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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.

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    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.2012-11-02

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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 $x$.

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    Ah, I see what I've done now. Thanks.2012-11-02
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Hint: What is

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

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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.