Cfrgtkky

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:

1. $K[[X]]$ is PID




2. 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
$$E:={ nin Bbb{N}: n=tau (f(X)), f(X)in I }subseteq Bbb{N}$$
where $tau(f(X))$ is the order of the formal power series $f(X)$.

Then, $Eneq emptyset$ (because $I neq {0_K } $) and from the Well-Ordering Principle, there is an element $n_0in E$ 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=0iff 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, $a_{n_0}neq 0_Kiff h(X) in U(K[[X]])$. 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,
$$Isubseteq langle X^{n_0} rangle tag{2}.$$

And now, from $(1),(2)$ we get $I = langle X^{n_0} rangle$. So, every non zero ideal $Itrianglelefteq K[[Χ]]$ 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 frome 2. that our $K[[X]]$ is Noetherian?

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
$$
E={tau(f):fin I,fne0}
$$
Let $n$ be the minimum of $E$, which is not empty because $Ine{0}$, and take $gin I$ such that $tau(g)=n$. Then $g=X^n g_0$, with $tau(g_0)=0$, so $g_0$ is invertible. Hence $X^nin I$ and so $langle X^nranglesubseteq I$. If $fin I$, then $tau(f)ge n$, which implies $f=X^nf_1$, so $Isubseteqlangle X^nrangle$.

Therefore $I=langle X^nrangle$.

It is obvious that $langle X^mranglesupseteqlangle X^nrangle$ whenever $nge m$, because $X^n=X^mX^{n-m}$.

If $mathcal{F}$ is a non empty set of ideals of $K[[X]]$, we have two cases:

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


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

With ascending chains, suppose $I_0subseteq I_1subseteqdots I_kdotsb$ is an ascending chain of ideals. It is not restrictive to assume at least one of these ideals is nonzero, say $I_kne{0}$. Then, for some $n$, $I_k=langle X^nrangle$. Then the chain must stabilize at most after $n+1$ steps: $I_{k+n+1}=I_{k+n+2}=dotsb$.