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Let $R$ be a commutative ring with a multiplicative id. The question reads:

Let $r$ be an element of a ring $R$. Show that, in the polynomial ring $R[X]$, the polynomial $1+rX$ is a unit if and only if $r$ is nilpotent. Is it possible for the polynomial $1 + X$ to be a product of two non-units?

I've done the first part, it wasn't too bad. But I'm quite stuck with the second, and any hints or pointers would be appreciated. Am I looking for a proof or counterexample? What kind of thing should I be looking at? Etc.

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  • $\begingroup$ Is there anything I can do to make the question clearer? $\endgroup$ – Shakespeare Feb 14 '15 at 10:47
  • $\begingroup$ You look for a proof for awhile, then look for a counterexample for awhile, etc. $\endgroup$ – Qiaochu Yuan Feb 14 '15 at 11:08
  • $\begingroup$ @QiaochuYuan Yes I didn't just post without trying... I've been at this for about an hour $\endgroup$ – Shakespeare Feb 14 '15 at 11:09
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Hint: Is there any ring where it is possible to write $1 + X = (1+aX)(1+bX)$?

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  • $\begingroup$ That's possible (without making $(1+aX)$ or $(1+bX)$ units) if and only if $a+b=1, ab=0$ which implies $b(1-b)=0$ or equivalently $b=b^2$ which means we are not in an integral domain... Does $\mathbb{Z}/6\mathbb{Z}$ with $b=3+\mathbb{Z},a=4+\mathbb{Z}$ work? (P.S. Thanks a lot :)) $\endgroup$ – Shakespeare Feb 14 '15 at 11:08
  • $\begingroup$ @Shakespeare That's exactly what I had in mind. You can use the first part to conclude that these factors are not units. $\endgroup$ – Slade Feb 14 '15 at 11:22
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    $\begingroup$ @Sha Indeed, $\ (1+ax)(1+(1\!-\!a)x)\, =\, 1+x\ $ in $\ \Bbb Z/a(a\!-\!1)\ \ $ $\endgroup$ – Bill Dubuque Feb 14 '15 at 19:30

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