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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

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The units in $k[[x]]$ are precisely those power series whose constant term is a unit. I hope this helps. –  Chris Leary Feb 29 '12 at 3:29
Thank you Chris. –  Rajesh Feb 29 '12 at 4:01

3 Answers 3

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

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$\bf Hint:$ $\sum_{n=0}^\infty a_nx^n$ is a unit iff $a_0\ne 0$.

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Thank you azarel, this makes sense. –  Rajesh Feb 29 '12 at 4:00
Could you give a hint as to why this is true? –  math1234567 May 10 '14 at 17:19

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.

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Thank you Brett. –  Rajesh Feb 29 '12 at 4:00

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