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Is there a simple proof/clarification of this statement?

The set of all formal power series in X with coefficients in a commutative ring R form another ring that is written R[[X]], and called the ring of formal power series in the variable X over R.

Thank You.

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See this: math.stackexchange.com/questions/18699/… –  wj32 Oct 24 '12 at 22:08

2 Answers 2

up vote 2 down vote accepted

As a first introduction to the idea of $R[[X]]$, as GilYoung Cheong suggests, just checking the ring axioms from a clear definition of $R[[X]]$ might be an interesting and useful exercise.

Indeed, this provides an example where things generalizing polynomials are not used to define functions in the sense of having point-wise values, but for some other purpose (e.g., generating-functions).

This may cause some cognitive dissonance if one is accustomed to things that either are "numbers" or produce numbers by evaluation.

Another useful, slightly more sophisticated, presentation of $R[[X]]$ which nearly eliminates "checking the axioms" is characterizing it as the _projective_limit_ of the quotient rings $R[X]/X^n$ as $n\rightarrow \infty$. Then the formulas for multiplication in terms of "formal" coefficients are deduced. (I won't write a polemic about projective limits here... partly since such things are widely available on-line, once one knows the keyword.)

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If you have the right definition of addition and multiplication, you need to check all the ring axioms. It shouldn't be difficult at all, but this ring is really useful since you don't have to care about the convergence of power series.

For example, note that $1/(1 - qX) = 1 + qX + (qX)^{2} + \cdots$ where $q \in R \setminus \{0\}$. This is exactly what we wanted from this ring (Hint: check $1 = (1 - qX)(1 + qX + (qX)^{2} + \cdots)$).

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