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Is there any way to calculate the multiplicative group of the units of power series ring $k[[x]]$, where $k$ is a field ?

user26857
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Rajesh
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4 Answers4

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Hint $\rm\displaystyle\quad 1\: =\: (a-xf)(b-xg)\ \Rightarrow\ \color{#c00}{ab=\bf 1}\ $ so scaling top & bottom below by $\rm \,b\,$ yields
$$\Rightarrow\ \ \displaystyle\rm\ \ \frac{1}{a-xf}\ =\ \frac{b}{\color{#c00}{\bf 1}-bxf}\ =\ b\:(1+bxf+(bxf)^2+(bxf)^3+\:\cdots\:)$$

Bill Dubuque
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    I still cannot see what to do next? how do you prove that the group of units are indeed power series with no-zero constant term? – Kran Jan 21 '17 at 21:13
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$\bf Hint:$ $\sum_{n=0}^\infty a_nx^n$ is a unit iff $a_0\ne 0$.

azarel
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The multiplicative group is $k[[x]]\backslash (x)$. Certainly those elements divisible by $x$ are not units. If an element is not divisble by $x$ (in other words, has nonzero constant term), you can construct the inverse term by term.

Brett Frankel
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It's possible to be much more specific about the structure of the unit group than has been done so far, as follows. So far we know that the units in $k[[x]]$ are those formal power series with nonzero constant term. Dividing by the constant term shows that this group is isomorphic to $k^{\times}$ (the constant term part) times the group of formal power series $1 + a_1 x + a_2 x^2 + \dots$ with constant term $1$.

Claim: If $k$ has characteristic $0$, then this group is isomorphic to $(k[[x]], +)$. In particular, it is a $\mathbb{Q}$-vector space.

(Note that this can't be true in characteristic $p$ since the group of units is not $p$-torsion: we have, for example, $(1 + x)^p = 1 + x^p \neq 1$.)

The isomorphism is given by the "exponential" map

$$k[[x]] \ni f \mapsto (1 + x)^f \in k[[x]]^{\times}.$$

This is defined using the identity

$$1 + x = \exp \log (1 + x) = \exp \sum_{n \ge 1} (-1)^{n-1} \frac{x^n}{n}$$

(which is where we need that $k$ has characteristic $0$), which allows us to define

$$(1 + x)^f = \exp f \log (1 + x)$$

as a formal power series. This has all of the standard properties of the exponential, and is in particular a homomorphism. The inverse of this isomorphism is the "logarithm" map

$$\log_{1 + x}(g) = \frac{\log g}{\log (1 + x)}$$

where $g$ has constant term $1$. It's formal to verify that these two operations are inverse to each other.

The answer is more interesting in positive characteristic: we get the underlying abelian group of the ring of Witt vectors $W(k)$. For example, when $k = \mathbb{F}_p$ this group is related to the group of $p$-adic integers $\mathbb{Z}_p$, and in fact there is a natural exponential $(1 + x)^f \in \mathbb{F}_p[[x]]$ where $f \in \mathbb{Z}_p$. Edit: See the comments!

Qiaochu Yuan
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  • I think the unit group of $\mathbb F_p[[X]]$ is isomorphic to $\mathbb F_p^\times\oplus\mathbb Z_p^{(\mathbb N)}$, where $\mathbb Z_p^{(\mathbb N)}$ denotes a countable direct sum of copies of the $p$-adic integers $\mathbb Z_p$. – user26857 Feb 28 '16 at 10:16
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    Just passing through and I noticed this answer. I think that Qiaochu's comment about the char p setting is inaccurate -- if $k=\mathbb{F}_p$ then the group is much much bigger than $\mathbb{Z}_p$ -- for example $(1+Xk[[X]]) / (1+X^Nk[[X]])$ is a finite $p$-group which is very far from being cyclic in general, and this sort of argument even shows that $1+Xk[[X]]$ is not finitely-generated. But I don't think @user26857's suggestion is right either -- as $1+Xk[[X]]$ is compact if $k$ is finite, whereas their suggestion does not seem to be (at least if I am interpreting it correctly). – Kevin Buzzard May 05 '17 at 14:52
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    I think that the unit group of $\mathbb{F}_p[[X]]$ is isomorphic to $\mathbb{F}_p^\times$ direct sum a countably infinite product (not sum) of $\mathbb{Z}_p$'s, but this comment box is too small for me to fit the proof in. – Kevin Buzzard May 08 '17 at 11:43
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    The structure of $\Bbb F_p[[X]]^{\times}$ is given in thm 4.4 here. – Watson May 30 '18 at 21:44
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    @KevinBuzzard is quite right: the units of $\Bbb F_p[[x]]$ that are $\equiv1\pmod x$ are a direct product of $\Bbb Z_p$’s, the indexing set being the positive integers prime to $p$. – Lubin Jun 04 '18 at 16:09
  • Does anyone know a reference for Kevin Buzzard's assertion that $\mathbb F_p[[x]]^\times\cong \mathbb F_p^\times\times\prod_{\mathbb N}\mathbb Z_p$? (The source indicated by Watson is no longer accessible.) – Peter Kropholler Jul 26 '23 at 08:22
  • @PeterKropholler: https://web.archive.org/web/20220526205303/https://www.math.arizona.edu/~ura-reports/063/McMurdie.Christopher/Final.pdf – Drew Nov 28 '23 at 09:02