# The non-units in $\mathbb{R}[[x]]$ form a principal ideal.

I'm having a bit of confusion regarding the ideal in $\mathbb{R}[[x]]$ consisting of non-units and I'm probably making some silly mistake somewhere. It's clear from order considerations that the units of this ring are the non-zero constants and so my intuition has suggested that the ideal of non-units is principal and generated by $x$. But, in this case, every element of $(x)$ is divisible by $x$. However, $1+x\in \mathbb{R}$ is not divisible by $x$ yet it is non-unit. Can someone point out where my error is?

Thank you.

-
Are you sure the set of non-units is an ideal? $1+x$ and $1-x$ are non-units, but... – wj32 Nov 2 '12 at 6:01
Whoa! Typographical error. I meant to write power series instead of polynomials. Correction soon to come. – Alexander Sibelius Nov 2 '12 at 6:07
$1+x$ is a unit! – Najib Idrissi Nov 2 '12 at 6:12
Alright, at this point it looks like a real power series is a unit if and only if it has a nonzero constant term. After I prove this, the fact that the non-units are principal will be trivial. Thanks again. – Alexander Sibelius Nov 2 '12 at 6:19

The units are not the non-zero constants. For example, $$(1-x)^{-1}=1+x+x^2+\cdots.$$ The ideal of non-units is indeed generated by $x$.

-
Ah, I see what I've done now. Thanks. – Alexander Sibelius Nov 2 '12 at 6:13

The units of this ring are the non-zero constants. $\;\;$ $\:1+x\:$ and $x$ are non-units, and $1$ is a unit.

$1+x+(-x) \: = \: 1+0 \: = \: 1$

The set of non-units in $\mathbb{R}[x]$ is not closed under addition,
therefore the set of non-units in $\mathbb{R}[x]$ is not an ideal.

-

Hint: What is

$$(1+x)\sum_{n=0}^\infty (-1)^n x^n ?$$

-