# What are significant differences between polynomial rings and the ring of formal power series?

I would like to know the difference between the polynomial ring and the ring of formal power series. We didn't mention it in our lectures so I have to understand it on my own. I saw some posts here and on the internet and I understand that:

Formal power series are a generalisation of polynomials, in that the coefficients can be infinite while in the polynomial they have to end at one point. Then I guess we can evaluate the polynomial and therefore have a function while this isn't so simple in power series since they do not always converge.

I feel like this is a difference between polynomials and power series, but what about them as rings? Does it make a difference? Are there other important differences?

Edit: I also saw on Wikipedia that there are differences when we consider topologies, but I haven't done any topological course until now. This is my very first algebra course.

• $\def\deg{\operatorname{deg}} \def\ord{\operatorname{ord}}$In $A[x]$, for $A$ a domain, degree satisfies $\deg fg=\deg f+\deg g$, for $f,g\in A[x]$, but in $A[[x]]$ this formula breaks, since $$(1-x)(1+x+x^2+\cdots)=1.$$ On the other hand, in the rings $A[x]\subset A[[x]]$ one can define the order of a polynomial or power series $f$ to be the biggest $n$ such that $f\in (x)^n$ (note that $\ord f=+\infty$ iff $f=0$). On this case, $\ord fg=\ord f+\ord g$ holds in $A[[x]]$ (hence in $A[x]$ as well). This formula allows to show that $A[[x]]$ is a domain if $A$ is a domain. Commented Sep 18, 2023 at 16:06

Well, an elementary but striking difference is that the polynomial $$1 -X$$ has no inverse in $$k[X]$$, but has an inverse in $$k[[X]]$$, namely the series $$\sum_{n \geqslant 0} X^n = 1 + X + X^2 + \cdots$$

$$\newcommand{\ZZ}{\mathbb{Z}} \newcommand{\Zx}{\ZZ\left[x\right]} \newcommand{\Zxx}{\ZZ\left[\left[x\right]\right]}$$A rather striking difference are the automorphisms.

For simplicity, I will compare the univariate polynomial ring $$\Zx$$ with the univariate FPS (= formal power series) ring $$\Zxx$$. A ring automorphism of $$\Zx$$ is a ring isomorphism $$\alpha : \Zx \to \Zx$$. By the universal property of polynomial rings, a ring morphism $$\alpha : \Zx \to \Zx$$ is uniquely determined by its value $$\alpha\left(x\right)$$. Moreover, it is easy to see that such a ring morphism $$\alpha$$ is an isomorphism if and only if $$\alpha\left(x\right) = \pm x + c$$ for some $$\pm$$ sign and some $$c \in \ZZ$$. Thus, there are countably many ring automorphisms of $$\Zx$$. In contrast, there are uncountably many continuous ring automorphisms of $$\Zxx$$: Indeed, if $$f = f_1 x + f_2 x^2 + f_3 x^3 + \cdots$$ is any FPS with zero constant term and with $$x^1$$-coefficient $$f_1 \in \left\{1,-1\right\}$$, then this FPS $$f$$ has a compositional inverse $$g$$, and the continuous ring morphism $$\beta : \Zxx \to \Zxx$$ that sends $$x$$ to $$f$$ thus has an inverse (namely, the continuous ring morphism $$\gamma : \Zxx \to \Zxx$$ that sends $$x$$ to $$g$$). Since there are uncountably many candidates for $$f$$, there are thus uncountably many continuous ring automorphisms of $$\Zxx$$. I'm not sure if there are additionally any discontinuous ones, but their existence cannot decrease the cardinality.

Another difference is that $$\Zx$$ is a free $$\ZZ$$-module, whereas $$\Zxx$$ is not (indeed, $$\Zxx$$ is isomorphic to $$\ZZ^{\mathbb{N}}$$ as a $$\ZZ$$-module, and it is well-known that the latter $$\ZZ$$-module is not free).

In practical terms, polynomials and FPS are indeed fairly similar in many aspects, except that FPS rings have to generally be considered as topological algebras (over their base rings) in order to be well-behaved, whereas polynomial rings are nice enough as (mere) algebras already. This is why I had to say "continuous ring morphism" above, for example -- as without the "continuous", a ring morphism from $$\Zxx$$ would not be uniquely determined by the image of $$x$$ (or so I think).

There are several interesting points in your question. Let me focus on one of them: you mention evaluation of polynomials, but you should note that polynomials and polynomial functions are not the same thing, in general. If you are working with coefficients in some ring $$A$$, you can associate the polynomial $$p = a_0 + a_1X + \ldots + a_nX^n \in R[X]$$ (viewed as a formal expression) with the polynomial function $$x \mapsto a_0 + a_1x + \ldots + a_n x^n \in A \to A$$. However this association may lose information. E.g., if $$R= \Bbb{Z}_2$$ (the field with two elements), $$p = X^2 - X$$ and $$q = 0$$ are distinct polynomials that give the same polynomial function. You can identify polynomials and polynomial functions for some useful classes of rings $$R$$, e.g., infinite fields, but not for all rings.

One obvious difference is that the polynomials are finitely generated as rings, while the formal power series are not.

• Hello! Thanks for the answer! Could you expand it please a little bit? Commented Jun 24, 2021 at 15:34

Even though this is not (by contemporary standards) an "elementary" explanation, I think it is much more functional and explanatory than most others.

(An example of a "bad explanation" is the idea that formal power series "are strings of symbols"... Seriously? Marks on the page?)

After my youthful dalliances with worse viewpoints, I seriously do think that the best way to describe "the formal power series ring $$R[[x]]$$" (for commutative ring $$R$$, surely with unit $$1$$) is as the projective limit of the quotients $$R[x]/\langle x^n\rangle$$. Whatever other structure the ring $$R$$ may have can be added to the "categorical" requirements.

So, yes, in any version of topology on $$R$$, in the formal power series ring $$x$$ is "very small", so small that any sum $$\sum_{n\ge 0} r_n x^n$$ converges, for any $$r_n\in R$$. But/and this pretense at metrization is not necessary, apart from an example of a construction, if/when we are content with the characterization of the formal power series ring as a projective limit.

(Namely, $$R$$-homs from another $$R$$-module to $$R[[x]]$$ are exactly compatible families of $$R$$-homs to the limitands $$R[x]/\langle x^n\rangle$$.)