Is there any known result of decomposing multivariable power series over $p$-adic field into product of single variable power series ?


For example, consider the following power series in $n$ variables:$$ f(x\_1,~x\_2, \cdots, x\_n)=\sum\_{j\_1,~j\_2,\cdots, j\_n=0}^{\infty} a\_{j\_1,~j\_2, \cdots, j\_n} \prod\_{k=1}^{n} (x\_k-c\_k)^{j\_k}.$$


Now we want to express $f(x\_1,~x\_2, \cdots, x\_n)$ in the following way:
$$ f(x\_1,~x\_2, \cdots, x\_n)=\left( \sum\_{i\_1=0}^{\infty} a\_{i\_1} x\_1^{i\_1}\right) \cdot \left( \sum\_{i\_2=0}^{\infty} a\_{i\_2} x\_2^{i\_2}\right) \cdots \left( \sum\_{i\_n=0}^{\infty} a\_{i\_n} x\_n^{i\_n}\right).$$
When and under which condition is it possible?


Is there any results or notes or resources available in this regard ?


Thanks,



When can we decompose a multivariable p-adic power series into product of single variable power series?
Some night thoughts on your question which are too long for a comment. For simplicity, I will look at the two variable case. Firstly, there is a very simple discrete criterion for when a function of two variable splits in the way you are interested in: Let $f$ be a function from $X \times Y$. Then is can be represented as a product of two functions of one variable if and only if $$ f(x\_1,y\_1)f(x\_2,y\_2)=f(x\_1,y\_2)f(x\_2,y\_1) $$ for all possible values of the variables.


This is purely set theoretical situation but one can ask the same question in various contexts (continuous, smooth functions, or power series) and a small additional argument is required to show that if a function splits in the above sense, then the factors automatically have the required smoothness.


With respect to a differential condition, I will cheat and suppose that $f$ is a smooth real-valued function on the plane. (I know nothing about the $p$-adic case but hope that this might give you some pointers). Then, as above,
$f f\_{xy}=f\_xf\_y$ is a necessary condition for splitting and the question is whether it is sufficient. This is certainly true (using logarithms) if $f$ has no zeroes. In cases (like yours) where it can only have isolated zeroes I imagine that one could use a localisation argument to prove the sufficiency but I haven’t examined the details.


