$\mathbb{Z}_p$ , $\mathbb{Q}_p$ , and the ring of Witt vectors

No.10:

DEFINITION 10.1   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:

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

We call $[a]$ the Teichmüller lift'' of $a$

LEMMA 10.2   Let $A$ be a commutative ring. Then:

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

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

$\qedsymbol$

PROPOSITION 10.3   Let $p$ be a prime number. Let $A$ be a ring of characteristic $p$ . Then:

1. If $n$ is a positive integer which is not divisible by $p$ , then $n$ is invertible in $\mathcal W_1(A)$ . To be more precise,
   $$\frac{1}{n}\boxdot [1]= (1-T)^{\frac{1}{n}}= 1+\sum_{j=1}^\infty \binom{\frac{1}{n}}{j} (-T)^j.$$

2. $p \boxdot : \mathcal W_1(A)\to \mathcal W_1(A)$ is an injection.
3. For any positive integer $n$ which is not divisible by $p$ , we define
   $$e_n=\frac{1}{n}\boxdot (1-T^n).$$
   as an element of $\mathcal W_1(A)$ .
4. For any positive integer $n$ , $e_n$ is an idempotent. (That means, $e_n^{\boxtimes 2}=e_n$ .)
5. If $n\vert m$ , then $e_n \succeq e_m$ in the order of idempotents. That means, $e_n \boxtimes e_m=e_m$ .

PROOF.. (1) follows from the next lemma. $\qedsymbol$

LEMMA 10.4   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..

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

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

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

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

SUBLEMMA 10.5   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. $$p^{c_{k,p}}\vert k!,\qquad p^{c_{k,p}+1} \not \vert k!$$

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). $\qedsymbol$

PROPOSITION 10.6   Let $p$ be a prime. Let $A$ be an integral domain of characteristic $p$ . Let us define an idempotent $f$ of $\mathcal W_1(A)$ as follows.

$$f= \bigvee _{\substack{ n>1\\ p\not \vert n }} e_n (=[1]\boxmi... ...{^\boxtimes } \prod_ {\substack {p \not \vert n\\ n>1 }} ([1]\boxminus e_n))$$

Then $f$ defines a direct product decomposition

$$\mathcal W_1(A) \cong \left ( f \boxtimes \mathcal W_1(A) \right ) \times \left( (1\boxminus f)\boxtimes \mathcal W_1(A) \right).$$

Furthermore, the factor algebra $(1\boxminus f)\boxtimes \mathcal W_1(A)$ is isomorphic to the ring $\mathcal W^{(p)}(A)$ of $p$ -adic Witt vectors.

The following proposition tells us the importance of the ring of $p$ -adic Witt vectors.

PROPOSITION 10.7   Let $p$ be a prime. Let $A$ be a commutative ring of characteristic $p$ . For each positive integer $k$ which is not divisible by $p$ , let us define an idempotent $f_k$ of $\mathcal W_1(A)$ as follows.

$$f_k=\bigvee_{ \substack{p\not \vert n n >1}} e_{k n } (=e_k\bo... ...\boxtimes }\prod _{\substack{p\not \vert n n >1}} (e_k\boxminus e_{k n}))$$

Then $f_k$ defines a direct product decomposition

$$e_k\mathcal W_1(A) \cong \left ( f_k \boxtimes \mathcal W_1(A) \right ) \times \left( (1\boxminus f_k)\boxtimes \mathcal W_1(A) \right).$$

Furthermore, the factor algebra $(1\boxminus f_k)\boxtimes \mathcal W_1(A)$ is isomorphic to the ring $\mathcal W^{(p)}(A)$ of $p$ -adic Witt vectors. Thus we have a direct product decomposition

$$\mathcal W_1(A) \cong \mathcal W^{(p)}(A)^{\mathbb{N}}.$$

To understand the mechanism which appears in the proposition above, it would be better to prove the following

LEMMA 10.8   Let $p$ be a prime number. Let $A$ be a ring of characteristic $p$ . Then for any $n$ which is not divisible by $p$ , a map

$$\frac{1}{n} \boxdot V_n : (\mathcal W_1(A),\boxplus,\boxtimes) \to (\mathcal W_1(A),\boxplus,\boxtimes)$$

is a ring homomorphism. Its image is equal to the range of the idempotent $e_n$ . That means,

$$\operatorname{Image}(\frac{1}{n}\boxdot V_n) =e_n \boxtimes \mathcal W_1(A) =\{ \sideset{}{^\boxplus}\sum_j (1-y_j T^{n j}); y_j \in A\}.$$

PROOF.. $V_n$ is already shown to be additive. The following calculation shows that $\frac{1}{n} \cdot V_n$ preserves the $\boxtimes$ -multiplication.

$$(\frac{1}{n}\boxdot V_n(1-x T^a)) \boxtimes (\frac{1}{n}\boxdot V_n(1-y T^b))$$

$$= (\frac{1}{n}\boxdot (1-x T^{a n})) \boxtimes (\frac{1}{n}\boxdot (1-y T^{b n}))$$

$$= \frac{1}{n^2}\boxdot(1-x^{m/a}y^{m/b} T^{nm} )^d$$

$$= \frac{1}{n} \boxdot((1-x T^a )\boxtimes (1-y T^b))$$

$\qedsymbol$

In preparing from No.7 to No.10 of this lecture, the following reference (especially its appendix) has been useful:

http://www.math.upenn.edu/~chai/course_notes/cartier_12_2004.pdf ARRAY(0x8b0c098)