,j j ∈Ek0 2 is in Ik + π R[x1, . . . , xn]. Computing modulo π we have X j0 akpk0,` − ak0 pk,` + hk0,`pk,j0 X j0 X j X j0 XX j0 j = −ak hk0,`zj0 + ak0 hk,`zj + hk0,`akzj0 − hk0,`hk,j0 zj X j X j0 j = ak0 hk,` − hk0,`hk,j0 zj

N+1 with Einstein summation convention. Combining with the above we see ak pk0,` is contained in the ideal generated by Ik and π in R[x1, . . . , xn, z1, . . . , zm]. Thus 0 pk ,` maps into π(Dk)ak . On the other hand, the equation X j0 πp 0 = −a 0 (f − πz ) + h 0 (f 0 − πz 0 ) k ,` k ` ` 0 k ,` j j j ∈Ek0

0 shows that πpk ,` is zero in (Dk)ak . Since we have assumed that AnnR(π) = 2 AnnR(π ) and since (Dk)q is smooth hence ﬂat over R we see that Ann (π) = k (Dk)qk 2 Ann (π ). We conclude that pk0,` maps to zero as well, hence Dq = (Dk)q (Dk)qk k and we win.

7. The desingularization lemma 07CQ Here is another fiendishly clever lemma. 07CR Lemma 7.1. Let R be a Noetherian ring. Let Λ be an R-algebra. Let π ∈ R and 2 assume that AnnΛ(π) = AnnΛ(π ). Let A → Λ be an R-algebra map with A of finite presentation. Assume (1) the image of π is strictly standard in A over R, and (2) there exists a section ρ : A/π4A → R/π4R which is compatible with the map to Λ/π4Λ. Then we can find R-algebra maps A → B → Λ with B of finite presentation such 2 that aB ⊂ HB/R where a = AnnR(AnnR(π )/AnnR(π)). Proof. Choose a presentation

A = R[x1, . . . , xn]/(f1, . . . , fm) and 0 ≤ c ≤ min(n, m) such that (2.3.3) holds for π and such that 2 (7.1.1)07CS πfc+j ∈ (f1, . . . , fc) + (f1, . . . , fm) SMOOTHING RING MAPS 19 for j = 1, . . . , m−c. Say ρ maps xi to the class of ri ∈ R. Then we can replace xi by 4 4 xi − ri. Hence we may assume ρ(xi) = 0 in R/π R. This implies that fj(0) ∈ π R 4 and that A → Λ maps xi to π λi for some λi ∈ Λ. Write X fj = fj(0) + rjixi + h.o.t. i=1,...,n

This implies that the constant term of ∂fj/∂xi is rji. Apply ρ to (2.3.3) for π and we see that X 4 π = rI det(rji)j=1,...,c, i∈I mod π R I⊂{1,...,n}, |I|=c for some rI ∈ R. Thus we have X uπ = rI det(rji)j=1,...,c, i∈I I⊂{1,...,n}, |I|=c for some u ∈ 1 + π3R. By Algebra, Lemma 15.5 this implies there exists a n × c matrix (sik) such that X uπδjk = rjisik for all j, k = 1, . . . , c i=1,...,n

(Kronecker delta). We introduce auxiliary variables v1, . . . , vc, w1, . . . , wn and we set 2 X 3 hi = xi − π sijvj − π wi j=1,...c In the following we will use that

R[x1, . . . , xn, v1, . . . , vc, w1, . . . , wn]/(h1, . . . , hn) = R[v1, . . . , vc, w1, . . . , wn] without further mention. In R[x1, . . . , xn, v1, . . . , vc, w1, . . . , wn]/(h1, . . . , hn) we have

fj = fj(x1 − h1, . . . , xn − hn) c n n 2 X X 3 X 4 = π rjisik vk + π rjiwi mod π k=1 i=1 i=1 n 3 3 X 4 = π vj + π rjiwi mod π i=1 for 1 ≤ j ≤ c. Hence we can choose elements gj ∈ R[v1, . . . , vc, w1, . . . , wn] P 3 such that gj = vj + rjiwi mod π and such that fj = π gj in the R-algebra R[x1, . . . , xn, v1, . . . , vc, w1, . . . , wn]/(h1, . . . , hn). We set

B = R[x1, . . . , xn, v1, . . . , vc, w1, . . . , wn]/(f1, . . . , fm, h1, . . . , hn, g1, . . . , gc). 4 The map A → B is clear. We deﬁne B → Λ by mapping xi → π λi, vi 7→ 0, and wi 7→ πλi. Then it is clear that the elements fj and hi are mapped to zero in Λ. 3 Moreover, it is clear that gi is mapped to an element t of πΛ such that π t = 0 3 (as fi = π gi modulo the ideal generated by the h’s). Hence our assumption that 2 AnnΛ(π) = AnnΛ(π ) implies that t = 0. Thus we are done if we can prove the statement about smoothness. ∼ Note that Bπ = Aπ[v1, . . . , vc] because the equations gi = 0 are implied by fi = 0. Hence Bπ is smooth over R as Aπ is smooth over R by the assumption that π is strictly standard in A over R, see Lemma 2.5. 0 P Set B = R[v1, . . . , vc, w1, . . . , wn]/(g1, . . . , gc). As gi = vi + rjiwi mod π we see 0 0 0 that B /πB = R/πR[w1, . . . , wn]. Hence R → B is smooth of relative dimension n at every point of V (π) by Algebra, Lemmas 136.11 and 137.17 (the ﬁrst lemma SMOOTHING RING MAPS 20

shows it is syntomic at those primes, in particular flat, whereupon the second lemma shows it is smooth). Let q ⊂ B be a prime with π ∈ q and for some r ∈ a, r 6∈ q. Denote q0 = B0 ∩ q. We 0 0 claim the surjection B → B induces an isomorphism of local rings (B )q0 → Bq. 0 This will conclude the proof of the lemma. Note that Bq is the quotient of (B )q0 by the ideal generated by fc+j, j = 1, . . . , m − c. We observe two things: first 0 2 the image of fc+j in (B )q0 is divisible by π and second the image of πfc+j in 0 0 P (B )q can be written as bj1j2 fc+j1 fc+j2 by (7.1.1). Thus we see that the image 2 of each πfc+j is contained in the ideal generated by the elements π fc+j0 . Hence 0 πfc+j = 0 in (B )q0 as this is a Noetherian local ring, see Algebra, Lemma 51.4. As 0 R → (B )q0 is flat we see that 2 0 2 Ann (π )/Ann (π) ⊗ (B ) 0 = Ann 0 (π )/Ann 0 (π) R R R q (B )q0 (B )q0 0 Because r ∈ a is invertible in (B )q0 we see that this module is zero. Hence we see 0 that the image of fc+j is zero in (B )q0 as desired. 07CT Lemma 7.2. Let R be a Noetherian ring. Let Λ be an R-algebra. Let π ∈ R 2 2 and assume that AnnR(π) = AnnR(π ) and AnnΛ(π) = AnnΛ(π ). Let A → Λ and D → Λ be R-algebra maps with A and D of finite presentation. Assume (1) π is strictly standard in A over R, and (2) there exists an R-algebra map A/π4A → D/π4D compatible with the maps to Λ/π4Λ. Then we can find an R-algebra map B → Λ with B of finite presentation and R-algebra maps A → B and D → B compatible with the maps to Λ such that HD/RB ⊂ HB/D and HD/RB ⊂ HB/R. Proof. We apply Lemma 7.1 to

D −→ A ⊗R D −→ Λ and the image of π in D. By Lemma 2.7 we see that π is strictly standard in 4 4 A ⊗R D over D. As our section ρ :(A ⊗R D)/π (A ⊗R D) → D/π D we take the map induced by the map in (2). Thus Lemma 7.1 applies and we obtain a factorization A ⊗R D → B → Λ with B of finite presentation and aB ⊂ HB/D where 2 a = AnnD(AnnD(π )/AnnD(π)). 2 For any prime q of D such that Dq is flat over R we have AnnDq (π )/AnnDq (π) = 0 because annihilators of elements commutes with flat base change and we assumed 2 2 AnnR(π) = AnnR(π ). Because D is Noetherian we see that AnnD(π )/AnnD(π) is a finite D-module, hence formation of its annihilator commutes with localization. Thus we see that a 6⊂ q. Hence we see that D → B is smooth at any prime of B lying over q. Since any prime of D where R → D is smooth is one where Dq is flat over R we conclude that HD/RB ⊂ HB/D. The final inclusion HD/RB ⊂ HB/R follows because compositions of smooth ring maps are smooth (Algebra, Lemma 137.14). 07F0 Lemma 7.3. Let R be a Noetherian ring. Let Λ be an R-algebra. Let π ∈ R and 2 2 assume that AnnR(π) = AnnR(π ) and AnnΛ(π) = AnnΛ(π ). Let A → Λ be an SMOOTHING RING MAPS 21

R-algebra map with A of finite presentation and assume π is strictly standard in A over R. Let A/π8A → C¯ → Λ/π8Λ be a factorization with C¯ of finite presentation. Then we can find a factorization A → B → Λ with B of finite presentation such that Rπ → Bπ is smooth and such that 8 q 8 HC/¯ (R/π8R) · Λ/π Λ ⊂ HB/RΛ mod π Λ.

Proof. Apply Lemma 6.1 to get R → D → Λ with a factorization C/π¯ 4C¯ → D/π4D → Λ/π4Λ such that R → D is smooth at any prime not containing π and at any prime lying over a prime of C/π¯ 4C¯ where R/π8R → C¯ is smooth. By Lemma 7.2 we can find a finitely presented R-algebra B and factorizations A → B → Λ and D → B → Λ such that HD/RB ⊂ HB/R. We omit the verification that this is a solution to the problem posed by the lemma.

8. Warmup: reduction to a base field 07F1 In this section we apply the lemmas in the previous sections to prove that it suffices to prove the main result when the base ring is a field, see Lemma 8.4. 07F2Situation 8.1. Here R → Λ is a regular ring map of Noetherian rings. Let R → Λ be as in Situation 8.1. We say PT holds for R → Λ if Λ is a filtered colimit of smooth R-algebras.

07F3 Lemma 8.2. Let Ri → Λi, i = 1, 2 be as in Situation 8.1. If PT holds for Ri → Λi, i = 1, 2, then PT holds for R1 × R2 → Λ1 × Λ2.

Proof. Omitted. Hint: A product of filtered colimits is a filtered colimit. 07F4 Lemma 8.3. Let R → A → Λ be ring maps with A of finite presentation over R. Let S ⊂ R be a multiplicative set. Let S−1A → B0 → S−1Λ be a factorization with B0 smooth over S−1R. Then we can find a factorization A → B → Λ such that some s ∈ S maps to an elementary standard element (Definition 2.3) in B over R. Proof. We first apply Lemma 3.4 to S−1R → B0. Thus we may assume B0 is −1 standard smooth over S R. Write A = R[x1, . . . , xn]/(g1, . . . , gt) and say xi 7→ λi 0 −1 in Λ. We may write B = S R[x1, . . . , xn+m]/(f1, . . . , fc) for some c ≥ n where 0 0 det(∂fj/∂xi)i,j=1,...,c is invertible in B and such that A → B is given by xi 7→ xi, see Lemma 3.6. After multiplying xi, i > n by an element of S and correspondingly 0 −1 modifying the equations fj we may assume B → S Λ maps xi to λi/1 for some λi ∈ Λ for i > n. Choose a relation X 1 = a0 det(∂fj/∂xi)i,j=1,...,c + ajfj j=1,...,c −1 0 for some aj ∈ S R[x1, . . . , xn+m]. Since each element of S is invertible in B we may (by clearing denominators) assume that fj, aj ∈ R[x1, . . . , xn+m] and that X s0 = a0 det(∂fj/∂xi)i,j=1,...,c + ajfj j=1,...,c −1 for some s0 ∈ S. Since gj maps to zero in S R[x1, . . . , xn+m]/(f1, . . . , xc) we can find elements sj ∈ S such that sjgj = 0 in R[x1, . . . , xn+m]/(f1, . . . , fc). Since fj SMOOTHING RING MAPS 22

−1 0 0 maps to zero in S Λ we can ﬁnd sj ∈ S such that sjfj(λ1, . . . , λn+m) = 0 in Λ. Consider the ring 0 0 B = R[x1, . . . , xn+m]/(s1f1, . . . , scfc, g1, . . . , gt)

and the factorization A → B → Λ with B → Λ given by xi 7→ λi. We claim that 0 0 s = s0s1 . . . sts1 . . . sc is elementary standard in B over R which finishes the proof. 0 0 Namely, sjgj ∈ (f1, . . . , fc) and hence sgj ∈ (s1f1, . . . , scfc). Finally, we have 0 X 0 ˆ0 0 0 0 0 a0 det(∂sjfj/∂xi)i,j=1,...,c + (s1 ... s . . . sc)ajsjfj = s0s1 . . . sc j=1,...,c j which divides s as desired. 07F5 Lemma 8.4. If for every Situation 8.1 where R is a field PT holds, then PT holds in general. Proof. Assume PT holds for any Situation 8.1 where R is a field. Let R → Λ be as in Situation 8.1 arbitrary. Note that R/I → Λ/IΛ is another regular ring map of Noetherian rings, see More on Algebra, Lemma 41.3. Consider the set of ideals I = {I ⊂ R | R/I → Λ/IΛ does not have PT} We have to show that I is empty. If this set is nonempty, then it contains a maximal element because R is Noetherian. Replacing R by R/I and Λ by Λ/I we obtain a situation where PT holds for R/I → Λ/IΛ for any nonzero ideal of R. In particular, we see by applying Proposition 5.3 that R is a reduced ring. Let A → Λ be an R-algebra homomorphism with A of finite presentation. We have to find a factorization A → B → Λ with B smooth over R, see Algebra, Lemma 127.4. Let S ⊂ R be the set of nonzerodivisors and consider the total ring of fractions −1 Q = S R of R. We know that Q = K1 ×...×Kn is a product of fields, see Algebra, Lemmas 25.4 and 31.6. By Lemma 8.2 and our assumption PT holds for the ring map S−1R → S−1Λ. Hence we can find a factorization S−1A → B0 → S−1Λ with B0 smooth over S−1R. We apply Lemma 8.3 and find a factorization A → B → Λ such that some π ∈ S is elementary standard in B over R. After replacing A by B we may assume that π is elementary standard, hence strictly standard in A. We know that R/π8R → Λ/π8Λ satisfies PT. Hence we can find a factorization R/π8R → A/π8A → C¯ → Λ/π8Λ with R/π8R → C¯ smooth. By Lemma 6.1 we can find an R-algebra map D → Λ with D smooth over R and a factorization R/π4R → A/π4A → D/π4D → Λ/π4Λ. By Lemma 7.2 we can find A → B → Λ with B smooth over R which finishes the proof.

9. Local tricks 07F6 07F7Situation 9.1. We are given a Noetherian ring R and an R-algebra map A → Λ and a prime q ⊂ Λ. We assume A is of ﬁnite presentation over R. In this situation p we denote hA = HA/RΛ. Let R → A → Λ ⊃ q be as in Situation 9.1. We say R → A → Λ ⊃ q can be resolved if there exists a factorization A → B → Λ with B of ﬁnite presentation SMOOTHING RING MAPS 23

and hA ⊂ hB 6⊂ q. In this case we will call the factorization A → B → Λ a resolution of R → A → Λ ⊃ q.

07F8 Lemma 9.2. Let R → A → Λ ⊃ q be as in Situation 9.1. Let r ≥ 1 and π1, . . . , πr ∈ R map to elements of q. Assume (1) for i = 1, . . . , r we have

2 Ann 8 8 (πi) = Ann 8 8 (π ) R/(π1 ,...,πi−1)R R/(π1 ,...,πi−1)R i and 2 Ann 8 8 (πi) = Ann 8 8 (π ) Λ/(π1 ,...,πi−1)Λ Λ/(π1 ,...,πi−1)Λ i

(2) for i = 1, . . . , r the element πi maps to a strictly standard element in A over R. Then, if

8 8 8 8 8 8 8 8 R/(π1, . . . , πr )R → A/(π1, . . . , πr )A → Λ/(π1, . . . , πr )Λ ⊃ q/(π1, . . . , πr )Λ

can be resolved, so can R → A → Λ ⊃ q.

Proof. We are going to prove this by induction on r.

8 The case r = 1. Here the assumption is that there exists a factorization A/π1 → ¯ 8 8 C → Λ/π1 which resolves the situation modulo π1. Conditions (1) and (2) are the assumptions needed to apply Lemma 7.3. Thus we can “lift” the resolution C¯ to a resolution of R → A → Λ ⊃ q.

The case r > 1. In this case we apply the induction hypothesis for r − 1 to the 8 8 8 8 situation R/π1 → A/π1 → Λ/π1 ⊃ q/π1Λ. Note that property (2) is preserved by Lemma 2.7.

07F9 Lemma 9.3. Let R → A → Λ ⊃ q be as in Situation 9.1. Let p = R ∩ q. Assume [Swa98, Lemma that q is minimal over hA and that Rp → Ap → Λq ⊃ qΛq can be resolved. Then 12.2] or [Pop85, there exists a factorization A → C → Λ with C of ﬁnite presentation such that Lemma 2] HC/RΛ 6⊂ q.

Proof. Let Ap → C → Λq be a resolution of Rp → Ap → Λq ⊃ qΛq. By our

assumption that q is minimal over hA this means that HC/Rp Λq = Λq. By Lemma 2.8 we may assume that C is smooth over Rp. By Lemma 3.4 we may assume that C is standard smooth over Rp. Write A = R[x1, . . . , xn]/(g1, . . . , gt) and say A → Λ is given by xi 7→ λi. Write C = Rp[x1, . . . , xn+m]/(f1, . . . , fc) for some c ≥ n such that A → C maps xi to xi and such that det(∂fj/∂xi)i,j=1,...,c is invertible in C, see Lemma 3.6. After clearing denominators we may assume f1, . . . , fc are elements of R[x1, . . . , xn+m]. Of course det(∂fj/∂xi)i,j=1,...,c is not invertible in R[x1, . . . , xn+m]/(f1, . . . , fc) but it becomes invertible after inverting some element s0 ∈ R, s0 6∈ p. As gj maps to zero under R[x1, . . . , xn] → A → C we can ﬁnd sj ∈ R, sj 6∈ p such that sjgj is zero in R[x1, . . . , xn+m]/(f1, . . . , fc). Write fj = Fj(x1, . . . , xn+m, 1) for some polynomial Fj ∈ R[x1, . . . , xn,Xn+1,...,Xn+m+1] ho- mogeneous in Xn+1,...,Xn+m+1. Pick λn+i ∈ Λ, i = 1, . . . , m+1 with λn+m+1 6∈ q SMOOTHING RING MAPS 24

such that xn+i maps to λn+i/λn+m+1 in Λq. Then λ λ deg(Fj ) n+1 n+m Fj(λ1, . . . , λn+m+1) = (λn+m+1) Fj(λ1, . . . , λn, ,..., , 1) λn+m+1 λn+m+1 λ λ deg(Fj ) n+1 n+m = (λn+m+1) fj(λ1, . . . , λn, ,..., ) λn+m+1 λn+m+1 = 0

in Λq. Thus we can ﬁnd λ0 ∈ Λ, λ0 6∈ q such that λ0Fj(λ1, . . . , λn+m+1) = 0 in Λ. Now we set B equal to

R[x0, . . . , xn+m+1]/(g1, . . . , gt, x0F1(x1, . . . , xn+m+1), . . . , x0Fc(x1, . . . , xn+m+1))

which we map to Λ by mapping xi to λi. Let b be the image of x0xn+m+1s0s1 . . . st in B. Then Bb is isomorphic to

Rs0s1...st [x0, x1, . . . , xn+m+1, 1/x0xn+m+1]/(f1, . . . , fc) which is smooth over R by construction. Since b does not map to an element of q, we win. 07FA Lemma 9.4. Let R → A → Λ ⊃ q be as in Situation 9.1. Let p = R ∩ q. Assume

(1) q is minimal over hA, (2) Rp → Ap → Λq ⊃ qΛq can be resolved, and (3) dim(Λq) = 0. Then R → A → Λ ⊃ q can be resolved.

N Proof. By (3) the ring Λq is Artinian local hence qΛq is nilpotent. Thus (hA) Λq = N 0 for some N > 0. Thus there exists a λ ∈ Λ, λ 6∈ q such that λ(hA) = 0 in Λ. N Say HA/R = (a1, . . . , ar) so that λai = 0 in Λ. By Lemma 9.3 we can ﬁnd a factorization A → C → Λ with C of ﬁnite presentation such that hC 6⊂ q. Write C = A[x1, . . . , xn]/(f1, . . . , fm). Set X B = A[x1, . . . , xn, y1, . . . , yr, z, tij]/(fj − yitij, zyi)

where tij is a set of rm variables. Note that there is a map B → C[yi, z]/(yiz) given by setting tij equal to zero. The map B → Λ is the composition B → C[yi, z]/(yiz) → Λ where C[yi, z]/(yiz) → Λ is the given map C → Λ, maps z to λ, N and maps yi to the image of ai in Λ.

We claim that B is a solution for R → A → Λ ⊃ q. First note that Bz is iso- −1 ∼ morphic to C[y1, . . . , yr, z, z ] and hence is smooth. On the other hand, By` = −1 A[xi, yi, y` , tij, i 6= `] which is smooth over A. Thus we see that z and a`y` (compositions of smooth maps are smooth) are all elements of HB/R. This proves the lemma. 10. Separable residue fields 07FB In this section we explain how to solve a local problem in the case of a separable residue field extension. 07FC Lemma 10.1 (Ogoma). Let A be a Noetherian ring and let M be a finite A- module. Let S ⊂ A be a multiplicative set. If π ∈ A and Ker(π : S−1M → S−1M) = Ker(π2 : S−1M → S−1M) then there exists an s ∈ S such that for any n > 0 we have Ker(snπ : M → M) = Ker((snπ)2 : M → M). SMOOTHING RING MAPS 25

Proof. Let K = Ker(π : M → M) and K0 = {m ∈ M | π2m = 0 in S−1M} and Q = K0/K. Note that S−1Q = 0 by assumption. Since A is Noetherian we see that Q is a ﬁnite A-module. Hence we can ﬁnd an s ∈ S such that s annihilates Q. Then s works.

07FD Lemma 10.2. Let Λ be a Noetherian ring. Let I ⊂ Λ be an ideal. Let I ⊂ q be n a prime. Let n, e be positive integers Assume that q Λq ⊂ IΛq and that Λq is a regular local ring of dimension d. Then there exists an n > 0 and π1, . . . , πd ∈ Λ such that

(1) (π1, . . . , πd)Λq = qΛq, n n (2) π1 , . . . , πd ∈ I, and (3) for i = 1, . . . , d we have

2 Ann e e (π ) = Ann e e (π ). Λ/(π1 ,...,πi−1)Λ i Λ/(π1 ,...,πi−1)Λ i

−1 Proof. Set S = Λ \ q so that Λq = S Λ. First pick π1, . . . , πd with (1) which is n possible as Λq is regular. By assumption πi ∈ IΛq. Thus we can ﬁnd s1, . . . , sd ∈ S n such that siπi ∈ I. Replacing πi by siπi we get (2). Note that (1) and (2) are preserved by further multiplying by elements of S. Suppose that (3) holds for i = 1, . . . , t for some t ∈ {0, . . . , d}. Note that π1, . . . , πd is a regular sequence −1 e e in S Λ, see Algebra, Lemma 106.3. In particular π1, . . . , πt , πt+1 is a regular −1 sequence in S Λ = Λq by Algebra, Lemma 68.9. Hence we see that

2 Ann −1 e e (πi) = Ann −1 e e (π ). S Λ/(π1 ,...,πi−1) S Λ/(π1 ,...,πi−1) i

Thus we get (3) for i = t+1 after replacing πt+1 by sπt+1 for some s ∈ S by Lemma 10.1. By induction on t this produces a sequence satisfying (1), (2), and (3).

07FE Lemma 10.3. Let k → A → Λ ⊃ q be as in Situation 9.1 where (1) k is a ﬁeld, (2) Λ is Noetherian, (3) q is minimal over hA, (4) Λq is a regular local ring, and (5) the ﬁeld extension k ⊂ κ(q) is separable. Then k → A → Λ ⊃ q can be resolved.

n Proof. Set d = dim Λq. Set R = k[x1, . . . , xd]. Choose n > 0 such that q Λq ⊂ hAΛq which is possible as q is minimal over hA. Choose generators a1, . . . , ar of HA/R. Set n X B = A[x1, . . . , xd, zij]/(xi − zijaj)

Each Baj is smooth over R it is a polynomial algebra over Aaj [x1, . . . , xd] and Aaj

is smooth over k. Hence Bxi is smooth over R. Let B → C be the R-algebra map constructed in Lemma 3.1 which comes with a R-algebra retraction C → B. In

particular a map C → Λ ﬁtting into the diagram above. By construction Cxi is a smooth R-algebra with Ω free. Hence we can ﬁnd c > 0 such that xc is Cxi /R i strictly standard in C/R, see Lemma 3.7. Now choose π1, . . . , πd ∈ Λ as in Lemma n P 10.2 where n = n, e = 8c, q = q and I = hA. Write πi = λijaj for some πij ∈ Λ. There is a map B → Λ given by xi 7→ πi and zij 7→ λij. Set R = k[x1, . . . , xd]. SMOOTHING RING MAPS 26

Diagram R / B O O

k / A / Λ Now we apply Lemma 9.2 to R → C → Λ ⊃ q and the sequence of elements c c x1, . . . , xd of R. Assumption (2) is clear. Assumption (1) holds for R by inspection 2 and for Λ by our choice of π1, . . . , πd. (Note that if AnnΛ(π) = AnnΛ(π ), then we c have AnnΛ(π) = AnnΛ(π ) for all c > 0.) Thus it suffices to resolve e e e e e e e e R/(x1, . . . , xd) → C/(x1, . . . , xd) → Λ/(π1, . . . , πd) ⊃ q/(π1, . . . , πd) for e = 8c. By Lemma 9.4 it suffices to resolve this after localizing at q. But since x1, . . . , xd map to a regular sequence in Λq we see that Rp → Λq is flat, see Algebra, Lemma 128.2. Hence e e e e Rp/(x1, . . . , xd) → Λq/(π1, . . . , πd) is a flat ring map of Artinian local rings. Moreover, this map induces a separable field extension on residue fields by assumption. Thus this map is a filtered colimit of smooth algebras by Algebra, Lemma 158.11 and Proposition 5.3. Existence of the desired solution follows from Algebra, Lemma 127.4. 11. Inseparable residue fields 07FF In this section we explain how to solve a local problem in the case of an inseparable residue field extension. 07FG Lemma 11.1. Let k be a field of characteristic p > 0. Let (Λ, m,K) be an Artinian local k-algebra. Assume that dim H1(LK/k) < ∞. Then Λ is a filtered colimit of Artinian local k-algebras A with each map A → Λ flat, with mAΛ = m, and with A essentially of finite type over k. Proof. Note that the flatness of A → Λ implies that A → Λ is injective, so the lemma really tells us that Λ is a directed union of these types of subrings A ⊂ Λ. Let n be the minimal integer such that mn = 0. We will prove this lemma by induction on n. The case n = 1 is clear as a field extension is a union of finitely generated field extensions.

Pick λ1, . . . , λd ∈ m which generate m. As K is formally smooth over Fp (see Algebra, Lemma 158.7) we can ﬁnd a ring map σ : K → Λ which is a section of the quotient map Λ → K. In general σ is not a k-algebra map. Given σ we deﬁne

Ψσ : K[x1, . . . , xd] −→ Λ

using σ on elements of K and mapping xi to λi. Claim: there exists a σ : K → Λ and a subfield k ⊂ F ⊂ K finitely generated over k such that the image of k in Λ is contained in Ψσ(F [x1, . . . , xd]). We will prove the claim by induction on the least integer n such that mn = 0. It is clear for n = 1. If n > 1 set I = mn−1 and Λ0 = Λ/I. By induction we may assume given σ0 : K → Λ0 and k ⊂ F 0 ⊂ K finitely generated such that the image of k → 0 0 0 0 0 Λ → Λ is contained in A = Ψσ0 (F [x1, . . . , xd]). Denote τ : k → A the induced map. Choose a lift σ : K → Λ of σ0 (this is possible by the formal smoothness of K/Fp we mentioned above). For later reference we note that we can change σ SMOOTHING RING MAPS 27

n to σ + D for some derivation D : K → I. Set A = F [x1, . . . , xd]/(x1, . . . , xd) . Then Ψσ induces a ring map Ψσ : A → Λ. The composition with the quotient map Λ → Λ0 induces a surjective map A → A0 with nilpotent kernel. Choose a lift 0 τ : k → A of τ (possible as k/Fp is formally smooth). Thus we obtain two maps k → Λ, namely Ψσ ◦ τ : k → Λ and the given map i : k → Λ. These maps agree modulo I, whence the diﬀerence is a derivation θ = i − Ψσ ◦ τ : k → I. Note that if we change σ into σ + D then we change θ into θ − D|k.

Choose a set of elements {yj}j∈J of k whose diﬀerentials dyj form a basis of Ωk/Fp . The Jacobi-Zariski sequence for Fp ⊂ k ⊂ K is

0 → H1(LK/k) → Ωk/Fp ⊗ K → ΩK/Fp → ΩK/k → 0

As dim H1(LK/k) < ∞ we can find a finite subset J0 ⊂ J such that the image of the first map is contained in L Kdy . Hence the elements dy , j ∈ J \ J map to j∈J0 j j 0

K-linearly independent elements of ΩK/Fp . Therefore we can choose a D : K → I such that θ − D|k = ξ ◦ d where ξ is a composition M M Ωk/Fp = kdyj −→ kdyj −→ I j∈J j∈J0

Let fj = ξ(dyj) ∈ I for j ∈ J0. Change σ into σ + D as above. Then we see that θ(a) = P a f for a ∈ k where da = P a dy in Ω . Note that I is j∈J0 j j j j k/Fp E e1 ed P generated by the monomials λ = λ1 . . . λd of total degree |E| = ei = n − 1 P E 0 0 in λ1, . . . , λd. Write fj = E cj,Eλ with cj,E ∈ K. Replace F by F = F (cj,E). Then the claim holds.

Choose σ and F as in the claim. The kernel of Ψσ is generated by finitely many polynomials g1, . . . , gt ∈ K[x1, . . . , xd] and we may assume their coefficients are in F after enlarging F by adjoining finitely many elements. In this case it is clear that the map A = F [x1, . . . , xd]/(g1, . . . , gt) → K[x1, . . . , xd]/(g1, . . . , gt) = Λ is flat. By the claim A is a k-subalgebra of Λ. It is clear that Λ is the filtered colimit of these algebras, as K is the filtered union of the subfields F . Finally, these algebras are essentially of finite type over k by Algebra, Lemma 54.4. 07FH Lemma 11.2. Let k be a field of characteristic p > 0. Let Λ be a Noetherian geometrically regular k-algebra. Let q ⊂ Λ be a prime ideal. Let n ≥ 1 be an n integer and let E ⊂ Λq/q Λq be a finite subset. Then we can find m ≥ 0 and ϕ : k[y1, . . . , ym] → Λ with the following properties −1 (1) setting p = ϕ (q) we have qΛq = pΛq and k[y1, . . . , ym]p → Λq is flat, (2) there is a factorization by homomorphisms of local Artinian rings n n k[y1, . . . , ym]p/p k[y1, . . . , ym]p → D → Λq/q Λq where the first arrow is essentially smooth and the second is flat, n (3) E is contained in D modulo q Λq. ¯ n Proof. Set Λ = Λq/q Λq. Note that dim H1(Lκ(q)/k) < ∞ by More on Algebra, Proposition 35.1. Pick A ⊂ Λ¯ containing E such that A is local Artinian, essentially ¯ of finite type over k, the map A → Λ is flat, and mA generates the maximal ideal of ¯ Λ, see Lemma 11.1. Denote F = A/mA the residue field so that k ⊂ F ⊂ K. Pick ¯ λ1, . . . , λt ∈ Λ which map to elements of A in Λ such that moreover the images 0 of dλ1,..., dλt form a basis of ΩF/k. Consider the map ϕ : k[y1, . . . , yt] → Λ 0 0 −1 sending yj to λj. Set p = (ϕ ) (q). By More on Algebra, Lemma 35.2 the SMOOTHING RING MAPS 28

0 ring map k[y1, . . . , yt]p0 → Λq is flat and Λq/p Λq is regular. Thus we can choose ¯ further elements λt+1, . . . , λm ∈ Λ which map into A ⊂ Λ and which map to 0 a regular system of parameters of Λq/p Λq. We obtain ϕ : k[y1, . . . , ym] → Λ n ¯ having property (1) such that k[y1, . . . , ym]p/p k[y1, . . . , ym]p → Λ factors through n A. Thus k[y1, . . . , ym]p/p k[y1, . . . , ym]p → A is flat by Algebra, Lemma 39.9. By construction the residue field extension κ(p) ⊂ F is finitely generated and ΩF/κ(p) = 0. Hence it is finite separable by More on Algebra, Lemma 34.1. Thus n k[y1, . . . , ym]p/p k[y1, . . . , ym]p → A is finite by Algebra, Lemma 54.4. Finally, we conclude that it is étale by Algebra, Lemma 143.7. Since an étale ring map is certainly essentially smooth we win.

07FI Lemma 11.3. Let ϕ : k[y1, . . . , ym] → Λ, n, q, p and n n k[y1, . . . , ym]p/p → D → Λq/q Λq be as in Lemma 11.2. Then for any λ ∈ Λ \ q there exists an integer q > 0 and a factorization n 0 n k[y1, . . . , ym]p/p → D → D → Λq/q Λq such that D → D0 is an essentially smooth map of local Artinian rings, the last arrow is flat, and λq is in D0. ¯ n ¯ ¯ Proof. Set Λ = Λq/q Λq. Let λ be the image of λ in Λ. Let α ∈ κ(q) be the image of λ in the residue field. Let k ⊂ F ⊂ κ(q) be the residue field of D. If α is in F then we can find an x ∈ D such that xλ¯ = 1 mod q. Hence (xλ¯)q = 1 mod (q)q if q is divisible by p. Hence λ¯q is in D. If α is transcendental over F , then we 0 ¯ ¯ can take D = (D[λ])m equal to the subring generated by D and λ localized at m = D[λ¯] ∩ qΛ¯. This works because D[λ¯] is in fact a polynomial algebra over D in this case. Finally, if λ mod q is algebraic over F , then we can find a p-power q such that αq is separable algebraic over F , see Fields, Section 28. Note that D and Λ¯ are henselian local rings, see Algebra, Lemma 153.10. Let D → D0 be a finite étale extension whose residue field extension is F ⊂ F (αq), see Algebra, Lemma 153.7. Since Λ¯ is henselian and F (αq) is contained in its residue field we can find a 0 factorization D0 → Λ¯. By the first part of the argument we see that λ¯qq ∈ D0 for 0 some q > 0. 07FJ Lemma 11.4. Let k → A → Λ ⊃ q be as in Situation 9.1 where (1) k is a field of characteristic p > 0, (2) Λ is Noetherian and geometrically regular over k, (3) q is minimal over hA. Then k → A → Λ ⊃ q can be resolved. Proof. The lemma is proven by the following steps in the given order. We will justify each of these steps below. N (1)07FKPick an integer N > 0 such that q Λq ⊂ HA/kΛq. (2)07FLPick generators a1, . . . , at ∈ A of the ideal HA/R. (3)07FMSet d = dim(Λq). 2N P (4)07FNSet B = A[x1, . . . , xd, zij]/(xi − zijaj). (5)07FPConsider B as a k[x1, . . . , xd]-algebra and let B → C be as in Lemma 3.1. We also obtain a section C → B. c (6)07FQ Choose c > 0 such that each xi is strictly standard in C over k[x1, . . . , xd]. SMOOTHING RING MAPS 29

(7)07FRSet n = N + dc and e = 8c. n (8)07FSLet E ⊂ Λq/q Λq be the images of generators of A as a k-algebra. (9)07FTChoose an integer m and a k-algebra map ϕ : k[y1, . . . , ym] → Λ and a factorization by local Artinian rings n n k[y1, . . . , ym]p/p k[y1, . . . , ym]p → D → Λq/q Λq such that the first arrow is essentially smooth, the second is flat, E is −1 contained in D, with p = ϕ (q) the map k[y1, . . . , ym]p → Λq is flat, and pΛq = qΛq. (10)07FUChoose π1, . . . , πd ∈ p which map to a regular system of parameters of k[y1, . . . , ym]p. (11)07FVLet R = k[y1, . . . , ym, t1, . . . , tm] and γi = πiti. (12)07FWIf necessary modify the choice of πi such that for i = 1, . . . , d we have 2 Ann e e (γ ) = Ann e e (γ ) R/(γ1 ,...,γi−1)R i R/(γ1 ,...,γi−1)R i 0 n (13)07FXThere exist δ1, . . . , δd ∈ Λ, δi 6∈ q and a factorization D → D → Λq/q Λq 0 0 0 n with D local Artinian, D → D essentially smooth, the map D → Λq/q Λq 0 flat such that, with πi = δiπi, we have for i = 1, . . . , d 0 2N P n 0 (a) (πi) = ajλij in Λ where λij mod q Λq is an element of D , 0 02 (b) Ann 0e 0e (π ) = Ann 0e 0e (π ), Λ/(π 1,...,π i−1) i Λ/(π 1,...,π i−1) i n 0 (c) δi mod q Λq is an element of D . 0 (14)07FYDefine B → Λ by sending xi to πi and zij to λij found above. Define C → Λ by composing the map B → Λ with the retraction C → B. (15)07FZMap R → Λ by ϕ on k[y1, . . . , ym] and by sending ti to δi. Further introduce a map

k[x1, . . . , xd] −→ R = k[y1, . . . , ym, t1, . . . , td]

by sending xi to γi = πiti. (16)07G0It suﬃces to resolve

R → C ⊗k[x1,...,xd] R → Λ ⊃ q e e (17)07G1Set I = (γ1, . . . , γd) ⊂ R. (18)07G2It suﬃces to resolve

R/I → C ⊗k[x1,...,xd] R/I → Λ/IΛ ⊃ q/IΛ

(19)07G3 We denote r ⊂ R = k[y1, . . . , ym, t1, . . . , td] the inverse image of q. (20)07G4It suﬃces to resolve

(R/I)r → C ⊗k[x1,...,xd] (R/I)r → Λq/IΛq ⊃ qΛq/IΛq e e (21)07G5Set J = (π1, . . . , πd) in k[y1, . . . , ym]. (22)07G6It suﬃces to resolve

(R/JR)p → C ⊗k[x1,...,xd] (R/JR)p → Λq/JΛq ⊃ qΛq/JΛq (23)07G7 It suﬃces to resolve n n n n (R/p R)p → C ⊗k[x1,...,xd] (R/p R)p → Λq/q Λq ⊃ qΛq/q Λq (24)07G8It suﬃces to resolve n n n n (R/p R)p → B ⊗k[x1,...,xd] (R/p R)p → Λq/q Λq ⊃ qΛq/q Λq SMOOTHING RING MAPS 30

0 n (25)07G9The ring D [t1, . . . , td] is given the structure of an Rp/p Rp-algebra by the n 0 given map k[y1, . . . , ym]p/p k[y1, . . . , ym]p → D and by sending ti to ti. It suﬃces to ﬁnd a factorization n 0 n B ⊗k[x1,...,xd] (R/p R)p → D [t1, . . . , td] → Λq/q Λq

where the second arrow sends ti to δi and induces the given homomorphism 0 n D → Λq/q Λq. (26)07GASuch a factorization exists by our choice of D0 above. We now give the justiﬁcation for each of the steps, except that we skip justifying the steps which just introduce notation. p Ad (1). This is possible as q is minimal over hA = HA/kΛ.

Ad (6). Note that Aai is smooth over k. Hence Baj , which is isomorphic to a polynomial algebra over Aaj [x1, . . . , xd], is smooth over k[x1, . . . , xd]. Thus Bxi is smooth over k[x1, . . . , xd]. By Lemma 3.1 we see that Cxi is smooth over k[x1, . . . , xd] with ﬁnite free module of diﬀerentials. Hence some power of xi is strictly standard in C over k[x1, . . . , xn] by Lemma 3.7. Ad (9). This follows by applying Lemma 11.2.

Ad (10). Since k[y1, . . . , ym]p → Λq is ﬂat and pΛq = qΛq by construction we see that dim(k[y1, . . . , ym]p) = d by Algebra, Lemma 112.7. Thus we can ﬁnd π1, . . . , πd ∈ Λ which map to a regular system of parameters in Λq.

Ad (12). By Algebra, Lemma 106.3 any permutation of the sequence π1, . . . , πd is a regular sequence in k[y1, . . . , ym]p. Hence γ1 = π1t1, . . . , γd = πdtd is a regular sequence in Rp = k[y1, . . . , ym]p[t1, . . . , td], see Algebra, Lemma 68.10. Let S = −1 k[y1, . . . , ym] \ p so that Rp = S R. Note that π1, . . . , πd and γ1, . . . , γd remain regular sequences if we multiply our πi by elements of S. Suppose that 2 Ann e e (γ ) = Ann e e (γ ) R/(γ1 ,...,γi−1)R i R/(γ1 ,...,γi−1)R i e e holds for i = 1, . . . , t for some t ∈ {0, . . . , d}. Note that γ1, . . . , γt , γt+1 is a regular sequence in S−1R by Algebra, Lemma 68.9. Hence we see that 2 Ann −1 e e (γi) = Ann −1 e e (γ ). S R/(γ1 ,...,γi−1) S R/(γ1 ,...,γi−1) i Thus we get 2 Ann e e (γ ) = Ann e e (γ ) R/(γ1 ,...,γt )R t+1 R/(γ1 ,...,γt )R t+1 after replacing πt+1 by sπt+1 for some s ∈ S by Lemma 10.1. By induction on t this produces the desired sequence. −1 ¯ n Ad (13). Let S = Λ \ q so that Λq = S Λ. Set Λ = Λq/q Λq. Suppose that 0 ¯ we have a t ∈ {0, . . . , d} and δ1, . . . , δt ∈ S and a factorization D → D → Λ as N in (13) such that (a), (b), (c) hold for i = 1, . . . , t. We have πt+1 ∈ HA/kΛq as N N ¯ N 0 0 ¯ q Λq ⊂ HA/kΛq by (1). Hence πt+1 ∈ HA/kΛ. Hence πt+1 ∈ HA/kD as D → Λ is faithfully ﬂat, see Algebra, Lemma 82.11. Recall that HA/k = (a1, . . . , at). Say N P 0 0 N P πt+1 = ajdj in D and choose cj ∈ Λq lifting dj ∈ D . Then πt+1 = cjaj + n n−N P 0 0 n−N with ∈ q Λq ⊂ q HA/kΛq. Write = ajcj for some cj ∈ q Λq. Hence 2N P N N 0 N 0 ¯ πt+1 = (πt+1cj + πt+1cj)aj. Note that πt+1cj maps to zero in Λ; this trivial but key observation will ensure later that (a) holds. Now we choose s ∈ S such that N N 0 2N there exist µt+1j ∈ Λ such that on the one hand πt+1cj + πt+1cj = µt+1j/s in −1 2N P S Λ and on the other (sπt+1) = µt+1jaj in Λ (minor detail omitted). We SMOOTHING RING MAPS 31 may further replace s by a power and enlarge D0 such that s maps to an element of 0 2N 0 D . With these choices µt+1j maps to s dj which is an element of D . Note that −1 π1, . . . , πd are a regular sequence of parameters in S Λ by our choice of ϕ. Hence π1, . . . , πd forms a regular sequence in Λq by Algebra, Lemma 106.3. It follows that 0e 0e −1 π 1, . . . , π t , sπt+1 is a regular sequence in S Λ by Algebra, Lemma 68.9. Thus we get 2 Ann −1 0e 0e (sπ ) = Ann −1 0e 0e ((sπ ) ). S Λ/(π 1,...,π t ) t+1 S Λ/(π 1,...,π t ) t+1 Hence we may apply Lemma 10.1 to ﬁnd an s0 ∈ S such that 0 q 0 q 2 Ann 0e 0e ((s ) sπ ) = Ann 0e 0e (((s ) sπ ) ). Λ/(π 1,...,π t ) t+1 Λ/(π 1,...,π t ) t+1 for any q > 0. By Lemma 11.3 we can choose q and enlarge D0 such that (s0)q maps 0 0 q to an element of D . Setting δt+1 = (s ) s and we conclude that (a), (b), (c) hold 0 2Nq for i = 1, . . . , t + 1. For (a) note that λt+1j = (s ) µt+1j works. By induction on t we win.

Ad (16). By construction the radical of H(C⊗ R)/RΛ contains hA. Namely, k[x1,...,xd] the elements aj ∈ HA/k map to elements of HB/k[x1,...,xn], hence map to elements of HC/k[x ,...,x ], hence aj ⊗ 1 map to elements of HC⊗ R/R. Moreover, if 1 n k[x1,...,xd] we have a solution C ⊗k[x1,...,xn] R → T → Λ of

R → C ⊗k[x1,...,xd] R → Λ ⊃ q then HT/R ⊂ HT/k as R is smooth over k. Hence T will also be a solution for the original situation k → A → Λ ⊃ q.

Ad (18). Follows on applying Lemma 9.2 to R → C ⊗k[x1,...,xd] R → Λ ⊃ q and the c c c sequence of elements γ1, . . . , γd. We note that since xi are strictly standard in C c over k[x1, . . . , xd] the elements γi are strictly standard in C ⊗k[x1,...,xd] R over R by Lemma 2.7. The other assumption of Lemma 9.2 holds by steps (12) and (13). Ad (20). Apply Lemma 9.4 to the situation in (18). In the rest of the arguments the target ring is local Artinian, hence we are looking for a factorization by a smooth algebra T over the source ring.

Ad (22). Suppose that C ⊗k[x1,...,xd] (R/JR)p → T → Λq/JΛq is a solution to

(R/JR)p → C ⊗k[x1,...,xd] (R/JR)p → Λq/JΛq ⊃ qΛq/JΛq

Then C ⊗k[x1,...,xd] (R/I)r → Tr → Λq/IΛq is a solution to the situation in (20). n Ad (23). Our n = N + dc is large enough so that p k[y1, . . . , ym]p ⊂ Jp and n n n q Λq ⊂ JΛq. Hence if we have a solution C ⊗k[x1,...,xd] (R/p R)p → T → Λq/q Λq of (22 then we can take T/JT as the solution for (23). Ad (24). This is true because we have a section C → B in the category of R- algebras. Ad (25). This is true because D0 is essentially smooth over the local Artinian ring n k[y1, . . . , ym]p/p k[y1, . . . , ym]p and n n Rp/p Rp = k[y1, . . . , ym]p/p k[y1, . . . , ym]p[t1, . . . , td]. 0 n Hence D [t1, . . . , td] is a ﬁltered colimit of smooth Rp/p Rp-algebras and B⊗k[x1,...,xd] n (Rp/p Rp) factors through one of these. SMOOTHING RING MAPS 32

Ad (26). The ﬁnal twist of the proof is that we cannot just use the map B → D0 0 0 0 which maps xi to the image of πi in D and zij to the image of λij in D because we need the diagram

0 B / D [t1, . . . , td] O O

n k[x1, . . . , xd] / Rp/p Rp

0 n to commute and we need the composition B → D [t1, . . . , td] → Λq/q Λq to be the 0 map of (14). This requires us to map xi to the image of πiti in D [t1, . . . , td]. Hence 2N 2N 0 we map zij to the image of λijti /δi in D [t1, . . . , td] and everything is clear.

12. The main theorem 07GB In this section we wrap up the discussion. 07GC Theorem 12.1 (Popescu). Any regular homomorphism of Noetherian rings is a filtered colimit of smooth ring maps. Proof. By Lemma 8.4 it suffices to prove this for k → Λ where Λ is Noetherian and geometrically regular over k. Let k → A → Λ be a factorization with A a finite type k-algebra. It suffices to construct a factorization A → B → Λ with B of finite type such that hB = Λ, see Lemma 2.8. Hence we may perform Noetherian induction on the ideal hA. Pick a prime q ⊃ hA such that q is minimal over hA. It now suffices to resolve k → A → Λ ⊃ q (as defined in the text following Situation 9.1). If the characteristic of k is zero, this follows from Lemma 10.3. If the characteristic of k is p > 0, this follows from Lemma 11.4.

13. The approximation property for G-rings 07QX Let R be a Noetherian local ring. In this case R is a G-ring if and only if the ring map R → R∧ is regular, see More on Algebra, Lemma 50.7. In this case it is true that the henselization Rh and the strict henselization Rsh of R are G-rings, see More on Algebra, Lemma 50.8. Moreover, any algebra essentially of ﬁnite type over a ﬁeld, over a complete local ring, over Z, or over a characteristic zero Dedekind ring is a G-ring, see More on Algebra, Proposition 50.12. This gives an ample supply of rings to which the result below applies.

Let R be a ring. Let f1, . . . , fm ∈ R[x1, . . . , xn]. Let S be an R-algebra. In this n situation we say a vector (a1, . . . , an) ∈ S is a solution in S if and only if

fj(a1, . . . , an) = 0 in S, for j = 1, . . . , m Of course an important question in algebraic geometry is to see when systems of polynomial equations have solutions. The following theorem tells us that having solutions in the completion of a local Noetherian ring is often enough to show there exist solutions in the henselization of the ring.

07QY Theorem 13.1. Let R be a Noetherian local ring. Let f1, . . . , fm ∈ R[x1, . . . , xn]. ∧ n ∧ Suppose that (a1, . . . , an) ∈ (R ) is a solution in R . If R is a henselian G-ring, n then for every integer N there exists a solution (b1, . . . , bn) ∈ R in R such that N ∧ ai − bi ∈ m R . SMOOTHING RING MAPS 33

N Proof. Let ci ∈ R be an element such that ai − ci ∈ m . Choose generators N P m = (d1, . . . , dM ). Write ai = ci + ai,ldl. Consider the polynomial ring R[xi,l] and the elements X X gj = fj(c1 + x1,ldl, . . . , cn + xn,ldn,l) ∈ R[xi,l]

The system of equations gj = 0 has the solution (ai,l). Suppose that we can show P that gj as a solution (bi,l) in R. Then it follows that bi = ci + bi,ldl is a solution N of fj = 0 which is congruent to ai modulo m . Thus it suffices to show that solvability over R∧ implies solvability over R. ∧ Let A ⊂ R be the R-subalgebra generated by a1, . . . , an. Since we’ve assumed R is a G-ring, i.e., that R → R∧ is regular, we see that there exists a factorization A → B → R∧ with B smooth over R, see Theorem 12.1. Denote κ = R/m the residue field. It is also the residue field of R∧, so we get a commutative diagram B / R0 O

R / κ Since the vertical arrow is smooth, More on Algebra, Lemma 9.14 implies that there exists an étale ring map R → R0 which induces an isomorphism R/m → R0/mR0 and an R-algebra map B → R0 making the diagram above commute. Since R is 0 henselian we see that R → R has a section, see Algebra, Lemma 153.3. Let bi ∈ R 0 be the image of ai under the ring maps A → B → R → R. Since all of these maps are R-algebra maps, we see that (b1, . . . , bn) is a solution in R. Given a Noetherian local ring (R, m), an étale ring map R → R0, and a maximal ideal m0 ⊂ R0 lying over m with κ(m) = κ(m0), then we have inclusions h ∧ R ⊂ Rm0 ⊂ R ⊂ R , by Algebra, Lemma 155.5 and More on Algebra, Lemma 45.3.

07QZ Theorem 13.2. Let R be a Noetherian local ring. Let f1, . . . , fm ∈ R[x1, . . . , xn]. ∧ n Suppose that (a1, . . . , an) ∈ (R ) is a solution. If R is a G-ring, then for every integer N there exist (1) an étale ring map R → R0, (2) a maximal ideal m0 ⊂ R0 lying over m 0 n 0 (3) a solution (b1, . . . , bn) ∈ (R ) in R 0 0 N ∧ such that κ(m) = κ(m ) and ai − bi ∈ (m ) R . Proof. We could deduce this theorem from Theorem 13.1 using that the henselization Rh is a G-ring by More on Algebra, Lemma 50.8 and writing Rh as a directed colimit of étale extension R0. Instead we prove this by redoing the proof of the previous theorem in this case. N N Let ci ∈ R be an element such that ai − ci ∈ m . Choose generators m = P (d1, . . . , dM ). Write ai = ci + ai,ldl. Consider the polynomial ring R[xi,l] and the elements X X gj = fj(c1 + x1,ldl, . . . , cn + xn,ldn,l) ∈ R[xi,l] SMOOTHING RING MAPS 34

The system of equations gj = 0 has the solution (ai,l). Suppose that we can show 0 0 that gj as a solution (bi,l) in R for some étale ring map R → R endowed with a 0 0 P maximal ideal m such that κ(m) = κ(m ). Then it follows that bi = ci + bi,ldl 0 N is a solution of fj = 0 which is congruent to ai modulo (m ) . Thus it suffices to show that solvability over R∧ implies solvability over some étale ring extension which induces a trivial residue field extension at some prime over m. ∧ Let A ⊂ R be the R-subalgebra generated by a1, . . . , an. Since we’ve assumed R is a G-ring, i.e., that R → R∧ is regular, we see that there exists a factorization A → B → R∧ with B smooth over R, see Theorem 12.1. Denote κ = R/m the residue field. It is also the residue field of R∧, so we get a commutative diagram B / R0 O

R / κ Since the vertical arrow is smooth, More on Algebra, Lemma 9.14 implies that there exists an étale ring map R → R0 which induces an isomorphism R/m → R0/mR0 0 0 and an R-algebra map B → R making the diagram above commute. Let bi ∈ R 0 be the image of ai under the ring maps A → B → R . Since all of these maps are 0 R-algebra maps, we see that (b1, . . . , bn) is a solution in R . 0A1WExample 13.3. Let (R, m) be a Noetherian local ring with henselization Rh. The map on completions R∧ → (Rh)∧ is an isomorphism, see More on Algebra, Lemma 45.3. Since also Rh is Noetherian (ibid.) we may think of Rh as a subring of its completion (because the completion is faithfully flat). In this way we see that we may identify Rh with a subring of R∧. Let us try to understand which elements of R∧ are in Rh. For simplicity we assume R is a domain with fraction field K. Clearly, every element f of Rh is algebraic over n R, in the sense that there exists an equation of the form anf + ... + a1f + a0 = 0 for some ai ∈ R with n > 0 and an 6= 0. ∧ Conversely, assume that f ∈ R , n ∈ N, and a0, . . . , an ∈ R with an 6= 0 such n that anf + ... + a1f + a0 = 0. If R is a G-ring, then, for every N > 0 there h n N ∧ exists an element g ∈ R with ang + ... + a1g + a0 = 0 and f − g ∈ m R , see Theorem 13.2. We’d like to conclude that f = g when N 0. If this is not true, then we find infinitely many roots g of P (T ) in Rh. This is impossible because h h h (1) R ⊂ R ⊗R K and (2) R ⊗R K is a finite product of field extensions of K. h h h Namely, R → K is injective and R → R is flat, hence R → R ⊗R K is injective and (2) follows from More on Algebra, Lemma 45.13. Conclusion: If R is a Noetherian local domain with fraction field K and a G-ring, then Rh ⊂ R∧ is the set of all elements which are algebraic over K. Here is another variant of the main theorem of this section. 0CAR Lemma 13.4. Let R be a Noetherian ring. Let p ⊂ R be a prime ideal. Let ∧ n f1, . . . , fm ∈ R[x1, . . . , xn]. Suppose that (a1, . . . , an) ∈ ((Rp) ) is a solution. If Rp is a G-ring, then for every integer N there exist (1) an étale ring map R → R0, SMOOTHING RING MAPS 35

(2) a prime ideal p0 ⊂ R0 lying over p 0 n 0 (3) a solution (b1, . . . , bn) ∈ (R ) in R 0 0 N 0 ∧ such that κ(p) = κ(p ) and ai − bi ∈ (p ) (Rp0 ) . 0 0 00 Proof. By Theorem 13.2 we can find a solution (b1, . . . , bn) in some ring R étale 00 00 over Rp which comes with a prime ideal p lying over p such that κ(p) = κ(p ) 0 00 N 00 ∧ 00 0 and ai − bi ∈ (p ) (Rp00 ) . We can write R = R ⊗R Rp for some étale R-algebra R0 (see Algebra, Lemma 143.3). After replacing R0 by a principal localization 0 0 0 if necessary we may assume (b1, . . . , bn) come from a solution (b1, . . . , bn) in R . 0 0 00 00 0 Setting p = R ∩ p we see that Rp00 = Rp0 which finishes the proof. 14. Approximation for henselian pairs 0AH4 We can generalize the discussion of Section 13 to the case of henselian pairs. Henselian pairs where defined in More on Algebra, Section 11. 0AH5 Lemma 14.1. Let (A, I) be a henselian pair with A Noetherian. Let A∧ be the I-adic completion of A. Assume at least one of the following conditions holds (1) A → A∧ is a regular ring map, (2) A is a Noetherian G-ring, or (3) (A, I) is the henselization (More on Algebra, Lemma 12.1) of a pair (B,J) where B is a Noetherian G-ring. ∧ Given f1, . . . , fm ∈ A[x1, . . . , xn] and aˆ1,..., aˆn ∈ A such that fj(â1,..., aˆn) = 0 N for j = 1, . . . , m, for every N ≥ 1 there exist a1, . . . , an ∈ A such that aî − ai ∈ I and such that fj(a1, . . . , an) = 0 for j = 1, . . . , m. Proof. By More on Algebra, Lemma 50.15 we see that (3) implies (2). By More on Algebra, Lemma 50.14 we see that (2) implies (1). Thus it suffices to prove the lemma in case A → A∧ is a regular ring map.

Let aˆ1,..., aˆn be as in the statement of the lemma. By Theorem 12.1 we can ∧ ﬁnd a factorization A → B → A with A → P smooth and b1, . . . , bn ∈ B with ∧ N fj(b1, . . . , bn) = 0 in B. Denote σ : B → A → A/I the composition. By More on Algebra, Lemma 9.14 we can ﬁnd an étale ring map A → A0 which induces an isomorphism A/IN → A0/IN A0 and an A-algebra map σ˜ : B → A0 lifting σ. Since (A, I) is henselian, there is an A-algebra map χ : A0 → A, see More on Algebra, Lemma 11.6. Then setting ai = χ(˜σ(bi)) gives a solution. 15. Other chapters

Preliminaries (11) Brauer Groups (1) Introduction (12) Homological Algebra (2) Conventions (13) Derived Categories (3) Set Theory (14) Simplicial Methods (4) Categories (15) More on Algebra (5) Topology (16) Smoothing Ring Maps (6) Sheaves on Spaces (17) Sheaves of Modules (7) Sites and Sheaves (18) Modules on Sites (8) Stacks (19) Injectives (9) Fields (20) Cohomology of Sheaves (10) Commutative Algebra (21) Cohomology on Sites SMOOTHING RING MAPS 36

(22) Differential Graded Algebra (66) Decent Algebraic Spaces (23) Divided Power Algebra (67) Cohomology of Algebraic Spaces (24) Differential Graded Sheaves (68) Limits of Algebraic Spaces (25) Hypercoverings (69) Divisors on Algebraic Spaces Schemes (70) Algebraic Spaces over Fields (71) Topologies on Algebraic Spaces (26) Schemes (72) Descent and Algebraic Spaces (27) Constructions of Schemes (73) Derived Categories of Spaces (28) Properties of Schemes (74) More on Morphisms of Spaces (29) Morphisms of Schemes (75) Flatness on Algebraic Spaces (30) Cohomology of Schemes (76) Groupoids in Algebraic Spaces (31) Divisors (77) More on Groupoids in Spaces (32) Limits of Schemes (78) Bootstrap (33) Varieties (79) Pushouts of Algebraic Spaces (34) Topologies on Schemes (35) Descent Topics in Geometry (36) Derived Categories of Schemes (80) Chow Groups of Spaces (37) More on Morphisms (81) Quotients of Groupoids (38) More on Flatness (82) More on Cohomology of Spaces (39) Groupoid Schemes (83) Simplicial Spaces (40) More on Groupoid Schemes (84) Duality for Spaces (41) Étale Morphisms of Schemes (85) Formal Algebraic Spaces (86) Algebraization of Formal Spaces Topics in Scheme Theory (87) Resolution of Surfaces Revisited (42) Chow Homology Deformation Theory (43) Intersection Theory (44) Picard Schemes of Curves (88) Formal Deformation Theory (45) Weil Cohomology Theories (89) Deformation Theory (46) Adequate Modules (90) The Cotangent Complex (47) Dualizing Complexes (91) Deformation Problems (48) Duality for Schemes Algebraic Stacks (49) Discriminants and Differents (92) Algebraic Stacks (50) de Rham Cohomology (93) Examples of Stacks (51) Local Cohomology (94) Sheaves on Algebraic Stacks (52) Algebraic and Formal Geometry (95) Criteria for Representability (53) Algebraic Curves (96) Artin’s Axioms (54) Resolution of Surfaces (97) Quot and Hilbert Spaces (55) Semistable Reduction (98) Properties of Algebraic Stacks (56) Derived Categories of Varieties (99) Morphisms of Algebraic Stacks (57) Fundamental Groups of Schemes (100) Limits of Algebraic Stacks (58) Étale Cohomology (101) Cohomology of Algebraic Stacks (59) Crystalline Cohomology (102) Derived Categories of Stacks (60) Pro-étale Cohomology (103) Introducing Algebraic Stacks (61) More Étale Cohomology (104) More on Morphisms of Stacks (62) The Trace Formula (105) The Geometry of Stacks Algebraic Spaces Topics in Moduli Theory (63) Algebraic Spaces (106) Moduli Stacks (64) Properties of Algebraic Spaces (107) Moduli of Curves (65) Morphisms of Algebraic Spaces Miscellany SMOOTHING RING MAPS 37

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