I have to show that
If $f(x)$ is an irreducible element in the ring of formal power series over $\mathbb{C}$ then $f(x)$ and $x$ are associates. Also the constant term has to be $0.$
I tried by writing$f(x) = g(x)h(x)$, where one of $g(x)$ or $h(x)$ is a unit which eventually gives that the constant term $b_{0}$ of $g(x)$ ( let's say $g(x)$ is a unit) is a unit in $\mathbb{C}.$
So $b_0 = 1 $ or $-1$ or $i$ or $ -i.$
Hence we have the constant term of $f(x)$ say $a_0 = b_{0}c_{0}. c_0$ is the constant term of $ h(x)$.
Again I was thinking that $\mathbb{C}[[x]]$ is a ufd, so every irreducible element will be prime and from there we can solve, but I could not get beyond that.
After this I have no clue. I think lost.
Anyone to help me out. Thank you.
Best Answer
First: to show $f(x)$ is irreducible you don’t start with $f(x)=g(x)h(x)$ and the assumption that one of $g$ and $h$ is a unit. You must show that if $f(x)=g(x)h(h)$, then one of $g$ or $h$ must be a unit, and that $f$ is itself not a unit.
Second: The units of $\mathbb{C}$ are the nonzero complex numbers, not just $1$, $-1$, $i$, or $-i$.
Since an element of the power series ring is a unit if and only if the constant term is a unit, the fact that $f(x)$ is irreducible already tells you that the constant term is $0$. That means you can write $f(x)$ as $f(x) = xh(x)$ for some power series $h(x)$. But $f(x)$ is irreducible, and $x$ is not a unit, so therefore...