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

No.20: Supplement.

LEMMA 20.1   Let $A$ be a commutative ring. Let $n$ be a positive integer. Then:

1. $V_n(\mathcal W_1(A))$ is an ideal of $\mathcal W_1(A)$.
2. If $n$ is invertible in $A$, then
   $$e_n\cdot \mathcal W_1(A)= V_n(\mathcal W_1(A)).$$

DEFINITION 20.2   For any commutative ring $A$, let us define $I(A)$ to be the submodule of $\mathcal W_1(A)$ generated by all the images $V_n(\mathcal W_1(A))$, where $n$ is a positive integer which is not divisible by $p$:

$$I(A)= (\sum_{ p \nmid n} V_n (\mathcal W_1(A)))$$

Let us denote by $\bar I(A)$ its closure. Then we define:

$$\mathcal W^{(p)}(A) \overset{\rm {def}}{=} \mathcal W_1(A) / \overline{I(A)}$$

PROPOSITION 20.3   Let us define a map

$$\Phi:A^{\mathbb{N}} \ni (a_i) \mapsto \sum_{i=0}^\infty (1-a_i T^{p^i})_W \in \mathcal W^{(p)}(A) .$$

Then $\Phi$ is a bijection.

PROOF.. Surjectivity: Every element $(f)_W$ of $\mathcal W_1(A)$ may be written as $(f)_W= \sum_j (1-c_j T^j)_W$. Knowing that $(1-c_j T^j)_W$ is an element of $I(A)$ whenever $j$ is not divisible by $p$, we see that $\Phi((c_{p^i})_{i=0}^\infty)=(f)_W$.

Injectivity: Assume $\Phi((a_i))=\Phi((b_i))$. Let $i_0$ be the smallest integer $i$ such that $a_i \neq b_i$. Then by subtraction we obtain an equation

$$\sum_{i\geq i_0} (1-a_i T^{p^i})_W = \sum_{i\geq i_0} (1-b_i T^{p^i})_W$$

in $\mathcal W^{(p)}(A)$.

$$(1-a_{i_0} T^{p^{i_0}}+$$higher order terms$$)_W =(1-b_{i_0} T^{p^{i_0}}+$$higher order terms$$)_W$$

Since we know that the terms of order $p^{i_0}$ are not affected by additions of elements of $\overline{I(A)}$, we see $a_{i_0}=b_{i_0}$, which is a contradiction.

$\qedsymbol$