If $u$ is a unit of a ring $R$, is it necessarily true that the additive inverse of $u$, say $-u$ is a unit?

Asked Feb 15 '25 at 15:50

Active Feb 15 '25 at 20:17

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I'm trying to show that "If $a \in R$ is a unit, then -a is also a unit."

I started the attempt of the proof like this:

Say $ab=1$, then consider

$\begin{align*} a+(-a) = 0\\ \implies b(a+(-a))=0\\ \implies ba+b(-a)=0\\ \implies 1+b(-a)=0\\ \implies b(-a)=-1\\ \implies (-1)b(-a)=(-1)(-1)\\ \implies -b(-a)=(-1)(-1) \ \ \end{align*}$

So, if (-1)(-1)=1, I can conclude my proof. Any help on proving the thing would also be appreciated!

edited Feb 15 '25 at 15:57

Martin Brandenburg

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asked Feb 15 '25 at 15:50

Amogh Kulkarni

2

Do you mean by "unity" the multiplicative identity? In an even, more generally, $(-a)(-b)=ab.$ – Thomas Andrews Feb 15 '25 at 15:56
Yeah, I mean the multiplicative identity. – Amogh Kulkarni Feb 15 '25 at 15:58
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I think you don't. I also fixed the title. (Before, the title was different from the question.) I found all the duplicates with https://approach0.xyz by the way. – Martin Brandenburg Feb 15 '25 at 16:02
Thanks for the duplicates. I did however mean multiplicative identity, because I wanted to know whether (-1)(-1)=1, since 1 is the unity and (-1) is the additive inverse of the unity I wrote the question as such. – Amogh Kulkarni Feb 15 '25 at 16:06
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This is answered in all the duplicates mentioned. Please click on them and read. – Martin Brandenburg Feb 15 '25 at 16:58

If $u$ is a unit of a ring $R$, is it necessarily true that the additive inverse of $u$, say $-u$ is a unit?

0 Answers0