It's a classical fact that the commutative power series ring $\mathbb{Z}\_p[[x\_1,x\_2]]$ is isomorphic to the completed group algebra $\mathbb{Z}\_p[[\mathbb{Z}\_p\times\mathbb{Z}\_p]]$, the isomorphism sending generators $a\_1,a\_2$ of $\mathbb{Z}\_p\times\mathbb{Z}\_p$ to $1+x\_1,1+x\_2$.


In $\mathbb{Z}\_p[[x\_1,x\_2]]$, we have the coset $1+(x\_1,x\_2)$ of the ideal $(x\_1,x\_2)$, which is a closed and open subgroup of its group of units, which certainly contain $1+x\_1,1+x\_2$, and hence contains the closed subgroup $\langle 1+x\_1,1+x\_2\rangle$ generated by $1+x\_1,1+x\_2$, which is isomorphic to $\mathbb{Z}\_p\times\mathbb{Z}\_p$, but is [strictly smaller than $1+(x\_1,x\_2)$](https://math.stackexchange.com/questions/2386126/is-the-subgroup-1-langle-x-y-rangle-subset-mathbbz-px-y-times-generate/2387081#2387081).


Is it possible to describe the image of the closed subgroup $\langle 1+x\_1,1+x\_2\rangle$ in $\mathbb{Z}\_p[[x\_1,x\_2]]$?


Ie, suppose we are given a power series $f\in 1+(x\_1,x\_2)\subset\mathbb{Z}\_p[[x\_1,x\_2]]$. How can we determine if $f$ lies in the closed subgroup $\langle 1+x\_1,1+x\_2\rangle$?


(I know this question isn't really well-defined. I'm just looking for certain criteria, ideally linked to algebraic properties of the power series, which can be used to identify the image of $\langle 1+x\_1,1+x\_2\rangle$).


It would also be nice if one could describe the quotient $(1+( x\_1,x\_2))/\langle 1+x\_1,1+x\_2\rangle$.



Which elements of $1+(x_1,x_2)\subset\mathbb{Z}_p[[x_1,x_2]]^\times$ are in $\langle 1+x_1,1+x_2\rangle$?
The usual power series for $\log(1+x)$ determines an injection from $1+(x\_1,x\_2)\subset\mathbb{Z}\_p[[x\_1,x\_2]]$ into $\mathbb{Q}\_p[[x\_1,x\_2]]$. A power series $f\in 1+(x\_1,x\_2)$ is in $\langle 1+x\_1,1+x\_2\rangle$ if and only if there are $\alpha\_1$, $\alpha\_2\in\mathbb{Z}\_p$ such that
\begin{align\*}
\log f(x)&=\alpha\_1\log(1+x\_1)+\alpha\_2\log(1+x\_2)\\
&=\alpha\_1\sum\_{n\geq 1}\frac{(-1)^{n-1}}{n}x\_1^n+\alpha\_2\sum\_{n\geq 1}\frac{(-1)^{n-1}}{n}x\_2^n.
\end{align\*}
Note that $\alpha\_1$ and $\alpha\_2$ can be read off from $\log f(x)$: they are the coefficients of $x\_1$ and $x\_2$, respectively. So if
$$
\log f(x)=\sum\_{i,j\geq 0}a\_{ij}x\_1^{i}x\_2^{j},
$$
a necessary and sufficient condition for $f(x)\in\langle 1+x\_1,1+x\_2\rangle$ is that $a\_{ij}=0$ if $i$, $j\geq 1$, and
$$
a\_{n,0}=\frac{(-1)^{n-1}}{n}a\_{1,0},
$$
$$
a\_{0,n}=\frac{(-1)^{n-1}}{n}a\_{0,1}.
$$


