# stacks/stacks-project

Fetching contributors…
Cannot retrieve contributors at this time
2263 lines (2038 sloc) 83.3 KB
 \input{preamble} % OK, start here. % \begin{document} \title{Divided Power Algebra} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we talk about divided power algebras and what you can do with them. A reference is the book \cite{Berthelot}. \section{Divided powers} \label{section-divided-powers} \noindent In this section we collect some results on divided power rings. We will use the convention $0! = 1$ (as empty products should give $1$). \begin{definition} \label{definition-divided-powers} 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 {\it divided power structure} on $I$ if for all $n \geq 0$, $m > 0$, $x, y \in I$, and $a \in A$ we have \begin{enumerate} \item $\gamma_1(x) = x$, we also set $\gamma_0(x) = 1$, \item $\gamma_n(x)\gamma_m(x) = \frac{(n + m)!}{n! m!} \gamma_{n + m}(x)$, \item $\gamma_n(ax) = a^n \gamma_n(x)$, \item $\gamma_n(x + y) = \sum_{i = 0, \ldots, n} \gamma_i(x)\gamma_{n - i}(y)$, \item $\gamma_n(\gamma_m(x)) = \frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x)$. \end{enumerate} \end{definition} \noindent 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 $I$. \begin{lemma} \label{lemma-silly} Let $A$ be a ring. Let $I$ be an ideal of $A$. \begin{enumerate} \item If $\gamma$ is a divided power structure\footnote{Here and in the following, $\gamma$ stands short for a sequence of maps $\gamma_1, \gamma_2, \gamma_3, \ldots$ from $I$ to $I$.} on $I$, then $n! \gamma_n(x) = x^n$ for $n \geq 1$, $x \in I$. \end{enumerate} Assume $A$ is torsion free as a $\mathbf{Z}$-module. \begin{enumerate} \item[(2)] A divided power structure on $I$, if it exists, is unique. \item[(3)] If $\gamma_n : I \to I$ are maps then $$\gamma\text{ is a divided power structure} \Leftrightarrow n! \gamma_n(x) = x^n\ \forall x \in I, n \geq 1.$$ \item[(4)] 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 $x_i^n \in (n!)I$. \end{enumerate} \end{lemma} \begin{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 $n! \gamma_n(x) = x^n$. \medskip\noindent Assume $A$ is torsion free as a $\mathbf{Z}$-module. Proof of (2). This is clear from (1). \medskip\noindent 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 \ref{definition-divided-powers} 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. \medskip\noindent 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. \end{proof} \begin{example} \label{example-ideal-generated-by-p} 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 \ref{definition-divided-powers} are satisfied. Alternatively, it is clear that the definition works for $A_0 = \mathbf{Z}_{(p)}$ and then the result follows from Lemma \ref{lemma-gamma-extends}. \end{example} \noindent 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 \ref{definition-divided-powers}, applied to $a=0$.) \begin{lemma} \label{lemma-check-on-generators} 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 \begin{enumerate} \item[(a)] (1), (3), and (4) of Definition \ref{definition-divided-powers} hold for all $x, y \in I$, and \item[(b)] properties (2) and (5) hold for $x$ in some set of generators of $I$ as an ideal. \end{enumerate} Then $\gamma$ is a divided power structure on $I$. \end{lemma} \begin{proof} The numbers (1), (2), (3), (4), (5) in this proof refer to the conditions listed in Definition \ref{definition-divided-powers}. 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 $x + y$. \medskip\noindent 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. \medskip\noindent 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. \end{proof} \begin{lemma} \label{lemma-two-ideals} 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 \begin{enumerate} \item $\gamma$ and $\delta$ agree on $IJ$, \item if $\gamma$ and $\delta$ agree on $I \cap J$ then they are the restriction of a unique divided power structure $\epsilon$ on $I + J$. \end{enumerate} \end{lemma} \begin{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 \ref{definition-divided-powers} it follows that they agree on all of $IJ$. \medskip\noindent 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 \ref{definition-divided-powers} 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 \ref{lemma-check-on-generators}. This is clear because $\gamma$ and $\delta$ are divided power structures. \medskip\noindent 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. \end{proof} \begin{lemma} \label{lemma-nil} 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 $A$. \end{lemma} \begin{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. \end{proof} \section{Divided power rings} \label{section-divided-power-rings} \noindent There is a category of divided power rings. Here is the definition. \begin{definition} \label{definition-divided-power-ring} A {\it 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 {\it 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 $n \geq 1$. \end{definition} \noindent 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 $(A, I, \gamma) \to (B, J, \delta)$. \begin{lemma} \label{lemma-limits} The category of divided power rings has all limits and they agree with limits in the category of rings. \end{lemma} \begin{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 \ref{categories-lemma-limits-products-equalizers}. \end{proof} \noindent The following lemma illustrates a very general category theoretic phenomenon in the case of divided power algebras. \begin{lemma} \label{lemma-a-version-of-brown} Let $\mathcal{C}$ be the category of divided power rings. Let $F : \mathcal{C} \to \textit{Sets}$ be a functor. Assume that \begin{enumerate} \item 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 \item $F$ commutes with limits. \end{enumerate} 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 $(A, I, \gamma)$. \end{lemma} \begin{proof} This is a special case of Categories, Lemma \ref{categories-lemma-a-version-of-brown}. \end{proof} \begin{lemma} \label{lemma-colimits} The category of divided power rings has all colimits. \end{lemma} \begin{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 \ref{lemma-a-version-of-brown} to see that $F$ is representable and we win. \end{proof} \begin{remark} \label{remark-forgetful} 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. Then in general the map $B \otimes_A B' \to B''$ isn't an isomorphism. (It is always surjective.) 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 $\delta'_2(2) = 0$ and all higher divided powers equal to zero. Then $(B'', J'', \delta'') = (\mathbf{F}_2, (0), \emptyset)$ which doesn't agree with the tensor product. However, note that it is always true that $$B''/J'' = B/J \otimes_{A/I} B'/J'$$ as can be seen from the universal property of the pushout by considering maps into divided power algebras of the form $(C, (0), \emptyset)$. \end{remark} \section{Extending divided powers} \label{section-extend} \noindent Here is the definition. \begin{definition} \label{definition-extends} Given a divided power ring $(A, I, \gamma)$ and a ring map $A \to B$ we say $\gamma$ {\it 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. \end{definition} \begin{lemma} \label{lemma-gamma-extends} 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 \begin{enumerate} \item $IB = 0$, \item $I$ is principal, or \item $A \to B$ is flat. \end{enumerate} Then $\gamma$ extends to $B$. \end{lemma} \begin{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 \ref{definition-divided-powers} 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. \medskip\noindent 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 $\gamma_n(I) \subseteq I$) and hence annihilated by $b - b'$. Next, we prove conditions (1) -- (5) of Definition \ref{definition-divided-powers}. 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. \medskip\noindent 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 \ref{algebra-theorem-lazard}) 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 $y_j = \sum x_ia_{ij} = \sum x'_{i'}a'_{i'j}$ holds\footnote{This can also be proven without recourse to Algebra, Theorem \ref{algebra-theorem-lazard}. Indeed, if $z = \sum x_ib_i$ and $z = \sum x'_{i'}b'_{i'}$, then $\sum x_ib_i - \sum x'_{i'}b'_{i'} = 0$ is a relation in the $A$-module $B$. Thus, Algebra, Lemma \ref{algebra-lemma-flat-eq} (applied to the $x_i$ and $x'_{i'}$ taking the place of the $f_i$, and the $b_i$ and $b'_{i'}$ taking the role of the $x_i$) yields the existence of the $c_1, \ldots, c_s \in B$ and $a_{ij}, a'_{i'j} \in A$ as required.}. 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 \ref{lemma-check-on-generators} that $\bar\gamma$ is a divided power structure on $IB$. \end{proof} \begin{lemma} \label{lemma-kernel} Let $(A, I, \gamma)$ be a divided power ring. \begin{enumerate} \item 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 $n \geq 1$. \item Let $\mathfrak a \subset A$ be an ideal and set $I' = I \cap \mathfrak a$. The following are equivalent \begin{enumerate} \item $I'$ is preserved by $\gamma_n$ for all $n > 0$, \item $\gamma$ extends to $A/\mathfrak a$, and \item there exist a set of generators $x_i$ of $I'$ as an ideal such that $\gamma_n(x_i) \in I'$ for all $n > 0$. \end{enumerate} \end{enumerate} \end{lemma} \begin{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 \ref{definition-divided-powers} (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. \end{proof} \begin{lemma} \label{lemma-sub-dp-ideal} 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 $\gamma_n(f)$, $n \geq 1$, $f \in E$. \end{lemma} \begin{proof} Follows immediately from Lemma \ref{lemma-kernel}. \end{proof} \begin{lemma} \label{lemma-extend-to-completion} Let $(A, I, \gamma)$ be a divided power ring. Let $p$ be a prime. If $p$ is nilpotent in $A/I$, then \begin{enumerate} \item the $p$-adic completion $A^\wedge = \lim_e A/p^eA$ surjects onto $A/I$, \item the kernel of this map is the $p$-adic completion $I^\wedge$ of $I$, and \item 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 $I^\wedge$. \end{enumerate} If moreover $A$ is a $\mathbf{Z}_{(p)}$-algebra, then \begin{enumerate} \item[(4)] 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. \end{enumerate} \end{lemma} \begin{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 \ref{algebra-lemma-completion-generalities}). 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 \ref{lemma-kernel}. The statement on limits is clear from the construction of limits in the proof of Lemma \ref{lemma-limits}. \end{proof} \section{Divided power polynomial algebras} \label{section-divided-power-polynomial-ring} \noindent A very useful example is the {\it 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 $A\langle x_1, \ldots, x_t \rangle_{+}$. \begin{lemma} \label{lemma-divided-power-polynomial-algebra} 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 \begin{enumerate} \item $\delta_n(x_i) = x_i^{[n]}$, and \item $(A, I, \gamma) \to (A\langle x_1, \ldots, x_t \rangle, J, \delta)$ is a homomorphism of divided power rings. \end{enumerate} Moreover, $(A\langle x_1, \ldots, x_t \rangle, J, \delta)$ has the following universal property: A homomorphism of divided power rings $\varphi : (A\langle x \rangle, J, \delta) \to (C, K, \epsilon)$ is the same thing as a homomorphism of divided power rings $A \to C$ and elements $k_1, \ldots, k_t \in K$. \end{lemma} \begin{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. \medskip\noindent 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 \ref{lemma-two-ideals} 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 \ref{lemma-gamma-extends} as $A \to A\langle x \rangle$ is flat. For the second, note that if $A$ is torsion free, then we can apply Lemma \ref{lemma-silly} (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 \ref{lemma-kernel} we see that the divided power structure on $R\langle x \rangle_{+}$ extends to $A\langle x \rangle$ as desired. \medskip\noindent 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. \end{proof} \begin{remark} \label{remark-divided-power-polynomial-algebra} Let $(A, I, \gamma)$ be a divided power ring. There is a variant of Lemma \ref{lemma-divided-power-polynomial-algebra} 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 \ref{lemma-divided-power-polynomial-algebra}. \end{remark} \noindent The following lemma can be found in \cite{BO}. \begin{lemma} \label{lemma-need-only-gamma-p} 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 \begin{enumerate} \item $p!\delta(x) = x^p$ for all $x \in I$, \item $\delta(ax) = a^p\delta(x)$ for all $a \in A$, $x \in I$, and \item $\delta(x + y) = \delta(x) + \sum\nolimits_{i + j = p, i,j \geq 1} \frac{1}{i!j!} x^i y^j + \delta(y)$ for all $x, y \in I$, \end{enumerate} then there exists a unique divided power structure $\gamma$ on $I$ such that $\gamma_p = \delta$. \end{lemma} \begin{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 \ref{definition-divided-powers}. 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. \end{proof} \section{Tate resolutions} \label{section-tate} \noindent In this section we briefly discuss the resolutions constructed in \cite{Tate-homology} which combine divided power structures with differential graded algebras. In this section we will use {\it 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. \medskip\noindent Let $R$ be a ring. 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. \begin{definition} \label{definition-divided-powers-graded} 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 {\it divided power structure} on $A$ if we have \begin{enumerate} \item $\gamma_n(x) \in A_{2nd}$ if $x \in A_{2d}$, \item $\gamma_1(x) = x$ for any $x$, we also set $\gamma_0(x) = 1$, \item $\gamma_n(x)\gamma_m(x) = \frac{(n + m)!}{n! m!} \gamma_{n + m}(x)$, \item $\gamma_n(xy) = x^n \gamma_n(y)$ for all $x \in A_{even}$ and $y \in A_{even, +}$, \item $\gamma_n(xy) = 0$ if $x, y \in A_{odd}$ homogeneous and $n > 1$ \item if $x, y \in A_{even, +}$ then $\gamma_n(x + y) = \sum_{i = 0, \ldots, n} \gamma_i(x)\gamma_{n - i}(y)$, \item $\gamma_n(\gamma_m(x)) = \frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x)$ for $x \in A_{even, +}$. \end{enumerate} \end{definition} \noindent 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 \ref{section-divided-powers}, \ref{section-divided-power-rings}, \ref{section-extend}, and \ref{section-divided-power-polynomial-ring}. 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. \begin{example}[Adjoining odd variable] \label{example-adjoining-odd} 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 $y \in A_{odd}$. \end{example} \begin{example}[Adjoining even variable] \label{example-adjoining-even} 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. \end{example} \noindent 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 $x \in A_{even, +}$. \begin{definition} \label{definition-divided-powers-dga} 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 {\it compatible with the differential graded structure} if $\text{d}(\gamma_n(x)) = \text{d}(x) \gamma_{n - 1}(x)$ for all $x \in A_{even, +}$. \end{definition} \noindent Warning: Let $(A, \text{d}, \gamma)$ be as in Definition \ref{definition-divided-powers-dga}. 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 insure this is the case, but we elect not to do so. \begin{lemma} \label{lemma-dpdga-good} Let $(A, \text{d}, \gamma)$ and $(B, \text{d}, \gamma)$ be as in Definition \ref{definition-divided-powers-dga}. Let $f : A \to B$ be a map of differential graded algebras compatible with divided power structures. Assume \begin{enumerate} \item $H_k(A) = 0$ for $k > 0$, and \item $f$ is surjective. \end{enumerate} Then $\gamma$ induces a divided power structure on the graded $R$-algebra $H(B)$. \end{lemma} \begin{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 $H(B)$. \end{proof} \begin{lemma} \label{lemma-base-change-div} Let $(A, \text{d}, \gamma)$ be as in Definition \ref{definition-divided-powers-dga}. 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 \ref{definition-divided-powers-dga}. \end{lemma} \begin{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 \ref{definition-extends}). Once we have shown $\gamma$ extends it follows easily that this extension has all the desired properties. \medskip\noindent Choose a polynomial $R$-algebra $P$ 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 \ref{lemma-gamma-extends}. 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 $f \in \Ker(P \to R')$. \end{proof} \begin{lemma} \label{lemma-extend-differential} Let $(A, \text{d}, \gamma)$ be as in Definition \ref{definition-divided-powers-dga}. 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 \ref{example-adjoining-odd} or \ref{example-adjoining-even}. 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 $A\langle T \rangle$. \end{lemma} \begin{proof} This is proved by a direct computation which is omitted. \end{proof} \noindent Here is the construction of Tate. \begin{lemma} \label{lemma-tate-resolution} Let $R$ be a Noetherian ring. Let $R \to S$ be of finite type. There exists a factorization $$R \to A \to S$$ with the following properties \begin{enumerate} \item $(A, \text{d}, \gamma)$ is as in Definition \ref{definition-divided-powers-dga}, \item $A \to S$ is a quasi-isomorphism (if we endow $S$ with the zero differential), \item $A_0 = R[x_1, \ldots, x_n] \to S$ is any surjection of a polynomial ring onto $S$, and \item $A$ is a graded divided power polynomial algebra over $R$ with finitely many variables in each degree. \end{enumerate} The last condition means that $A$ is constructed out of $A_0$ by successively adjoining variables $T$ of degree $> 0$ as in Examples \ref{example-adjoining-odd} and \ref{example-adjoining-even}. \end{lemma} \begin{proof} 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. \medskip\noindent Choose generators $f_1, \ldots, f_m \in R[x_1, \ldots, x_m]$ for the kernel of the given map $A(0) = R[x_1, \ldots, x_m] \to S$. We apply Examples \ref{example-adjoining-odd} $m$ 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 \ref{lemma-extend-differential}. \medskip\noindent 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 $m = 1$. \medskip\noindent 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 \ref{example-adjoining-even} or \ref{example-adjoining-odd} $t$ 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 $A(1)$ is a divided power polynomial algebra over $A(0)$ 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. \medskip\noindent 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 $A = \colim A(m)$. \end{proof} \begin{lemma} \label{lemma-tate-resoluton-pseudo-coherent-ring-map} Let $R \to S$ be a pseudo-coherent ring map (More on Algebra, Definition \ref{more-algebra-definition-pseudo-coherent-perfect}). Then Lemma \ref{lemma-tate-resolution} holds. \end{lemma} \begin{proof} This is proved in exactly the same way as Lemma \ref{lemma-tate-resolution}. 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 \ref{more-algebra-lemma-complex-pseudo-coherent-modules}. Since the complex is exact in (homological) degrees $> m$ we conclude that $H_m$ is a finite $R$-module by More on Algebra, Lemma \ref{more-algebra-lemma-finite-cohomology}. \end{proof} \begin{lemma} \label{lemma-uniqueness-tate-resolution} Let $R$ be a ring. Suppose that $(A, \text{d}, \gamma)$ and $(B, \text{d}, \gamma)$ are as in Definition \ref{definition-divided-powers-dga}. Let $\overline{\varphi} : H_0(A) \to H_0(B)$ be an $R$-algebra map. Assume \begin{enumerate} \item $A$ is a graded divided power polynomial algebra over $R$ with finitely many variables in each degree, \item $H_k(B) = 0$ for $k > 0$. \end{enumerate} Then there exists a map $\varphi : A \to B$ of differential graded $R$-algebras compatible with divided powers lifting $\varphi$. \end{lemma} \begin{proof} Since $A$ is obtained from $R$ by adjoining divided power variables, we obtain filtrations $R \subset A(0) \subset A(1) \subset \ldots$ such that $A(m + 1)$ is obtained from $A(m)$ by adjoining finitely many divided power variables of degree $m + 1$. Then $A(0) \to S$ is a surjection from a (usual) polynomial algebra over $R$ onto $S$. Thus we can lift $\overline{\varphi}$ to an $R$-algebra map $\varphi(0) : A(0) \to B(0)$. \medskip\noindent Write $A(1) = A(0)\langle T_1, \ldots, T_m\rangle$ for some 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 we find a lift $\varphi(1) : A(1) \to B$ of $\varphi(0)$ mapping $T_j$ to $f_j$. \medskip\noindent 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. \end{proof} \begin{lemma} \label{lemma-divided-powers-on-tor} Let $R$ be a Noetherian ring. Let $R \to S$ and $R \to T$ be finite type ring maps. There exists a canonical structure of a divided power graded $R$-algebra on $$\text{Tor}_*^R(S, T)$$ \end{lemma} \begin{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 $R[y_1, \ldots, y_k] \to T$. Then we see that $B$ is a quotient of the differential graded algebra $A[y_1, \ldots, y_k]$ whose homology sits in degree $0$ (it is equal to $S[y_1, \ldots, y_k]$). By Lemma \ref{lemma-base-change-div} the differential graded algebras $B$ and $A[y_1, \ldots, y_k]$ have divided power structures compatible with the differentials. Hence we obtain our divided power structure on $H(B)$ by Lemma \ref{lemma-dpdga-good}. \medskip\noindent 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 \ref{lemma-uniqueness-tate-resolution} implies there is a map $A \to A'$ preserving all structure and the augmentations towards $S$. Then the induced map $B = A \otimes_R T \to A' \otimes_R T' = B'$ is likewise and is a quasi-isomorphism. The induced isomorphism of Tor algebras is therefore compatible with all multiplication and divided powers. \end{proof} \section{Application to complete intersections} \label{section-application-ci} \noindent Let $R$ be a ring. Let $(A, \text{d}, \gamma)$ be as in Definition \ref{definition-divided-powers-dga}. A {\it derivation} of degree $2$ is an $R$-linear map $\theta : A \to A$ with the following properties \begin{enumerate} \item $\theta(A_d) \subset A_{d - 2}$, \item $\theta(xy) = \theta(x)y + x\theta(y)$, \item $\theta$ commutes with $\text{d}$, \item $\theta(\gamma_n(x)) = \theta(x) \gamma_{n - 1}(x)$ for all $x \in A_{2d}$ all $d$. \end{enumerate} In the following lemma we construct a derivation. \begin{lemma} \label{lemma-get-derivation} Let $R$ be a ring. Let $(A, \text{d}, \gamma)$ be as in Definition \ref{definition-divided-powers-dga}. 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 $R$. \end{lemma} \begin{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 \begin{enumerate} \item $\text{d}(xy) = \text{d}(x)y + (-1)^{\deg(x)}x \text{d}(y)$ for $x, y \in A'$ homogeneous, and \item $\text{d}(\gamma_n(x)) = \text{d}(x) \gamma_{n - 1}(x)$ for $x \in A'_{even, +}$. \end{enumerate} 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. \end{proof} \begin{lemma} \label{lemma-compute-theta} Assumption and notation as in Lemma \ref{lemma-get-derivation}. 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 $S/IS$. \end{lemma} \begin{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 $H_2(A/IA)$. \medskip\noindent The construction of $\theta$ in the proof of Lemma \ref{lemma-get-derivation} 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. \end{proof} \noindent The method of proof of the following lemma is apparently due to Gulliksen. \begin{lemma} \label{lemma-not-finite-pd} 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 $R[x_1, \ldots, x_m]$. Assume that \begin{enumerate} \item each $r_j$ has coefficients in the annihilator $I$ of $f$ in $R$, \item 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 $S$. \end{enumerate} Then $S$ does not have finite tor dimension over $R$ (i.e., $S$ is not a perfect $R$-algebra). \end{lemma} \begin{proof} Choose a Tate resolution $R \to A \to S$ as in Lemma \ref{lemma-tate-resolution}. Let $\xi \in H_2(A/IA)$ and $\theta : A/IA \to A/IA$ be the element and derivation found in Lemmas \ref{lemma-get-derivation} and \ref{lemma-compute-theta}. Observe that $$\theta^n(\gamma_n(\xi)) = g^n$$ Hence if $g$ is not nilpotent, 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. \end{proof} \noindent The following result can be found in \cite{Rodicio}. \begin{lemma} \label{lemma-injective} 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. \end{lemma} \begin{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 \ref{lemma-not-finite-pd} (with the ideal $I$ of that lemma equal to $\mathfrak m_R$) and the lemma is proved. \end{proof} \begin{lemma} \label{lemma-regular-sequence} Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset J \subset A$ be proper ideals. Assume \begin{enumerate} \item $A/J$ has finite tor dimension over $A/I$, and \item $J$ is generated by a regular sequence. \end{enumerate} Then $I$ is generated by a regular sequence and $J/I$ is generated by a regular sequence. \end{lemma} \begin{proof} By Lemma \ref{lemma-injective} 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 \ref{more-algebra-lemma-independence-of-generators} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. \end{proof} \begin{lemma} \label{lemma-perfect-map-ci} 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 \begin{enumerate} \item $J$ is generated by a regular sequence and $S/J$ has finite tor dimension as a module over $R/I$, \item $J$ is generated by a regular sequence and $\text{Tor}^{R/I}_p(S/J, R/\mathfrak m_R)$ is nonzero for only finitely many $p$, \item $I$ is generated by a regular sequence and $J/IS$ is generated by a regular sequence in $S/IS$. \end{enumerate} \end{lemma} \begin{proof} If (3) holds, then $J$ is generated by a regular sequence, see for example More on Algebra, Lemmas \ref{more-algebra-lemma-join-koszul-regular-sequences} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. 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 \ref{more-algebra-lemma-flat-push-tor-amplitude}. Thus (3) implies (1). \medskip\noindent The implication (1) $\Rightarrow$ (2) is trivial. Assume (2). By More on Algebra, Lemma \ref{more-algebra-lemma-perfect-over-regular-local-ring} we find that $S/J$ has finite tor dimension over $S/IS$. Thus we can apply Lemma \ref{lemma-regular-sequence} 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 \ref{more-algebra-lemma-independence-of-generators} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. Thus $f_1, \ldots, f_r$ is a regular sequence in $R$ by Algebra, Lemma \ref{algebra-lemma-flat-increases-depth}. \end{proof} \section{Local complete intersection rings} \label{section-lci} \noindent Let $(A, \mathfrak m)$ be a Noetherian complete local ring. By the Cohen structure theorem (see Algebra, Theorem \ref{algebra-theorem-cohen-structure-theorem}) we can write $A$ as the quotient of a regular Noetherian complete local ring $R$. Let us say that $A$ is a {\it 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. \begin{lemma} \label{lemma-ci-well-defined} Let $(A, \mathfrak m)$ be a Noetherian complete local ring. The following are equivalent \begin{enumerate} \item 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 \item 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. \end{enumerate} \end{lemma} \begin{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 \ref{algebra-definition-cohen-ring}) with residue field $k$ (Algebra, Lemma \ref{algebra-lemma-cohen-rings-exist}). 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 \ref{more-algebra-lemma-power-series-ring-over-Cohen-fs}. Fix a surjection $\Lambda[[x_1, \ldots, x_n]] \to A$ as in the Cohen structure theorem (Algebra, Theorem \ref{algebra-theorem-cohen-structure-theorem}). \medskip\noindent 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 \ref{algebra-lemma-flat-increases-depth} and \ref{algebra-lemma-completion-flat}. 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 \ref{algebra-lemma-completion-tensor}). Finally, the ring $R^\wedge$ is regular since $R$ is regular (More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}). Hence we may assume $R$ is complete. \medskip\noindent 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 \ref{more-algebra-lemma-lift-continuous}. 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 \ref{algebra-lemma-ci-well-defined} 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. \end{proof} \noindent The following two lemmas are sanity checks on the definition given above. \begin{lemma} \label{lemma-quotient-regular-ring-by-regular-sequence} 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. \end{lemma} \begin{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 \ref{algebra-lemma-completion-tensor}). The image of the sequence $f_1, \ldots, f_r$ in $R_\mathfrak p$ is a regular sequence by Algebra, Lemmas \ref{algebra-lemma-completion-flat} and \ref{algebra-lemma-flat-increases-depth}. Moreover, $R_\mathfrak p^\wedge$ is a regular local ring by More on Algebra, Lemma \ref{more-algebra-lemma-completion-regular}. Hence the result holds by our definition of complete intersection for complete local rings. \end{proof} \noindent The following lemma is the analogue of Algebra, Lemma \ref{algebra-lemma-lci}. \begin{lemma} \label{lemma-quotient-regular-ring} 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 \begin{enumerate} \item the completion of $A$ is a complete intersection in the sense above, \item $I_\mathfrak p \subset R_\mathfrak p$ is generated by a regular sequence, \item the module $(I/I^2)_\mathfrak p$ can be generated by $\dim(R_\mathfrak p) - \dim(A)$ elements, \item add more here. \end{enumerate} \end{lemma} \begin{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 \ref{algebra-lemma-completion-tensor}). \medskip\noindent 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 \ref{algebra-lemma-completion-flat} and \ref{algebra-lemma-flat-increases-depth}. Thus (1) implies (2). \medskip\noindent 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 \ref{algebra-lemma-NAK}) we see that $I = (f_1, \ldots, f_c)$. Since $R$ is regular and hence Cohen-Macaulay (Algebra, Proposition \ref{algebra-proposition-CM-module}) we see that $f_1, \ldots, f_c$ is a regular sequence by Algebra, Proposition \ref{algebra-proposition-CM-module}. Thus (3) implies (2). Finally, (2) implies (1) by Lemma \ref{lemma-quotient-regular-ring-by-regular-sequence}. \end{proof} \noindent The following result is due to Avramov, see \cite{Avramov}. \begin{proposition} \label{proposition-avramov} Let $A \to B$ be a flat local homomorphism of Noetherian local rings. Then the following are equivalent \begin{enumerate} \item $B^\wedge$ is a complete intersection, \item $A^\wedge$ and $(B/\mathfrak m_A B)^\wedge$ are complete intersections. \end{enumerate} \end{proposition} \begin{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 \ref{algebra-lemma-completion-faithfully-flat}) we conclude that the right vertical arrow is flat (for example by Algebra, Lemma \ref{algebra-lemma-criterion-flatness-fibre-Noetherian}). Moreover, we have $(B/\mathfrak m_A B)^\wedge = B^\wedge/\mathfrak m_{A^\wedge} B^\wedge$ by Algebra, Lemma \ref{algebra-lemma-completion-tensor}. Thus we may assume $A$ and $B$ are complete local Noetherian rings. \medskip\noindent 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 \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings}. 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 $J/I \otimes_R R/\mathfrak m_R \to J/J \cap \mathfrak m_R S$ is an isomorphism. Hence a minimal system of generators of $J/I$ 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. \medskip\noindent 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 \ref{lemma-perfect-map-ci} applies and we deduce that $I$ and $J/I$ 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 \ref{algebra-lemma-dimension-base-fibre-equals-total}). Thus $J/I$ is generated by $$\dim(S/J) - \dim(S/IS) = \dim(S/\mathfrak m_R S) - \dim(B/\mathfrak m_A B)$$ elements (Algebra, Lemma \ref{algebra-lemma-one-equation}). 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 \ref{lemma-quotient-regular-ring}. In this way we see that (2) holds. \medskip\noindent 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 \ref{algebra-lemma-grothendieck-regular-sequence}) we find that $J/I$ is generated by a regular sequence in $S/IS$. Thus Lemma \ref{lemma-perfect-map-ci} tells us that $J$ is generated by a regular sequence, whence (1) holds. \end{proof} \begin{definition} \label{definition-lci} Let $A$ be a Noetherian ring. \begin{enumerate} \item If $A$ is local, then we say $A$ is a {\it complete intersection} if its completion is a complete intersection in the sense above. \item In general we say $A$ is a {\it local complete intersection} if all of its local rings are complete intersections. \end{enumerate} \end{definition} \noindent We will check below that this does not conflict with the terminology introduced in Algebra, Definitions \ref{algebra-definition-lci-field} and \ref{algebra-definition-lci-local-ring}. 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. \begin{lemma} \label{lemma-ci-good} 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. \end{lemma} \begin{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 \ref{algebra-lemma-completion-faithfully-flat}). Then $A_\mathfrak p \to (A^\wedge)_\mathfrak q$ is a flat local ring homomorphism. Thus by Proposition \ref{proposition-avramov} 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). \medskip\noindent 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 \ref{lemma-quotient-regular-ring-by-regular-sequence}. \end{proof} \begin{lemma} \label{lemma-check-lci-at-maximal-ideals} 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 $A$. \end{lemma} \begin{proof} This follows immediately from Lemma \ref{lemma-ci-good} and the definitions. \end{proof} \begin{lemma} \label{lemma-check-lci-agrees} Let $S$ be a finite type algebra over a field $k$. \begin{enumerate} \item for a prime $\mathfrak q \subset S$ the local ring $S_\mathfrak q$ is a complete intersection in the sense of Algebra, Definition \ref{algebra-definition-lci-local-ring} if and only if $S_\mathfrak q$ is a complete intersection in the sense of Definition \ref{definition-lci}, and \item $S$ is a local complete intersection in the sense of Algebra, Definition \ref{algebra-definition-lci-field} if and only if $S$ is a local complete intersection in the sense of Definition \ref{definition-lci}. \end{enumerate} \end{lemma} \begin{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 \ref{algebra-proposition-finite-gl-dim-polynomial-ring}. Hence the equivalence of the two notions in (1) follows by combining Lemma \ref{lemma-quotient-regular-ring} with Algebra, Lemma \ref{algebra-lemma-lci-local}. \medskip\noindent Having proved (1) the equivalence in (2) follows from the definition and Algebra, Lemma \ref{algebra-lemma-lci-global}. \end{proof} \begin{lemma} \label{lemma-avramov} Let $A \to B$ be a flat local homomorphism of Noetherian local rings. Then the following are equivalent \begin{enumerate} \item $B$ is a complete intersection, \item $A$ and $B/\mathfrak m_A B$ are complete intersections. \end{enumerate} \end{lemma} \begin{proof} Now that the definition makes sense this is a trivial reformulation of the (nontrivial) Proposition \ref{proposition-avramov}. \end{proof} \section{Local complete intersection maps} \label{section-lci-homomorphisms} \noindent 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 \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings}. Let us (temporarily) say $A \to S \to B$ is a {\it 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 {\it 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. \begin{lemma} \label{lemma-ci-map-well-defined} Let $A \to B$ be a local homomorphism of Noetherian complete local rings. The following are equivalent \begin{enumerate} \item for some good factorization $A \to S \to B$ the kernel of $S \to B$ is generated by a regular sequence, and \item for every good factorization $A \to S \to B$ the kernel of $S \to B$ is generated by a regular sequence. \end{enumerate} \end{lemma} \begin{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 \ref{algebra-lemma-completion-flat}), 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 \ref{more-algebra-lemma-completion-regular}). 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 \ref{algebra-lemma-flat-increases-depth}. 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 \ref{algebra-lemma-completion-tensor}). 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. \medskip\noindent 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 \ref{more-algebra-lemma-dominate-two-surjections} the ring $S \times_B S'$ is a Noetherian complete local ring. Hence, using More on Algebra, Lemma \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings} 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. \medskip\noindent 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 \ref{algebra-lemma-regular-quotient-regular}). 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 \ref{algebra-lemma-grothendieck-regular-sequence}. \medskip\noindent 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 \ref{more-algebra-lemma-join-koszul-regular-sequences} and \ref{more-algebra-lemma-noetherian-finite-all-equivalent}. \medskip\noindent 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. \end{proof} \noindent 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 $A$. \begin{proposition} \label{proposition-avramov-map} Let $A \to B$ be a local homomorphism of Noetherian local rings. Then the following are equivalent \begin{enumerate} \item $B$ is a complete intersection and $\text{Tor}^A_p(B, A/\mathfrak m_A)$ is nonzero for only finitely many $p$, \item $A$ is a complete intersection and $A^\wedge \to B^\wedge$ is a complete intersection homomorphism in the sense defined above. \end{enumerate} \end{proposition} \begin{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 \ref{algebra-lemma-completion-tensor} and \ref{algebra-lemma-completion-flat}). 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). \medskip\noindent 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 \ref{more-algebra-lemma-embed-map-Noetherian-complete-local-rings}. Let $I = \Ker(R \to A)$ and $J = \Ker(S \to B)$. The proposition now follows from Lemma \ref{lemma-perfect-map-ci}. \end{proof} \begin{remark} \label{remark-no-good-ci-map} 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 {\bf 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 \cite{Rodicio-ci}. We will develop this theory in Dualizing Complexes, Section \ref{dualizing-section-formal-fibres}. \end{remark} \noindent To finish of this section we compare the notion defined above with the notion introduced in More on Algebra, Section \ref{section-lci}. \begin{lemma} \label{lemma-well-defined-if-you-can-find-good-factorization} 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 \begin{enumerate} \item $\Ker(S \to B)$ is generated by a regular sequence, and \item $A^\wedge \to B^\wedge$ is a complete intersection homomorphism as defined above. \end{enumerate} \end{lemma} \begin{proof} Omitted. Hint: the proof is identical to the argument given in the first paragraph of the proof of Lemma \ref{lemma-ci-map-well-defined}. \end{proof} \begin{lemma} \label{lemma-finite-type-lci-map} Let $A$ be a Noetherian ring. Let $A \to B$ be a finite type ring map. The following are equivalent \begin{enumerate} \item $A \to B$ is a local complete intersection in the sense of More on Algebra, Definition \ref{more-algebra-definition-local-complete-intersection}, \item 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. \end{enumerate} \end{lemma} \begin{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 \ref{lemma-well-defined-if-you-can-find-good-factorization} and Algebra, Lemma \ref{algebra-lemma-regular-sequence-in-neighbourhood}. 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 \ref{more-algebra-lemma-noetherian-finite-all-equivalent} we find that $I$ is locally generated by a regular sequence. Hence (2) hold by Lemma \ref{lemma-well-defined-if-you-can-find-good-factorization}. Some details omitted. \end{proof} \begin{lemma} \label{lemma-avramov-map-finite-type} 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 \begin{enumerate} \item $B$ is a complete intersection and $A \to B$ has finite tor dimension, \item $A$ is a complete intersection and $A \to B$ is a complete intersection in the sense of More on Algebra, Definition \ref{more-algebra-definition-local-complete-intersection}. \end{enumerate} \end{lemma} \begin{proof} This is a reformulation of Proposition \ref{proposition-avramov-map} via Lemma \ref{lemma-finite-type-lci-map}. We omit the details. \end{proof} \input{chapters} \bibliography{my} \bibliographystyle{amsalpha} \end{document}