# An element of formal power series over a commutative ring

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.

• 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? – Mohan Sep 5 '19 at 21:46

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.

• @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. – Pavel Čoupek Sep 8 '19 at 20:46
• @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 $x+y \in N_f$. – Pavel Čoupek Sep 9 '19 at 11:55
• Now I get it! Thanks for your patience. Deleting my comments now. – Matemáticos Chibchas Sep 9 '19 at 14:34