Proof that in a ring $R$, $(-1_R)(-1_R)=1_R$

Question

I saw another proof on here using only the fundamental ring axioms, but I was curious if my method is close to being correct.

We know that $1_R$ is a unit in $R$ with inverse $(1_R)^{-1}=1_R$.

Left-multiplying by $(-1_R)$ we then get, $(-1_R)(1_R)^{-1}=(-1_R)(1_R)$

On the right side by the property of the identity element, we get $(-1_R)$, but I wasn't sure what can be done with the left side. I was hoping to manufacture that $(-1_R)^{-1}=-1_R$ so then I could say $(-1_R)(-1_R)=1_R$ by the property of the inverse.

Is this a correct way of approaching this proof? If not, what am I missing?

In your work, you haven't used any property of $-1_R$, so it cannot possibly show that $(-1_R)^{-1}=-1_R$, as $x^{-1}=1_R$ is not necessarily true for all $x\in R$. — Brian Moehring, Oct 22 '21 at 18:04
To correct a typo in my previous comment: I meant $x^{-1}=x$ is not generally true for every invertible $x\in R$. — Brian Moehring, Oct 22 '21 at 18:51
Despite the dupe asking for using "field axioms" it really only uses "ring axioms" — rschwieb, Oct 22 '21 at 20:20

score 0 · Answer 1 · answered Oct 22 '21 at 17:52

0

You must be missing something. It will be like this... $$ (-1_R)+1_R=0_R $$ and... $$ (-1_R)(-1_R)+(-1_R)=(-1_R)(-1_R)+1_R(-1_R)=((-1_R)+1_R)(-1_R)=0_R(-1_R)=0_R $$

answered Oct 22 '21 at 17:52

MH.Lee

5,568

Proof that in a ring $R$, $(-1_R)(-1_R)=1_R$

1 Answers1