# Rabinoff on Witt Vectors

I have sometimes thought of writing a post on the ring of Witt vectors. But now I see that there is no need, because Joe Rabinoff has written a superb guide. Every time that I thought “this is pretty good, but it would be clearer if he pointed out $X$“, the next paragraph was an explanation of $X$!

There is one little thing that I could think to add, so I’ll do that here. I think you will get the most out of this paper if you approach each result as an exercise and try to give your own proof. However, right near the beginning is a theorem — Theorem 1.2, part 2 — which is too hard to be an exercise and where failure will be frustrating rather than illuminating. So I’m going to give you a hint.

Here is the result: Let $K$ be a ring of characteristic $p$ for which the map $x \mapsto x^p$ is bijective. Let $R$ be a ring in which $p$ is not a zero divisor, which is complete and Hausdorff in the $p$adic topology*, and such that $R/pR \cong K$. Theorem/Exercise There is a unique lift $\tau: K \to R$ such that $\tau(x)$ is congruent to $x$ modulo $p$ and such that $\tau(x) \tau(y) = \tau(xy)$.

Here is the hint: $\tau(x)$ can be characterized by the fact that it is the unique lift of $x$ such that $\tau(x)^{1/p^i}$ exists for every $i$.

That’s the only improvement I have. Go and enjoy!

* If the statement that $R$ is complete and Hausdorff is weird for you, here is a restatement: The Hausdorff condition says that, for $x \in R$, if $p^i$ divides $x$ for every $i$, then $x=0$. The complete condition says this: Suppose we have a sequence $x_n$ in $R$ such that, for any $i$, the images of $x_n$ in $R/p^i$ are eventually constant. Then there is an element $x \in R$ such that, for every $i$ the image of $x$ in $R/p^i$ assumes the above-mentioned constant value.

I would advise, however, that you learn to think about these conditions topologically before attempting the Theorem/Exercise.

## 4 thoughts on “Rabinoff on Witt Vectors”

1. Allen Knutson says:

Let K be a ring of characteristic p for which the map x \mapsto x^p is bijective.

Surely such a thing should be called a perfect ring.

As David knows, I’ve run into these recently. If R is a ring, and one wants to classify Frobenius splittings on R[x_1,…,x_n], it’s much easier to do if R is a perfect ring. Unfortunately perfect & Noetherian => field.

2. A minor point: you are probably also assuming that R is local here.
(A product of perfect rings will again be perfect, so e.g. a product of perfect fields is also possible. I agree that this is nit-picking.)

3. Thanks for the kind post! I’m happy that someone found those notes useful. Of course this also prompted me to take a look at them to make sure I didn’t say anything too stupid. So I just posted an updated version with a few small changes of varying importance — e.g. I was horrified to find that I didn’t even mention that my indexing of Witt components is nonstandard (for number theorists anyway). As always, comments are welcome!

4. James says:

Nonstandard, but the right one!