algebra

# Divided Power Structures 1

At some point in the distant future we may want to work with Divided Power structures if I ever get around to crystalline cohomology, so why not start writing about it now? Basically this is how we are going to be able to talk about things that require division when working in positive characteristic. Today we’ll just quickly give the definition and then a bunch of easy examples.

Suppose ${A}$ is a commutative ring and ${I}$ an ideal. Divided powers on ${I}$ are a collection of maps ${\gamma_i: I\rightarrow A}$ for all integers ${i\geq 0}$ for which we have the following five properties:

1. For all ${x\in I}$, ${\gamma_0(x)=1}$, ${\gamma_1(x)=x}$ and ${\gamma_i(x)\in I}$
2. For ${x, y\in I}$ we have ${\gamma_k(x+y)=\sum_{i+j=k}\gamma_i(x)\gamma_j(y)}$
3. For ${\lambda\in A}$, ${x\in I}$, we have ${\gamma_k(\lambda x)=\lambda^k\gamma_k(x)}$
4. For ${x\in I}$ we have ${\gamma_i(x)\gamma_j(x)=((i,j))\gamma_{i+j}(x)}$, where ${((i,j))=\frac{(i+j)!}{(i!)(j!)}}$
5. ${\gamma_p(\gamma_q(x))=C_{p,q}\gamma_{pq}(x)}$ where ${C_{p,q}=\frac{(pq)!}{p!(q!)^p}}$, the number of partitions of a set with ${pq}$ elements into ${p}$ subsets with ${q}$ each.

You may have noticed that these conditions seem to just be a formal encoding of the power map ${\gamma_n(x)=x^n}$ from characteristic ${0}$. You’d be wrong, but basically correct. A close examination of the fourth condition actually gives that the map must be ${\gamma_n(x)=\frac{x^n}{n!}}$ when dividing makes sense (i.e. ${A}$ is a ${\mathbb{Q}}$-algebra).

We say ${(I, \gamma)}$ is a P.D. ideal, ${(A, I, \gamma)}$ is a P.D. ring and ${\gamma}$ is a P.D. structure on ${I}$. Of course, P.D. stands for “Divided Power”. OK, not really, it stands for “Puissances Divisees” which is just french for divided powers. We may want to form this into a category, so we’ll say a P.D. morphism ${f:(A, I, \gamma)\rightarrow (B, J, \delta)}$ is a ring map ${f:A\rightarrow B}$ with the property ${f(I)\subset J}$ and it commutes with the divided powers ${f\circ \gamma_n(x)=\delta_n\circ f(x)}$ for all ${x\in I}$.

We already gave the ${\mathbb{Q}}$-algebra example. In fact, in that case every ideal has that as its unique P.D. structure. For any ring, ${\{0\}}$ together with ${\gamma_0(0)=1}$ and ${\gamma_i(0)=0}$ is a P.D. structure. The first interesting example is to let ${V}$ be a DVR of unequal characteristic ${(p,0)}$ with uniformizing parameter ${\pi}$. Recall that if ${p=u\pi^e}$, then ${e}$ is the absolute ramification index of ${V}$. We get that ${\frak{m}}$ has a P.D. structure if and only if ${e\leq p-1}$. In particular, if ${k}$ is perfect, then since ${W(k)}$ is absolutely unramified ${pW(k)}$ has a unique P.D. structure on it.

We can also define subobjects in the obvious way. If ${(I, \gamma)}$ is a P.D. ideal, then another ideal ${J\subset I}$ is a sub P.D. ideal if ${\gamma_i(x)\in J}$ for any ${x\in J}$. In other words, the P.D. structure restricts to be a P.D. structure on the smaller ideal.

Let’s end with a nice little lemma for how find sub P.D. ideals. Suppose ${(A, I, \gamma)}$ is a P.D. algebra, ${S\subset I}$ a subset and ${J}$ the ideal generated by ${S}$. We have that ${J}$ is a sub P.D. ideal if and only if ${\gamma_n(s)\in J}$ for all ${s\in S}$.

Here is the proof. By definition if it is a sub P.D. ideal, then ${\gamma_n(s)\in J}$. That direction is done. Now suppose ${\gamma_n(s)\in J}$ for all ${s\in S}$. Let ${J'}$ be the subset of ${J}$ of the ${x}$ for which ${\gamma_n(x)\in J}$ for all ${n\geq 1}$. By assumption and construction ${S\subset J' \subset J}$. By definition of generation, if ${J'}$ is an ideal, then ${J'=J}$ and we are done. Choose ${x,y \in J'}$ and fix ${n\geq 1}$. Now ${\gamma_n(x+y)=\sum_{i+j=n}\gamma_i(x)\gamma_j(y)\in J}$ since ${J}$ is an ideal. Thus ${x+y\in J'}$. Lastly, suppose ${\lambda\in A}$, then using ${J}$ an ideal again ${\gamma_n(\lambda x)=\lambda^n\gamma_n(x)\in J}$, so ${\lambda x\in J'}$. Thus ${J'}$ is an ideal which proves the lemma.