# The group of $k$-automorphisms of $k[[x,y]]$, $k$ is a field

Let $k$ be a field. Is the group of $k$-automorphisms of $k[[x,y]]$ known? ($k[[x,y]]$ is the ring of formal power series in two variables, see Wikipedia.)

A somewhat relevant question is this question, which deals with $k[[x]]$, with $k$ any commutative ring.

Thanks for any hints and comments.

• I meant for $k[x,y]$ apply to $(x,y,1)$ the invertible $k$-linear maps $k^2 \to k^2$. For $k[[x,y]]$ you need to find which elements are invertible, Commented Apr 9, 2017 at 4:11
• I know that in $k[[x]]$ an element $\sum_{i=0} a_i x^i$ is invertible iff its constant coefficient $a_0$ is invertible in $k$ (=namely in $k-0$). I guess a 'similar' result holds in $k[[x,y]]$. Commented Apr 9, 2017 at 4:17
• In $k[[x,y]]$: Let $A=\sum a_{ij}x^iy^j$. For $A$ to be invertible in $k[[x,y]]$, it is clear that it is necessary that $a_{00}$ is invertible in $k$. Is $a_{00}$ invertible a sufficient condition for invertibility of $A$? Commented Apr 9, 2017 at 4:26
• @user237522 Yes, a power series with invertible constant term (you may as well make it 1) is invertible. It's essentially the geometric series: $(1 - f)^{-1} = 1 + f + f^2 + f^3 + ...$ Commented Apr 9, 2017 at 4:29
• A recursive formula for the coefficients of the inverse $B$ can be obtained similarly to the one variable case? Commented Apr 9, 2017 at 4:31

It is simple to describe all automorphisms of $R=k[[x,y]]$ as a $k$-algebra. First, note that $R$ is a local ring with maximal ideal $m=(x,y)$, so any automorphism must map $m$ to itself and thus be continuous in the $m$-adic topology. Since $k[x,y]$ is dense in $R$ in the $m$-adic topology, this means that an automorphism is determined by where it sends $x$ and $y$. Moreover, given $f,g\in m$, there is a unique continuous $k$-algebra homomorphism $\varphi:R\to R$ such that $\varphi(x)=f$ and $\varphi(y)=g$ (given by "evaluating" power series by plugging in $f$ for $x$ and $g$ for $y$). So the only question is what conditions on $f$ and $g$ guarantee that this $\varphi$ is an automorphism.
The answer is simple: you just need the images of $f$ and $g$ in the $k$-vector space $m/m^2$ to be linearly independent. Concretely, this just means that the linear homogeneous parts of $f$ and $g$ are linearly independent. Clearly this condition is necessary, since the images of $x$ and $y$ in $m/m^2$ are linearly independent. Conversely, if the linear homogeneous parts of $f$ and $g$ are linearly independent, it is easy to show $\varphi$ is surjective (for any homogeneous polynomial $p$ of degree $n$, you can find a homogeneous polynomial $q$ of degree $n$ such that $p$ is the degree $n$ part of $\varphi(q)$, and then you can use this to build a power series whose image under $\varphi$ has any desired value, one degree at a time). A surjective homomorphism from a Noetherian ring to itself is automatically injective, so it follows that $\varphi$ is an automorphism.
So automorphisms of $R$ are in bijection with pairs of power series with no constant term whose linear parts are linearly independent. Beware that the group operation is very complicated from this description--to compose two automorphisms, you need to compose the power series (that is, substitute the power series of one automorphism for the variables $x$ and $y$ in the power series of the other automorphism).
• I guess it is possible to replace $x,y$ by $x_1,\ldots,x_n$ and get a similar answer to the above? Commented Apr 9, 2017 at 4:52
• Please, there is something that I do not understand: According to your answer $(x,y) \mapsto (x+1,y+2)$ is not an automorphism, but it is (since it is surjective). What am I missing? Thanks. Commented Apr 12, 2017 at 2:15
• That isn't even a well-defined map. To be able to evaluate a power series with $f$ and $g$ substituted for $x$ and $y$, $f$ and $g$ must be in $m$. For instance, try defining where your map would send $\sum_n x^n$. Commented Apr 12, 2017 at 2:17