Exercise 25, § 2

Let A be a ring, commutative or otherwise, and let $\sigma$ be an endomorphism of A. On the additive product group $E = A^\mathbf{N}$ an internal law of composition is defined by writing $(\alpha_n)(\beta_n) = (\gamma_n)$, where $\gamma_n = \sum_{p+q=n} \alpha_p \sigma^p(\beta_q)$ (with the convention $\sigma^0(\xi) = \xi$).

(a) Show that this law defines, with addition on E, a ring structure on E, admitting as unit element $e$ the sequence $(\alpha_n)$ where $\alpha_0 = 1$ and $\alpha_n = 0$ for $n \geq 1$. The mapping which associates with every $\xi \in A$ the element $(\alpha_n) \in E$, such that $\alpha_0 = \xi, \alpha_n = 0$ for $n \geq 1$, is an isomorphism of A onto a subring of E, with which it is identified. Then we write $\sum_{n=0}^\infty \alpha_n X^n$ instead of $(\alpha_n)$ and $X^p \beta = \sigma^p(\beta) X^p$ for all $\beta \in A$; if $u = \sum_{n=0}^\infty \alpha_n X^n \neq 0$, the smallest integer $n$ such that $\alpha_n \neq 0$ is called the order of $u$ and denoted by $\omega(u)$.

(b) If A is a ring with no divisor of 0 and $\sigma$ is an isomorphism of A onto a subring of A, show that E is a ring with no divisor of 0 and that

for $u \neq 0$ and $v \neq 0$.

(c) For $u = \sum_{n=0}^\infty \alpha_n X^n$ to be invertible, it is necessary and sufficient that $\alpha_0$ be invertible in A.

(d) Suppose that A is a field and $\sigma$ an automorphism of A. Show that E admits a field of left fractions (I, § 9, Exercise 15) and that in this field F every element $\neq 0$ can be written uniquely in the form $u X^{-h}$, where $u$ is an element of E of order 0.