# Ring of formal power series over a field is a principal ideal domain

If $K$ is a field, any non-zero ideal in the ring of formal power series $A=K[[X]]$ is of the form $AX^n$ with $n\geq 0$, so $A=K[[X]]$ is a principal ideal ring.

I can't see why. Usually when we consider the ring of polynomials $B=K[X]$, the ideals are the multiples of any fixed polynomial $P(x)\in K[X]$.

Here, for formal power series, why can we restrict it to $AX^n$?

The ring of a formal power series over a field is even more than a PID--it's a discrete valuation ring (hence a local ring). This follows from the lemma that a formal power series $f(X) = \sum_{n \geq 0} a_n X^n \in K[[X]]$ is a unit iff $a_0 \neq 0$. (More generally over a ring $R$, $f(X) \in R[[X]]^\times$ iff $a_0 \in R^\times$.) This shows that $(X)$ is the unique maximal ideal of $K[[X]]$. Thus the only ideals of $K[[X]]$ are $(X^n)$, so $f(X) = u(X) X^n$ for some $n \in \mathbb{Z}_{\geq 0}$ and $u(X) \in K[[X]]^\times$.