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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,

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    $\begingroup$ A necessary condition is given by $$\frac{\partial f}{\partial x_i} \frac{\partial f}{\partial x_j} = \left( \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_j} f \right) f$$ for all $i \neq j$. My guess is that this is also sufficient (at least in characteristic zero, which you are assuming anyway). $\endgroup$
    – RP_
    Aug 28 '20 at 13:35
  • $\begingroup$ @RP_, Thank you very much. Yes , I checked for small degrees, your condition comes true. But I didn't see how did you get the necessary condition ? Can you be more explicit about the derivation of the necessary condition ? $\endgroup$
    – Why
    Aug 28 '20 at 14:20
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    $\begingroup$ Well, just write $f = f_1(x_1)f_2(x_2) \cdots f_n(x_n)$ and compute both sides of my equation. For simplicity, let $i=1, j=2$. Then both sides are clearly equal to $f'_1(x_1)f_1(x_1)f'_2(x_2)f_2(x_2)f_3(x_3)^2 f_4(x_4)^2 \ldots f_n(x_n)^2$. $\endgroup$
    – RP_
    Aug 28 '20 at 14:28
  • $\begingroup$ @RP_, Thanks. So it seems that it is also sufficient condition as you said in p-adic case (char $0$). $\endgroup$
    – Why
    Aug 28 '20 at 14:31
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    $\begingroup$ Just to be sure: by "necessary condition" I meant that if the power series can be written as a product, then it must satisfy the equation (so if it doesn't satisfy it, it is not a product of $n$ univariate power series). From what I wrote above I can't draw the conclusion that the condition is also sufficient. I have thought about it but I don't see how to prove that it is sufficient. I thought I could maybe prove it but unfortunately I was too optimistic. $\endgroup$
    – RP_
    Aug 28 '20 at 19:57
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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.

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  • $\begingroup$ Thank you very much . In the last para 3rd line, what do mean by $\text{as above}$ ? Are you referring the comment by @RP_ ? $\endgroup$
    – Why
    Aug 29 '20 at 11:28
  • $\begingroup$ Yes, indeed I was. $\endgroup$
    – user131781
    Aug 29 '20 at 11:55
  • $\begingroup$ ok,thank you very much $\endgroup$
    – Why
    Aug 29 '20 at 13:02
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    $\begingroup$ If you use the tranformation $g= \ln f$, the differential equation takes on a very simple form. For the second question, one can use this fact to get a factorisation in a neighbourhood of any point where $f$ is non zero, then piece them together to get one away from the zeroes. But as I said, I haven’t thought this through so if you want to use this “fact” anywhere you will have to bite the bullet and find out whether it can be solidified to a precise proof. $\endgroup$
    – user131781
    Aug 29 '20 at 13:46
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    $\begingroup$ Sorry—I looked but don’t have the expertise to be able to help you. $\endgroup$
    – user131781
    Sep 6 '20 at 9:39

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