$\Lambda(A)$ for arbitrary commutative ring $A$

In the previous lecture we defined the ring structure on $\Lambda(A)$ for $A=\Omega$, a field of characteristic 0. Now we want to define the structure for arbitrary commutative ring $A$. Note that addition is already known:

$$(f)_W+(g)_W=(fg)_W$$

We would like to know the product $(f)_W(g)_W$. Before doing that, we consider “universal” power serieses:

$$a(T)=1+a_1 T + a_2 T^2 +a_3 T^3+\dots,$$

$$b(T)=1+b_1 T + b_2 T^2 +b_3 T^3+\dots,$$

with $a_1,a_2,\dots, b_1, b_2,b_3,\dots$ be all independent variables. We need a fairly large field $\Omega$, namely,

$$\Omega=\overline{\mathbb{Q}(a_1,a_2,\dots, b_1,b_2,\dots)},$$

the algebraic closure of an infinite trancendent extension of $\mathbb{Q}$. We find:

$$(a(T))_W (b(T))_W=(m_{a,b}(T))W$$

where

$$m_{a,b}(T) =1+ m_{a,b;1}T +m_{a,b;2}T^2 +m_{a,b;3}T^3 +\dots$$

with $m_{a,b;k} \in \Omega$.

We also see:

• For fixed $a$, $m_{a,b,k}$ only depend on $b_1,b_2,b_3,\dots, b_k$. (In other words, it is an element of $\overline{\mathbb{Q}(a_1,\dots,a_k,b_1,b_2,\dots,b_k)}$.
• By using a Galois-theoretic arguments (or by using arguments on symmetric polynomials,) we see that $m_{a,b,k}$ actually lie in $\mathbb{Q}(a_1,a_2,\dots,a_k,b_1,b_2,\dots, b_k)$.
• $m_{a,b,k}$ is integral over the polynomial ring $\mathbb{Z}[a_1,a_2,\dots,a_k,b_1,b_2,\dots, b_k]$. It is thus itself belongs to the ring $\mathbb{Z}[a_1,a_2,\dots,a_k,b_1,b_2,\dots, b_k]$.
• The fact that $\Lambda(\Omega)$ obeys each of the axioms of ring, such as
  $$((a)_W (b)_W) (c)_W =(a)_W( (b)_W (c)_W)$$
  (associativity), gives a set of polynomial identities in $a,b,c,m_{ab;k},m_{bc,k}$. Such identities in term guarantees that for any ring $A$, $\Lambda(A)$ satisfy such axiom.

PROPOSITION 06.1   For any commutative ring $A$, $\Lambda(A)$ carries the structure of a ring.