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Let $R$ be a commutative ring with 1 and let $R[[x]]$ be the formal power series ring over $R$. Now let $f\in R[[x]]$ with the property that if $g, h\in R[[x]]$ and $f=g+h$ then either $\langle h \rangle\subseteq \langle g\rangle $ or $\langle g \rangle\subseteq \langle h\rangle $, where $ \langle a\rangle $ if the ideal generated by $a$ in the ring $R[[x]]$ for $a\in R[[x]]$. Is there any characterization for such an element? or can we deduced that $f$ is only a single term?

Example: Let $K$ be a field. Then $K[[x]]$ is a valuation ring and hence every element of $K[[x]]$ has this property.

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    $\begingroup$ Since the property carries over to $R$. shouldn't you be first looking at $R$ which has your property rather than the power series ring? $\endgroup$ – Mohan Sep 5 at 21:46
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Let me give a longer comment about the more general form of the question suggested in comments.

So let $0 \neq f \in A$ be an element such that $f=a+b$ implies $a$ divides $b$ or $b$ divides $a$. Equivalently, $f=a+b$ implies $a$ divides $f$ or $b$ divides $f$ (this is because $a$ dividing $b$ implies that $a$ divides $f=a+b$, and $a$ dividing $f$ implies $a$ divides $b=f-a$).

This is further equivalent to:

The set $N$ of all non-divisors of $f$ forms an ideal.

The point is that the set of non-divisors of $f$ is always stable under multiplication by elements of $A$, but in the presence of the condition, it is also stable under addition (if $x, y$ do not divide $f$, then neither does $x+y$, since $(x+y)z=f$ implies that either $xz$ or $yz$ divides $f$, so either $x$ or $y$ divides $f$).

A consequence:

$(f)$ contains a unique maximal proper sub-ideal.

To see this, one looks at $A/N$. Here $f$ is nonzero, and all the proper cosets are represented by elements that divide $f$ in $A$. That is, $f$ is in $A/N$ a nonzero element contained in every nonzero principal, hence every nonzero ideal. Thus, the quotient $N+(f)/N \simeq (f) / (f) \cap N$ is simple. The fact that $(f)\setminus (f) \cap N$ consists of elements associated to $f$ (in the sense that they give the same principal ideal) shows that $(f) \cap N$ is a unique such sub-ideal.

A consequence when $f$ is a non-zero divisor:

If such an element exists and is a non-zero divisor, $A$ is a local ring (hence, all the principal ideals contain a unique maximal sub-ideal).

This holds since $f\cdot -:A \rightarrow (f)$ is in this case a module isomorphism.

So in the particular case when $A=R[[X]]$ is a domain, even existence of such nonzero $f$ implies that $R[[X]]$, hence $R$, has to be a local ring.

One final remark: The initial condition, i.e. $N$ forming an ideal, seems a lot stronger than these consequences. I wonder whether among domains $A$, the only example where this can happen are actually valuation rings.

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  • $\begingroup$ @MatemáticosChibchas You almost convinced me, but on second thought I don't see the problem: Do you claim that the forward implication does not hold? If $f$ has the property in question and $x, y$ are non-divisors of $f$, then assuming that $(x+y)$ divides $f$ leads to $(x+y)z=f$. By the ssumption on $f$, this implies that either $xz,$ or $yz$ divides $f$. But then one gets that either $x$, or $y$ has to be divisor of $f$ also, which is a contradiction. $\endgroup$ – Pavel Čoupek Sep 8 at 20:46
  • $\begingroup$ @MatemáticosChibchas the fact that $(*)$ is equivalent to the discussed property of $f$ is immediate, as it is just a formal restatement of the property "$f=a+b$ implies that $a$ or $b$ is a divisor of $f$". As you say, $(**)$ implies $(*)$ is trivial. So the part that requires an argument is $(*)$ implies $(**)$, and it is the argument given: if you assume $(*)$, i.e. equivalently the given property of $f$, and have $x, y \in N_f$, the reasoning in the above comment shows by contradiction that $x+y$ cannot be a divisor of $f$. That is, we also have to have $x+y \in N_f$. $\endgroup$ – Pavel Čoupek Sep 9 at 11:55
  • $\begingroup$ Now I get it! Thanks for your patience. Deleting my comments now. $\endgroup$ – Matemáticos Chibchas Sep 9 at 14:34

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