# 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:

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 $$E:=\{ n\in \Bbb{N}: n=\tau (f(X)),\ f(X)\in I \}\subseteq \Bbb{N}$$ 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?

• Surely $\langle X^0 \rangle = K[[X]]$ is an ideal too.
– lhf
Nov 13, 2018 at 23:13
• see math.stackexchange.com/questions/1208609/… and note that all nonzero elements in a field are units Nov 14, 2018 at 0:05
• Thank you both for your comments. Ok, $U(K)=K^*=K\backslash \{0_K\}$. Nov 14, 2018 at 0:18

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:

1. if $$\{0\}$$ is the only element in $$\mathcal{F}$$, it is a maximal element of $$\mathcal{F}$$;
2. otherwise take $$F=\{n:\langle X^n\rangle\in\mathcal{F}\}$$ 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: $$I_{k+n+1}=I_{k+n+2}=\dotsb$$.

• Thank you for your answer and sorry for the delay. 1) So, you used the set $\mathcal{F}$ of all ideals of $K[[X]]$, you proved that $\mathcal{F}$ has maximal element (upper bound?). And then, from this fact you said that our ascending sequence stabilizes for some $n\in \mathbb{N}$. Right? 2) Could we use directly the lemma; " If $R$ is a PID, then every ascending sequence of ideals of $R$ stabilizes" in order to claim that $K[[X]]$ is Noetherian? Nov 15, 2018 at 1:36
• @Chris Not “the set of all ideals”, but “any nonempty set of ideals”. Using the theorem that a PID is Noetherian is obviously possible, but it requires more tools. Here the set of all ideals is order antiisomorphic to the ordinal $\omega+1$, so every nonempty subset has a maximal element. Nov 15, 2018 at 8:57
• Yes, I mean "any nonempty set of ideals". What is order antiisomorphic? Nov 16, 2018 at 22:28
• @Chris Two partially ordered set $(X,\le)$ and $(Y,\le)$ are order antiisomorphic if there exists a bijective map $f\colon X\to Y$ such that, for all $a,b\in X$, $a\le b$ if and only if $f(b)\le f(a)$. In other words, if in one of the sets we take the opposite order, then the two posets are order isomorphic. Nov 16, 2018 at 22:33
• @Chris Yes, I should have added $f\ne0$ in the definition of $E$. But if we define $\tau(0)=\infty$, then nothing really changes, because the ideal $I$ is by assumption nonzero, so an element with finite order exists. Nov 16, 2018 at 23:19