$K[[x]]$ has only one prime element - what is it?

Question

$K[[x]]$ is the ring of formal power series over a field $K$. What can we say about the prime elements? In the exercise I should prove there exists only one prime element (to be precise one class of prime elements which are related to each other, i.e. if two elements $r,s$ are from this class than one can express the other one as a product with a unit $u,r=us$). I already know that the units are the elements where the constant part is not zero. That every ideal looks like $<x^n>$ and that for every nonzero element $f$ there exists a $x^n$ such that $f=ux^n$, where $u$ is a unit again.

How can I start?

Answer 1

You have to look for some element $p \ne 0$ which is not a unit and has the property that if $p \mid (f\cdot g)$, then $p \mid f$ or $p \mid g$. In this condition we may of course restrict to the case $f, g \ne 0$.

Certainly $p = x$ is the simplest non-zero power series which is no unit. Now assume $x \mid (f \cdot g)$ with $f, g \ne 0$. Then $f = ux^n, g = vx^m$ with units $u,v$. If $n > 0$, then $x \mid f$, and if $m > 0$, then $x \mid g$. Is it possible that $m = n = 0$? In that case the constant term of $f \cdot g$ would be non-zero so that $x$ could not divide $f \cdot g$.

Thus $x$ is a prime element.

Now let $p$ be an arbitrary prime element. It has the form $p = ux^n$ with a unit $u$ and $n > 0$ (note that $n = 0$ is impossible since then $p$ would be a unit). We conclude that $x^n$ is a prime element. But $x^n \mid (x \cdot x^{n-1})$ so that $x^n \mid x$ or $x^n \mid x^{n-1}$. Certainly $x^n \not\mid x^{n-1}$, thus $x^n \mid x$ which shows that $n = 1$.

Thus $p = ux$ which finishes the proof.