Let $k$ be a perfect field of characteristic $p>0$ and denote by $W_n(k)$ the ring of Witt vectors over $k$ of length $n$. In their article on the decomposition of the de Rham complex, Deligne and Illusie claim that $W_n(k)$ is the unique flat lifting of $k$ over $\mathbf{Z}/p^n$ (1.3). I know how to justify that $W_n(k)$ satisfies these properties, but it is not clear to me why it should be (up to isomorphism) the only one satisfying them. Translating this into commutative algebra, we need to prove that if $A$ is a flat $\mathbf{Z}/p^n$algebra such that its reduction modulo $p$ is isomorphic to $k$, then we have $A \cong W_n(k)$. Could someone shed a light on this?
1 Answers
$\def\ZZ{\mathbb{Z}}$You can almost find a detailed proof in Rabinoff's notes on Witt vectors. Specifically, Rabinoff proves that, if $k$ is a perfect field of char $p$, then there is a unique $\ZZ_p$algebra $R$ such that $R$ is complete and Hausdorff in the $p$adic topology, $p$ is not a zero divisor and $R/pR \cong k$. I will sketch how to adapt this to work with $\ZZ/p^n$ rather than the $p$adics. This is "folklore" that lots of people knew before Rabinoff, but he is the best reference I know for it.
Let $A$ and $B$ be two $\ZZ/p^n$ algebras, which are flat over $\ZZ/p^n$ and have $A/pA \cong B/pB \cong k$. For $x \in k$, choose a sequence $a_0$, $a_1$, $a_2$, … of elements of $A$ with $a_i^{p^i} \equiv x \bmod pA$; such a sequence exists since $k$ is perfect. Then $a_i^{p^i}$ is eventually constant modulo $p^n$. (Proof: We have $a_{i} \equiv a_{i+1}^p \bmod p A$, so $a_i^{p^i} \equiv a_{i+1}^{p^{i+1}} \bmod p^{i+1} A$ by Rabinoff's Lemma 1.4.) Define $\alpha(x)$ to be the eventual limit of $a_i^{p^i}$ in $A$, and similarly define $\beta(x)$ to be the analogous limit in $B$. These are the analogues of Teichmuller lifts in our setting.
Then $\{ \alpha(x) \}_{x \in k}$ are a set of representatives for $A/pA$, and $p^n=0$ in $A$, so any element of $A$ can be written as $\sum_{i=0}^{n1} \alpha(x_i) p^{i}$ for some sequence $x_i$. Flatness shows that this expression is unique. So we have a bijection $A \longleftrightarrow k^n$ by $\sum_{i=0}^{n1} \alpha(x_i) p^{i} \longleftrightarrow (x_0, x_1, \ldots, x_{n1})$. We similarly have a bijection $k^n \longleftrightarrow B$. We need to show that the composite is an isomorphism of rings. I.e. we need to show that $$\sum_{i=0}^{n1} \alpha(x_i) p^i + \sum_{i=0}^{n1} \alpha(y_i) p^i = \sum_{i=0}^{n1} \alpha(s_i) p^i \ \Longleftrightarrow \ \sum_{i=0}^{n1} \beta(x_i) p^i + \sum_{i=0}^{n1} \beta(y_i) p^i = \sum_{i=0}^{n1} \beta(s_i) p^i$$ and $$\left( \sum_{i=0}^{n1} \alpha(x_i) p^i \right) \left( \sum_{i=0}^{n1} \alpha(y_i) p^i \right) = \sum_{i=0}^{n1} \alpha(z_i) p^i \ \Longleftrightarrow \ \left( \sum_{i=0}^{n1} \beta(x_i) p^i \right) \left( \sum_{i=0}^{n1} \beta(y_i) p^i \right) = \sum_{i=0}^{n1} \beta(z_i) p^i$$
The point is that both of sides of the equivalence these are equivalent to $s_i$ (respectively $z_i$) being given by certain universal polynomials in the $x_i$ and $y_i$; see Theorem 1.5 in Rabinoff.
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Thank you very, very much for this wellthoughtout answer. I was already aware of Rabinoff's notes and of your [recommendation](http://sbseminar.wordpress.com/2009/05/06/rabinoffonwittvectors/) about them, but, due to my lack of time, I opted to just read (some parts of) Serre's local fields. Sometime in the future I'll try to read them. Please give me some time to understand your explanation before accepting your answer. – Nuno Aug 26 '14 at 17:48

Am I missing something obvious or is it possible to prove the same result stated in my initial post assuming only that $k$ is a *perfect domain*? Also, is there some kind of structure theorem for such rings? – Nuno Aug 28 '14 at 15:43

Assuming that perfect domain means integral domain where $p=0$ and $x \mapsto x^p$ is bijective, I think you are right. I can't think of any interesting examples that aren't fields, though. – David E Speyer Aug 28 '14 at 17:06

Thanks for the attention. Since there are other kind of rings called perfect, I should've made that clear. But you guessed it right. By perfect domain I meant an integral domain of characteristic $p>0$ such that the Frobenius endomorphism is an automorphism. – Nuno Aug 30 '14 at 12:37