Playing with idempotents in the ring of Witt vectors

DEFINITION 8.7   Let $A$ be a commutative ring. For any $a\in A$, we denote by $[a]$ the element of $\mathcal W_1(A)$ defined as follows:

$$[a]=(1-a T)_W$$

We call $[a]$ the “Teichmüller lift” of $a$.

LEMMA 8.8   Let $A$ be a commutative ring. Then:

1. $\mathcal W_1(A)$ is a commutative ring with the zero element $[0]$ and the unity $[1]$.

2. For any $a,b \in A$, we have
   $$[a]\cdot [b]=[a b]$$

$\qedsymbol$

PROPOSITION 8.9   Let $A$ be a commutative ring. If $n$ is a positive integer which is invertible in $A$, then $n$ is invertible in $\mathcal W_1(A)$. To be more precise, we have

$$\frac{1}{n}\cdot [1]= \left( (1-T)^{\frac{1}{n}}\right)_W = \left((1+\sum_{j=1}^\infty \binom{\frac{1}{n}}{j} (-T)^j \right)_W.$$

PROOF.. It is easy to find out, by using iterative approximation, an element $x$ of $A[[T]]$ such that

$$(1+x)^n= (1-T).$$

It also follows from the next lemma. $\qedsymbol$

LEMMA 8.10   Let $n$ be a positive integer. Let $k$ be a non negative integer. Then we have always

$$\binom{\frac{1}{n}}{k}\in \mathbb{Z}\left[\frac{1}{n}\right].$$

PROOF..

$$\binom{\frac{1}{n}}{k} = \frac{\frac{1}{n}(\frac{1}{n}-1)\cdots (\frac{1}{n}-(k-1))}{k!}$$

$$= \frac{1}{n^k} \frac{(1(1-n)(1-2n)\dots (1-(k-1)n)}{k!}$$

So the result follows from the next sublemma. $\qedsymbol$

SUBLEMMA 8.11   Let $n$ be a positive integer. Let $k$ be a non negative integer. Let $\{a_j\}_{j=1}^k\subset \mathbb{Z}$ be an arithmetic progression of common difference $n$. Then:

1. For any positive integer $m$ which is relatively prime to $n$, we have
   $$\char93 \{j;\ m \vert a_j\ \} \geq \left\lfloor \frac{k}{m} \right\rfloor$$

2. For any prime $p$ which does not divide $n$, let us define
   $$c_{k,p}= \sum_{i=1}^\infty \lfloor \frac{k}{p^i}\rfloor$$
   (which is evidently a finite sum in practice.) Then
   $$p^{c_{k,p}} \vert \prod_{j=1}^k a_j$$

3. $$p^{c_{k,p}}\vert k!,\qquad p^{c_{k,p}+1} \nmid k!$$

4. $$\frac{\prod_{j=1}^k a_j }{k!} \in \mathbb{Z}_{(p)}$$

PROOF.. (1) Let us put $t=\lfloor\frac{k}{m}\rfloor$. Then we divide the set of first $kt$-terms of the sequence $\{a_j\}$ into disjoint sets in the following way.

$$S_0=\{a_1,a_2,\dots, a_m\},$$

$$S_1=\{a_{m+1},a_{m+2},a_{m+m}\},$$

$$S_2=\{a_{2m+1},a_{2m+2},a_{2m+m}\},$$

$$\dots$$

$$S_{t-1}=\{a_{(t-1)m+1},a_{(t-1)m+2},\dots, a_{(t-1)m+m}\}$$

Since $m$ is coprime to $n$, we see that each of the $S_u$ gives a complete representative of $\mathbb{Z}/n\mathbb{Z}$.

(2): Apply (1) to the cases where $m=p,p^2,p^3,\dots$ and count the powers of $p$ which appear in $\prod a_j$.

(3): Easy. (4) is a direct consequence of (2),(3). $\qedsymbol$

DEFINITION 8.12   For any positive integer $n$ which is invertible in a commutative ring $A$, we define an element $e_n$ as follows:

$$e_n=\frac{1}{n} \cdot (1-T^n)_W.$$

LEMMA 8.13   Let $A$ be a commutative ring. Then for any positive integer $n$ which is invertible in $A$, we have:

1. $e_n$ is an idempotent.

2. $$e_n=(1-\frac{1}{n} T^n+$$   (higher order terms)$$)_W$$

3. If $n\vert m$, with $m$ invertible in $A$, then $e_n \geq e_m$ in the order of idempotents.

PROOF.. if $n\vert m$, then we have

$$e_n \cdot e_m= e_m.$$

$\qedsymbol$

It should be important to note that the range of the projection $e_n$ is easy to describe.

PROPOSITION 8.14   Let $n$ be an integer invertible in $A$. $e_n \cdot \mathcal W_1(A) = \{ (f)_W \vert f \in 1+ T^n A[[T^n]] \}$

PROOF.. Easy. Compare with Lemma below.

$\qedsymbol$

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