In an arbitrary ring, I have that $ab = 1$. I'm pretty certain that $a(-b) = -1$, but I'm not sure how to prove that the $-$ sign commutes. Is this necessarily true? Could someone point me in the direction of a proof or counterexample?
Showing that multiplying by $-1$ commutes with a unit and its inverse in an arbitrary ring.
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$\begingroup$
abstract-algebra
ring-theory
commutative-algebra
noncommutative-algebra
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2It's a general fact that $a(-b)=(-a)b=-(ab)$. – 2017-01-16
1 Answers
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Hint: You want to prove $a(-b)=-(ab)$. That is, you want to prove $a(-b)$ is the additive inverse of $ab$. So try adding $a(-b)$ and $ab$ and see if you can show the sum is $0$.
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0$ab + a(-b) = a(b-b) = 0$. Got it! Thank you. – 2017-01-16
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0@Alex $\,a(-b)\,$ and $\,(-a)b\,$ are both inverses of $\,ab\,$ so are equal by **uniqueness of inverses.** which yields a proof of the [Law of Signs.](http://math.stackexchange.com/a/9940/242) – 2017-01-16