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.

  • 5
    $\begingroup$ 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, 2019 at 21:46

1 Answer 1


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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