§ 1. Witt vectors

Throughout this paragraph, $p$ denotes a prime number.

1. Witt polynomials

For every integer $n \geqslant 0$ , one calls the $n$ -th Witt polynomial the element $\Phi_n$ of $\mathbf{Z}[X_0, ..., X_n]$ defined by

\Phi_n(X_0, ..., X_n) = \sum_{i=0}^n p^i X_i^{p^{n-i}} = X_0^{p^n} + p X_1^{p^{n-1}} + \cdots + p^n X_n .

One has obviously $\Phi_0 = X_0$ and the recurrence relations

\Phi_{n+1}(X_0, ..., X_{n+1}) = \Phi_n(X_0^p, ..., X_n^p) + p^{n+1} X_{n+1}

\Phi_{n+1}(X_0, ..., X_{n+1}) = X_0^{p^{n+1}} + p \Phi_n(X_1, ..., X_{n+1}) .

When one assigns to $X_i$ the weight $p^i$ , the polynomial $\Phi_n$ is isobaric of weight $p^n$ (A, IV, p. 3).

Proposition 1

Let A be a filtered ring and $(\mathbf{J}_n)_{n \in \mathbf{Z}}$ its filtration. Assume that one has $\mathbf{J}_0 = \mathbf{A}$ and $p.1_A \in \mathbf{J}_1$ . Let m and n be integers such that $m \geqslant 1$ and $n \geqslant 0$ , and let $a_0, ..., a_n, b_0, ..., b_n$ be elements of A.

a) If one has $a_i \equiv b_i \mod \mathbf{J}_m$ for $0 \leqslant i \leqslant n$ , then one has

\Phi_i(a_0, ..., a_i) \equiv \Phi_i(b_0, ..., b_i) \mod \mathbf{J}_{m+i} \quad \text{for } 0 \leqslant i \leqslant n .

b) Suppose that, for every integer $k \geqslant 1$ , and every $x \in \mathbf{A}$ , the relation $p.x \in \mathbf{J}_{k+1}$ entails $x \in \mathbf{J}_k$ . If one has $\Phi_i(a_0, ..., a_i) \equiv \Phi_i(b_0, ..., b_i) \mod \mathbf{J}_{m+i}$ for $0 \leqslant i \leqslant n$ , then one has $a_i \equiv b_i \mod \mathbf{J}_m$ for $0 \leqslant i \leqslant n$ .

Lemma 1

If x and y are two elements of A congruent modulo $\mathbf{J}_m$ , one has

x^{p^n} \equiv y^{p^n} \mod \mathbf{J}_{m+n} .

By induction on n, one reduces to the case where

n = 1

. Let P denote the polynomial

\sum_{i=0}^{p-1} X^i Y^{p-1-i}

\mathbf{Z}[X, Y]

. In view of the assumption made on x and y, one has

P(x, y) \equiv P(x, x) \equiv p.x^{p-1} \mod \mathbf{J}_m

. Now one has

\mathbf{J}_m + p.\mathbf{A} \subset \mathbf{J}_1

, whence

P(x, y) \in \mathbf{J}_1

. Finally,

x^p - y^p = (x - y) P(x, y)

belongs to

\mathbf{J}_m \mathbf{J}_1 \subset \mathbf{J}_{m+1}

Let us prove a) by induction on n. The case $n = 0$ is immediate. Suppose $n \geqslant 1$ . Under the hypotheses of a), one has (4) $a_i^p \equiv b_i^p \mod \mathbf{J}_{m+1}$ for $0 \leqslant i \leqslant n-1$ according to Lemma 1 , (5) $\Phi_{n-1}(a_0^p, ..., a_{n-1}^p) \equiv \Phi_{n-1}(b_0^p, ..., b_{n-1}^p) \mod \mathbf{J}_{m+n}$ according to the induction hypothesis applied to the elements $a_0^p, ..., a_{n-1}^p, b_0^p, ..., b_{n-1}^p$ of A, and (6) $\Phi_n(a_0, ..., a_n) - p^n.a_n \equiv \Phi_n(b_0, ..., b_n) - p^n.b_n \mod \mathbf{J}_{m+n}$ according to formulas (2) and (5). Since $a_n - b_n$ belongs to $\mathbf{J}_m$ , the element $p^n.a_n - p^n.b_n$ belongs to $\mathbf{J}_{m+n}$ and one deduces from (6) the congruence

\Phi_n(a_0, ..., a_n) \equiv \Phi_n(b_0, ..., b_n) \mod \mathbf{J}_{m+n} ,

whence a).

Let us prove b) by induction on n. The case $n = 0$ is immediate. Suppose $n \geqslant 1$ . Under the assumptions of b), we have $a_i \equiv b_i \mod \mathbf{J}_m$ for $0 \leqslant i \leqslant n-1$ by the induction hypothesis, and we deduce from this, as previously, the congruences (4), (5) and (6). But by assumption $\Phi_n(a_0, ..., a_n)$ and $\Phi_n(b_0, ..., b_n)$ are congruent mod. $\mathbf{J}_{m+n}$ , and hence we have $p^n.(a_n - b_n) \in \mathbf{J}_{m+n}$ . Since the relation $p.x \in \mathbf{J}_{k+1}$ implies $x \in \mathbf{J}_k$ for every $x \in \mathbf{A}$ and every $k \geqslant 1$ , we have $a_n - b_n \in \mathbf{J}_m$ , which completes the proof.

2. The mappings $f, v$ and $\Phi$

Let A be a ring. Endow $\mathbf{A}^\mathbf{N}$ with the product ring structure. Denote by $f_A$ , or simply by $f$ , the endomorphism $(a_n)_{n \in \mathbf{N}} \mapsto (a_{n+1})_{n \in \mathbf{N}}$ of $\mathbf{A}^\mathbf{N}$ . Denote by $v_A$ , or simply by $v$ , the endomorphism of the additive group underlying $\mathbf{A}^{\mathbf{N}}$ which associates $(0, p \cdot a_0, p \cdot a_1, ...)$ with $(a_n)_{n \in \mathbf{N}}$ .

For every integer $m \geqslant 0$ , denote by $\Phi_m$ the mapping from $\mathbf{A}^{\mathbf{N}}$ into $\mathbf{A}$ which associates $\Phi_m(a_0, ..., a_m)$ with $a = (a_n)_{n \in \mathbf{N}}$ . We denote by $\Phi_A$ , or simply by $\Phi$ , the mapping $a \mapsto (\Phi_n(a))_{n \in \mathbf{N}}$ from $\mathbf{A}^{\mathbf{N}}$ into itself.

Lemma 2

Let $\mathbf{A}$ be a ring endowed with an endomorphism $\sigma$ satisfying $\sigma(a) \equiv a^p \mod. p \cdot \mathbf{A}$ for every $a \in \mathbf{A}$ . Let $n \geqslant 1$ be an integer and let $a_0, ..., a_{n-1}$ be elements of $\mathbf{A}$ . Put $u_i = \Phi_i(a_0, ..., a_i)$ for $0 \leqslant i \leqslant n - 1$ . Let $u_n$ be an element of $\mathbf{A}$ . The following conditions are equivalent:

a) There exists $a_n \in \mathbf{A}$ such that $u_n = \Phi_n(a_0, ..., a_n)$ .

b) One has $\sigma(u_{n-1}) \equiv u_n \mod. p^n \cdot \mathbf{A}$ .

For $0 \leqslant i \leqslant n - 1$ , one has $\sigma(a_i) \equiv a_i^p \mod. p \cdot \mathbf{A}$ . According to prop. 1 of no. 1 applied to the case where $J_k = p^k \cdot \mathbf{A}$ (for $k \in \mathbf{N}$ ) and where $m = 1$ , one has the congruence

\Phi_{n-1}(\sigma(a_0), ..., \sigma(a_{n-1})) \equiv \Phi_{n-1}(a_0^p, ..., a_{n-1}^p) \mod. p^n \cdot \mathbf{A},

that is to say

\sigma(u_{n-1}) \equiv \Phi_{n-1}(a_0^p, ..., a_{n-1}^p) \mod. p^n \cdot \mathbf{A}.

Now, according to formula (2), the relation $u_n = \Phi_n(a_0, ..., a_n)$ is equivalent to

u_n = \Phi_{n-1}(a_0^p, ..., a_{n-1}^p) + p^n \cdot a_n.

The lemma follows.

Proposition 2

Let $\mathbf{A}$ be a ring.

(a) If $p \cdot 1_A$ is not a divisor of 0 in $\mathbf{A}$ , the mapping $\Phi_A$ is injective.

(b) If $p \cdot 1_A$ is invertible in $\mathbf{A}$ , the mapping $\Phi_A$ is bijective.

(c) If $\sigma$ is an endomorphism of the ring $\mathbf{A}$ , satisfying $\sigma(a) \equiv a^p \mod. p \cdot \mathbf{A}$ for every $a \in \mathbf{A}$ , the image $\mathbf{A}'$ of $\Phi_A$ is a subring of $\mathbf{A}^{\mathbf{N}}$ , stable under $f_A$ and $v_A$ . It is the set of elements $(u_n)_{n \in \mathbf{N}}$ of $\mathbf{A}^{\mathbf{N}}$ such that $\sigma(u_n) \equiv u_{n+1} \mod. p^{n+1} \cdot \mathbf{A}$ for every $n \in \mathbf{N}$ .

If $a = (a_n)_{n \in \mathbf{N}}$ and $u = (u_n)_{n \in \mathbf{N}}$ are elements of $\mathbf{A}^{\mathbf{N}}$ , the relation $\Phi_A(a) = u$ is equivalent, according to formula (2), to the equalities

\begin{cases} u_0 = a_0, \\ u_n = \Phi_{n-1}(a_0^p, ..., a_{n-1}^p) + p^n \cdot a_n & \text{for all } n \geqslant 1. \end{cases}

Let $u = (u_n)_{n \in \mathbf{N}}$ be in $\mathbf{A}^{\mathbf{N}}$ . When $p \cdot 1_A$ is not a divisor of 0 in $\mathbf{A}$ (resp. when $p \cdot 1_A$ is invertible in $\mathbf{A}$ ), there exists at most one sequence $(a_n)_{n \in \mathbf{N}}$ in $\mathbf{A}$ (resp. exactly one sequence $(a_n)_{n \in \mathbf{N}}$ in $\mathbf{A}$ ) satisfying the equalities (10), whence a) and b).

Let us prove c). According to Lemma 2, the image $\mathbf{A}'$ of $\mathbf{A}^{\mathbf{N}}$ by $\Phi_A$ is the set of $u = (u_n)_{n \in \mathbf{N}}$ in $\mathbf{A}^{\mathbf{N}}$ such that $\sigma(u_n) \equiv u_{n+1} \mod. p^{n+1} \cdot \mathbf{A}$ for every $n \in \mathbf{N}$ . It follows immediately that $\mathbf{A}'$ is a subring of $\mathbf{A}^{\mathbf{N}}$ , stable under $f_A$ and $v_A$ .

Remark

Let $a = (a_n)_{n \in \mathbf{N}}$ and $u = (u_n)_{n \in \mathbf{N}}$ be elements of $A^\mathbf{N}$ such that $u = \Phi_A(a)$ , and let $m$ be an integer $\geqslant 0$ . From (10) one deduces the following assertions:

If the $u_n$ , for $0 \leqslant n \leqslant m$ , belong to a subring $B$ of $A$ and if, for every $x \in A$ , the relation $p.x \in B$ implies $x \in B$ , then the $a_n$ , for $0 \leqslant n \leqslant m$ , belong to $B$ .

If $A$ is endowed with a graduation of type $\mathbf{N}$ , if $p.1_A$ is not a divisor of 0 in $A$ , if $d \in \mathbf{N}$ and if $u_n$ is homogeneous of degree $dp^n$ for $0 \leqslant n \leqslant m$ , then $a_n$ is homogeneous of degree $dp^n$ for $0 \leqslant n \leqslant m$ .

3. Construction of polynomials

Let $A$ be the ring $\mathbf{Z}[X, Y]$ of polynomials with integer coefficients in two families of indeterminates $X = (X_n)_{n \in \mathbf{N}}$ and $Y = (Y_n)_{n \in \mathbf{N}}$ . Let $\theta$ be the endomorphism of $A$ defined by $\theta(X_n) = X_n^p$ and $\theta(Y_n) = Y_n^p$ for all $n \in \mathbf{N}$ . Then $p$ is not a divisor of 0 in $A$ and the set of the $a$ in $A$ such that $\theta(a) \equiv a^p \mod. p.A$ is a subring of $A$ containing the $X_n$ and the $Y_n$ , hence equal to $A$ itself.

By the prop. 2, a) and c) of no 2, there exist elements $S = (S_n)_{n \in \mathbf{N}}, P = (P_n)_{n \in \mathbf{N}}, I = (I_n)_{n \in \mathbf{N}}$ and $F = (F_n)_{n \in \mathbf{N}}$ of $A^\mathbf{N}$ characterized respectively by the equalities

\left\{ \begin{array}{l} \Phi_A(S) = \Phi_A(X) + \Phi_A(Y) \\ \Phi_A(P) = \Phi_A(X) \Phi_A(Y) \\ \Phi_A(I) = - \Phi_A(X) \\ \Phi_A(F) = f_A(\Phi_A(X)) \end{array} \right.

The elements $S_n, P_n, I_n$ and $F_n$ of $A$ are therefore characterized by the following formulas (where $n$ runs through $\mathbf{N}$ ):

(12)

\Phi_n(S_0, ..., S_n) = \Phi_n(X_0, ..., X_n) + \Phi_n(Y_0, ..., Y_n),

(13)

\Phi_n(P_0, ..., P_n) = \Phi_n(X_0, ..., X_n) \Phi_n(Y_0, ..., Y_n),

(14)

\Phi_n(I_0, ..., I_n) = - \Phi_n(X_0, ..., X_n),

(15)

\Phi_n(F_0, ..., F_n) = \Phi_{n+1}(X_0, ..., X_{n+1}).

Assign $X_n$ and $Y_n$ the weight $p^n$ for every $n \in \mathbf{N}$ . We deduce from the remark of no 2 the following assertions:

a) We have $S_n \in \mathbf{Z}[X_0, ..., X_n, Y_0, ..., Y_n]$ and $S_n$ is isobaric of weight $p^n$ . b) We have $P_n \in \mathbf{Z}[X_0, ..., X_n, Y_0, ..., Y_n]$ and $P_n$ is isobaric of weight $p^n$ in each of the families $(X_0, ..., X_n)$ and $(Y_0, ..., Y_n)$ . c) We have $I_n \in \mathbf{Z}[X_0, ..., X_n]$ and $I_n$ is isobaric of weight $p^n$ . d) We have $F_n \in \mathbf{Z}[X_0, ..., X_{n+1}]$ and $F_n$ is isobaric of weight $p^{n+1}$ .

Formula (2) makes it possible in practice to determine the polynomials $S_n, P_n, I_n$ and $F_n$ successively.

Example 1

We have

S_0 = X_0 + Y_0

S_1 = X_1 + Y_1 - \sum_{i=1}^{p-1} \frac{1}{p} \binom{p}{i} X_0^i Y_0^{p-i}.

Moreover, $S_n - X_n - Y_n$ belongs to the ring $\mathbf{Z}[X_0, ..., X_{n-1}, Y_0, ..., Y_{n-1}]$ .

Example 2

We have

P_0 = X_0 Y_0

P_1 = p X_1 Y_1 + X_0^p Y_1 + X_1 Y_0^p.

Example 3

When $p \neq 2$ , we have $I_n = - X_n$ . For $p = 2$ , we have

I_0 = - X_0

I_1 = - (X_0^2 + X_1)

I_2 = - X_0^4 - X_0^2 X_1 - X_1^2 - X_2.

Example 4

We have

F_0 = X_0^p + p X_1

F_1 = X_1^p + p X_2 - \sum_{i=0}^{p-1} \binom{p}{i} p^{p-i-1} X_0^{pi} X_1^{p-i}.

Since we have $\Phi_n(F_0, ..., F_n) \equiv \Phi_n(X_0^p, ..., X_n^p) \mod. p^{n+1}$ . A for every $n \in \mathbf{N}$ (formulas (2) and (15)), it follows from prop. 1, b) that we have $F_n \equiv X_n^p \mod. p$ . A for every $n \in \mathbf{N}$ .

Remark

Let $J$ be the set of integers $j \geq 1$ . For every element $j$ of $J$ , define the polynomial $\varphi_j$ of $\mathbf{Z}[(X_j)_{j \in J}]$ by the formula

\varphi_j = \sum_d d X_d^{j/d},

where the sum is taken over the elements of $J$ which divide $j$ . For every integer $n \geq 0$ , we have

\varphi_{p^n} = \Phi_n(X_{p^0}, ..., X_{p^n}).

For every ring $A$ and every element $m$ of $J$ , we denote by $\varphi_m$ the mapping from $A^J$ into $A$ which associates to $(a_j)_{j \in J}$ $\varphi_m((a_j)_{j \in J})$ ; we denote by $\varphi_A$ , or simply $\varphi$ , the mapping from $A^J$ into itself which associates to $a = (a_j)_{j \in J}$ $(\varphi_m(a))_{m \in J}$ .

Let $\mathscr{A} = \mathbf{Z}[(X_j)_{j \in J}, (Y_j)_{j \in J}]$ be the ring of polynomials with integer coefficients in the two families of indeterminates $X = (X_j)_{j \in J}$ and $Y = (Y_j)_{j \in J}$ . One can show (p. 51, exerc. 34) that there exist in $\mathscr{A}$ elements

s = (s_j)_{j \in J}, \quad p = (p_j)_{j \in J} \quad \text{and} \quad i = (i_j)_{j \in J},

characterized by the following equalities:

\varphi_{\mathscr{A}}(s) = \varphi_{\mathscr{A}}(\mathbf{X}) + \varphi_{\mathscr{A}}(\mathbf{Y})

\varphi_{\mathscr{A}}(p) = \varphi_{\mathscr{A}}(\mathbf{X}) \varphi_{\mathscr{A}}(\mathbf{Y})

\varphi_{\mathscr{A}}(i) = - \varphi_{\mathscr{A}}(\mathbf{X}) .

4. The ring W(A) of Witt vectors

Let A be a ring. If $a = (a_n)_{n \in \mathbf{N}}$ and $b = (b_n)_{n \in \mathbf{N}}$ are elements of $A^\mathbf{N}$ , we shall denote by $S_A(a, b)$ (resp. $P_A(a, b)$ , resp. $I_A(a)$ ) or simply $S(a, b)$ (resp. $P(a, b)$ , resp. $I(a)$ ) the sequence $(S_n(a_0, ..., a_n; b_0, ..., b_n))_{n \in \mathbf{N}}$ (resp. $(P_n(a_0, ..., a_n; b_0, ..., b_n))_{n \in \mathbf{N}}$ , resp. $(I_n(a_0, ..., a_n))_{n \in \mathbf{N}}$ ). By substituting $a_n$ for $X_n$ and $b_n$ for $Y_n$ , for every $n \in \mathbf{N}$ , in formulas (12), (13) and (14), one obtains the equalities

(16)

\Phi_A(S_A(a, b)) = \Phi_A(a) + \Phi_A(b)

(17)

\Phi_A(P_A(a, b)) = \Phi_A(a) \Phi_A(b)

(18)

\Phi_A(I_A(a)) = - \Phi_A(a) .

We shall denote by W(A) the set $A^\mathbf{N}$ endowed with the composition laws $S_A$ and $P_A$ .

Let $\rho : B \to A$ be a ring homomorphism. We shall denote by $\rho^\mathbf{N}$ or also W( $\rho$ ) the mapping from $B^\mathbf{N}$ into $A^\mathbf{N}$ which associates to the element $b = (b_n)_{n \in \mathbf{N}}$ of $B^\mathbf{N}$ $(\rho(b_n))_{n \in \mathbf{N}}$ . It follows immediately from the definitions that one has

(19)

W(\rho) \circ S_B = S_A \circ (W(\rho) \times W(\rho))

(20)

W(\rho) \circ P_B = P_A \circ (W(\rho) \times W(\rho))

(21)

W(\rho) \circ I_B = I_A \circ W(\rho)

(22)

\rho^\mathbf{N} \circ \Phi_B = \Phi_A \circ W(\rho) .

Lemma 3

Let A be a ring. There exists a surjective homomorphism of rings $\rho : B \to A$ , where B is a ring satisfying the following conditions: p is not a divisor of 0 in B, and there exists an endomorphism $\sigma$ of B such that $\sigma(b) \equiv b^p \mod. p.B$ for all $b \in B$ .

It suffices indeed to set $B = \mathbf{Z}[(X_a)_{a \in A}]$ , to take for $\sigma$ the endomorphism of B defined by $\sigma(X_a) = X_a^p$ for all $a \in A$ , and for $\rho$ the homomorphism of B into A defined by $\rho(X_a) = a$ for all $a \in A$ .

Theorem 1

a) Let A be a (commutative) ring. Endowed with the addition $S_A$ and the multiplication $P_A$ , W(A) is a (commutative) ring. The identity element for the addition is the sequence $0_A$ all of whose terms are zero; the identity element for the multiplication is the sequence $1_A$ all of whose terms are zero except the one of index 0 which is equal to $1_A$ . The opposite of an element a of W(A) is $I_A(a)$ .

b) Let $\rho : B \to A$ be a homomorphism of rings. Then $W(\rho) : W(B) \to W(A)$ is a homomorphism of rings.

c) Let $A$ be a ring. The mapping $\Phi_A$ is a homomorphism of rings from $W(A)$ into the product ring $A^\mathbf{N}$ . In particular, for all $n \in \mathbf{N}$ , the mapping $\Phi_n : a \mapsto \Phi_n(a_0, ..., a_n)$ is a homomorphism of rings from $W(A)$ into $A$ .

Taking into account formulas (16), (17), (19) and (20), it suffices to prove assertion $a$ .

Let $\rho : B \to A$ be a homomorphism of rings satisfying the conditions of Lemma 3. Let $B'$ be the subring of $B^\mathbf{N}$ formed by the elements $(b_n)_{n \in \mathbf{N}}$ such that $\sigma(b_n) \equiv b_{n+1} \mod. p^{n+1}$ .B for all $n \in \mathbf{N}$ . According to Prop. 2 of No. 2, $\Phi_B$ induces a bijection $\Phi'_B$ from $W(B)$ onto $B'$ . Viewed from formulas (16) to (18) and the relations $\Phi_n(0_B) = 0$ and $\Phi_n(1_B) = 1_B$ ( $n \in \mathbf{N}$ ), one sees by transport of structure that $W(B)$ is a ring, with identity element $0_B$ for the addition, $1_B$ for the multiplication, the opposite of $b$ being $I_B(b)$ .

The mapping $W(\rho) : W(B) \to W(A)$ is surjective. According to formulas (19) and (20), the equivalence relation R on $W(B)$ associated with the mapping $W(\rho)$ is compatible with the ring structure of $W(B)$ . Since $W(\rho)$ induces a bijection $\Psi$ from the quotient ring $W(B)/R$ onto $W(A)$ , compatible with the laws of addition and multiplication, assertion $a$ follows from this by transport of structure.

Definition 1

Let $A$ be a ring. The ring $W(A)$ is called the ring of Witt vectors with coefficients in $A$ .

For $a$ in $W(A)$ and $n$ in $\mathbf{N}$ , the element $\Phi_n(a) = \Phi_n(a_0, ..., a_n)$ is sometimes called the ghost component of index $n$ of $a$ .

Remark

Let us take up again the notations of the remark of No. 3. Let $A$ be a ring. If $a$ and $b$ are elements of $A^J$ and $r = (r_j)_{j \in J}$ is an element of $A^J$ , we denote by $r_A(a, b)$ the element $(r_j(a, b))_{j \in J}$ of $A^J$ . Let us denote by $U(A)$ the set $A^J$ endowed with the composition laws $s_A$ and $p_A$ . One can show (p. 52, Exercise 35) that, endowed with the addition $s_A$ and the multiplication $p_A$ , $U(A)$ is a (commutative) ring; it is called the universal Witt ring of $A$ . The identity element for the addition is the element of $U(A)$ all of whose components are zero; the identity element for the multiplication is the element of $U(A)$ all of whose components are zero except that of index 1 which is equal to $1_A$ ; the opposite of an element $a$ of $U(A)$ is $i_A(a)$ . The mapping $\varphi_A$ is a ring homomorphism of $U(A)$ into the product ring $A^J$ .

Let $\rho : B \to A$ be a ring homomorphism; we denote by $U(\rho)$ the mapping from $B^J$ into $A^J$ which associates with the element $(b_j)_{j \in J}$ of $B^J$ the element $(\rho(b_j))_{j \in J}$ of $A^J$ . One can show (loc. cit.) that $U(\rho)$ is a ring homomorphism of $U(B)$ into $U(A)$ .

5. The homomorphism F and the shift V

Let $A$ be a ring. In the sequel of this paragraph, we denote respectively by $+$ and $\times$ the laws of addition and multiplication in $W(A)$ . We shall also write $0$ for $0_A$ and $1$ for $1_A$ . We define $^1$ two mappings $F_A$ and $V_A$ (also simply denoted by $F$ and $V$ ) from $W(A)$ into itself by the formulas

(23) \quad F_A(a) = (F_n(a_0, ..., a_{n+1}))_{n \in \mathbf{N}},

(24) \quad V_A(a) = (0, a_0, a_1, ...)

(pour $a = (a_n)_{n \in \mathbf{N}}$ dans $W(A)$ ). The mapping $V_A$ is called the shift.

The formula

(25) \quad \Phi_n(F_0(a), ..., F_n(a)) = \Phi_{n+1}(a_0, ..., a_{n+1}) \quad (n \in \mathbf{N})

follows immediately from (15). It can also be written in the form

(26) \quad \Phi_A \circ F_A = f_A \circ \Phi_A .

The formula

(27) \quad \Phi_A \circ V_A = v_A \circ \Phi_A

follows from relation (3).

Let $\rho : B \to A$ be a homomorphism of rings. The relations

(28) \quad W(\rho) \circ F_B = F_A \circ W(\rho)

(29) \quad W(\rho) \circ V_B = V_A \circ W(\rho)

follow immediately from the definitions.

Proposition 3

Let A be a ring.* a) The mapping $F_A$ is an endomorphism of the ring $W(A)$ . b) The mapping $V_A$ is an endomorphism of the additive group underlying the ring $W(A)$ . c) For every $a$ in $W(A)$ , one has $F_A(V_A(a)) = p.a$ (sum in $W(A)$ of $p$ terms equal to $a$ ). d) Whatever $a$ and $b$ in $W(A)$ may be, one has

(30) \quad V_A(a \times F_A(b)) = V_A(a) \times b

(31) \quad V_A(a) \times V_A(b) = p.V_A(a \times b)

(sum in

W(A)

p

terms equal to

V_A(a \times b)

). e) Put $\mu = V_A(1) = (0, 1, 0, ...)$ . For every $b$ in $W(A)$ , one has

(32) \quad V_A(F_A(b)) = \mu \times b .

$^1$ The letter $F$ is the initial of the name Frobenius, and the letter $V$ that of the German word Verschiebung.

f) For every element $a$ of $W(A)$ denote by $a^{*p}$ the product in $W(A)$ of $p$ elements equal to $a$ . Then one has

(33) \quad F_A(a) \equiv a^{*p} \mod. p.W(A) \quad (\text{ideal of } W(A) \text{ generated by } p.\mathbf{1}) .

Let $\rho : B \to A$ be a homomorphism of rings satisfying the conditions of Lemma 3 of No. 4. Then $W(\rho) : W(B) \to W(A)$ is a surjective homomorphism of rings, and $\Phi_B : W(B) \to B^N$ is an injective homomorphism of rings. Moreover, $f_B : B^N \to B^N$ is a homomorphism of rings. According to the formulas (26) and (28), one has

\Phi_B \circ F_B = f_B \circ \Phi_B, \quad W(\rho) \circ F_B = F_A \circ W(\rho),

whence immediately assertion $a$ . Assertion $b$ follows in an analogous manner from the formulas (27) and (29) and from the fact that $v_B$ is an endomorphism of the additive group underlying $B^N$ .

Let $a$ be an element of $W(A)$ , and choose an element $x$ of $W(B)$ which $W(\rho)$ maps onto $a$ . Put $\xi = \Phi_B(x)$ . It follows immediately from the definitions of $f_B$ and $v_B$ that one has $f_B(v_B(\xi)) = p.\xi$ (sum in $B^N$ of $p$ terms equal to $\xi$ ). According to formulas (26) and (27) (where A is replaced by B), the elements $F_B(V_B(x))$ and $p.x$ of $W(B)$ therefore have the same image $p.\xi$ by the injective mapping $\Phi_B$ , and hence are equal. The formula $F_A(V_A(a)) = p.a$ then follows from relations (28) and (29). This proves $c$ .

Reasoning in an analogous manner, we reduce the proof of formula (30) to that of the relation

v_B(\xi f_B(\eta)) = v_B(\xi) \eta

for $\xi, \eta$ in $B^N$ . Now this results from the equalities

\xi f_B(\eta) = (\xi_0 \eta_1, \xi_1 \eta_2, ...)

v_B(\xi) \eta = (0, p\xi_0 \eta_1, p\xi_1 \eta_2, ...)

Taking into account $b$ and $c$ , formula (31) results from formula (30), where $b$ is replaced by $V_A(b)$ . Formula (32) is the particular case $a = 1$ of formula (30).

In an analogous manner, we reduce the proof of formula (33) to that of the relation

f_B(\xi) \equiv \xi^p \mod. p.\Phi_B(B^N),

where $\xi^p$ denotes the product in $B^N$ of $p$ elements equal to $\xi$ . By Prop. 2, c) of No. 2, this is equivalent to the fact that for every $n \geqslant 0$ , one has

\sigma(\xi_{n+1} - \xi_n^p) \equiv \xi_{n+2} - \xi_{n+1}^p \mod. p^{n+2}B .

Now, for every $n \geqslant 0$ , one has, by loc. cit.,

\sigma(\xi_n) \equiv \xi_{n+1} \mod. p^{n+1}B

since $\xi = \Phi_B(x)$ ; one deduces from this, thanks to Lemma 1 of No. 1,

\sigma(\xi_n)^p \equiv \xi_{n+1}^p \mod. p^{n+2}\mathbf{B}.

This proves the desired relation.

Remark

For the definition of mappings analogous to the mappings F and V, in the case of the universal Witt ring, see exercises 36, 37 and 38, p. 52 and following.

6. Filtration and topology of the ring W(A)

Lemma 4

Let A be a ring and m $\geqslant 1$ an integer. One has

(34)

a = (a_0, ..., a_{m-1}, 0, ...) + (\underbrace{0, ..., 0}_{m \text{ terms}}, a_m, a_{m+1}, ...)

for every

a

in W(A).

Let $\rho : B \to A$ be a ring homomorphism satisfying the conditions of Lemma 3 of No. 4. Then $W(\rho) : W(B) \to W(A)$ is a surjective homomorphism of rings, and $\Phi_B : W(B) \to B^N$ is an injective homomorphism. It is therefore enough to prove that one has

(35)

\Phi_n(b) = \Phi_n(b_0, ..., b_{m-1}, 0, ...) + \Phi_n(0, ..., 0, b_m, b_{m+1}, ...)

whatever

b

in W(B) and the integers

m \geqslant 1, n \geqslant 0

may be. Now one has

\Phi_n(b_0, ..., b_{m-1}, ...) = \Phi_n(b_0, ..., b_n) \quad \text{if} \quad 0 \leqslant n < m

= \sum_{i=0}^{m-1} p^i \cdot b_i^{p^{n-i}} \quad \text{if} \quad m \leqslant n

\Phi_n(0, ..., 0, b_m, b_{m+1}, ...) = 0 \quad \text{if} \quad 0 \leqslant n < m

= \sum_{i=m}^n p^i \cdot b_i^{p^{n-i}} \quad \text{if} \quad m \leqslant n,

whence formula (35).

Soit A a ring. For every integer $m \geqslant 0$ , denote by $V_m(A)$ the set of Witt vectors $a = (a_n)_{n \in \mathbf{N}}$ such that $a_n = 0$ for $0 \leqslant n < m$ . It is the image of the m-th power $V^m$ of the mapping $V_A$ . The formulas

(36)

V^m(a + b) = V^m(a) + V^m(b)

(37)

V^m(a) \times b = V^m(a \times F^m(b))

result from prop. 3 of No. 5 by induction on

m

. They imply that

V_m(A)

is an ideal of W(A).

In what follows, we shall endow $W(A)$ with the topology $\mathscr{T}$ associated with the filtration $(V_m(A))_{m \in \mathbf{Z}}$ . Since $V_m(A)$ is an ideal of $W(A)$ for every $m \in \mathbf{Z}$ , the topology $\mathscr{T}$ is compatible with the ring structure of $W(A)$ (TG, III, p. 49, example 3). Let $a \in W(A)$ ; the sets $a + V_m(A)$ , where $m$ runs through $\mathbf{N}$ , form a fundamental system of neighbourhoods of $a$ for $\mathscr{T}$ . Now, it follows from lemma 4 that $a + V_m(A)$ consists of the Witt vectors $b$ such that $a_i = b_i$ for $0 \leq i < m$ . Consequently, $\mathscr{T}$ is none other than the product topology on $A^\mathbf{N}$ of the discrete topology on each of the factors, and $W(A)$ is therefore a separated and complete topological ring (TG, II, p. 17, prop. 10 and TG, III, p. 22, prop. 4).

Let us denote by $\tau_A$ (or simply $\tau$ ) the mapping of $A$ into $W(A)$ which associates with an element $a$ of $A$ $(a, 0, 0, ...)$ . We have $\Phi_n(\tau(a)) = a^{p^n}$ for every $n \in \mathbf{N}$ . For every ring homomorphism $\rho : B \to A$ , we have $W(\rho) \circ \tau_B = \tau_A \circ \rho$ .

Proposition 4

Let $a$ and $b$ be in $A$ and let $x = (x_n)_{n \in \mathbf{N}}$ be an element of $W(A)$ .

a) We have the formulas

\tau(ab) = \tau(a) \times \tau(b)

\tau(a) \times x = (a^{p^n} x_n)_{n \in \mathbf{N}}.

b) The series with general term $V^n(\tau(x_n))$ is convergent in $W(A)$ , with sum $x$ . Let $n$ be a positive integer. The polynomial $P_n(X_0, ..., X_n; Y_0, ..., Y_n)$ introduced in No. 3 is isobaric of weight $p^n$ in the family $(X_0, ..., X_n)$ when $X_i$ is assigned the weight $p^i$ . Hence

P_n(X_0, 0, ..., 0; Y_0, ..., Y_n) = X_0^{p^n} P_n(1, 0, ..., 0; Y_0, ..., Y_n).

Since

1 = (1, 0, 0, ...)

is the unit element of the ring of Witt vectors with coefficients in

\mathbf{Z}[(X_n)_{n \in \mathbf{N}}, (Y_n)_{n \in \mathbf{N}}]

, we have

P_n(1, 0, ..., 0; Y_0, ..., Y_n) = Y_n.

By substituting

a

for

X_0

and

x_i

for

Y_i

, we deduce from (40) and (41) the relation:

P_n(a, 0, ..., 0; x_0, ..., x_n) = a^{p^n} x_n.

According to the definition of the multiplication in

W(A)

, we have proved (39); formula (38) is a particular case of (39).

Let us prove b). By definition, $V^n(\tau(x_n))$ is the sequence all of whose components are zero, except the one with index $n$ , which is equal to $x_n$ . It follows from Lemma 4, by induction on $m$ , that we have

\sum_{n=0}^m V^n(\tau(x_n)) = (x_0, ..., x_m, 0, 0, ...)

for every integer $m \geq 0$ ; we deduce b) by passing to the limit since the topology $\mathscr{T}$ on $W(A)$ is the product of the discrete topologies of the factors $A$ .

7. The rings $W_n(A)$ of Witt vectors of finite length

Definition 2

Let $A$ be a ring and let $n \geq 1$ be an integer. We denote by $W_n(A)$ the quotient ring $W(A)/V_n(A)$ .

Given elements $a_0, ..., a_{n-1}$ of $A$ , we denote by $[a_0, ..., a_{n-1}]$ or $[a_i]_{0 \leq i < n}$ the class modulo $V_n(A)$ of the element $(a_0, ..., a_{n-1}, 0, 0, ...)$ of $W(A)$ . According to Lemma 4 of No. 6, the mapping $(a_0, ..., a_{n-1}) \mapsto [a_0, ..., a_{n-1}]$ from $A^n$ into $W_n(A)$ is a bijection. For this reason, we say that the elements of $W_n(A)$ are the Witt vectors of length $n$ ; by analogy, one sometimes qualifies the elements of $W(A)$ as Witt vectors of infinite length.

We denote by $\pi_n$ the canonical homomorphism from $W(A)$ into $W_n(A)$ . According to Lemma 4 of No. 6, we have

\pi_n(a) = [a_0, ..., a_{n-1}]

for every

a = (a_n)_{n \in \mathbf{N}}

W(A)

According to the definition of the operations in $W(A)$ , we have the following description of the operations in $W_n(A)$ :

[a_0, ..., a_{n-1}] + [b_0, ..., b_{n-1}] = [S_i(a_0, ..., a_i; b_0, ..., b_i)]_{0 \leq i < n}

[a_0, ..., a_{n-1}] \times [b_0, ..., b_{n-1}] = [P_i(a_0, ..., a_i; b_0, ..., b_i)]_{0 \leq i < n}

- [a_0, ..., a_{n-1}] = [I_i(a_0, ..., a_i)]_{0 \leq i < n}.

Moreover, the identity element of the addition in $W_n(A)$ is $[0, ..., 0]$ and that of the multiplication is $[1, 0, ..., 0]$ .

Let $i$ be an integer such that $0 \leq i \leq n$ . By passing to the quotient, the homomorphism $\Phi_i$ from $W(A)$ into $A$ defines a homomorphism $\Phi_i$ from $W_n(A)$ into $A$ . This associates with the Witt vector $[a_0, ..., a_{n-1}]$ the element $\Phi_i(a_0, ..., a_i)$ of $A$ (also called the ghost component of index $i$ of $[a_0, ..., a_{n-1}]$ ).

Let $\rho : B \to A$ be a ring homomorphism. By passing to the quotients, the homomorphism $W(\rho)$ from $W(B)$ into $W(A)$ defines a homomorphism $W_n(\rho)$ from $W_n(B)$ into $W_n(A)$ . It is described by the formula

W_n(\rho)[b_0, ..., b_{n-1}] = [\rho(b_0), ..., \rho(b_{n-1})]

for every

[b_0, ..., b_{n-1}]

W_n(B)

Let $m$ and $n$ be two integers such that $1 \leq n \leq m$ . We have $V_n(A) \supset V_m(A)$ , whence a canonical homomorphism from $W_m(A) = W(A)/V_m(A)$ onto $W_n(A) = W(A)/V_n(A)$ ; this homomorphism will be denoted by $\pi_{n,m}$ . Explicitly, we have

\pi_{n,m}[a_0, ..., a_{m-1}] = [a_0, ..., a_{n-1}]

for $[a_0, ..., a_{m-1}]$ in $W_m(A)$ . The family $(W_n(A), \pi_{n,m})$ is a projective system of rings and the mapping $\pi : a \mapsto (\pi_n(a))_{n \geq 1}$ is a ring homomorphism from $W(A)$ into $\lim_{\leftarrow} W_n(A)$ , called canonical. Since $W(A)$ is separated and complete for the filtration $(V_n(A))_{n \in \mathbf{Z}}$ (cf. No. 6), the canonical homomorphism $\pi$ is an isomorphism of topological rings, when $W_n(A)$ is endowed with the discrete topology for every integer $n \geq 1$ (III, § 2, No. 6).

Henceforth, the homomorphisms $\pi_n$ and $\pi_{n,m}$ will be called projection homomorphisms from $W(A)$ into $W_n(A)$ , and from $W_m(A)$ into $W_n(A)$ respectively.

Example 1

The homomorphism $\Phi_0 : W_1(A) \to A$ is an isomorphism.

Example 2

Let us make explicit the operations in $W_2(A)$ . We have

[a_0, a_1] + [b_0, b_1] = \left[ a_0 + b_0, a_1 + b_1 - \sum_{i=1}^{p-1} \frac{1}{p} \binom{p}{i} a_0^i b_0^{p-i} \right]

[a_0, a_1] \times [b_0, b_1] = [a_0 b_0, a_0^p b_1 + a_1 b_0^p + p \cdot a_1 b_1]

for $[a_0, a_1]$ and $[b_0, b_1]$ in $W_2(A)$ . The ghost components of $[a_0, a_1]$ are $a_0$ and $a_0^p + p \cdot a_1$ .

Example 3

Let $n \geq 1$ be an integer. If $a_0, ..., a_{n-1}, b_0, ..., b_{n-1}$ are integers such that $a_i \equiv b_i \bmod p$ for $0 \leq i < n$ , one has (No. 1, prop. 1)

\Phi_{n-1}(a_0, ..., a_{n-1}) \equiv \Phi_{n-1}(b_0, ..., b_{n-1}) \bmod p^n.

Consequently, $\Phi_{n-1}$ defines by passing to the quotients a ring homomorphism $\varphi_n : W_n(\mathbf{Z}/p\mathbf{Z}) \to \mathbf{Z}/p^n\mathbf{Z}$ . The image of $\varphi_n$ is a subgroup of $\mathbf{Z}/p^n\mathbf{Z}$ containing 1, hence $\varphi_n$ is surjective. Since the finite sets $W_n(\mathbf{Z}/p\mathbf{Z})$ and $\mathbf{Z}/p^n\mathbf{Z}$ have the same cardinal $p^n$ , $\varphi_n$ is an isomorphism.

Let $m$ and $n$ be integers such that $1 \leq n \leq m$ . There exists a unique ring homomorphism $\alpha_{n,m} : \mathbf{Z}/p^m\mathbf{Z} \to \mathbf{Z}/p^n\mathbf{Z}$ ; consequently the diagram

\begin{array}{ccc} \mathbf{Z}/p^m\mathbf{Z} & \xrightarrow{\alpha_{n,m}} & \mathbf{Z}/p^n\mathbf{Z} \\ \varphi_m \uparrow & & \varphi_n \uparrow \\ W_m(\mathbf{Z}/p\mathbf{Z}) & \xrightarrow{\pi_{n,m}} & W_n(\mathbf{Z}/p\mathbf{Z}) \end{array}

is commutative. It follows that $\varphi = \lim \varphi_n$ is an isomorphism of topological rings from $W(\mathbf{Z}/p\mathbf{Z}) = \lim_{\leftarrow} W_n(\mathbf{Z}/p\mathbf{Z})$ onto $\mathbf{Z}_p = \lim_{\leftarrow} \mathbf{Z}/p^n\mathbf{Z}$ (III, § 2, No. 12, example 3).

Let $m$ and $n$ be two integers $\geq 1$ . By construction, one has an exact sequence of additive groups

(E)

0 \longrightarrow W(A) \xrightarrow{\nu^m} W(A) \xrightarrow{\pi_m} W_m(A) \longrightarrow 0.

By passing to the quotients, the endomorphism $V^n$ of the additive group of $W(A)$ defines a homomorphism $V^n_m$ of the additive group of $W_m(A)$ into that of $W_{m+n}(A)$ . In other words, one has a commutative diagram

\begin{array}{ccc} W(A) & \xrightarrow{V^n} & W(A) \\ \pi_m \downarrow & & \pi_{n+m} \downarrow \\ W_m(A) & \xrightarrow{V^n_m} & W_{n+m}(A) \end{array}

By passing to the quotients, one deduces from the exact sequence (E) an exact sequence

(E')

0 \longrightarrow W_m(A) \xrightarrow{V^n_m} W_{n+m}(A) \xrightarrow{\pi_{n,n+m}} W_n(A) \longrightarrow 0 .

One has

(45)

V^n_m[a_0, ..., a_{m-1}] = [\underbrace{0, ..., 0}_{n \text{ fois}}, a_0, ..., a_{m-1}],

for every element $[a_0, ..., a_{m-1}]$ of $W_m(A)$ .

According to prop. 3, c) of No. 5, one has $FV^{m+1}(a) = p.V^m(a)$ for every $a$ in $W(A)$ and one consequently has $F(V_{m+1}(A)) \subset V_m(A)$ . By induction on $n$ , one deduces that $F^n$ maps $V_{n+m}(A)$ into $V_m(A)$ , and hence defines, by passing to the quotients, a ring homomorphism $F^n_m : W_{n+m}(A) \to W_m(A)$ . By construction, one has a commutative diagram

\begin{array}{ccc} W(A) & \xrightarrow{F^n} & W(A) \\ \pi_{n+m} \downarrow & & \pi_m \downarrow \\ W_{n+m}(A) & \xrightarrow{F^n_m} & W_m(A) \end{array}

Recall (No. 3) that the polynomial $F_i$ belongs to $\mathbf{Z}[X_0, ..., X_{i+1}]$ for every integer $i \geqslant 0$ ; the homomorphism $F^1_m$ from $W_{m+1}(A)$ into $W_m(A)$ is therefore made explicit as follows:

(46)

F^1_m[a_0, ..., a_m] = [F_i(a_0, ..., a_{i+1})]_{0 \leqslant i < m}.

Let $a \in W_m(A)$ , $a' \in W_m(A)$ and $b \in W_{m+1}(A)$ . The following formulas result by passing to the quotients from prop. 3 of No. 5:

(47)

F^1_m(V^1_m(a)) = p.a

(48)

V^1_m(a \times F^1_m(b)) = V^1_m(a) \times b

(49)

V^1_m(a) \times V^1_m(a') = p.V^1_m(a \times a')

(50)

V^1_m(F^1_m(b)) = \mu_{m+1} \times b

(avec $\mu_{m+1} = [0, 1, 0, ..., 0]$ ).

8. The ring of Witt vectors with coefficients in a ring of characteristic $p$

Proposition 5

Let $A$ be a ring of characteristic $p$ (A, V, p. 2). Whatever the elements $a$ and $b$ of $W(A)$ , and the positive integers $m, n$ , one has, if $a = (a_n)_{n \in \mathbf{N}}$ ,

(51) \quad F(a) = (a_n^p)_{n \in \mathbf{N}}

(52) \quad p.a = VF(a) = FV(a) = (0, a_0^p, a_1^p, ...)

(53) \quad V^m(a) \times V^n(b) = V^{m+n}(F^n(a) \times F^m(b)) .

Formula (51) results from Example 4 of No. 3. One immediately deduces from this the equality

VF(a) = FV(a) = (0, a_0^p, a_1^p, ...) ,

and the equality $p.a = FV(a)$ has been proved (No. 5, prop. 3), whence (52).

Let us prove (53). According to formula (37) (where one substitutes $V^n(b)$ for $b$ ), one has

(54) \quad V^m(a) \times V^n(b) = V^m(a \times F^m(V^n(b))) .

From formula (37), one also deduces

(55) \quad V^n(F^m(b)) \times a = V^n(F^m(b) \times F^n(a)) .

Formula (53) then results from (54) and (55) and from the relation $F^m \circ V^n = V^n \circ F^m$ , itself a consequence of (51).

Corollary

If $m$ and $n$ are two positive integers, one has

V_m(A) \times V_n(A) \subset V_{m+n}(A) .

This results from formula (53), since $V_m(A)$ is the image of $V^m : W(A) \to W(A)$ .

Proposition 6

Let $A$ be a ring.

a) For every integer $k \geqslant 1$ , one has $(V_1(A))^k = p^{k-1} \cdot V_1(A)$ .

b) Suppose that $A$ is a ring of characteristic $p$ . On the ring $W(A)$ , the $V_1(A)$ -adic topology and the $p$ -adic topology coincide, and they are finer than the product topology $\mathscr{C}$ (cf. No. 6). The ring $W(A)$ is separated and complete for the $p$ -adic topology.

Let us prove a) by induction on $k$ . The case $k = 1$ is evident. Suppose $k \geqslant 2$ . By the induction hypothesis, one has $V_1(A)^{k-1} = p^{k-2} \cdot V_1(A)$ and consequently $V_1(A)^k = p^{k-2} \cdot (V_1(A))^2$ . But it follows from prop. 3, d), formula (31), of No. 5 that one has $(V_1(A))^2 = p \cdot V_1(A)$ , whence a).

Suppose now that $A$ is of characteristic $p$ . As one has

p \cdot W(A) = VF(W(A)) \subset V_1(A) \quad (\text{formula (52)}) ,

one deduces from a) the inclusions $p^k.W(A) \subset (V_1(A))^k \subset p^{k-1}.W(A)$ , and from the corollary to prop. 5 the inclusion $(V_1(A))^k \subset V_k(A)$ , for every integer $k \geqslant 1$ . The first assertion of b) follows.

Let $k$ be an integer $\geqslant 1$ . According to formula (52), the ideal $p^k.W(A)$ of $W(A)$ is the set of elements $a = (a_n)_{n \in \mathbf{N}}$ of $W(A)$ such that one has $a_n = 0$ for $n < k$ and $a_n \in A^{p^k}$ for $n \geqslant k$ . It is therefore closed for the topology $\mathscr{G}$ . Since $W(A)$ is separated and complete for the topology $\mathscr{G}$ (no 6) and since the ideals $p^k.W(A)$ of $W(A)$ , for $k \geqslant 1$ , form a basis of neighbourhoods of $0$ in $W(A)$ for the $p$ -adic topology, the ring $W(A)$ is separated and complete for the $p$ -adic topology (TG, III, p. 26, cor. 1 to prop. 10).

Proposition 7

Let A be a perfect ring of characteristic p.

a) For every element $a = (a_n)_{n \in \mathbf{N}}$ of $W(A)$ , the series with general term $p^n \tau(a_n^{p^{-n}})$ is convergent in $W(A)$ , with sum $a$ .

b) On $W(A)$ , the $V_1(A)$ -adic topology, the $p$ -adic topology and the topology $\mathscr{G}$ coincide. More precisely, one has $V_n(A) = p^n.W(A) = (V_1(A))^n$ for every integer $n \geqslant 0$ . In particular $\Phi_0$ defines an isomorphism of $W(A)/p.W(A)$ onto $A$ .

By definition (A, V, p. 5), the mapping $a \mapsto a^p$ is an automorphism of the ring $A$ . According to prop. 5, F is therefore an automorphism of the ring $W(A)$ , and one has, for every $n \in \mathbf{N}$ ,

p^n.W(A) = V^n F^n(W(A)) = V^n(W(A)) = V_n(A).

In particular, one has $(V_1(A))^n = (p.W(A))^n = p^n.W(A)$ . The assertion b) follows from this.

According to prop. 5, one has

p^n.\tau(a_n^{p^{-n}}) = V^n F^n \tau(a_n^{p^{-n}}) = V^n \tau(a_n),

and the assertion a) follows from prop. 4 of no 6.

Proposition 8

Let A be a field of characteristic p. The ring $W(A)$ is a separated and complete integral local ring, with maximal ideal $V_1(A)$ and residue field isomorphic to A. If the field A is perfect, the ring $W(A)$ is a discrete valuation ring, and its maximal ideal is $p.W(A)$ .

The homomorphism $\Phi_0$ defines an isomorphism of $W(A)/V_1(A)$ onto $A$ (no 7, example 1). The ideal $V_1(A)$ of $W(A)$ is therefore maximal. Since the ring $W(A)$ is separated and complete for the $V_1(A)$ -adic topology (prop. 6, $b$ ), it is a local ring, with maximal ideal $V_1(A)$ (III, § 2, no 13, prop. 19).

Let $a$ and $b$ be two nonzero elements of $W(A)$ . There exist integers $m \geqslant 0$ and $n \geqslant 0$ , and elements $a' = (a'_n)_{n \in \mathbf{N}}$ and $b' = (b'_n)_{n \in \mathbf{N}}$ of $W(A)$ such that $a = V^m(a')$ , $b = V^n(b')$ and that the elements $a'_0$ and $b'_0$ of $A$ are nonzero. Then the component of index $m + n$ of $a \times b$ is equal to the component of index 0 of $F^n(a') \times F^m(b')$ (formula (53)), that is to say to ${a'_0}^{p^n} {b'_0}^{p^m}$ (formula (51) and No. 3, Example 2). Consequently $a \times b$ is nonzero and $W(A)$ is an integral domain.

If the field $A$ is perfect, the maximal ideal $V_1(A)$ of $W(A)$ is equal to $p.W(A)$ (Prop. 7, b)) and consequently W(A) is a discrete valuation ring (VI, § 3, No. 6, Prop. 9, c)).

Remark 1

Let A be a field of characteristic p. One can now prove that the ring W(A) is noetherian if and only if A is perfect (p. 43, Exercise 9).

Remark 2

Let A be a ring of characteristic p. According to Prop. 5, one has the formulas

F_m^n[a_0, ..., a_{n+m-1}] = [a_0^{p^n}, ..., a_{m-1}^{p^n}]

p^n.[a_0, ..., a_{n+m-1}] = [\underbrace{0, ..., 0}_{n \text{ fois}}, a_0^{p^n}, ..., a_{m-1}^{p^n}]

for every Witt vector $[a_0, ..., a_{n+m-1}]$ of length $n + m$ .

In fact, the mapping $F : W(A) \to W(A)$ allows one, by passing to the quotients by $V_m(A)$ , to define a mapping $\overline{F}_m : W_m(A) \to W_m(A)$ . One has the formula

\overline{F}_m[a_0, ..., a_{m-1}] = [a_0^p, ..., a_{m-1}^p].

The mappings $V_m^1 \circ \overline{F}_m$ and $\overline{F}_{m+1} \circ V_m^1$ from $W_m(A)$ into $W_{m+1}(A)$ are equal and are obtained, by passing to the quotient, from multiplication by $p$ in $W_{m+1}(A)$ .

Exercises

In Exercises 1 to 27, $p$ is a fixed prime number. If $A$ is a ring, the ring of Witt vectors $W(A)$ is the one attached to the prime number $p$ .

§ 1. Witt vectors

1. Witt polynomials

Proposition 1

Lemma 1

2. The mappings f,vf, vf,v and Φ\PhiΦ

Lemma 2

Proposition 2

Remark

3. Construction of polynomials

Example 1

Example 2

Example 3

Example 4

Remark

4. The ring W(A) of Witt vectors

Lemma 3

Theorem 1

Definition 1

Remark

5. The homomorphism F and the shift V

Proposition 3

Remark

6. Filtration and topology of the ring W(A)

Lemma 4

Proposition 4

7. The rings Wn(A)W_n(A)Wn​(A) of Witt vectors of finite length

Definition 2

Example 1

Example 2

Example 3

8. The ring of Witt vectors with coefficients in a ring of characteristic ppp

Proposition 5

Corollary

Proposition 6

Proposition 7

Proposition 8

Remark 1

Remark 2

Exercises

2. The mappings $f, v$ and $\Phi$

7. The rings $W_n(A)$ of Witt vectors of finite length

8. The ring of Witt vectors with coefficients in a ring of characteristic $p$