Help Understanding Proof Involving Formal Power Series

Question

Let $R$ be a ring, and let $f, g \in R[[t]]$.

As part of a proof that I am reading, we have the following setup:

$$ \begin{align} f(0) &= 1\\ g(0) &= 0\\ p f(t) &= f(g(t))\cdot g'(t) \end{align} $$

for some prime, $p$.

The author then concludes, "hence, $g'(t) \in pR[[t]]$". How is this conclusion reached - is it really as obvious as the author makes it seem?

[I should add that I'm not sure how much of the information in the setup is required for the conclusion.]

Answer 1

The easiest way to see the result is to note that any formal power series of the form $(1 + c_1t + c_2t^2 + \cdots)$ is invertible$^\dagger$.

In particular, in your example, $f(g(t))$ is invertible, with inverse, $h(t)$, say. Then, $g'(t) = pf(t)h(t) \in pR[[t]]$.

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A more explicit way to show the result is as follows. Let

\begin{align} f(t) &= 1 + a_1t + a_2 t^2 + \cdots \\ g'(t) &= b_0 + b_1t + b_2t^2 + \cdots \end{align}

Then, substituting this into the equality given in the question, we get

$$ pf(t) = (1 + a_1t + a_2t^2 + \cdots)(b_0 + b_1t + b_2t^2 + \cdots) $$

We can now proceed to show that $p|b_n$ by induction on $n$:

• Base case, $n=0$: The constant term on the RHS is $b_0$ so must be divisible by $p$ (since every coefficient on the LHS is obviously divisible by $p$)
• Suppose $p|b_0, ..., b_k$: The coefficient of $t^{k+1}$ on the RHS is $(b_0 a_k + b_1 a_{k-1} + \cdots + b_{k-1}a_1 + b_k$) and must be divisible by $p$. By our induction hypothesis, this implies $p|b_k$.

$\therefore$ By induction, $p|b_n$ for each $n \in \mathbb{N}$, and hence, $g'(t) \in pR[[t]]$.

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($\dagger$): In fact, a power series in $R[[t]]$ is invertible iff its constant term is invertible in $R$.