# Automorphism of a formal power series ring

Is the following theorem true? If yes, how would you prove it?

Theorem Let $A$ be a commutative ring. Let $A[[x]]$ be the ring of formal power series in one variable. Let $\mathfrak{m}$ be the ideal of $A[[x]]$ generated by $x$. Let $u$ be an invertible element of $A$. Let $f(x) = ux + g(x)$, where $g(x) \in \mathfrak{m}^2$. Then there exists a unique automorphism $\psi$ of $A[[x]]$ fixing every element of $A$ such that $\psi(x) = f$.

-
I presume you meant "a unique automorphism $\psi$". Isomorphism asks for a domain and a codomain; automorphism assumes they are equal. – M Turgeon Jul 20 '12 at 22:47
I edited my question regarding automorphism. Thanks. – Makoto Kato Jul 20 '12 at 22:50

## 2 Answers

According to Eisenbud Theorem 7.16a, there is a unique homomorphism $\psi$ defined as in the question. To see that $\psi$ is an isomorphism, it suffices to check that $\ker\psi=0.$ But this is immediate, since, assuming that $p$ is a power series we have $0=\psi(p(x))=p(\psi(x))$ which implies the coefficient $u=0.$

-
I wonder where you use the assumption that $f(x) = ux + g(x)$ and $u$ is an invertible element of $A$. Regards. – Makoto Kato Jul 20 '12 at 23:06
The assumption that $f(x)=ux+g(x)$ implies in particular that $f(x)\in \langle x\rangle\subset A[[x]],$ where $\langle x\rangle$ is the ideal with respect to which $A[[x]]$ is complete. The assumption that $u$ is invertible allows us to conclude that the lowest degree term of $\psi(p(x))$ for $p\neq 0$ must have a non-zero coefficient, namely $u.$ – Andrew Jul 20 '12 at 23:10
Is it immediate that $\psi$ is surjective? – Dylan Moreland Jul 20 '12 at 23:12
Hmm, maybe that's not obvious. But I think it should be true. Consider $q\in A[[x]].$ Then we match the first non-zero coefficient to give a first approximation of a $p$ such that $\psi(p)=q.$ Then all terms of $\psi(p)-q$ have higher degree, and we can again solve for the coefficients of higher order of $p$, continuing ad infinitum. – Andrew Jul 20 '12 at 23:17
On surjectivity: In dealing with power series, one should be fully aware that they are completely computable. By hand! To say that the associated homomorphism is invertible is exactly the same as saying that the power series itself is invertible in the sense of substitution of power series. Once you've done a couple of hand computations of power-series substitution, you see that if the first-degree coefficient has a reciprocal, the power series is invertible. – Lubin Jul 21 '12 at 0:08

$\psi$ is "evaluate at $f(x)$" is a homomorphism is a standard (and not very enlightening) manipulation of power series, so I'll omit it.

To check that it's invertible, it suffices to find $f^{-1}(x)$. To solve $f(y) = x$, or $x = uy + g(y)$:

$$y = u^{-1} x + u^{-1} g(y) = u^{-1} x + u^{-1} g\left(u^{-1} x + u^{-1} g(y)\right) = \cdots$$

Because $g(x) \equiv 0 \mod{x^2}$, it's not hard to see that this series converges. The inverse of $\psi$ is then "Evaluate at $y$".

It might be easier to use Newton's method, starting with the initial approximation

$$y \approx u^{-1} x$$

I do not know a proof that $\psi$ is the only ring homomorphism (over $A$) that sends $x$ to $f(x)$, though it's obviously the only one that's continuous.

-