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Let $R$ be a commutative ring with $1$, $R[[x]]$ be the power series ring over $R$ and $A$ be an (prime) ideal of $R[[x]]$ with $x\not\in A$ and $\{f_i\}_{i=1}^\infty$ be a sequence of element of $A$. Now I have two questions:

  1. Is $f\mathrel{:=}f_1+xf_2+x^2f_3+\dotsb+x^nf_{n+1}+\dotsb$ a well defined element of $R[[x]]$? (Since we can find the coefficient of $x^n$ in $f$ for each $n$, it seems that $f$ is well defined.)

  2. If (1) is true is $f\in A$? (If (1) is true and (2) is not true, under what conditions is (2) true?)

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    $\begingroup$ The usual definition of the topology on $R[[x]]$ sets its ideals $x^n R[[x]]$ as a neighbourhood base at the identity. According to that definition, the answer to (1) is obviously yes. For (2), of course the answer is that $A$ should be closed, but I don't know if there's a nice characterisation of closed ideals in $R[[x]]$. $\endgroup$
    – LSpice
    Commented Jul 21, 2020 at 16:20
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    $\begingroup$ Here is at least some characterization (not sure it can be called "nice"): An ideal $I$ is in $R[[x]]$ is closed in the $x$-adic topology if and only if $\bigcap_{n \geq 1} I + (x)^n = I$. $\endgroup$
    – AlexIvanov
    Commented Jul 23, 2020 at 21:00

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