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.


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?

  • 1
    $\begingroup$ Surely $\langle X^0 \rangle = K[[X]]$ is an ideal too. $\endgroup$
    – lhf
    Nov 13, 2018 at 23:13
  • 2
    $\begingroup$ see math.stackexchange.com/questions/1208609/… and note that all nonzero elements in a field are units $\endgroup$
    – Will Jagy
    Nov 14, 2018 at 0:05
  • $\begingroup$ Thank you both for your comments. Ok, $U(K)=K^*=K\backslash \{0_K\}$. $\endgroup$
    – Chris
    Nov 14, 2018 at 0:18

1 Answer 1


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: $I_{k+n+1}=I_{k+n+2}=\dotsb$.

  • $\begingroup$ 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? $\endgroup$
    – Chris
    Nov 15, 2018 at 1:36
  • 1
    $\begingroup$ @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. $\endgroup$
    – egreg
    Nov 15, 2018 at 8:57
  • $\begingroup$ Yes, I mean "any nonempty set of ideals". What is order antiisomorphic? $\endgroup$
    – Chris
    Nov 16, 2018 at 22:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – egreg
    Nov 16, 2018 at 22:33
  • 1
    $\begingroup$ @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. $\endgroup$
    – egreg
    Nov 16, 2018 at 23:19

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