# formal power series over field

Theorem.  If $K$ is a field, then the ring $K[[X]]$ of formal power series is a discrete valuation ring with $(X)$ its unique maximal ideal.

Proof.  We show first that an arbitrary ideal $I$ of $K[[X]]$ is a principal ideal.  If  $I=(0)$,  the thing is ready.  Therefore, let  $I\neq(0)$.  Take an element

 $f(X)\;:=\;\sum_{i=0}^{\infty}a_{i}X^{i}$

of $I$ such that it has the least possible amount of successive zero coefficients in its beginning; let its first non-zero coefficient be $a_{k}$.  Then

 $f(X)\;=\;X^{k}(a_{k}\!+\!a_{k+1}X\!+\ldots).$

Here we have in the parentheses an invertible formal power series $g(X)$, whence get the equation

 $X^{k}\;=\;f(X)[g(X)]^{-1}$

implying  $X^{k}\in I$  and consequently  $(X^{k})\subseteq I$.
For obtaining the reverse inclusion, suppose that

 $h(X)\;:=\;b_{n}X^{n}\!+\!b_{n+1}X^{n+1}\!+\ldots$

is an arbitrary nonzero element of $I$ where  $b_{n}\neq 0$.  Because  $n\geq k$,  we may write

 $h(X)\;=\;X^{k}(b_{n}X^{n-k}\!+\!b_{n+1}X^{n-k+1}\!+\ldots).$

This equation says that  $h(X)\in(X^{k})$,  whence  $I\subseteq(X^{k})$.
Thus we have seen that $I$ is the principal ideal $(X^{k})$, so that $K[[X]]$ is a principal ideal domain.
Now, all ideals of the ring $K[[X]]$ form apparently the strictly descending chain

 $(X)\;\supset\;(X^{2})\;\supset\;(X^{3})\;\supset\;\ldots\;\supset\;(0),$

whence the ring has the unique maximal ideal $(X)$.  A principal ideal domain with only one maximal ideal is a discrete valuation ring.

Title formal power series over field FormalPowerSeriesOverField 2015-10-19 9:13:35 2015-10-19 9:13:35 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 13H05 msc 13J05 msc 13F25