Divided Power Algebra

Introduction

In this chapter we talk about divided power algebras and what you can do with them. A reference is the book [Berthelot].

Divided powers

In this section we collect some results on divided power rings. We will use the convention \(0! = 1\) (as empty products should give \(1\)).

Definition

Let \(A\) be a ring. Let \(I\) be an ideal of \(A\). A collection of maps \(\gamma_n : I \to I\), \(n > 0\) is called a divided power structure on \(I\) if for all \(n \geq 0\), \(m > 0\), \(x, y \in I\), and \(a \in A\) we have

1. \(\gamma_1(x) = x\), we also set \(\gamma_0(x) = 1\),

2. \(\gamma_n(x)\gamma_m(x) = \frac{(n + m)!}{n! m!} \gamma_{n + m}(x)\),

3. \(\gamma_n(ax) = a^n \gamma_n(x)\),

4. \(\gamma_n(x + y) = \sum_{i = 0, \ldots, n} \gamma_i(x)\gamma_{n - i}(y)\),

5. \(\gamma_n(\gamma_m(x)) = \frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x)\).

Note that the rational numbers \(\frac{(n + m)!}{n! m!}\) and \(\frac{(nm)!}{n! (m!)^n}\) occurring in the definition are in fact integers; the first is the number of ways to choose \(n\) out of \(n + m\) and the second counts the number of ways to divide a group of \(nm\) objects into \(n\) groups of \(m\). We make some remarks about the definition which show that \(\gamma_n(x)\) is a replacement for \(x^n/n!\) in \(I\).

Lemma

Let \(A\) be a ring. Let \(I\) be an ideal of \(A\).

1. If \(\gamma\) is a divided power structure¹ on \(I\), then \(n! \gamma_n(x) = x^n\) for \(n \geq 1\), \(x \in I\).

Assume \(A\) is torsion free as a \(\mathbf{Z}\)-module.

1. A divided power structure on \(I\), if it exists, is unique.

2. If \(\gamma_n : I \to I\) are maps then \[\gamma\text{ is a divided power structure} \Leftrightarrow n! \gamma_n(x) = x^n\ \forall x \in I, n \geq 1.\]

3. The ideal \(I\) has a divided power structure if and only if there exists a set of generators \(x_i\) of \(I\) as an ideal such that for all \(n \geq 1\) we have \(x_i^n \in (n!)I\).

Proof

Proof of (1). If \(\gamma\) is a divided power structure, then condition (2) (applied to \(1\) and \(n-1\) instead of \(n\) and \(m\)) implies that \(n \gamma_n(x) = \gamma_1(x)\gamma_{n - 1}(x)\). Hence by induction and condition (1) we get \(n! \gamma_n(x) = x^n\).

Assume \(A\) is torsion free as a \(\mathbf{Z}\)-module. Proof of (2). This is clear from (1).

Proof of (3). Assume that \(n! \gamma_n(x) = x^n\) for all \(x \in I\) and \(n \geq 1\). Since \(A \subset A \otimes_{\mathbf{Z}} \mathbf{Q}\) it suffices to prove the axioms (1) – (5) of Definition 07GL in case \(A\) is a \(\mathbf{Q}\)-algebra. In this case \(\gamma_n(x) = x^n/n!\) and it is straightforward to verify (1) – (5); for example, (4) corresponds to the binomial formula \[(x + y)^n = \sum_{i = 0, \ldots, n} \frac{n!}{i!(n - i)!} x^iy^{n - i}\] We encourage the reader to do the verifications to make sure that we have the coefficients correct.

Proof of (4). Assume we have generators \(x_i\) of \(I\) as an ideal such that \(x_i^n \in (n!)I\) for all \(n \geq 1\). We claim that for all \(x \in I\) we have \(x^n \in (n!)I\). If the claim holds then we can set \(\gamma_n(x) = x^n/n!\) which is a divided power structure by (3). To prove the claim we note that it holds for \(x = ax_i\). Hence we see that the claim holds for a set of generators of \(I\) as an abelian group. By induction on the length of an expression in terms of these, it suffices to prove the claim for \(x + y\) if it holds for \(x\) and \(y\). This follows immediately from the binomial theorem.

Example

Let \(p\) be a prime number. Let \(A\) be a ring such that every integer \(n\) not divisible by \(p\) is invertible, i.e., \(A\) is a \(\mathbf{Z}_{(p)}\)-algebra. Then \(I = pA\) has a canonical divided power structure. Namely, given \(x = pa \in I\) we set \[\gamma_n(x) = \frac{p^n}{n!} a^n\] The reader verifies immediately that \(p^n/n! \in p\mathbf{Z}_{(p)}\) for \(n \geq 1\) (for instance, this can be derived from the fact that the exponent of \(p\) in the prime factorization of \(n!\) is \(\left\lfloor n/p \right\rfloor + \left\lfloor n/p^2 \right\rfloor + \left\lfloor n/p^3 \right\rfloor + \ldots\)), so that the definition makes sense and gives us a sequence of maps \(\gamma_n : I \to I\). It is a straightforward exercise to verify that conditions (1) – (5) of Definition 07GL are satisfied. Alternatively, it is clear that the definition works for \(A_0 = \mathbf{Z}_{(p)}\) and then the result follows from Lemma 07H1.

We notice that \(\gamma_n\left(0\right) = 0\) for any ideal \(I\) of \(A\) and any divided power structure \(\gamma\) on \(I\). (This follows from axiom (3) in Definition 07GL, applied to \(a=0\).)

Lemma

Let \(A\) be a ring. Let \(I\) be an ideal of \(A\). Let \(\gamma_n : I \to I\), \(n \geq 1\) be a sequence of maps. Assume

1. (1), (3), and (4) of Definition 07GL hold for all \(x, y \in I\), and

2. properties (2) and (5) hold for \(x\) in some set of generators of \(I\) as an ideal.

Then \(\gamma\) is a divided power structure on \(I\).

Proof

The numbers (1), (2), (3), (4), (5) in this proof refer to the conditions listed in Definition 07GL. Applying (3) we see that if (2) and (5) hold for \(x\) then (2) and (5) hold for \(ax\) for all \(a \in A\). Hence we see (b) implies (2) and (5) hold for a set of generators of \(I\) as an abelian group. Hence, by induction of the length of an expression in terms of these it suffices to prove that, given \(x, y \in I\) such that (2) and (5) hold for \(x\) and \(y\), then (2) and (5) hold for \(x + y\).

Proof of (2) for \(x + y\). By (4) we have \[\gamma_n(x + y)\gamma_m(x + y) = \sum\nolimits_{i + j = n,\ k + l = m} \gamma_i(x)\gamma_k(x)\gamma_j(y)\gamma_l(y)\] Using (2) for \(x\) and \(y\) this equals \[\sum \frac{(i + k)!}{i!k!}\frac{(j + l)!}{j!l!} \gamma_{i + k}(x)\gamma_{j + l}(y)\] Comparing this with the expansion \[\gamma_{n + m}(x + y) = \sum \gamma_a(x)\gamma_b(y)\] we see that we have to prove that given \(a + b = n + m\) we have \[\sum\nolimits_{i + k = a,\ j + l = b,\ i + j = n,\ k + l = m} \frac{(i + k)!}{i!k!}\frac{(j + l)!}{j!l!} = \frac{(n + m)!}{n!m!}.\] Instead of arguing this directly, we note that the result is true for the ideal \(I = (x, y)\) in the polynomial ring \(\mathbf{Q}[x, y]\) because \(\gamma_n(f) = f^n/n!\), \(f \in I\) defines a divided power structure on \(I\). Hence the equality of rational numbers above is true.

Proof of (5) for \(x + y\) given that (1) – (4) hold and that (5) holds for \(x\) and \(y\). We will again reduce the proof to an equality of rational numbers. Namely, using (4) we can write \(\gamma_n(\gamma_m(x + y)) = \gamma_n(\sum \gamma_i(x)\gamma_j(y))\). Using (4) we can write \(\gamma_n(\gamma_m(x + y))\) as a sum of terms which are products of factors of the form \(\gamma_k(\gamma_i(x)\gamma_j(y))\). If \(i > 0\) then \[\begin{align*} \gamma_k(\gamma_i(x)\gamma_j(y)) & = \gamma_j(y)^k\gamma_k(\gamma_i(x)) \\ & = \frac{(ki)!}{k!(i!)^k} \gamma_j(y)^k \gamma_{ki}(x) \\ & = \frac{(ki)!}{k!(i!)^k} \frac{(kj)!}{(j!)^k} \gamma_{ki}(x) \gamma_{kj}(y) \end{align*}\] using (3) in the first equality, (5) for \(x\) in the second, and (2) exactly \(k\) times in the third. Using (5) for \(y\) we see the same equality holds when \(i = 0\). Continuing like this using all axioms but (5) we see that we can write \[\gamma_n(\gamma_m(x + y)) = \sum\nolimits_{i + j = nm} c_{ij}\gamma_i(x)\gamma_j(y)\] for certain universal constants \(c_{ij} \in \mathbf{Z}\). Again the fact that the equality is valid in the polynomial ring \(\mathbf{Q}[x, y]\) implies that the coefficients \(c_{ij}\) are all equal to \((nm)!/n!(m!)^n\) as desired.

Lemma

Let \(A\) be a ring with two ideals \(I, J \subset A\). Let \(\gamma\) be a divided power structure on \(I\) and let \(\delta\) be a divided power structure on \(J\). Then

1. \(\gamma\) and \(\delta\) agree on \(IJ\),

2. if \(\gamma\) and \(\delta\) agree on \(I \cap J\) then they are the restriction of a unique divided power structure \(\epsilon\) on \(I + J\).

Proof

Let \(x \in I\) and \(y \in J\). Then \[\gamma_n(xy) = y^n\gamma_n(x) = n! \delta_n(y) \gamma_n(x) = \delta_n(y) x^n = \delta_n(xy).\] Hence \(\gamma\) and \(\delta\) agree on a set of (additive) generators of \(IJ\). By property (4) of Definition 07GL it follows that they agree on all of \(IJ\).

Assume \(\gamma\) and \(\delta\) agree on \(I \cap J\). Let \(z \in I + J\). Write \(z = x + y\) with \(x \in I\) and \(y \in J\). Then we set \[\epsilon_n(z) = \sum \gamma_i(x)\delta_{n - i}(y)\] for all \(n \geq 1\). To see that this is well defined, suppose that \(z = x' + y'\) is another representation with \(x' \in I\) and \(y' \in J\). Then \(w = x - x' = y' - y \in I \cap J\). Hence \[\begin{align*} \sum\nolimits_{i + j = n} \gamma_i(x)\delta_j(y) & = \sum\nolimits_{i + j = n} \gamma_i(x' + w)\delta_j(y) \\ & = \sum\nolimits_{i' + l + j = n} \gamma_{i'}(x')\gamma_l(w)\delta_j(y) \\ & = \sum\nolimits_{i' + l + j = n} \gamma_{i'}(x')\delta_l(w)\delta_j(y) \\ & = \sum\nolimits_{i' + j' = n} \gamma_{i'}(x')\delta_{j'}(y + w) \\ & = \sum\nolimits_{i' + j' = n} \gamma_{i'}(x')\delta_{j'}(y') \end{align*}\] as desired. Hence, we have defined maps \(\epsilon_n : I + J \to I + J\) for all \(n \geq 1\); it is easy to see that \(\epsilon_n \mid_{I} = \gamma_n\) and \(\epsilon_n \mid_{J} = \delta_n\). Next, we prove conditions (1) – (5) of Definition 07GL for the collection of maps \(\epsilon_n\). Properties (1) and (3) are clear. To see (4), suppose that \(z = x + y\) and \(z' = x' + y'\) with \(x, x' \in I\) and \(y, y' \in J\) and compute \[\begin{align*} \epsilon_n(z + z') & = \sum\nolimits_{a + b = n} \gamma_a(x + x')\delta_b(y + y') \\ & = \sum\nolimits_{i + i' + j + j' = n} \gamma_i(x) \gamma_{i'}(x')\delta_j(y)\delta_{j'}(y') \\ & = \sum\nolimits_{k = 0, \ldots, n} \sum\nolimits_{i+j=k} \gamma_i(x)\delta_j(y) \sum\nolimits_{i'+j'=n-k} \gamma_{i'}(x')\delta_{j'}(y') \\ & = \sum\nolimits_{k = 0, \ldots, n}\epsilon_k(z)\epsilon_{n-k}(z') \end{align*}\] as desired. Now we see that it suffices to prove (2) and (5) for elements of \(I\) or \(J\), see Lemma 07GP. This is clear because \(\gamma\) and \(\delta\) are divided power structures.

The existence of a divided power structure \(\epsilon\) on \(I+J\) whose restrictions to \(I\) and \(J\) are \(\gamma\) and \(\delta\) is thus proven; its uniqueness is rather clear.

Lemma

Let \(p\) be a prime number. Let \(A\) be a ring, let \(I \subset A\) be an ideal, and let \(\gamma\) be a divided power structure on \(I\). Assume \(p\) is nilpotent in \(A/I\). Then \(I\) is locally nilpotent if and only if \(p\) is nilpotent in \(A\).

Proof

If \(p^N = 0\) in \(A\), then for \(x \in I\) we have \(x^{pN} = (pN)!\gamma_{pN}(x) = 0\) because \((pN)!\) is divisible by \(p^N\). Conversely, assume \(I\) is locally nilpotent. We’ve also assumed that \(p\) is nilpotent in \(A/I\), hence \(p^r \in I\) for some \(r\), hence \(p^r\) nilpotent, hence \(p\) nilpotent.

Divided power rings

There is a category of divided power rings. Here is the definition.

Definition

A divided power ring is a triple \((A, I, \gamma)\) where \(A\) is a ring, \(I \subset A\) is an ideal, and \(\gamma = (\gamma_n)_{n \geq 1}\) is a divided power structure on \(I\). A homomorphism of divided power rings \(\varphi : (A, I, \gamma) \to (B, J, \delta)\) is a ring homomorphism \(\varphi : A \to B\) such that \(\varphi(I) \subset J\) and such that \(\delta_n(\varphi(x)) = \varphi(\gamma_n(x))\) for all \(x \in I\) and \(n \geq 1\).

We sometimes say “let \((B, J, \delta)\) be a divided power algebra over \((A, I, \gamma)\)” to indicate that \((B, J, \delta)\) is a divided power ring which comes equipped with a homomorphism of divided power rings \((A, I, \gamma) \to (B, J, \delta)\).

Lemma

The category of divided power rings has all limits and they agree with limits in the category of rings.

Proof

The empty limit is the zero ring (that’s weird but we need it). The product of a collection of divided power rings \((A_t, I_t, \gamma_t)\), \(t \in T\) is given by \((\prod A_t, \prod I_t, \gamma)\) where \(\gamma_n((x_t)) = (\gamma_{t, n}(x_t))\). The equalizer of \(\alpha, \beta : (A, I, \gamma) \to (B, J, \delta)\) is just \(C = \{a \in A \mid \alpha(a) = \beta(a)\}\) with ideal \(C \cap I\) and induced divided powers. It follows that all limits exist, see Categories, Lemma 002N.

The following lemma illustrates a very general category theoretic phenomenon in the case of divided power algebras.

Lemma

Let \(\mathcal{C}\) be the category of divided power rings. Let \(F : \mathcal{C} \to \textit{Sets}\) be a functor. Assume that

1. there exists a cardinal \(\kappa\) such that for every \(f \in F(A, I, \gamma)\) there exists a morphism \((A', I', \gamma') \to (A, I, \gamma)\) of \(\mathcal{C}\) such that \(f\) is the image of \(f' \in F(A', I', \gamma')\) and \(|A'| \leq \kappa\), and

2. \(F\) commutes with limits.

Then \(F\) is representable, i.e., there exists an object \((B, J, \delta)\) of \(\mathcal{C}\) such that \[F(A, I, \gamma) = \Hom_\mathcal{C}((B, J, \delta), (A, I, \gamma))\] functorially in \((A, I, \gamma)\).

Proof

This is a special case of Categories, Lemma 0AHN.

Lemma

The category of divided power rings has all colimits.

Proof

The empty colimit is \(\mathbf{Z}\) with divided power ideal \((0)\). Let’s discuss general colimits. Let \(\mathcal{C}\) be a category and let \(c \mapsto (A_c, I_c, \gamma_c)\) be a diagram. Consider the functor \[F(B, J, \delta) = \lim_{c \in \mathcal{C}} Hom((A_c, I_c, \gamma_c), (B, J, \delta))\] Note that any \(f = (f_c)_{c \in C} \in F(B, J, \delta)\) has the property that all the images \(f_c(A_c)\) generate a subring \(B'\) of \(B\) of bounded cardinality \(\kappa\) and that all the images \(f_c(I_c)\) generate a divided power sub ideal \(J'\) of \(B'\). And we get a factorization of \(f\) as a \(f'\) in \(F(B')\) followed by the inclusion \(B' \to B\). Also, \(F\) commutes with limits. Hence we may apply Lemma 07GW to see that \(F\) is representable and we win.

Remark

Let \[\xymatrix{ (B, J, \delta) \ar[r] & (B'', J'', \delta'') \\ (A, I, \gamma) \ar[r] \ar[u] & (B', J', \delta') \ar[u] }\] be a pushout in the category of divided power rings (pushouts exist by Lemma 07GX). Then

1. \(B''/J'' = B/J \otimes_{A/I} B'/J'\),

2. the map \(\varphi : B \otimes_A B' \to B''\) is surjective, and

3. the map \(J \otimes_A B' + B \otimes_A J' \to J''\) induced by \(\varphi\) is surjective.

To see (1) consider maps \((B'', J'', \delta'') \to (C, (0), \emptyset)\) and apply the definition of a pushout. To see (2) consider the image \(\tilde B = \varphi(B \otimes_A B')\) and the ideal \(\tilde J = \varphi(J \otimes_A B' + B \otimes_A J')\). Then it is clear that \(\delta''_n(\tilde J) \subset \tilde J\) for all \(n \geq 1\) because of the compatibility of \(\delta''\) with \(\delta\) and \(\delta'\). Hence we must have \(\tilde B = B''\) and \(\tilde J = J''\) by the universality of the pushout.

Remark

The forgetful functor \((A, I, \gamma) \mapsto A\) does not commute with colimits. For example, let \[\xymatrix{ (B, J, \delta) \ar[r] & (B'', J'', \delta'') \\ (A, I, \gamma) \ar[r] \ar[u] & (B', J', \delta') \ar[u] }\] be a pushout in the category of divided power rings as discussed in Remark 0H86. Then in general the map \(B \otimes_A B' \to B''\) isn’t an isomorphism. An explicit example is given by \((A, I, \gamma) = (\mathbf{Z}, (0), \emptyset)\), \((B, J, \delta) = (\mathbf{Z}/4\mathbf{Z}, 2\mathbf{Z}/4\mathbf{Z}, \delta)\), and \((B', J', \delta') = (\mathbf{Z}/4\mathbf{Z}, 2\mathbf{Z}/4\mathbf{Z}, \delta')\) where \(\delta_2(2) = 2\) and \(\delta'_2(2) = 0\). More precisely, using Lemma 07GS we let \(\delta\), resp. \(\delta'\) be the unique divided power structure on \(J\), resp. \(J'\) such that \(\delta_2 : J \to J\), resp. \(\delta'_2 : J' \to J'\) is the map \(0 \mapsto 0, 2 \mapsto 2\), resp. \(0 \mapsto 0, 2 \mapsto 0\). Then \((B'', J'', \delta'') = (\mathbf{F}_2, (0), \emptyset)\) which doesn’t agree with the tensor product.

Extending divided powers

Here is the definition.

Definition

Given a divided power ring \((A, I, \gamma)\) and a ring map \(A \to B\) we say \(\gamma\) extends to \(B\) if there exists a divided power structure \(\bar \gamma\) on \(IB\) such that \((A, I, \gamma) \to (B, IB, \bar\gamma)\) is a homomorphism of divided power rings.

Lemma

Let \((A, I, \gamma)\) be a divided power ring. Let \(A \to B\) be a ring map. If \(\gamma\) extends to \(B\) then it extends uniquely. Assume (at least) one of the following conditions holds

1. \(IB = 0\),

2. \(I\) is principal, or

3. \(A \to B\) is flat.

Then \(\gamma\) extends to \(B\).

Proof

Any element of \(IB\) can be written as a finite sum \(\sum\nolimits_{i=1}^t b_ix_i\) with \(b_i \in B\) and \(x_i \in I\). If \(\gamma\) extends to \(\bar\gamma\) on \(IB\) then \(\bar\gamma_n(x_i) = \gamma_n(x_i)\). Thus, conditions (3) and (4) in Definition 07GL imply that \[\bar\gamma_n(\sum\nolimits_{i=1}^t b_ix_i) = \sum\nolimits_{n_1 + \ldots + n_t = n} \prod\nolimits_{i = 1}^t b_i^{n_i}\gamma_{n_i}(x_i)\] Thus we see that \(\bar\gamma\) is unique if it exists.

If \(IB = 0\) then setting \(\bar\gamma_n(0) = 0\) works. If \(I = (x)\) then we define \(\bar\gamma_n(bx) = b^n\gamma_n(x)\). This is well defined: if \(b'x = bx\), i.e., \((b - b')x = 0\) then \[\begin{align*} b^n\gamma_n(x) - (b')^n\gamma_n(x) & = (b^n - (b')^n)\gamma_n(x) \\ & = (b^{n - 1} + \ldots + (b')^{n - 1})(b - b')\gamma_n(x) = 0 \end{align*}\] because \(\gamma_n(x)\) is divisible by \(x\) (since \(\gamma_n(I) \subset I\)) and hence annihilated by \(b - b'\). Next, we prove conditions (1) – (5) of Definition 07GL. Parts (1), (2), (3), (5) are obvious from the construction. For (4) suppose that \(y, z \in IB\), say \(y = bx\) and \(z = cx\). Then \(y + z = (b + c)x\) hence \[\begin{align*} \bar\gamma_n(y + z) & = (b + c)^n\gamma_n(x) \\ & = \sum \frac{n!}{i!(n - i)!}b^ic^{n -i}\gamma_n(x) \\ & = \sum b^ic^{n - i}\gamma_i(x)\gamma_{n - i}(x) \\ & = \sum \bar\gamma_i(y)\bar\gamma_{n -i}(z) \end{align*}\] as desired.

Assume \(A \to B\) is flat. Suppose that \(b_1, \ldots, b_r \in B\) and \(x_1, \ldots, x_r \in I\). Then \[\bar\gamma_n(\sum b_ix_i) = \sum b_1^{e_1} \ldots b_r^{e_r} \gamma_{e_1}(x_1) \ldots \gamma_{e_r}(x_r)\] where the sum is over \(e_1 + \ldots + e_r = n\) if \(\bar\gamma_n\) exists. Next suppose that we have \(c_1, \ldots, c_s \in B\) and \(a_{ij} \in A\) such that \(b_i = \sum a_{ij}c_j\). Setting \(y_j = \sum a_{ij}x_i\) we claim that \[\sum b_1^{e_1} \ldots b_r^{e_r} \gamma_{e_1}(x_1) \ldots \gamma_{e_r}(x_r) = \sum c_1^{d_1} \ldots c_s^{d_s} \gamma_{d_1}(y_1) \ldots \gamma_{d_s}(y_s)\] in \(B\) where on the right hand side we are summing over \(d_1 + \ldots + d_s = n\). Namely, using the axioms of a divided power structure we can expand both sides into a sum with coefficients in \(\mathbf{Z}[a_{ij}]\) of terms of the form \(c_1^{d_1} \ldots c_s^{d_s}\gamma_{e_1}(x_1) \ldots \gamma_{e_r}(x_r)\). To see that the coefficients agree we note that the result is true in \(\mathbf{Q}[x_1, \ldots, x_r, c_1, \ldots, c_s, a_{ij}]\) with \(\gamma\) the unique divided power structure on \((x_1, \ldots, x_r)\). By Lazard’s theorem (Algebra, Theorem 058G) we can write \(B\) as a directed colimit of finite free \(A\)-modules. In particular, if \(z \in IB\) is written as \(z = \sum x_ib_i\) and \(z = \sum x'_{i'}b'_{i'}\), then we can find \(c_1, \ldots, c_s \in B\) and \(a_{ij}, a'_{i'j} \in A\) such that \(b_i = \sum a_{ij}c_j\) and \(b'_{i'} = \sum a'_{i'j}c_j\) such that \(y_j = \sum x_ia_{ij} = \sum x'_{i'}a'_{i'j}\) holds². Hence the procedure above gives a well defined map \(\bar\gamma_n\) on \(IB\). By construction \(\bar\gamma\) satisfies conditions (1), (3), and (4). Moreover, for \(x \in I\) we have \(\bar\gamma_n(x) = \gamma_n(x)\). Hence it follows from Lemma 07GP that \(\bar\gamma\) is a divided power structure on \(IB\).

Lemma

Let \((A, I, \gamma)\) be a divided power ring.

1. If \(\varphi : (A, I, \gamma) \to (B, J, \delta)\) is a homomorphism of divided power rings, then \(\Ker(\varphi) \cap I\) is preserved by \(\gamma_n\) for all \(n \geq 1\).

2. Let \(\mathfrak a \subset A\) be an ideal and set \(I' = I \cap \mathfrak a\). The following are equivalent

   1. \(I'\) is preserved by \(\gamma_n\) for all \(n > 0\),

   2. \(\gamma\) extends to \(A/\mathfrak a\), and

   3. there exist a set of generators \(x_i\) of \(I'\) as an ideal such that \(\gamma_n(x_i) \in I'\) for all \(n > 0\).

Proof

Proof of (1). This is clear. Assume (2)(a). Define \(\bar\gamma_n(x \bmod I') = \gamma_n(x) \bmod I'\) for \(x \in I\). This is well defined since \(\gamma_n(x + y) = \gamma_n(x) \bmod I'\) for \(y \in I'\) by Definition 07GL (4) and the fact that \(\gamma_j(y) \in I'\) by assumption. It is clear that \(\bar\gamma\) is a divided power structure as \(\gamma\) is one. Hence (2)(b) holds. Also, (2)(b) implies (2)(a) by part (1). It is clear that (2)(a) implies (2)(c). Assume (2)(c). Note that \(\gamma_n(x) = a^n\gamma_n(x_i) \in I'\) for \(x = ax_i\). Hence we see that \(\gamma_n(x) \in I'\) for a set of generators of \(I'\) as an abelian group. By induction on the length of an expression in terms of these, it suffices to prove \(\forall n : \gamma_n(x + y) \in I'\) if \(\forall n : \gamma_n(x), \gamma_n(y) \in I'\). This follows immediately from the fourth axiom of a divided power structure.

Lemma

Let \((A, I, \gamma)\) be a divided power ring. Let \(E \subset I\) be a subset. Then the smallest ideal \(J \subset I\) preserved by \(\gamma\) and containing all \(f \in E\) is the ideal \(J\) generated by \(\gamma_n(f)\), \(n \geq 1\), \(f \in E\).

Proof

Follows immediately from Lemma 07H2.

Lemma

Let \((A, I, \gamma)\) be a divided power ring. Let \(p\) be a prime. If \(p\) is nilpotent in \(A/I\), then

1. the \(p\)-adic completion \(A^\wedge = \lim_e A/p^eA\) surjects onto \(A/I\),

2. the kernel of this map is the \(p\)-adic completion \(I^\wedge\) of \(I\), and

3. each \(\gamma_n\) is continuous for the \(p\)-adic topology and extends to \(\gamma_n^\wedge : I^\wedge \to I^\wedge\) defining a divided power structure on \(I^\wedge\).

If moreover \(A\) is a \(\mathbf{Z}_{(p)}\)-algebra, then

1. for \(e\) large enough the ideal \(p^eA \subset I\) is preserved by the divided power structure \(\gamma\) and \[(A^\wedge, I^\wedge, \gamma^\wedge) = \lim_e (A/p^eA, I/p^eA, \bar\gamma)\] in the category of divided power rings.

Proof

Let \(t \geq 1\) be an integer such that \(p^tA/I = 0\), i.e., \(p^tA \subset I\). The map \(A^\wedge \to A/I\) is the composition \(A^\wedge \to A/p^tA \to A/I\) which is surjective (for example by Algebra, Lemma 0315). As \(p^eI \subset p^eA \cap I \subset p^{e - t}I\) for \(e \geq t\) we see that the kernel of the composition \(A^\wedge \to A/I\) is the \(p\)-adic completion of \(I\). The map \(\gamma_n\) is continuous because \[\gamma_n(x + p^ey) = \sum\nolimits_{i + j = n} p^{je}\gamma_i(x)\gamma_j(y) = \gamma_n(x) \bmod p^eI\] by the axioms of a divided power structure. It is clear that the axioms for divided power structures are inherited by the maps \(\gamma_n^\wedge\) from the maps \(\gamma_n\). Finally, to see the last statement say \(e > t\). Then \(p^eA \subset I\) and \(\gamma_1(p^eA) \subset p^eA\) and for \(n > 1\) we have \[\gamma_n(p^ea) = p^n \gamma_n(p^{e - 1}a) = \frac{p^n}{n!} p^{n(e - 1)}a^n \in p^e A\] as \(p^n/n! \in \mathbf{Z}_{(p)}\) and as \(n \geq 2\) and \(e \geq 2\) so \(n(e - 1) \geq e\). This proves that \(\gamma\) extends to \(A/p^eA\), see Lemma 07H2. The statement on limits is clear from the construction of limits in the proof of Lemma 07GV.

Divided power polynomial algebras

A very useful example is the divided power polynomial algebra. Let \(A\) be a ring. Let \(t \geq 1\). We will denote \(A\langle x_1, \ldots, x_t \rangle\) the following \(A\)-algebra: As an \(A\)-module we set \[A\langle x_1, \ldots, x_t \rangle = \bigoplus\nolimits_{n_1, \ldots, n_t \geq 0} A x_1^{[n_1]} \ldots x_t^{[n_t]}\] with multiplication given by \[x_i^{[n]}x_i^{[m]} = \frac{(n + m)!}{n!m!}x_i^{[n + m]}.\] We also set \(x_i = x_i^{[1]}\). Note that \(1 = x_1^{[0]} \ldots x_t^{[0]}\). There is a similar construction which gives the divided power polynomial algebra in infinitely many variables. There is an canonical \(A\)-algebra map \(A\langle x_1, \ldots, x_t \rangle \to A\) sending \(x_i^{[n]}\) to zero for \(n > 0\). The kernel of this map is denoted \(A\langle x_1, \ldots, x_t \rangle_{+}\).

Lemma

Let \((A, I, \gamma)\) be a divided power ring. There exists a unique divided power structure \(\delta\) on \[J = IA\langle x_1, \ldots, x_t \rangle + A\langle x_1, \ldots, x_t \rangle_{+}\] such that

1. \(\delta_n(x_i) = x_i^{[n]}\), and

2. \((A, I, \gamma) \to (A\langle x_1, \ldots, x_t \rangle, J, \delta)\) is a homomorphism of divided power rings.

Moreover, \((A\langle x_1, \ldots, x_t \rangle, J, \delta)\) has the following universal property: A homomorphism of divided power rings \(\varphi : (A\langle x_1, \ldots, x_t \rangle, J, \delta) \to (C, K, \epsilon)\) is the same thing as a homomorphism of divided power rings \(A \to C\) and elements \(k_1, \ldots, k_t \in K\).

Proof

We will prove the lemma in case of a divided power polynomial algebra in one variable. The result for the general case can be argued in exactly the same way, or by noting that \(A\langle x_1, \ldots, x_t\rangle\) is isomorphic to the ring obtained by adjoining the divided power variables \(x_1, \ldots, x_t\) one by one.

Let \(A\langle x \rangle_{+}\) be the ideal generated by \(x, x^{[2]}, x^{[3]}, \ldots\). Note that \(J = IA\langle x \rangle + A\langle x \rangle_{+}\) and that \[IA\langle x \rangle \cap A\langle x \rangle_{+} = IA\langle x \rangle \cdot A\langle x \rangle_{+}\] Hence by Lemma 07GQ it suffices to show that there exist divided power structures on the ideals \(IA\langle x \rangle\) and \(A\langle x \rangle_{+}\). The existence of the first follows from Lemma 07H1 as \(A \to A\langle x \rangle\) is flat. For the second, note that if \(A\) is torsion free, then we can apply Lemma 07GM (4) to see that \(\delta\) exists. Namely, choosing as generators the elements \(x^{[m]}\) we see that \((x^{[m]})^n = \frac{(nm)!}{(m!)^n} x^{[nm]}\) and \(n!\) divides the integer \(\frac{(nm)!}{(m!)^n}\). In general write \(A = R/\mathfrak a\) for some torsion free ring \(R\) (e.g., a polynomial ring over \(\mathbf{Z}\)). The kernel of \(R\langle x \rangle \to A\langle x \rangle\) is \(\bigoplus \mathfrak a x^{[m]}\). Applying criterion (2)(c) of Lemma 07H2 we see that the divided power structure on \(R\langle x \rangle_{+}\) extends to \(A\langle x \rangle\) as desired.

Proof of the universal property. Given a homomorphism \(\varphi : A \to C\) of divided power rings and \(k_1, \ldots, k_t \in K\) we consider \[A\langle x_1, \ldots, x_t \rangle \to C,\quad x_1^{[n_1]} \ldots x_t^{[n_t]} \longmapsto \epsilon_{n_1}(k_1) \ldots \epsilon_{n_t}(k_t)\] using \(\varphi\) on coefficients. The only thing to check is that this is an \(A\)-algebra homomorphism (details omitted). The inverse construction is clear.

Remark

Let \((A, I, \gamma)\) be a divided power ring. There is a variant of Lemma 07H5 for infinitely many variables. First note that if \(s < t\) then there is a canonical map \[A\langle x_1, \ldots, x_s \rangle \to A\langle x_1, \ldots, x_t\rangle\] Hence if \(W\) is any set, then we set \[A\langle x_w: w \in W\rangle = \colim_{E \subset W} A\langle x_e:e \in E\rangle\] (colimit over \(E\) finite subset of \(W\)) with transition maps as above. By the definition of a colimit we see that the universal mapping property of \(A\langle x_w: w \in W\rangle\) is completely analogous to the mapping property stated in Lemma 07H5.

The following lemma can be found in [BO].

Lemma

Let \(p\) be a prime number. Let \(A\) be a ring such that every integer \(n\) not divisible by \(p\) is invertible, i.e., \(A\) is a \(\mathbf{Z}_{(p)}\)-algebra. Let \(I \subset A\) be an ideal. Two divided power structures \(\gamma, \gamma'\) on \(I\) are equal if and only if \(\gamma_p = \gamma'_p\). Moreover, given a map \(\delta : I \to I\) such that

1. \(p!\delta(x) = x^p\) for all \(x \in I\)³,

2. \(\delta(ax) = a^p\delta(x)\) for all \(a \in A\), \(x \in I\), and

3. \(\delta(x + y) = \delta(x) + \sum\nolimits_{i + j = p, i,j \geq 1} \frac{1}{i!j!} x^i y^j + \delta(y)\) for all \(x, y \in I\),

then there exists a unique divided power structure \(\gamma\) on \(I\) such that \(\gamma_p = \delta\).

Proof

If \(n\) is not divisible by \(p\), then \(\gamma_n(x) = c x \gamma_{n - 1}(x)\) where \(c\) is a unit in \(\mathbf{Z}_{(p)}\). Moreover, \[\gamma_{pm}(x) = c \gamma_m(\gamma_p(x))\] where \(c\) is a unit in \(\mathbf{Z}_{(p)}\). Thus the first assertion is clear. For the second assertion, we can, working backwards, use these equalities to define all \(\gamma_n\). More precisely, if \(n = a_0 + a_1p + \ldots + a_e p^e\) with \(a_i \in \{0, \ldots, p - 1\}\) then we set \[\gamma_n(x) = c_n x^{a_0} \delta(x)^{a_1} \ldots \delta^e(x)^{a_e}\] for \(c_n \in \mathbf{Z}_{(p)}\) defined by \[c_n = {(p!)^{a_1 + a_2(1 + p) + \ldots + a_e(1 + \ldots + p^{e - 1})}}/{n!}.\] Now we have to show the axioms (1) – (5) of a divided power structure, see Definition 07GL. We observe that (1) and (3) are immediate. Verification of (2) and (5) is by a direct calculation which we omit. Let \(x, y \in I\). We claim there is a ring map \[\varphi : \mathbf{Z}_{(p)}\langle u, v \rangle \longrightarrow A\] which maps \(u^{[n]}\) to \(\gamma_n(x)\) and \(v^{[n]}\) to \(\gamma_n(y)\). By construction of \(\mathbf{Z}_{(p)}\langle u, v \rangle\) this means we have to check that \[\gamma_n(x)\gamma_m(x) = \frac{(n + m)!}{n!m!} \gamma_{n + m}(x)\] in \(A\) and similarly for \(y\). This is true because (2) holds for \(\gamma\). Let \(\epsilon\) denote the divided power structure on the ideal \(\mathbf{Z}_{(p)}\langle u, v\rangle_{+}\) of \(\mathbf{Z}_{(p)}\langle u, v\rangle\). Next, we claim that \(\varphi(\epsilon_n(f)) = \gamma_n(\varphi(f))\) for \(f \in \mathbf{Z}_{(p)}\langle u, v\rangle_{+}\) and all \(n\). This is clear for \(n = 0, 1, \ldots, p - 1\). For \(n = p\) it suffices to prove it for a set of generators of the ideal \(\mathbf{Z}_{(p)}\langle u, v\rangle_{+}\) because both \(\epsilon_p\) and \(\gamma_p = \delta\) satisfy properties (1) and (3) of the lemma. Hence it suffices to prove that \(\gamma_p(\gamma_n(x)) = \frac{(pn)!}{p!(n!)^p}\gamma_{pn}(x)\) and similarly for \(y\), which follows as (5) holds for \(\gamma\). Now, if \(n = a_0 + a_1p + \ldots + a_e p^e\) is an arbitrary integer written in \(p\)-adic expansion as above, then \[\epsilon_n(f) = c_n f^{a_0} \gamma_p(f)^{a_1} \ldots \gamma_p^e(f)^{a_e}\] because \(\epsilon\) is a divided power structure. Hence we see that \(\varphi(\epsilon_n(f)) = \gamma_n(\varphi(f))\) holds for all \(n\). Applying this for \(f = u + v\) we see that axiom (4) for \(\gamma\) follows from the fact that \(\epsilon\) is a divided power structure.

Remark

With \(A \supset I\) as in Lemma 07GS, let \(\delta : I \to I\) be a map satisfying (2) and (3) of the lemma. Then (1) holds too. First, we prove \(\delta(nx) = n\delta(x) + \frac{n^p - n}{p!}x^p\) for \(n \geq 1\) by induction on \(n\). The case \(n = 1\) is OK. Assume that it is OK for some \(n\), then \(\delta((n + 1)x) = \delta(nx) + \delta(x) + (\sum_{i = 1}^{p - 1} \frac{n^i}{i!(p - i)!})x^p = (n + 1)\delta(x) + \left( \frac{n^p - n}{p!} + \frac{(n + 1)^p}{p!} - \frac{n^p + 1}{p!} \right)x^p = (n + 1)\delta(x) + \frac{(n + 1)^p - (n + 1)}{p!}x^p\), so it is also OK for \(n + 1\). On the other hand, we have \(\delta(nx) = n^p\delta(x)\), hence \((n^p - n)\delta(x) = \frac{n^p - n}{p!}x^p\). But for any \(p\) there exists \(n\) such that \(p^2 \not \mid n^p - n\). Namely, we have \((\mathbf{Z}/p^2\mathbf{Z})^\times \cong \mathbf{Z}/p(p-1)\mathbf{Z}\). So there exists \(n\) such that \(n^{p-1} \not \equiv 1 \mod p^2\). Hence we get \(p!\delta(x)=x^p\)

Tate resolutions

In this section we briefly discuss the resolutions constructed in [Tate-homology] and [AH] which combine divided power structures with differential graded algebras. In this section we will use homological notation for differential graded algebras. Our differential graded algebras will sit in nonnegative homological degrees. Thus our differential graded algebras \((A, \text{d})\) will be given as chain complexes \[\ldots \to A_2 \to A_1 \to A_0 \to 0 \to \ldots\] endowed with a multiplication.

Let \(R\) be a ring (commutative, as usual). In this section we will often consider graded \(R\)-algebras \(A = \bigoplus_{d \geq 0} A_d\) whose components are zero in negative degrees. We will set \(A_+ = \bigoplus_{d > 0} A_d\). We will write \(A_{even} = \bigoplus_{d \geq 0} A_{2d}\) and \(A_{odd} = \bigoplus_{d \geq 0} A_{2d + 1}\). Recall that \(A\) is graded commutative if \(x y = (-1)^{\deg(x)\deg(y)} y x\) for homogeneous elements \(x, y\). Recall that \(A\) is strictly graded commutative if in addition \(x^2 = 0\) for homogeneous elements \(x\) of odd degree. Finally, to understand the following definition, keep in mind that \(\gamma_n(x) = x^n/n!\) if \(A\) is a \(\mathbf{Q}\)-algebra.

Definition

Let \(R\) be a ring. Let \(A = \bigoplus_{d \geq 0} A_d\) be a graded \(R\)-algebra which is strictly graded commutative. A collection of maps \(\gamma_n : A_{even, +} \to A_{even, +}\) defined for all \(n > 0\) is called a divided power structure on \(A\) if we have

1. \(\gamma_n(x) \in A_{2nd}\) if \(x \in A_{2d}\),

2. \(\gamma_1(x) = x\) for any \(x\), we also set \(\gamma_0(x) = 1\),

3. \(\gamma_n(x)\gamma_m(x) = \frac{(n + m)!}{n! m!} \gamma_{n + m}(x)\),

4. \(\gamma_n(xy) = x^n \gamma_n(y)\) for all \(x \in A_{even}\) and \(y \in A_{even, +}\),

5. \(\gamma_n(xy) = 0\) if \(x, y \in A_{odd}\) homogeneous and \(n > 1\)

6. if \(x, y \in A_{even, +}\) then \(\gamma_n(x + y) = \sum_{i = 0, \ldots, n} \gamma_i(x)\gamma_{n - i}(y)\),

7. \(\gamma_n(\gamma_m(x)) = \frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x)\) for \(x \in A_{even, +}\).

Observe that conditions (2), (3), (4), (6), and (7) imply that \(\gamma\) is a “usual” divided power structure on the ideal \(A_{even, +}\) of the (commutative) ring \(A_{even}\), see Sections 07GK, 07GT, 07GZ, and 07H4. In particular, we have \(n! \gamma_n(x) = x^n\) for all \(x \in A_{even, +}\). Condition (1) states that \(\gamma\) is compatible with grading and condition (5) tells us \(\gamma_n\) for \(n > 1\) vanishes on products of homogeneous elements of odd degree. But note that it may happen that \[\gamma_2(z_1 z_2 + z_3 z_4) = z_1z_2z_3z_4\] is nonzero if \(z_1, z_2, z_3, z_4\) are homogeneous elements of odd degree.

Example

Let \(R\) be a ring. Let \((A, \gamma)\) be a strictly graded commutative graded \(R\)-algebra endowed with a divided power structure as in the definition above. Let \(d > 0\) be an odd integer. In this setting we can adjoin a variable \(T\) of degree \(d\) to \(A\). Namely, set \[A\langle T \rangle = A \oplus AT\] with grading given by \(A\langle T \rangle_m = A_m \oplus A_{m - d}T\). We claim there is a unique divided power structure on \(A\langle T \rangle\) compatible with the given divided power structure on \(A\). Namely, we set \[\gamma_n(x + yT) = \gamma_n(x) + \gamma_{n - 1}(x)yT\] for \(x \in A_{even, +}\) and \(y \in A_{odd}\).

Example

Let \(R\) be a ring. Let \((A, \gamma)\) be a strictly graded commutative graded \(R\)-algebra endowed with a divided power structure as in the definition above. Let \(d > 0\) be an even integer. In this setting we can adjoin a variable \(T\) of degree \(d\) to \(A\). Namely, set \[A\langle T \rangle = A \oplus AT \oplus AT^{(2)} \oplus AT^{(3)} \oplus \ldots\] with multiplication given by \[T^{(n)} T^{(m)} = \frac{(n + m)!}{n!m!} T^{(n + m)}\] and with grading given by \[A\langle T \rangle_m = A_m \oplus A_{m - d}T \oplus A_{m - 2d}T^{(2)} \oplus \ldots\] We claim there is a unique divided power structure on \(A\langle T \rangle\) compatible with the given divided power structure on \(A\) such that \(\gamma_n(T^{(i)}) = T^{(ni)}\). To define the divided power structure we first set \[\gamma_n\left(\sum\nolimits_{i > 0} x_i T^{(i)}\right) = \sum \prod\nolimits_{n = \sum e_i} x_i^{e_i} T^{(ie_i)}\] if \(x_i\) is in \(A_{even}\). If \(x_0 \in A_{even, +}\) then we take \[\gamma_n\left(\sum\nolimits_{i \geq 0} x_i T^{(i)}\right) = \sum\nolimits_{a + b = n} \gamma_a(x_0)\gamma_b\left(\sum\nolimits_{i > 0} x_iT^{(i)}\right)\] where \(\gamma_b\) is as defined above.

Remark

We can also adjoin a set (possibly infinite) of exterior or divided power generators in a given degree \(d > 0\), rather than just one as in Examples 09PH and 09PI. Namely, following Remark 07H6: for \((A,\gamma)\) as above and a set \(J\), let \(A\langle T_j:j\in J\rangle\) be the directed colimit of the algebras \(A\langle T_j:j\in S\rangle\) over all finite subsets \(S\) of \(J\). It is immediate that this algebra has a unique divided power structure, compatible with the given structure on \(A\) and on each generator \(T_j\).

At this point we tie in the definition of divided power structures with differentials. To understand the definition note that \(\text{d}(x^n/n!) = \text{d}(x) x^{n - 1}/(n - 1)!\) if \(A\) is a \(\mathbf{Q}\)-algebra and \(x \in A_{even, +}\).

Definition

Let \(R\) be a ring. Let \(A = \bigoplus_{d \geq 0} A_d\) be a differential graded \(R\)-algebra which is strictly graded commutative. A divided power structure \(\gamma\) on \(A\) is compatible with the differential graded structure if \(\text{d}(\gamma_n(x)) = \text{d}(x) \gamma_{n - 1}(x)\) for all \(x \in A_{even, +}\).

Warning: Let \((A, \text{d}, \gamma)\) be as in Definition 09PJ. It may not be true that \(\gamma_n(x)\) is a boundary, if \(x\) is a boundary. Thus \(\gamma\) in general does not induce a divided power structure on the homology algebra \(H(A)\). In some papers the authors put an additional compatibility condition in order to ensure that this is the case, but we elect not to do so.

Lemma

Let \((A, \text{d}, \gamma)\) and \((B, \text{d}, \gamma)\) be as in Definition 09PJ. Let \(f : A \to B\) be a map of differential graded algebras compatible with divided power structures. Assume

1. \(H_k(A) = 0\) for \(k > 0\), and

2. \(f\) is surjective.

Then \(\gamma\) induces a divided power structure on the graded \(R\)-algebra \(H(B)\).

Proof

Suppose that \(x\) and \(x'\) are homogeneous of the same degree \(2d\) and define the same cohomology class in \(H(B)\). Say \(x' - x = \text{d}(w)\). Choose a lift \(y \in A_{2d}\) of \(x\) and a lift \(z \in A_{2d + 1}\) of \(w\). Then \(y' = y + \text{d}(z)\) is a lift of \(x'\). Hence \[\gamma_n(y') = \sum \gamma_i(y) \gamma_{n - i}(\text{d}(z)) = \gamma_n(y) + \sum\nolimits_{i < n} \gamma_i(y) \gamma_{n - i}(\text{d}(z))\] Since \(A\) is acyclic in positive degrees and since \(\text{d}(\gamma_j(\text{d}(z))) = 0\) for all \(j\) we can write this as \[\gamma_n(y') = \gamma_n(y) + \sum\nolimits_{i < n} \gamma_i(y) \text{d}(z_i)\] for some \(z_i\) in \(A\). Moreover, for \(0 < i < n\) we have \[\text{d}(\gamma_i(y) z_i) = \text{d}(\gamma_i(y))z_i + \gamma_i(y)\text{d}(z_i) = \text{d}(y) \gamma_{i - 1}(y) z_i + \gamma_i(y)\text{d}(z_i)\] and the first term maps to zero in \(B\) as \(\text{d}(y)\) maps to zero in \(B\). Hence \(\gamma_n(x')\) and \(\gamma_n(x)\) map to the same element of \(H(B)\). Thus we obtain a well defined map \(\gamma_n : H_{2d}(B) \to H_{2nd}(B)\) for all \(d > 0\) and \(n > 0\). We omit the verification that this defines a divided power structure on \(H(B)\).

Lemma

Let \((A, \text{d}, \gamma)\) be as in Definition 09PJ. Let \(R \to R'\) be a ring map. Then \(\text{d}\) and \(\gamma\) induce similar structures on \(A' = A \otimes_R R'\) such that \((A', \text{d}, \gamma)\) is as in Definition 09PJ.

Proof

Observe that \(A'_{even} = A_{even} \otimes_R R'\) and \(A'_{even, +} = A_{even, +} \otimes_R R'\). Hence we are trying to show that the divided powers \(\gamma\) extend to \(A'_{even}\) (terminology as in Definition 07H0). Once we have shown \(\gamma\) extends it follows easily that this extension has all the desired properties.

Choose a polynomial \(R\)-algebra \(P\) (on any set of generators) and a surjection of \(R\)-algebras \(P \to R'\). The ring map \(A_{even} \to A_{even} \otimes_R P\) is flat, hence the divided powers \(\gamma\) extend to \(A_{even} \otimes_R P\) uniquely by Lemma 07H1. Let \(J = \Ker(P \to R')\). To show that \(\gamma\) extends to \(A \otimes_R R'\) it suffices to show that \(I' = \Ker(A_{even, +} \otimes_R P \to A_{even, +} \otimes_R R')\) is generated by elements \(z\) such that \(\gamma_n(z) \in I'\) for all \(n > 0\). This is clear as \(I'\) is generated by elements of the form \(x \otimes f\) with \(x \in A_{even, +}\) and \(f \in \Ker(P \to R')\).

Lemma

Let \((A, \text{d}, \gamma)\) be as in Definition 09PJ. Let \(d \geq 1\) be an integer. Let \(A\langle T \rangle\) be the graded divided power polynomial algebra on \(T\) with \(\deg(T) = d\) constructed in Example 09PH or 09PI. Let \(f \in A_{d - 1}\) be an element with \(\text{d}(f) = 0\). There exists a unique differential \(\text{d}\) on \(A\langle T\rangle\) such that \(\text{d}(T) = f\) and such that \(\text{d}\) is compatible with the divided power structure on \(A\langle T \rangle\).

Proof

This is proved by a direct computation which is omitted.

In Lemma 0GZ6 we will compute the cohomology of \(A\langle T \rangle\) in some special cases. Here is Tate’s construction, as extended by Avramov and Halperin.

Lemma

Let \(R \to S\) be a homomorphism of commutative rings. There exists a factorization \[R \to A \to S\] with the following properties:

1. \((A, \text{d}, \gamma)\) is as in Definition 09PJ,

2. \(A \to S\) is a quasi-isomorphism (if we endow \(S\) with the zero differential),

3. \(A_0 = R[x_j: j\in J] \to S\) is any surjection of a polynomial ring onto \(S\), and

4. \(A\) is a graded divided power polynomial algebra over \(R\).

The last condition means that \(A\) is constructed out of \(A_0\) by successively adjoining a set of variables \(T\) in each degree \(> 0\) as in Example 09PH or 09PI. Moreover, if \(R\) is Noetherian and \(R\to S\) is of finite type, then \(A\) can be taken to have only finitely many generators in each degree.

Proof

We write out the construction for the case that \(R\) is Noetherian and \(R\to S\) is of finite type. Without those assumptions, the proof is the same, except that we have to use some set (possibly infinite) of generators in each degree.

Start of the construction: Let \(A(0) = R[x_1, \ldots, x_n]\) be a (usual) polynomial ring and let \(A(0) \to S\) be a surjection. As grading we take \(A(0)_0 = A(0)\) and \(A(0)_d = 0\) for \(d \not = 0\). Thus \(\text{d} = 0\) and \(\gamma_n\), \(n > 0\), is zero as well.

Choose generators \(f_1, \ldots, f_m \in R[x_1, \ldots, x_n]\) for the kernel of the given map \(A(0) = R[x_1, \ldots, x_n] \to S\). We apply Example 09PH \(m\) times to get \[A(1) = A(0)\langle T_1, \ldots, T_m\rangle\] with \(\deg(T_i) = 1\) as a graded divided power polynomial algebra. We set \(\text{d}(T_i) = f_i\). Since \(A(1)\) is a divided power polynomial algebra over \(A(0)\) and since \(\text{d}(f_i) = 0\) this extends uniquely to a differential on \(A(1)\) by Lemma 09PM.

Induction hypothesis: Assume we are given factorizations \[R \to A(0) \to A(1) \to \ldots \to A(m) \to S\] where \(A(0)\) and \(A(1)\) are as above and each \(R \to A(m') \to S\) for \(2 \leq m' \leq m\) satisfies properties (1) and (4) of the statement of the lemma and (2) replaced by the condition that \(H_i(A(m')) \to H_i(S)\) is an isomorphism for \(m' > i \geq 0\). The base case is \(m = 1\).

Induction step: Assume we have \(R \to A(m) \to S\) as in the induction hypothesis. Consider the group \(H_m(A(m))\). This is a module over \(H_0(A(m)) = S\). In fact, it is a subquotient of \(A(m)_m\) which is a finite type module over \(A(m)_0 = R[x_1, \ldots, x_n]\). Thus we can pick finitely many elements \[e_1, \ldots, e_t \in \Ker(\text{d} : A(m)_m \to A(m)_{m - 1})\] which map to generators of this module. Applying Example 09PH or 09PI \(t\) times we get \[A(m + 1) = A(m)\langle T_1, \ldots, T_t\rangle\] with \(\deg(T_i) = m + 1\) as a graded divided power algebra. We set \(\text{d}(T_i) = e_i\). Since \(A(m+1)\) is a divided power polynomial algebra over \(A(m)\) and since \(\text{d}(e_i) = 0\) this extends uniquely to a differential on \(A(m + 1)\) compatible with the divided power structure. Since we’ve added only material in degree \(m + 1\) and higher we see that \(H_i(A(m + 1)) = H_i(A(m))\) for \(i < m\). Moreover, it is clear that \(H_m(A(m + 1)) = 0\) by construction.

To finish the proof we observe that we have shown there exists a sequence of maps \[R \to A(0) \to A(1) \to \ldots \to A(m) \to A(m + 1) \to \ldots \to S\] and to finish the proof we set \(A = \colim A(m)\).

Lemma

Let \(R \to S\) be a pseudo-coherent ring map (More on Algebra, Definition 067H). Then Lemma 09PN holds, with the resolution \(A\) of \(S\) having finitely many generators in each degree.

Proof

This is proved in exactly the same way as Lemma 09PN. The only additional twist is that, given \(A(m) \to S\) we have to show that \(H_m = H_m(A(m))\) is a finite \(R[x_1, \ldots, x_m]\)-module (so that in the next step we need only add finitely many variables). Consider the complex \[\ldots \to A(m)_{m - 1} \to A(m)_m \to A(m)_{m - 1} \to \ldots \to A(m)_0 \to S \to 0\] Since \(S\) is a pseudo-coherent \(R[x_1, \ldots, x_n]\)-module and since \(A(m)_i\) is a finite free \(R[x_1, \ldots, x_n]\)-module we conclude that this is a pseudo-coherent complex, see More on Algebra, Lemma 064Y. Since the complex is exact in (homological) degrees \(> m\) we conclude that \(H_m\) is a finite \(R\)-module by More on Algebra, Lemma 064S.

Lemma

Let \(R\) be a commutative ring. Suppose that \((A, \text{d}, \gamma)\) and \((B, \text{d}, \gamma)\) are as in Definition 09PJ. Let \(\overline{\varphi} : H_0(A) \to H_0(B)\) be an \(R\)-algebra map. Assume

1. \(A\) is a graded divided power polynomial algebra over \(R\).

2. \(H_k(B) = 0\) for \(k > 0\).

Then there exists a map \(\varphi : A \to B\) of differential graded \(R\)-algebras compatible with divided powers that lifts \(\overline{\varphi}\).

Proof

The assumption means that \(A\) is obtained from \(R\) by successively adjoining some set of polynomial generators in degree zero, exterior generators in positive odd degrees, and divided power generators in positive even degrees. So we have a filtration \(R \subset A(0) \subset A(1) \subset \ldots\) of \(A\) such that \(A(m + 1)\) is obtained from \(A(m)\) by adjoining generators of the appropriate type (which we simply call “divided power generators”) in degree \(m + 1\). In particular, \(A(0) \to H_0(A)\) is a surjection from a (usual) polynomial algebra over \(R\) onto \(H_0(A)\). Thus we can lift \(\overline{\varphi}\) to an \(R\)-algebra map \(\varphi(0) : A(0) \to B_0\).

Write \(A(1) = A(0)\langle T_j:j\in J\rangle\) for some set \(J\) of divided power variables \(T_j\) of degree \(1\). Let \(f_j \in B_0\) be \(f_j = \varphi(0)(\text{d}(T_j))\). Observe that \(f_j\) maps to zero in \(H_0(B)\) as \(\text{d}T_j\) maps to zero in \(H_0(A)\). Thus we can find \(b_j \in B_1\) with \(\text{d}(b_j) = f_j\). By the universal property of divided power polynomial algebras from Lemma 07H5, we find a lift \(\varphi(1) : A(1) \to B\) of \(\varphi(0)\) mapping \(T_j\) to \(f_j\).

Having constructed \(\varphi(m)\) for some \(m \geq 1\) we can construct \(\varphi(m + 1) : A(m + 1) \to B\) in exactly the same manner. We omit the details.

Lemma

Let \(R\) be a commutative ring. Let \(S\) and \(T\) be commutative \(R\)-algebras. Then there is a canonical structure of a strictly graded commutative \(R\)-algebra with divided powers on \[\operatorname{Tor}_*^R(S, T).\]

Proof

Choose a factorization \(R \to A \to S\) as above. Since \(A \to S\) is a quasi-isomorphism and since \(A_d\) is a free \(R\)-module, we see that the differential graded algebra \(B = A \otimes_R T\) computes the Tor groups displayed in the lemma. Choose a surjection \(R[y_j:j\in J] \to T\). Then we see that \(B\) is a quotient of the differential graded algebra \(A[y_j:j\in J]\) whose homology sits in degree \(0\) (it is equal to \(S[y_j:j\in J]\)). By Lemma 09PL the differential graded algebras \(B\) and \(A[y_j:j\in J]\) have divided power structures compatible with the differentials. Hence we obtain our divided power structure on \(H(B)\) by Lemma 09PK.

The divided power algebra structure constructed in this way is independent of the choice of \(A\). Namely, if \(A'\) is a second choice, then Lemma 09PP implies there is a map \(A \to A'\) preserving all structure and the augmentations towards \(S\). Then the induced map \(B = A \otimes_R T \to A' \otimes_R T' = B'\) also preserves all structure and is a quasi-isomorphism. The induced isomorphism of Tor algebras is therefore compatible with products and divided powers.

Application to complete intersections

Let \(R\) be a ring. Let \((A, \text{d}, \gamma)\) be as in Definition 09PJ. A derivation of degree \(2\) is an \(R\)-linear map \(\theta : A \to A\) with the following properties

1. \(\theta(A_d) \subset A_{d - 2}\),

2. \(\theta(xy) = \theta(x)y + x\theta(y)\),

3. \(\theta\) commutes with \(\text{d}\),

4. \(\theta(\gamma_n(x)) = \theta(x) \gamma_{n - 1}(x)\) for all \(x \in A_{2d}\) all \(d\).

In the following lemma we construct a derivation.

Lemma

Let \(R\) be a ring. Let \((A, \text{d}, \gamma)\) be as in Definition 09PJ. Let \(R' \to R\) be a surjection of rings whose kernel has square zero and is generated by one element \(f\). If \(A\) is a graded divided power polynomial algebra over \(R\) with finitely many variables in each degree, then we obtain a derivation \(\theta : A/IA \to A/IA\) where \(I\) is the annihilator of \(f\) in \(R\).

Proof

Since \(A\) is a divided power polynomial algebra, we can find a divided power polynomial algebra \(A'\) over \(R'\) such that \(A = A' \otimes_{R'} R\). Moreover, we can lift \(\text{d}\) to an \(R\)-linear operator \(\text{d}\) on \(A'\) such that

1. \(\text{d}(xy) = \text{d}(x)y + (-1)^{\deg(x)}x \text{d}(y)\) for \(x, y \in A'\) homogeneous, and

2. \(\text{d}(\gamma_n(x)) = \text{d}(x) \gamma_{n - 1}(x)\) for \(x \in A'_{even, +}\).

We omit the details (hint: proceed one variable at the time). However, it may not be the case that \(\text{d}^2\) is zero on \(A'\). It is clear that \(\text{d}^2\) maps \(A'\) into \(fA' \cong A/IA\). Hence \(\text{d}^2\) annihilates \(fA'\) and factors as a map \(A \to A/IA\). Since \(\text{d}^2\) is \(R\)-linear we obtain our map \(\theta : A/IA \to A/IA\). The verification of the properties of a derivation is immediate.

Lemma

Assumption and notation as in Lemma 09PS. Suppose \(S = H_0(A)\) is isomorphic to \(R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)\) for some \(n\), \(m\), and \(f_j \in R[x_1, \ldots, x_n]\). Moreover, suppose given a relation \[\sum r_j f_j = 0\] with \(r_j \in R[x_1, \ldots, x_n]\). Choose \(r'_j, f'_j \in R'[x_1, \ldots, x_n]\) lifting \(r_j, f_j\). Write \(\sum r'_j f'_j = gf\) for some \(g \in R/I[x_1, \ldots, x_n]\). If \(H_1(A) = 0\) and all the coefficients of each \(r_j\) are in \(I\), then there exists an element \(\xi \in H_2(A/IA)\) such that \(\theta(\xi) = g\) in \(S/IS\).

Proof

Let \(A(0) \subset A(1) \subset A(2) \subset \ldots\) be the filtration of \(A\) such that \(A(m)\) is gotten from \(A(m - 1)\) by adjoining divided power variables of degree \(m\). Then \(A(0)\) is a polynomial algebra over \(R\) equipped with an \(R\)-algebra surjection \(A(0) \to S\). Thus we can choose a map \[\varphi : R[x_1, \ldots, x_n] \to A(0)\] lifting the augmentations to \(S\). Next, \(A(1) = A(0)\langle T_1, \ldots, T_t \rangle\) for some divided power variables \(T_i\) of degree \(1\). Since \(H_0(A) = S\) we can pick \(\xi_j \in \sum A(0)T_i\) with \(\text{d}(\xi_j) = \varphi(f_j)\). Then \[\text{d}\left(\sum \varphi(r_j) \xi_j\right) = \sum \varphi(r_j) \varphi(f_j) = \sum \varphi(r_jf_j) = 0\] Since \(H_1(A) = 0\) we can pick \(\xi \in A_2\) with \(\text{d}(\xi) = \sum \varphi(r_j) \xi_j\). If the coefficients of \(r_j\) are in \(I\), then the same is true for \(\varphi(r_j)\). In this case \(\text{d}(\xi)\) dies in \(A_1/IA_1\) and hence \(\xi\) defines a class in \(H_2(A/IA)\).

The construction of \(\theta\) in the proof of Lemma 09PS proceeds by successively lifting \(A(i)\) to \(A'(i)\) and lifting the differential \(\text{d}\). We lift \(\varphi\) to \(\varphi' : R'[x_1, \ldots, x_n] \to A'(0)\). Next, we have \(A'(1) = A'(0)\langle T_1, \ldots, T_t\rangle\). Moreover, we can lift \(\xi_j\) to \(\xi'_j \in \sum A'(0)T_i\). Then \(\text{d}(\xi'_j) = \varphi'(f'_j) + f a_j\) for some \(a_j \in A'(0)\). Consider a lift \(\xi' \in A'_2\) of \(\xi\). Then we know that \[\text{d}(\xi') = \sum \varphi'(r'_j)\xi'_j + \sum fb_iT_i\] for some \(b_i \in A(0)\). Applying \(\text{d}\) again we find \[\theta(\xi) = \sum \varphi'(r'_j)\varphi'(f'_j) + \sum f \varphi'(r'_j) a_j + \sum fb_i \text{d}(T_i)\] The first term gives us what we want. The second term is zero because the coefficients of \(r_j\) are in \(I\) and hence are annihilated by \(f\). The third term maps to zero in \(H_0\) because \(\text{d}(T_i)\) maps to zero.

The method of proof of the following lemma is apparently due to Gulliksen.

Lemma

Let \(R' \to R\) be a surjection of Noetherian rings whose kernel has square zero and is generated by one element \(f\). Let \(S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)\). Let \(\sum r_j f_j = 0\) be a relation in \(R[x_1, \ldots, x_n]\). Assume that

1. each \(r_j\) has coefficients in the annihilator \(I\) of \(f\) in \(R\),

2. for some lifts \(r'_j, f'_j \in R'[x_1, \ldots, x_n]\) we have \(\sum r'_j f'_j = gf\) where \(g\) is not nilpotent in \(S/IS\).

Then \(S\) does not have finite tor dimension over \(R\) (i.e., \(S\) is not a perfect \(R\)-algebra).

Proof

Choose a Tate resolution \(R \to A \to S\) as in Lemma 09PN. Let \(\xi \in H_2(A/IA)\) and \(\theta : A/IA \to A/IA\) be the element and derivation found in Lemmas 09PS and 09PT. Observe that \[\theta^n(\gamma_n(\xi)) = g^n\] in \(H_0(A/IA) = S/IS\). Hence if \(g\) is not nilpotent in \(S/IS\), then \(\xi^n\) is nonzero in \(H_{2n}(A/IA)\) for all \(n > 0\). Since \(H_{2n}(A/IA) = \text{Tor}^R_{2n}(S, R/I)\) we conclude.

The following result can be found in [Rodicio].

Lemma

Let \((A, \mathfrak m)\) be a Noetherian local ring. Let \(I \subset J \subset A\) be proper ideals. If \(A/J\) has finite tor dimension over \(A/I\), then \(I/\mathfrak m I \to J/\mathfrak m J\) is injective.

Proof

Let \(f \in I\) be an element mapping to a nonzero element of \(I/\mathfrak m I\) which is mapped to zero in \(J/\mathfrak mJ\). We can choose an ideal \(I'\) with \(\mathfrak mI \subset I' \subset I\) such that \(I/I'\) is generated by the image of \(f\). Set \(R = A/I\) and \(R' = A/I'\). Let \(J = (a_1, \ldots, a_m)\) for some \(a_j \in A\). Then \(f = \sum b_j a_j\) for some \(b_j \in \mathfrak m\). Let \(r_j, f_j \in R\) resp. \(r'_j, f'_j \in R'\) be the image of \(b_j, a_j\). Then we see we are in the situation of Lemma 09PU (with the ideal \(I\) of that lemma equal to \(\mathfrak m_R\)) and the lemma is proved.

Lemma

Let \((A, \mathfrak m)\) be a Noetherian local ring. Let \(I \subset J \subset A\) be proper ideals. Assume

1. \(A/J\) has finite tor dimension over \(A/I\), and

2. \(J\) is generated by a regular sequence.

Then \(I\) is generated by a regular sequence and \(J/I\) is generated by a regular sequence.

Proof

By Lemma 09PV we see that \(I/\mathfrak m I \to J/\mathfrak m J\) is injective. Thus we can find \(s \leq r\) and a minimal system of generators \(f_1, \ldots, f_r\) of \(J\) such that \(f_1, \ldots, f_s\) are in \(I\) and form a minimal system of generators of \(I\). The lemma follows as any minimal system of generators of \(J\) is a regular sequence by More on Algebra, Lemmas 066A and 09CC.

Lemma

Let \(R \to S\) be a local ring map of Noetherian local rings. Let \(I \subset R\) and \(J \subset S\) be ideals with \(IS \subset J\). If \(R \to S\) is flat and \(S/\mathfrak m_RS\) is regular, then the following are equivalent

1. \(J\) is generated by a regular sequence and \(S/J\) has finite tor dimension as a module over \(R/I\),

2. \(J\) is generated by a regular sequence and \(\text{Tor}^{R/I}_p(S/J, R/\mathfrak m_R)\) is nonzero for only finitely many \(p\),

3. \(I\) is generated by a regular sequence and \(J/IS\) is generated by a regular sequence in \(S/IS\).

Proof

If (3) holds, then \(J\) is generated by a regular sequence, see for example More on Algebra, Lemmas 0669 and 09CC. Moreover, if (3) holds, then \(S/J = (S/I)/(J/I)\) has finite projective dimension over \(S/IS\) because the Koszul complex will be a finite free resolution of \(S/J\) over \(S/IS\). Since \(R/I \to S/IS\) is flat, it then follows that \(S/J\) has finite tor dimension over \(R/I\) by More on Algebra, Lemma 066J. Thus (3) implies (1).

The implication (1) \(\Rightarrow\) (2) is trivial. Assume (2). By More on Algebra, Lemma 09PC we find that \(S/J\) has finite tor dimension over \(S/IS\). Thus we can apply Lemma 09PW to conclude that \(IS\) and \(J/IS\) are generated by regular sequences. Let \(f_1, \ldots, f_r \in I\) be a minimal system of generators of \(I\). Since \(R \to S\) is flat, we see that \(f_1, \ldots, f_r\) form a minimal system of generators for \(IS\) in \(S\). Thus \(f_1, \ldots, f_r \in R\) is a sequence of elements whose images in \(S\) form a regular sequence by More on Algebra, Lemmas 066A and 09CC. Thus \(f_1, \ldots, f_r\) is a regular sequence in \(R\) by Algebra, Lemma 00LM.

Local complete intersection rings

Let \((A, \mathfrak m)\) be a Noetherian complete local ring. By the Cohen structure theorem (see Algebra, Theorem 032A) we can write \(A\) as the quotient of a regular Noetherian complete local ring \(R\). Let us say that \(A\) is a complete intersection if there exists some surjection \(R \to A\) with \(R\) a regular local ring such that the kernel is generated by a regular sequence. The following lemma shows this notion is independent of the choice of the surjection.

Lemma

Let \((A, \mathfrak m)\) be a Noetherian complete local ring. The following are equivalent

1. for every surjection of local rings \(R \to A\) with \(R\) a regular local ring, the kernel of \(R \to A\) is generated by a regular sequence, and

2. for some surjection of local rings \(R \to A\) with \(R\) a regular local ring, the kernel of \(R \to A\) is generated by a regular sequence.

Proof

Let \(k\) be the residue field of \(A\). If the characteristic of \(k\) is \(p > 0\), then we denote \(\Lambda\) a Cohen ring (Algebra, Definition 0327) with residue field \(k\) (Algebra, Lemma 0328). If the characteristic of \(k\) is \(0\) we set \(\Lambda = k\). Recall that \(\Lambda[[x_1, \ldots, x_n]]\) for any \(n\) is formally smooth over \(\mathbf{Z}\), resp. \(\mathbf{Q}\) in the \(\mathfrak m\)-adic topology, see More on Algebra, Lemma 07NL. Fix a surjection \(\Lambda[[x_1, \ldots, x_n]] \to A\) as in the Cohen structure theorem (Algebra, Theorem 032A).

Let \(R \to A\) be a surjection from a regular local ring \(R\). Let \(f_1, \ldots, f_r\) be a minimal sequence of generators of \(\Ker(R \to A)\). We will use without further mention that an ideal in a Noetherian local ring is generated by a regular sequence if and only if any minimal set of generators is a regular sequence. Observe that \(f_1, \ldots, f_r\) is a regular sequence in \(R\) if and only if \(f_1, \ldots, f_r\) is a regular sequence in the completion \(R^\wedge\) by Algebra, Lemmas 00LM and 00MB. Moreover, we have \[R^\wedge/(f_1, \ldots, f_r)R^\wedge = (R/(f_1, \ldots, f_n))^\wedge = A^\wedge = A\] because \(A\) is \(\mathfrak m_A\)-adically complete (first equality by Algebra, Lemma 00MA). Finally, the ring \(R^\wedge\) is regular since \(R\) is regular (More on Algebra, Lemma 07NY). Hence we may assume \(R\) is complete.

If \(R\) is complete we can choose a map \(\Lambda[[x_1, \ldots, x_n]] \to R\) lifting the given map \(\Lambda[[x_1, \ldots, x_n]] \to A\), see More on Algebra, Lemma 07NJ. By adding some more variables \(y_1, \ldots, y_m\) mapping to generators of the kernel of \(R \to A\) we may assume that \(\Lambda[[x_1, \ldots, x_n, y_1, \ldots, y_m]] \to R\) is surjective (some details omitted). Then we can consider the commutative diagram \[\xymatrix{ \Lambda[[x_1, \ldots, x_n, y_1, \ldots, y_m]] \ar[r] \ar[d] & R \ar[d] \\ \Lambda[[x_1, \ldots, x_n]] \ar[r] & A }\] By Algebra, Lemma 00SE we see that the condition for \(R \to A\) is equivalent to the condition for the fixed chosen map \(\Lambda[[x_1, \ldots, x_n]] \to A\). This finishes the proof of the lemma.

The following two lemmas are sanity checks on the definition given above.

Lemma

Let \(R\) be a regular ring. Let \(\mathfrak p \subset R\) be a prime. Let \(f_1, \ldots, f_r \in \mathfrak p\) be a regular sequence. Then the completion of \[A = (R/(f_1, \ldots, f_r))_\mathfrak p = R_\mathfrak p/(f_1, \ldots, f_r)R_\mathfrak p\] is a complete intersection in the sense defined above.

Proof

The completion of \(A\) is equal to \(A^\wedge = R_\mathfrak p^\wedge/(f_1, \ldots, f_r)R_\mathfrak p^\wedge\) because completion for finite modules over the Noetherian ring \(R_\mathfrak p\) is exact (Algebra, Lemma 00MA). The image of the sequence \(f_1, \ldots, f_r\) in \(R_\mathfrak p\) is a regular sequence by Algebra, Lemmas 00MB and 00LM. Moreover, \(R_\mathfrak p^\wedge\) is a regular local ring by More on Algebra, Lemma 07NY. Hence the result holds by our definition of complete intersection for complete local rings.

The following lemma is the analogue of Algebra, Lemma 00SC.

Lemma

Let \(R\) be a regular ring. Let \(\mathfrak p \subset R\) be a prime. Let \(I \subset \mathfrak p\) be an ideal. Set \(A = (R/I)_\mathfrak p = R_\mathfrak p/I_\mathfrak p\). The following are equivalent

1. the completion of \(A\) is a complete intersection in the sense above,

2. \(I_\mathfrak p \subset R_\mathfrak p\) is generated by a regular sequence,

3. the module \((I/I^2)_\mathfrak p\) can be generated by \(\dim(R_\mathfrak p) - \dim(A)\) elements,

4. add more here.

Proof

We may and do replace \(R\) by its localization at \(\mathfrak p\). Then \(\mathfrak p = \mathfrak m\) is the maximal ideal of \(R\) and \(A = R/I\). Let \(f_1, \ldots, f_r \in I\) be a minimal sequence of generators. The completion of \(A\) is equal to \(A^\wedge = R^\wedge/(f_1, \ldots, f_r)R^\wedge\) because completion for finite modules over the Noetherian ring \(R_\mathfrak p\) is exact (Algebra, Lemma 00MA).

If (1) holds, then the image of the sequence \(f_1, \ldots, f_r\) in \(R^\wedge\) is a regular sequence by assumption. Hence it is a regular sequence in \(R\) by Algebra, Lemmas 00MB and 00LM. Thus (1) implies (2).

Assume (3) holds. Set \(c = \dim(R) - \dim(A)\) and let \(f_1, \ldots, f_c \in I\) map to generators of \(I/I^2\). by Nakayama’s lemma (Algebra, Lemma 00DV) we see that \(I = (f_1, \ldots, f_c)\). Since \(R\) is regular and hence Cohen-Macaulay (Algebra, Proposition 00N6) we see that \(f_1, \ldots, f_c\) is a regular sequence by Algebra, Proposition 00N6. Thus (3) implies (2). Finally, (2) implies (1) by Lemma 09Q0.

The following result is due to Avramov, see [Avramov].

Proposition

Let \(A \to B\) be a flat local homomorphism of Noetherian local rings. Then the following are equivalent

1. \(B^\wedge\) is a complete intersection,

2. \(A^\wedge\) and \((B/\mathfrak m_A B)^\wedge\) are complete intersections.

Proof

Consider the diagram \[\xymatrix{ B \ar[r] & B^\wedge \\ A \ar[u] \ar[r] & A^\wedge \ar[u] }\] Since the horizontal maps are faithfully flat (Algebra, Lemma 00MC) we conclude that the right vertical arrow is flat (for example by Algebra, Lemma 00MP). Moreover, we have \((B/\mathfrak m_A B)^\wedge = B^\wedge/\mathfrak m_{A^\wedge} B^\wedge\) by Algebra, Lemma 00MA. Thus we may assume \(A\) and \(B\) are complete local Noetherian rings.

Assume \(A\) and \(B\) are complete local Noetherian rings. Choose a diagram \[\xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] }\] as in More on Algebra, Lemma 07NN. Let \(I = \Ker(R \to A)\) and \(J = \Ker(S \to B)\). Note that since \(R/I = A \to B = S/J\) is flat the map \(J/IS \otimes_R R/\mathfrak m_R \to J/J \cap \mathfrak m_R S\) is an isomorphism. Hence a minimal system of generators of \(J/IS\) maps to a minimal system of generators of \(\Ker(S/\mathfrak m_R S \to B/\mathfrak m_A B)\). Finally, \(S/\mathfrak m_R S\) is a regular local ring.

Assume (1) holds, i.e., \(J\) is generated by a regular sequence. Since \(A = R/I \to B = S/J\) is flat we see Lemma 09PX applies and we deduce that \(I\) and \(J/IS\) are generated by regular sequences. We have \(\dim(B) = \dim(A) + \dim(B/\mathfrak m_A B)\) and \(\dim(S/IS) = \dim(A) + \dim(S/\mathfrak m_R S)\) (Algebra, Lemma 00ON). Thus \(J/IS\) is generated by \[\dim(S/IS) - \dim(S/J) = \dim(S/\mathfrak m_R S) - \dim(B/\mathfrak m_A B)\] elements (Algebra, Lemma 00KW). It follows that \(\Ker(S/\mathfrak m_R S \to B/\mathfrak m_A B)\) is generated by the same number of elements (see above). Hence \(\Ker(S/\mathfrak m_R S \to B/\mathfrak m_A B)\) is generated by a regular sequence, see for example Lemma 09Q1. In this way we see that (2) holds.

If (2) holds, then \(I\) and \(J/J \cap \mathfrak m_RS\) are generated by regular sequences. Lifting these generators (see above), using flatness of \(R/I \to S/IS\), and using Grothendieck’s lemma (Algebra, Lemma 00MG) we find that \(J/IS\) is generated by a regular sequence in \(S/IS\). Thus Lemma 09PX tells us that \(J\) is generated by a regular sequence, whence (1) holds.

Definition

Let \(A\) be a Noetherian ring.

1. If \(A\) is local, then we say \(A\) is a complete intersection if its completion is a complete intersection in the sense above.

2. In general we say \(A\) is a local complete intersection if all of its local rings are complete intersections.

We will check below that this does not conflict with the terminology introduced in Algebra, Definitions 00S9 and 00SD. But first, we show this “makes sense” by showing that if \(A\) is a Noetherian local complete intersection, then \(A\) is a local complete intersection, i.e., all of its local rings are complete intersections.

Lemma

Let \((A, \mathfrak m)\) be a Noetherian local ring. Let \(\mathfrak p \subset A\) be a prime ideal. If \(A\) is a complete intersection, then \(A_\mathfrak p\) is a complete intersection too.

Proof

Choose a prime \(\mathfrak q\) of \(A^\wedge\) lying over \(\mathfrak p\) (this is possible as \(A \to A^\wedge\) is faithfully flat by Algebra, Lemma 00MC). Then \(A_\mathfrak p \to (A^\wedge)_\mathfrak q\) is a flat local ring homomorphism. Thus by Proposition 09Q2 we see that \(A_\mathfrak p\) is a complete intersection if and only if \((A^\wedge)_\mathfrak q\) is a complete intersection. Thus it suffices to prove the lemma in case \(A\) is complete (this is the key step of the proof).

Assume \(A\) is complete. By definition we may write \(A = R/(f_1, \ldots, f_r)\) for some regular sequence \(f_1, \ldots, f_r\) in a regular local ring \(R\). Let \(\mathfrak q \subset R\) be the prime corresponding to \(\mathfrak p\). Observe that \(f_1, \ldots, f_r \in \mathfrak q\) and that \(A_\mathfrak p = R_\mathfrak q/(f_1, \ldots, f_r)R_\mathfrak q\). Hence \(A_\mathfrak p\) is a complete intersection by Lemma 09Q0.

Lemma

Let \(A\) be a Noetherian ring. Then \(A\) is a local complete intersection if and only if \(A_\mathfrak m\) is a complete intersection for every maximal ideal \(\mathfrak m\) of \(A\).

Proof

This follows immediately from Lemma 09Q4 and the definitions.

Lemma

Let \(S\) be a finite type algebra over a field \(k\).

1. for a prime \(\mathfrak q \subset S\) the local ring \(S_\mathfrak q\) is a complete intersection in the sense of Algebra, Definition 00SD if and only if \(S_\mathfrak q\) is a complete intersection in the sense of Definition 09Q3, and

2. \(S\) is a local complete intersection in the sense of Algebra, Definition 00S9 if and only if \(S\) is a local complete intersection in the sense of Definition 09Q3.

Proof

Proof of (1). Let \(k[x_1, \ldots, x_n] \to S\) be a surjection. Let \(\mathfrak p \subset k[x_1, \ldots, x_n]\) be the prime ideal corresponding to \(\mathfrak q\). Let \(I \subset k[x_1, \ldots, x_n]\) be the kernel of our surjection. Note that \(k[x_1, \ldots, x_n]_\mathfrak p \to S_\mathfrak q\) is surjective with kernel \(I_\mathfrak p\). Observe that \(k[x_1, \ldots, x_n]\) is a regular ring by Algebra, Proposition 00OQ. Hence the equivalence of the two notions in (1) follows by combining Lemma 09Q1 with Algebra, Lemma 00SF.

Having proved (1) the equivalence in (2) follows from the definition and Algebra, Lemma 00SH.

Lemma

Let \(A \to B\) be a flat local homomorphism of Noetherian local rings. Then the following are equivalent

1. \(B\) is a complete intersection,

2. \(A\) and \(B/\mathfrak m_A B\) are complete intersections.

Proof

Now that the definition makes sense this is a trivial reformulation of the (nontrivial) Proposition 09Q2.

Local complete intersection maps

Let \(A \to B\) be a local homomorphism of Noetherian complete local rings. A consequence of the Cohen structure theorem is that we can find a commutative diagram \[\xymatrix{ S \ar[r] & B \\ & A \ar[lu] \ar[u] }\] of Noetherian complete local rings with \(S \to B\) surjective, \(A \to S\) flat, and \(S/\mathfrak m_A S\) a regular local ring. This follows from More on Algebra, Lemma 07NN. Let us (temporarily) say \(A \to S \to B\) is a good factorization of \(A \to B\) if \(S\) is a Noetherian local ring, \(A \to S \to B\) are local ring maps, \(S \to B\) surjective, \(A \to S\) flat, and \(S/\mathfrak m_AS\) regular. Let us say that \(A \to B\) is a complete intersection homomorphism if there exists some good factorization \(A \to S \to B\) such that the kernel of \(S \to B\) is generated by a regular sequence. The following lemma shows this notion is independent of the choice of the diagram.

Lemma

Let \(A \to B\) be a local homomorphism of Noetherian complete local rings. The following are equivalent

1. for some good factorization \(A \to S \to B\) the kernel of \(S \to B\) is generated by a regular sequence, and

2. for every good factorization \(A \to S \to B\) the kernel of \(S \to B\) is generated by a regular sequence.

Proof

Let \(A \to S \to B\) be a good factorization. As \(B\) is complete we obtain a factorization \(A \to S^\wedge \to B\) where \(S^\wedge\) is the completion of \(S\). Note that this is also a good factorization: The ring map \(S \to S^\wedge\) is flat (Algebra, Lemma 00MB), hence \(A \to S^\wedge\) is flat. The ring \(S^\wedge/\mathfrak m_A S^\wedge = (S/\mathfrak m_A S)^\wedge\) is regular since \(S/\mathfrak m_A S\) is regular (More on Algebra, Lemma 07NY). Let \(f_1, \ldots, f_r\) be a minimal sequence of generators of \(\Ker(S \to B)\). We will use without further mention that an ideal in a Noetherian local ring is generated by a regular sequence if and only if any minimal set of generators is a regular sequence. Observe that \(f_1, \ldots, f_r\) is a regular sequence in \(S\) if and only if \(f_1, \ldots, f_r\) is a regular sequence in the completion \(S^\wedge\) by Algebra, Lemma 00LM. Moreover, we have \[S^\wedge/(f_1, \ldots, f_r)R^\wedge = (S/(f_1, \ldots, f_n))^\wedge = B^\wedge = B\] because \(B\) is \(\mathfrak m_B\)-adically complete (first equality by Algebra, Lemma 00MA). Thus the kernel of \(S \to B\) is generated by a regular sequence if and only if the kernel of \(S^\wedge \to B\) is generated by a regular sequence. Hence it suffices to consider good factorizations where \(S\) is complete.

Assume we have two factorizations \(A \to S \to B\) and \(A \to S' \to B\) with \(S\) and \(S'\) complete. By More on Algebra, Lemma 09Q8 the ring \(S \times_B S'\) is a Noetherian complete local ring. Hence, using More on Algebra, Lemma 07NN we can choose a good factorization \(A \to S'' \to S \times_B S'\) with \(S''\) complete. Thus it suffices to show: If \(A \to S' \to S \to B\) are comparable good factorizations, then \(\Ker(S \to B)\) is generated by a regular sequence if and only if \(\Ker(S' \to B)\) is generated by a regular sequence.

Let \(A \to S' \to S \to B\) be comparable good factorizations. First, since \(S'/\mathfrak m_R S' \to S/\mathfrak m_R S\) is a surjection of regular local rings, the kernel is generated by a regular sequence \(\overline{x}_1, \ldots, \overline{x}_c \in \mathfrak m_{S'}/\mathfrak m_R S'\) which can be extended to a regular system of parameters for the regular local ring \(S'/\mathfrak m_R S'\), see (Algebra, Lemma 00NR). Set \(I = \Ker(S' \to S)\). By flatness of \(S\) over \(R\) we have \[I/\mathfrak m_R I = \Ker(S'/\mathfrak m_R S' \to S/\mathfrak m_R S) = (\overline{x}_1, \ldots, \overline{x}_c).\] Choose lifts \(x_1, \ldots, x_c \in I\). These lifts form a regular sequence generating \(I\) as \(S'\) is flat over \(R\), see Algebra, Lemma 00MG.

We conclude that if also \(\Ker(S \to B)\) is generated by a regular sequence, then so is \(\Ker(S' \to B)\), see More on Algebra, Lemmas 0669 and 09CC.

Conversely, assume that \(J = \Ker(S' \to B)\) is generated by a regular sequence. Because the generators \(x_1, \ldots, x_c\) of \(I\) map to linearly independent elements of \(\mathfrak m_{S'}/\mathfrak m_{S'}^2\) we see that \(I/\mathfrak m_{S'}I \to J/\mathfrak m_{S'}J\) is injective. Hence there exists a minimal system of generators \(x_1, \ldots, x_c, y_1, \ldots, y_d\) for \(J\). Then \(x_1, \ldots, x_c, y_1, \ldots, y_d\) is a regular sequence and it follows that the images of \(y_1, \ldots, y_d\) in \(S\) form a regular sequence generating \(\Ker(S \to B)\). This finishes the proof of the lemma.

In the following proposition observe that the condition on vanishing of Tor’s applies in particular if \(B\) has finite tor dimension over \(A\) and thus in particular if \(B\) is flat over \(A\).

Proposition

Let \(A \to B\) be a local homomorphism of Noetherian local rings. Then the following are equivalent

1. \(B\) is a complete intersection and \(\text{Tor}^A_p(B, A/\mathfrak m_A)\) is nonzero for only finitely many \(p\),

2. \(A\) is a complete intersection and \(A^\wedge \to B^\wedge\) is a complete intersection homomorphism in the sense defined above.

Proof

Let \(F_\bullet \to A/\mathfrak m_A\) be a resolution by finite free \(A\)-modules. Observe that \(\text{Tor}^A_p(B, A/\mathfrak m_A)\) is the \(p\)th homology of the complex \(F_\bullet \otimes_A B\). Let \(F_\bullet^\wedge = F_\bullet \otimes_A A^\wedge\) be the completion. Then \(F_\bullet^\wedge\) is a resolution of \(A^\wedge/\mathfrak m_{A^\wedge}\) by finite free \(A^\wedge\)-modules (as \(A \to A^\wedge\) is flat and completion on finite modules is exact, see Algebra, Lemmas 00MA and 00MB). It follows that \[F_\bullet^\wedge \otimes_{A^\wedge} B^\wedge = F_\bullet \otimes_A B \otimes_B B^\wedge\] By flatness of \(B \to B^\wedge\) we conclude that \[\text{Tor}^{A^\wedge}_p(B^\wedge, A^\wedge/\mathfrak m_{A^\wedge}) = \text{Tor}^A_p(B, A/\mathfrak m_A) \otimes_B B^\wedge\] In this way we see that the condition in (1) on the local ring map \(A \to B\) is equivalent to the same condition for the local ring map \(A^\wedge \to B^\wedge\). Thus we may assume \(A\) and \(B\) are complete local Noetherian rings (since the other conditions are formulated in terms of the completions in any case).

Assume \(A\) and \(B\) are complete local Noetherian rings. Choose a diagram \[\xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] }\] as in More on Algebra, Lemma 07NN. Let \(I = \Ker(R \to A)\) and \(J = \Ker(S \to B)\). The proposition now follows from Lemma 09PX.

Remark

It appears difficult to define an good notion of “local complete intersection homomorphisms” for maps between general Noetherian rings. The reason is that, for a local Noetherian ring \(A\), the fibres of \(A \to A^\wedge\) are not local complete intersection rings. Thus, if \(A \to B\) is a local homomorphism of local Noetherian rings, and the map of completions \(A^\wedge \to B^\wedge\) is a complete intersection homomorphism in the sense defined above, then \((A_\mathfrak p)^\wedge \to (B_\mathfrak q)^\wedge\) is in general not a complete intersection homomorphism in the sense defined above. A solution can be had by working exclusively with excellent Noetherian rings. More generally, one could work with those Noetherian rings whose formal fibres are complete intersections, see [Rodicio-ci]. We will develop this theory in Dualizing Complexes, Section 0BJM.

To finish of this section we compare the notion defined above with the notion introduced in More on Algebra, Section 09PY.

Lemma

Consider a commutative diagram \[\xymatrix{ S \ar[r] & B \\ & A \ar[lu] \ar[u] }\] of Noetherian local rings with \(S \to B\) surjective, \(A \to S\) flat, and \(S/\mathfrak m_A S\) a regular local ring. The following are equivalent

1. \(\Ker(S \to B)\) is generated by a regular sequence, and

2. \(A^\wedge \to B^\wedge\) is a complete intersection homomorphism as defined above.

Proof

Omitted. Hint: the proof is identical to the argument given in the first paragraph of the proof of Lemma 09QA.

Lemma

Let \(A\) be a Noetherian ring. Let \(A \to B\) be a finite type ring map. The following are equivalent

1. \(A \to B\) is a local complete intersection in the sense of More on Algebra, Definition 07D0,

2. for every prime \(\mathfrak q \subset B\) and with \(\mathfrak p = A \cap \mathfrak q\) the ring map \((A_\mathfrak p)^\wedge \to (B_\mathfrak q)^\wedge\) is a complete intersection homomorphism in the sense defined above.

Proof

Choose a surjection \(R = A[x_1, \ldots, x_n] \to B\). Observe that \(A \to R\) is flat with regular fibres. Let \(I\) be the kernel of \(R \to B\). Assume (2). Then we see that \(I\) is locally generated by a regular sequence by Lemma 09QD and Algebra, Lemma 061L. In other words, (1) holds. Conversely, assume (1). Then after localizing on \(R\) and \(B\) we can assume that \(I\) is generated by a Koszul regular sequence. By More on Algebra, Lemma 09CC we find that \(I\) is locally generated by a regular sequence. Hence (2) hold by Lemma 09QD. Some details omitted.

Lemma

Let \(A\) be a Noetherian ring. Let \(A \to B\) be a finite type ring map such that the image of \(\Spec(B) \to \Spec(A)\) contains all closed points of \(\Spec(A)\). Then the following are equivalent

1. \(B\) is a complete intersection and \(A \to B\) has finite tor dimension,

2. \(A\) is a complete intersection and \(A \to B\) is a local complete intersection in the sense of More on Algebra, Definition 07D0.

Proof

This is a reformulation of Proposition 09QB via Lemma 09QE. We omit the details.

Smooth ring maps and diagonals

In this section we use the material above to characterize smooth ring maps as those whose diagonal is perfect.

Lemma

Let \(A \to B\) be a local ring homomorphism of Noetherian local rings such that \(B\) is flat and essentially of finite type over \(A\). If \[B \otimes_A B \longrightarrow B\] is a perfect ring map, i.e., if \(B\) has finite tor dimension over \(B \otimes_A B\), then \(B\) is the localization of a smooth \(A\)-algebra.

Proof

As \(B\) is essentially of finite type over \(A\), so is \(B \otimes_A B\) and in particular \(B \otimes_A B\) is Noetherian. Hence the quotient \(B\) of \(B \otimes_A B\) is pseudo-coherent over \(B \otimes_A B\) (More on Algebra, Lemma 066E) which explains why perfectness of the ring map (More on Algebra, Definition 067H) agrees with the condition of finite tor dimension.

We may write \(B = R/K\) where \(R\) is the localization of \(A[x_1, \ldots, x_n]\) at a prime ideal and \(K \subset R\) is an ideal. Denote \(\mathfrak m \subset R \otimes_A R\) the maximal ideal which is the inverse image of the maximal ideal of \(B\) via the surjection \(R \otimes_A R \to B \otimes_A B \to B\). Then we have surjections \[(R \otimes_A R)_\mathfrak m \to (B \otimes_A B)_\mathfrak m \to B\] and hence ideals \(I \subset J \subset (R \otimes_A R)_\mathfrak m\) as in Lemma 09PV. We conclude that \(I/\mathfrak m I \to J/\mathfrak m J\) is injective.

Let \(K = (f_1, \ldots, f_r)\) with \(r\) minimal. We may and do assume that \(f_i \in R\) is the image of an element of \(A[x_1, \ldots, x_n]\) which we also denote \(f_i\). Observe that \(I\) is generated by \(f_1 \otimes 1, \ldots, f_r \otimes 1\) and \(1 \otimes f_1, \ldots, 1 \otimes f_r\). We claim that this is a minimal set of generators of \(I\). Namely, if \(\kappa\) is the common residue field of \(R\), \(B\), \((R \otimes_A R)_\mathfrak m\), and \((B \otimes_A B)_\mathfrak m\) then we have a map \(R \otimes_A R \to R \otimes_A \kappa \oplus \kappa \otimes_A R\) which factors through \((R \otimes_A R)_\mathfrak m\). Since \(B\) is flat over \(A\) and since we have the short exact sequence \(0 \to K \to R \to B \to 0\) we see that \(K \otimes_A \kappa \subset R \otimes_A \kappa\), see Algebra, Lemma 00HL. Thus restricting the map \((R \otimes_A R)_\mathfrak m \to R \otimes_A \kappa \oplus \kappa \otimes_A R\) to \(I\) we obtain a map \[I \to K \otimes_A \kappa \oplus \kappa \otimes_A K \to K \otimes_B \kappa \oplus \kappa \otimes_B K.\] The elements \(f_1 \otimes 1, \ldots, f_r \otimes 1, 1 \otimes f_1, \ldots, 1 \otimes f_r\) map to a basis of the target of this map, since by Nakayama’s lemma (Algebra, Lemma 00DV) \(f_1, \ldots, f_r\) map to a basis of \(K \otimes_B \kappa\). This proves our claim.

The ideal \(J\) is generated by \(f_1 \otimes 1, \ldots, f_r \otimes 1\) and the elements \(x_1 \otimes 1 - 1 \otimes x_1, \ldots, x_n \otimes 1 - 1 \otimes x_n\) (for the proof it suffices to see that these elements are contained in the ideal \(J\)). Now we can write \[f_i \otimes 1 - 1 \otimes f_i = \sum g_{ij} (x_j \otimes 1 - 1 \otimes x_j)\] for some \(g_{ij}\) in \((R \otimes_A R)_\mathfrak m\). This is a general fact about elements of \(A[x_1, \ldots, x_n]\) whose proof we omit. Denote \(a_{ij} \in \kappa\) the image of \(g_{ij}\). Another computation shows that \(a_{ij}\) is the image of \(\partial f_i / \partial x_j\) in \(\kappa\). The injectivity of \(I/\mathfrak m I \to J/\mathfrak m J\) and the remarks made above forces the matrix \((a_{ij})\) to have maximal rank \(r\). Set \[C = A[x_1, \ldots, x_n]/(f_1, \ldots, f_r)\] and consider the naive cotangent complex \(\NL_{C/A} \cong (C^{\oplus r} \to C^{\oplus n})\) where the map is given by the matrix of partial derivatives. Thus \(\NL_{C/A} \otimes_A B\) is isomorphic to a free \(B\)-module of rank \(n - r\) placed in degree \(0\). Hence \(C_g\) is smooth over \(A\) for some \(g \in C\) mapping to a unit in \(B\), see Algebra, Lemma 07BU. This finishes the proof.

Lemma

Let \(A \to B\) be a flat finite type ring map of Noetherian rings. If \[B \otimes_A B \longrightarrow B\] is a perfect ring map, i.e., if \(B\) has finite tor dimension over \(B \otimes_A B\), then \(B\) is a smooth \(A\)-algebra.

Proof

This follows from Lemma 0FCW and general facts about smooth ring maps, see Algebra, Lemmas 07BU and 00TC. Alternatively, the reader can slightly modify the proof of Lemma 0FCW to prove this lemma.

Freeness of the conormal module

Tate resolutions and derivations on them can be used to prove (stronger) versions of the results in this section, see [Iyengar]. Two more elementary references are [Vasconcelos] and [Ferrand-lci].

Lemma

Let \(R\) be a Noetherian local ring. Let \(I \subset R\) be an ideal of finite projective dimension over \(R\). If \(F \subset I/I^2\) is a direct summand isomorphic to \(R/I\), then there exists a nonzerodivisor \(x \in I\) such that the image of \(x\) in \(I/I^2\) generates \(F\).

Proof

By assumption we may choose a finite free resolution \[0 \to R^{\oplus n_e} \to R^{\oplus n_{e-1}} \to \ldots \to R^{\oplus n_1} \to R \to R/I \to 0\] Then \(\varphi_1 : R^{\oplus n_1} \to R\) has rank \(1\) and we see that \(I\) contains a nonzerodivisor \(y\) by Algebra, Proposition 00N1. Let \(\mathfrak p_1, \ldots, \mathfrak p_n\) be the associated primes of \(R\), see Algebra, Lemma 00LC. Let \(I^2 \subset J \subset I\) be an ideal such that \(J/I^2 = F\). Then \(J \not \subset \mathfrak p_i\) for all \(i\) as \(y^2 \in J\) and \(y^2 \not \in \mathfrak p_i\), see Algebra, Lemma 00LD. By Nakayama’s lemma (Algebra, Lemma 00DV) we have \(J \not \subset \mathfrak m J + I^2\). By Algebra, Lemma 00DS we can pick \(x \in J\), \(x \not \in \mathfrak m J + I^2\) and \(x \not \in \mathfrak p_i\) for \(i = 1, \ldots, n\). Then \(x\) is a nonzerodivisor and the image of \(x\) in \(I/I^2\) generates (by Nakayama’s lemma) the summand \(J/I^2 \cong R/I\).

Lemma

Let \(R\) be a Noetherian local ring. Let \(I \subset R\) be an ideal of finite projective dimension over \(R\). If \(F \subset I/I^2\) is a direct summand free of rank \(r\), then there exists a regular sequence \(x_1, \ldots, x_r \in I\) such that \(x_1 \bmod I^2, \ldots, x_r \bmod I^2\) generate \(F\).

Proof

If \(r = 0\) there is nothing to prove. Assume \(r > 0\). We may pick \(x \in I\) such that \(x\) is a nonzerodivisor and \(x \bmod I^2\) generates a summand of \(F\) isomorphic to \(R/I\), see Lemma 0FJQ. Consider the ring \(R' = R/(x)\) and the ideal \(I' = I/(x)\). Of course \(R'/I' = R/I\). The short exact sequence \[0 \to R/I \xrightarrow{x} I/xI \to I' \to 0\] splits because the map \(I/xI \to I/I^2\) sends \(xR/xI\) to a direct summand. Now \(I/xI = I \otimes_R^\mathbf{L} R'\) has finite projective dimension over \(R'\), see More on Algebra, Lemmas 066Q and 066W. Hence the summand \(I'\) has finite projective dimension over \(R'\). On the other hand, we have the short exact sequence \(0 \to xR/xI \to I/I^2 \to I'/(I')^2 \to 0\) and we conclude \(I'/(I')^2\) has the free direct summand \(F' = F/(R/I \cdot x)\) of rank \(r - 1\). By induction on \(r\) we may we pick a regular sequence \(x'_2, \ldots, x'_r \in I'\) such that there congruence classes freely generate \(F'\). If \(x_1 = x\) and \(x_2, \ldots, x_r\) are any elements lifting \(x'_1, \ldots, x'_r\) in \(R\), then we see that the lemma holds.

Proposition

Let \(R\) be a Noetherian ring. Let \(I \subset R\) be an ideal which has finite projective dimension and such that \(I/I^2\) is finite locally free over \(R/I\). Then \(I\) is a regular ideal (More on Algebra, Definition 07CV).

Proof

By Algebra, Lemma 061L it suffices to show that \(I_\mathfrak p \subset R_\mathfrak p\) is generated by a regular sequence for every \(\mathfrak p \supset I\). Thus we may assume \(R\) is local. If \(I/I^2\) has rank \(r\), then by Lemma 0FJR we find a regular sequence \(x_1, \ldots, x_r \in I\) generating \(I/I^2\). By Nakayama (Algebra, Lemma 00DV) we conclude that \(I\) is generated by \(x_1, \ldots, x_r\).

For any local complete intersection homomorphism \(A \to B\) of rings, the naive cotangent complex \(\NL_{B/A}\) is perfect of tor-amplitude in \([-1, 0]\), see More on Algebra, Lemma 0FV0. Using the above, we can show that this sometimes characterizes local complete intersection homomorphisms.

Lemma

Let \(A \to B\) be a perfect (More on Algebra, Definition 067H) ring homomorphism of Noetherian rings. Then the following are equivalent

1. \(\NL_{B/A}\) has tor-amplitude in \([-1, 0]\),

2. \(\NL_{B/A}\) is a perfect object of \(D(B)\) with tor-amplitude in \([-1, 0]\), and

3. \(A \to B\) is a local complete intersection (More on Algebra, Definition 07D0).

Proof

Write \(B = A[x_1, \ldots, x_n]/I\). Then \(\NL_{B/A}\) is represented by the complex \[I/I^2 \longrightarrow \bigoplus B \text{d}x_i\] of \(B\)-modules with \(I/I^2\) placed in degree \(-1\). Since the term in degree \(0\) is finite free, this complex has tor-amplitude in \([-1, 0]\) if and only if \(I/I^2\) is a flat \(B\)-module, see More on Algebra, Lemma 0653. Since \(I/I^2\) is a finite \(B\)-module and \(B\) is Noetherian, this is true if and only if \(I/I^2\) is a finite locally free \(B\)-module (Algebra, Lemma 00NX). Thus the equivalence of (1) and (2) is clear. Moreover, the equivalence of (1) and (3) also follows if we apply Proposition 0FJS (and the observation that a regular ideal is a Koszul regular ideal as well as a quasi-regular ideal, see More on Algebra, Section 07CU).

Lemma

Let \(A \to B\) be a flat ring map of finite presentation. Then the following are equivalent

1. \(\NL_{B/A}\) has tor-amplitude in \([-1, 0]\),

2. \(\NL_{B/A}\) is a perfect object of \(D(B)\) with tor-amplitude in \([-1, 0]\),

3. \(A \to B\) is syntomic (Algebra, Definition 00SL), and

4. \(A \to B\) is a local complete intersection (More on Algebra, Definition 07D0).

Proof

The equivalence of (3) and (4) is More on Algebra, Lemma 07D3.

If \(A \to B\) is syntomic, then we can find a cocartesian diagram \[\xymatrix{ B_0 \ar[r] & B \\ A_0 \ar[r] \ar[u] & A \ar[u] }\] such that \(A_0 \to B_0\) is syntomic and \(A_0\) is Noetherian, see Algebra, Lemmas 00R1 and 0C33. By Lemma 0FJT we see that \(\NL_{B_0/A_0}\) is perfect of tor-amplitude in \([-1, 0]\). By More on Algebra, Lemma 0FJU we conclude the same thing is true for \(\NL_{B/A} = \NL_{B_0/A_0} \otimes_{B_0}^\mathbf{L} B\) (see also More on Algebra, Lemmas 066L and 066W). This proves that (3) implies (2).

Assume (1). By More on Algebra, Lemma 0FJU for every ring map \(A \to k\) where \(k\) is a field, we see that \(\NL_{B \otimes_A k/k}\) has tor-amplitude in \([-1, 0]\) (see More on Algebra, Lemma 066L). Hence by Lemma 0FJT we see that \(k \to B \otimes_A k\) is a local complete intersection homomorphism. Thus \(A \to B\) is syntomic by definition. This proves (1) implies (3) and finishes the proof.

Koszul complexes and Tate resolutions

In this section we “lift” the result of More on Algebra, Lemma 0921 to the category of differential graded algebras endowed with divided powers compatible with the differential graded structure (beware that in this section we represent Koszul complexes as chain complexes whereas in locus citatus we use cochain complexes).

Let \(R\) be a ring. Let \(I \subset R\) be an ideal generated by \(f_1, \ldots, f_r \in R\). For \(n \geq 1\) we denote \[K_n = K_{n, \bullet} = R\langle \xi_1, \ldots, \xi_r\rangle\] the differential graded Koszul algebra with \(\xi_i\) in degree \(1\) and \(\text{d}(\xi_i) = f_i^n\). There exists a unique divided power structure on this (as in Definition 09PJ), see Example 09PH. For \(m > n\) the transition map \(K_m \to K_n\) is the differential graded algebra map compatible with divided powers given by sending \(\xi_i\) to \(f_i^{m - n}\xi_i\).

Lemma

In the situation above, if \(R\) is Noetherian, then for every \(n\) there exists an \(N \geq n\) and maps \[K_N \to A \to R/(f_1^N, \ldots, f_r^N)\quad\text{and}\quad A \to K_n\] with the following properties

1. \((A, \text{d}, \gamma)\) is as in Definition 09PJ,

2. \(A \to R/(f_1^N, \ldots, f_r^N)\) is a quasi-isomorphism,

3. the composition \(K_N \to A \to R/(f_1^N, \ldots, f_r^N)\) is the canonical map,

4. the composition \(K_N \to A \to K_n\) is the transition map,

5. \(A_0 = R \to R/(f_1^N, \ldots, f_r^N)\) is the canonical surjection,

6. \(A\) is a graded divided power polynomial algebra over \(R\) with finitely many generators in each degree, and

7. \(A \to K_n\) is a homomorphism of differential graded \(R\)-algebras compatible with divided powers which induces the canonical map \(R/(f_1^N, \ldots, f_r^N) \to R/(f_1^n, \ldots, f_r^n)\) on homology in degree \(0\).

Condition (4) means that \(A\) is constructed out of \(A_0\) by successively adjoining a finite set of variables \(T\) in each degree \(> 0\) as in Example 09PH or 09PI.

Proof

Fix \(n\). If \(r = 0\), then we can just pick \(A = R\). Assume \(r > 0\). By More on Algebra, Lemma 0921 (translated into the language of chain complexes) we can choose \[n_{r} > n_{r - 1} > \ldots > n_1 > n_0 = n\] such that the transition maps \(K_{n_{i + 1}} \to K_{n_i}\) on Koszul algebras (see above) induce the zero map on homology in degrees \(> 0\). We will prove the lemma with \(N = n_r\).

We will construct \(A\) exactly as in the statement and proof of Lemma 09PN. Thus we will have \[A = \colim A(m),\quad\text{and}\quad A(0) \to A(1) \to A(2) \to \ldots \to R/(f_1^N, \ldots, f_r^N)\] This will immediately give us properties (1), (2), (5), and (6). To finish the proof we will construct the \(R\)-algebra maps \(K_N \to A \to K_n\). To do this we will construct

1. an isomorphism \(A(1) \to K_N = K_{n_r}\),

2. a map \(A(2) \to K_{n_{r - 1}}\),

3. \(\ldots\)

4. a map \(A(r) \to K_{n_1}\),

5. a map \(A(r + 1) \to K_{n_0} = K_n\), and

6. a map \(A \to K_n\).

In each of these steps the map constructed will be between differential graded algebras compatibly endowed with divided powers and each of the maps will be compatible with the previous one via the transition maps between the Koszul algebras and each of the maps will induce the obvious canonical map on homology in degree \(0\).

Recall that \(A(0) = R\). For \(m = 1\), the proof of Lemma 09PN chooses \(A(1) = R\langle T_1, \ldots, T_r\rangle\) with \(T_i\) of degree \(1\) and with \(\text{d}(T_i) = f_i^N\). Namely, the \(f_i^N\) are generators of the kernel of \(A(0) \to R/(f_1^N, \ldots, f_r^N)\). Thus for \(A(1) \to K_N = K_{n_r}\) we use the map \[\varphi_1 : A(1) \longrightarrow K_{n_r},\quad T_i \longmapsto \xi_i\] which is an isomorphism.

For \(m = 2\), the construction in the proof of Lemma 09PN chooses generators \(e_1, \ldots, e_t \in \Ker(\text{d} : A(1)_1 \to A(1)_0)\). The construction proceeds by taking \(A(2) = A(1)\langle T_1, \ldots, T_t\rangle\) as a divided power polynomial algebra with \(T_i\) of degree \(2\) and with \(\text{d}(T_i) = e_i\). Since \(\varphi_1(e_i)\) is a cocycle in \(K_{n_r}\) we see that its image in \(K_{n_{r - 1}}\) is a coboundary by our choice of \(n_r\) and \(n_{r - 1}\) above. Hence we can construct the following commutative diagram \[\xymatrix{ A(1) \ar[d] \ar[r]_{\varphi_1} & K_{n_r} \ar[d] \\ A(2) \ar[r]^{\varphi_2} & K_{n_{r - 1}} }\] by sending \(T_i\) to an element in degree \(2\) whose boundary is the image of \(\varphi_1(e_i)\). The map \(\varphi_2\) exists and is compatible with the differential and the divided powers by the universal of the divided power polynomial algebra.

The algebra \(A(m)\) and the map \(\varphi_m : A(m) \to K_{n_{r + 1 - m}}\) are constructed in exactly the same manner for \(m = 2, \ldots, r\).

Given the map \(A(r) \to K_{n_1}\) we see that the composition \(H_r(A(r)) \to H_r(K_{n_1}) \to H_r(K_{n_0}) \subset (K_{n_0})_r\) is zero, hence we can extend this to \(A(r + 1) \to K_{n_0} = K_n\) by sending the new polynomial generators of \(A(r + 1)\) to zero.

Having constructed \(A(r + 1) \to K_{n_0} = K_n\) we can simply extend to \(A(r + 2), A(r + 3), \ldots\) in the only possible way by sending the new polynomial generators to zero. This finishes the proof.

Remark

In the situation above, if \(R\) is Noetherian, we can inductively choose a sequence of integers \(1 = n_0 < n_1 < n_2 < \ldots\) such that for \(i = 1, 2, 3, \ldots\) we have maps \(K_{n_i} \to A_i \to R/(f_1^{n_i}, \ldots, f_r^{n_i})\) and \(A_i \to K_{n_{i - 1}}\) as in Lemma 0GZ4. Denote \(A_{i + 1} \to A_i\) the composition \(A_{i + 1} \to K_{n_i} \to A_i\). Then the diagram \[\xymatrix{ K_{n_1} \ar[d] & K_{n_2} \ar[d] \ar[l] & K_{n_3} \ar[d] \ar[l] & \ldots \ar[l] \\ A_1 \ar[d] & A_2 \ar[l] \ar[d] & A_3 \ar[l] \ar[d] & \ldots \ar[l] \\ K_1 & K_{n_1} \ar[l] & K_{n_2} \ar[l] & \ldots \ar[l] }\] commutes. In this way we see that the inverse systems \((K_n)\) and \((A_n)\) are pro-isomorphic in the category of differential graded \(R\)-algebras with compatible divided powers.

Lemma

Let \((A, \text{d}, \gamma)\), \(d \geq 1\), \(f \in A_{d - 1}\), and \(A\langle T \rangle\) be as in Lemma 09PM.

1. If \(d = 1\), then there is a long exact sequences \[\ldots \to H_0(A) \xrightarrow{f} H_0(A) \to H_0(A\langle T \rangle) \to 0\]

2. For \(d = 2\) there is a bounded spectral sequence \((E_1)_{i, j} = H_{j - i}(A) \cdot T^{[i]}\) converging to \(H_{i + j}(A\langle T \rangle)\). The differential \((d_1)_{i, j} : H_{j - i}(A) \cdot T^{[i]} \to H_{j - i + 1}(A) \cdot T^{[i - 1]}\) sends \(\xi \cdot T^{[i]}\) to the class of \(f \xi \cdot T^{[i - 1]}\).

3. Add more here for other degrees as needed.

Proof

For \(d = 1\), we have a short exact sequence of complexes \[0 \to A \to A\langle T \rangle \to A \cdot T \to 0\] and the result (1) follows easily from this. For \(d = 2\) we view \(A\langle T \rangle\) as a filtered chain complex with subcomplexes \[F^pA\langle T \rangle = \bigoplus\nolimits_{i \leq p} A \cdot T^{[i]}\] Applying the spectral sequence of Homology, Section 012K (translated into chain complexes) we obtain (2).

The following lemma will be needed later.

Lemma

In the situation above, for all \(n \geq t \geq 1\) there exists an \(N > n\) and a map \[K_t \longrightarrow K_n \otimes_R K_t\] in the derived category of left differential graded \(K_N\)-modules whose composition with the multiplication map is the transition map (in either direction).

Proof

We first prove this for \(r = 1\). Set \(f = f_1\). Write \(K_t = R\langle x \rangle\), \(K_n = R\langle y \rangle\), and \(K_N = R\langle z \rangle\) with \(x\), \(y\), \(z\) of degree \(1\) and \(\text{d}(x) = f^t\), \(\text{d}(y) = f^n\), and \(\text{d}(z) = f^N\). For all \(N > t\) we claim there is a quasi-isomorphism \[B_{N, t} = R\langle x, z, u \rangle \longrightarrow K_t,\quad x \mapsto x,\quad z \mapsto f^{N - t}x,\quad u \mapsto 0\] Here the left hand side denotes the divided power polynomial algebra in variables \(x\) and \(z\) of degree \(1\) and \(u\) of degree \(2\) with \(\text{d}(x) = f^t\), \(\text{d}(z) = f^N\), and \(\text{d}(u) = z - f^{N - t}x\). To prove the claim, we observe that the following three submodules of \(H_*(R\langle x, z\rangle)\) are the same

1. the kernel of \(H_*(R\langle x, z\rangle) \to H_*(K_t)\),

2. the image of \(z - f^{N - t}x : H_*(R\langle x, z\rangle) \to H_*(R\langle x, z\rangle)\), and

3. the kernel of \(z - f^{N - t}x : H_*(R\langle x, z\rangle) \to H_*(R\langle x, z\rangle)\).

This observation is proved by a direct computation⁴ which we omit. Then we can apply Lemma 0GZ6 part (2) to see that the claim is true.

Via the homomorphism \(K_N \to B_{N, t}\) of differential graded \(R\)-algebras sending \(z\) to \(z\), we may view \(B_{N, t} \to K_t\) as a quasi-isomorphism of left differential graded \(K_N\)-modules. To define the arrow in the statement of the lemma we use the homomorphism \[B_{N, t} = R\langle x, z, u \rangle \to K_n \otimes_R K_t,\quad x \mapsto 1 \otimes x,\quad z \mapsto f^{N - n}y \otimes 1,\quad u \mapsto - f^{N - n - t}y \otimes x\] This makes sense as long as we assume \(N \geq n + t\). It is a pleasant computation to show that the (pre or post) composition with the multiplication map is the transition map.

For \(r > 1\) we proceed by writing each of the Koszul algebras as a tensor product of Koszul algebras in \(1\) variable and we apply the previous construction. In other words, we write \[K_t = R\langle x_1, \ldots, x_r\rangle = R\langle x_1\rangle \otimes_R \ldots \otimes_R R\langle x_r\rangle\] where \(x_i\) is in degree \(1\) and \(\text{d}(x_i) = f_i^t\). In the case \(r > 1\) we then use \[B_{N, t} = R\langle x_1, z_1, u_1 \rangle \otimes_R \ldots \otimes_R R\langle x_r, z_r, u_r \rangle\] where \(x_i, z_i\) have degree \(1\) and \(u_i\) has degree \(2\) and we have \(\text{d}(x_i) = f_i^t\), \(\text{d}(z_i) = f_i^N\), and \(\text{d}(u_i) = z_i - f_i^{N - t}x_i\). The tensor product map \(B_{N, t} \to K_t\) will be a quasi-isomorphism as it is a tensor product of quasi-isomorphisms between bounded above complexes of free \(R\)-modules. Finally, we define the map \[B_{N, t} \to K_n \otimes_R K_t = R\langle y_1, \ldots, y_r\rangle \otimes_R R\langle x_1, \ldots, x_r\rangle\] as the tensor product of the maps constructed in the case of \(r = 1\) or simply by the rules \(x_i \mapsto 1 \otimes x_i\), \(z_i \mapsto f_i^{N - n}y_i \otimes 1\), and \(u_i \mapsto - f_i^{N - n - t}y_i \otimes x_i\) which makes sense as long as \(N \geq n + t\). We omit the details.