# Divided Powers 3

Today we’ll think about notions of compatibility for P.D. structures. The general setup is the following. We have ${(A,I,\gamma)}$ a P.D. ring and ${B}$ an ${A}$-algebra. We’ll say that ${\gamma}$ extends to ${B}$ if there is a P.D. structure ${\overline{\gamma}}$ on ${IB}$ such that ${(A,I, \gamma)\rightarrow (B, IB, \overline{\gamma})}$ is a P.D. morphism. From what we talked about last time we immediately get that ${\gamma}$ extends if and only if there is a P.D. ideal ${(J, \delta)}$ of ${B}$ such that ${(A, I, \gamma)\rightarrow (B, J, \delta)}$ is a P.D. morphism since we’ve checked that ${IB}$ is a sub P.D. ideal of ${J}$.

It also quickly follows from what we’ve done that if there is some extension, then that extension is unique. We should check that this definition isn’t useless by exhibiting an instance where it doesn’t extend. Just pick rings for which ${B=A/J}$ and ${I\cap J}$ is not a sub P.D. ideal of ${I}$ (which necessarily means we picked ${J}$ to have no P.D. structure).

We can guarantee that ${\gamma}$ extends in certain nice situations. For instance, let’s prove that ${\gamma}$ extends to any ${B}$ when our ideal ${I}$ is principal. Suppose ${I=(t)}$, then we need to define ${\overline{\gamma}}$ on ${IB}$ or in other words we need to understand how to define ${\overline{\gamma}_n(tb)}$ for any ${b\in B}$. First, let’s check that the natural guess is well-defined. Define ${\overline{\gamma}_n(tb)=b^n\gamma_n(t)}$. Suppose ${tb=tb'}$, then ${b^n\gamma_n(t)-b'^n\gamma_n(t)=\sum_{i=1}^{n-1} b^ib'^{n-i-1}(b-b')\gamma_n(t)}$. Now since ${\gamma_n(t)\in I}$ we get that it is just another multiple of ${t}$ which shows that ${(b-b')\gamma_n(t)=0}$ making the whole sum ${0}$ and hence ${\overline{\gamma}_n(tb)=\overline{\gamma}_n(tb')}$. Now that we have well-defined, we can just use the fact that ${\gamma}$ is a P.D. structure to check the properties for ${\overline{\gamma}}$.

Here we have the main definition of today. Suppose ${(A, I, \gamma)}$ is a P.D. ring and ${B}$ and ${A}$-algebra with ${(J, \delta)}$ a P.D. ideal of ${B}$. We say that ${\gamma}$ and ${\delta}$ are compatible if any of the following equivalent conditions are met.

1. ${\gamma}$ extends to ${B}$ and ${\overline{\gamma}=\delta}$ on ${IB\cap J}$.

2. The ideal ${K=IB+J}$ has a P.D. structure ${\overline{\delta}}$ such that ${(A, I, \gamma)\rightarrow (B, K, \overline{\delta})}$ and ${(B, J, \delta)\rightarrow (B, K, \overline{\delta})}$ are P.D. morphisms.

3. There is an ideal ${K\subset IB+J}$ with a P.D. structure ${\delta ' }$ such that ${(A, I, \gamma)\rightarrow (B, K', \delta')}$ and ${(B, J, \delta)\rightarrow (B, K, \delta')}$ are P.D. morphisms.

The equivalence is straightforward to check using the if and only if condition listed in the first part of the post for extending a P.D. structure. Let’s go back to thinking geometrically for a second. Then what’s going on is we have some affine variety, and possibly some thickening of it. The original variety has a P.D. structure, and so does the thickening. Compatibility is just saying that the original P.D. structure can be extended over the whole thickening (meaning there is a P.D structure on the thickening that restricts to the original one) and it can be done in such a way that restricting to where the ${\delta}$ P.D. structure is defined also agrees. So as opposed to just an extension, we have what could be thought of as a simultaneous extension of both P.D. structures.

Next time we’ll start talking about some more complicated things by introducing what could be thought of as formal completions of P.D. structures. This shouldn’t be hard to guess what it is if you are comfortable with formal completions of affine varieties.