Show that $\ I=(x^n) $ for some $n$, an element of $\Bbb N\cup\{0\}$.

Question

Assume $k$ is a field, $k[[x]]$ is the ring of formal power series. Let $I$ be an ideal of $k[[x]]$. Show that $\ I=(x^n) $ for some n, an element of $\Bbb N\cup\{0\}$.

Answer 1

Hint:

A formal power series: $\;u=a_0+a_1c+a_2x^2+\dots\;$ is a unit in $k[[x]]$ if and only if $a_0\ne 0$.

Deduce that if $n$ is the least order of non-zero formal power series in $I$, $I$ is generated by $x^n$.