Question

let R be a commutative Ring with identity a formal power series f(X) is invertible in R[[x]] if and only if the constant term f0 has an inverse in R

Answer 1

Answer on Question #82698 – Math – Abstract Algebra

Question

Let $R$ be a commutative ring with identity. A formal power series $f(X)$ is invertible in $R[[x]]$ if and only if the constant term $f_0$ has an inverse in $R$.

Solution

1. Necessity.

If $f(X)$ is invertible, then exists $g(X) \in R[[x]]$, such that $f(X)g(X) = 1$, and the constant term of $f(X)g(X)$ is equal to $f_0g_0$, so $f_0g_0 = 1$, so $f_0$ is invertible.

2. Sufficiency.

Suppose $f_0$ is invertible, so $\frac{1}{f_0}$ exists in $R$.

Let's define $g_{n}$ recursively, by induction: $g_0 \coloneqq \frac{1}{f_0}, g_n \coloneqq -\frac{1}{f_0} \left( \sum_{i=1}^{n} f_i g_{n-i} \right)$. It can be defined because $(n - i) < n$, so all $g_{n-i}$ are already defined.

We will prove that $g(x) = g_0 + g_1x + g_2x^2 + \cdots$ is the inverse of $f(x)$, so $f(x)g(x) = 1$.

Let $h_i$ be the coefficient of the $n$-th power of the $x$ in the $f(x)g(x)$. Let's prove that

$h_0 = 1$ and $h_i = 0$ for all $i \in \mathbb{N}$:

$$h_0 = f_0 g_0 = f_0 \cdot \frac{1}{f_0} = 1$$ $$h_n = \sum_{i=0}^{n} f_i g_{n-i} = f_0 g_n + \sum_{k=1}^{n} f_i g_{n-i} = f_0 \cdot \left( -\frac{1}{f_0} \right) \left( \sum_{i=1}^{n} f_i g_{n-i} \right) + \sum_{i=1}^{n} f_i g_{n-i} = 0$$

So, $f(x)g(x) = 1$.

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