Why $K[[X]]$ is PID and what's the form of the ring's ideals? Let $K$ be a field. We write $K[[X]]$ for the ring of all formal power series with coefficients from the field $K$. Then, we will try to prove the next theorem.

Theorem. If $K$ is a field then:
  
  
*
  
*$K[[X]]$ is PID
  
*The ideals of $K[[X]]$ have the form $\langle X^k \rangle$, $k=1,2,3,...$ and in particular, it is
  $$\langle X \rangle \supset \langle X^2 \rangle \supset \langle X^3 \rangle \supset \langle X^4 \rangle \supset \cdots \ .$$

Proof. 1. Let's take a non-zero ideal $\{0_K \} \neq I \trianglelefteq K[[X]]$ of $K[[X]]$. We define the subset
$$E:=\{ n\in \Bbb{N}: n=\tau (f(X)),\ f(X)\in I  \}\subseteq \Bbb{N}$$
where $\tau(f(X))$ is the order of the formal power series $f(X)$.
Then, $E\neq \emptyset$ (because $I \neq \{0_K \} $) and from the Well-Ordering Principle, there is an element $n_0\in E$ such that $n_0=\tau (g(X))$, for some $g(X)\in I$.
Claim. We will show that $I=\langle X^{n_0} \rangle $.
Trivial Case: If $n_0=0\iff \tau(g(X))=0 \iff $ the fixed term of $g(X)$ is non-zero $\iff g(X)\in U(K[[X]]).$ So, $I$ contains an invertible element $\iff I=K[[X]]=\langle 1_K \rangle $.
If $n_0>0 $, we can write
\begin{align}
g(X) &:=a_{n_0}X^{n_0}+a_{n_1}X^{n_1}+... && \in I \trianglelefteq K[[X]]\\
&= X^{n_0}\cdot (a_{n_0}+a_{n_0+1}X+...)  && \in I \trianglelefteq  K[[X]] \tag{♠}
\end{align}
where the term $a_{n_0} \neq 0_K$ and we set $h(X):=a_{n_0}+a_{n_0+1}X+... \in K[[X]] $. But then, $a_{n_0}\neq 0_K\iff h(X) \in U(K[[X]])$. So, $(♠)$ could be written in the form
\begin{align} 
X^{n_{0}}&=g(X)\cdot h(X)^{-1}\in I \implies \\ 
\langle X^{n_{0}}  \rangle & \subseteq  I. \tag{1}
\end{align}
On the other hand, if we take an element $f(X)\in I$, then 
\begin{align}
f(x) & \in I && \implies \\
\tau(f(X)):&=n_1 \geq n_0 && \implies \\ 
f(X) & =X^{n_0}\cdot \ell (X),\ \ell (X) \in K[[X]] &&  \implies \\
f(X) & \in \langle X^{n_{0}} \rangle .
\end{align}
So, 
$$I\subseteq \langle X^{n_0} \rangle \tag{2}.$$
And now, from $(1),(2)$ we get $I = \langle X^{n_0} \rangle$. So, every non zero ideal $I\trianglelefteq K[[Χ]]$ is principal.
Questions.
1) Is this proof completely right?
2) Why do we get from 1. this decreasing sequence of ideals?
3) Can we conclude from 2. that our $K[[X]]$ is Noetherian?
 A: The proof seems essentially right, but it is rather clumsy. You also fail to say what's $n_0$ (it is the minimum of $E$, of course).
If $I$ is a nonzero ideal of $K[[X]]$, then we can define
$$
E=\{\tau(f):f\in I,f\ne0\}
$$
Let $n$ be the minimum of $E$, which is not empty because $I\ne\{0\}$, and take $g\in I$ such that $\tau(g)=n$. Then $g=X^n g_0$, with $\tau(g_0)=0$, so $g_0$ is invertible. Hence $X^n\in I$ and so $\langle X^n\rangle\subseteq I$. If $f\in I$, then $\tau(f)\ge n$, which implies $f=X^nf_1$, so $I\subseteq\langle X^n\rangle$.
Therefore $I=\langle X^n\rangle$.
It is obvious that $\langle X^m\rangle\supseteq\langle X^n\rangle$ whenever $n\ge m$, because $X^n=X^mX^{n-m}$.
If $\mathcal{F}$ is a non empty set of ideals of $K[[X]]$, we have two cases:


*

*if $\{0\}$ is the only element in $\mathcal{F}$, it is a maximal element of $\mathcal{F}$;

*otherwise take $F=\{n:\langle X^n\rangle\in\mathcal{F}\}$ and let $m$ be the minimum of $F$; then $\langle X^m\rangle$ is a maximal element in $\mathcal{F}$.


With ascending chains, suppose $I_0\subseteq I_1\subseteq\dots I_k\dotsb$ is an ascending chain of ideals. It is not restrictive to assume at least one of these ideals is nonzero, say $I_k\ne\{0\}$. Then, for some $n$, $I_k=\langle X^n\rangle$. Then the chain must stabilize at most after $n+1$ steps: $I_{k+n+1}=I_{k+n+2}=\dotsb$.
