Hi! I need help with this exercise :

1)Let R be a ring with 1 .Prove that $\displaystyle (-1)^2=1 $ in R

another question , if I have to "u" is a unit in R, "-u" is a unit ? how i prove this ?

Thanks

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- Aug 21st 2012, 06:58 PMresident911Ring
Hi! I need help with this exercise :

1)Let R be a ring with 1 .Prove that $\displaystyle (-1)^2=1 $ in R

another question , if I have to "u" is a unit in R, "-u" is a unit ? how i prove this ?

Thanks - Aug 22nd 2012, 03:30 AMHallsofIvyRe: Ring
"-1" is defined, in a ring, as the additive inverse of the multiplicative identity. In particular, 1+ (-1)= 0 so, by the distributive law, $\displaystyle (1+ (-1))^2= 1^2+ (1)(-1)+ (-1)(1)+ (-1)^2= 0$. Because 1 is the multiplicative identity, $\displaystyle 1^2= 1$, of course, and $\displaystyle 1(-1)= (-1)(1)= -1$ so that $\displaystyle (1+ (-1))^2= 1- 2+ (-1)^2= -1+ (-1)^2= 0$. Since -1 is the additive inverse of 1, and this equation says it is the additive inverse of $\displaystyle (-1)^2$, it follows that $\displaystyle (-1)^2= 1$.

A 'unit' in a ring is a member that has a multiplicative inverse. If u is a unit, there exist "v" such that uv= vu= 1. But we have also that (-u)(-v)= uv= 1 (use the distributive law to prove that) so that the multiplicative inverse of -u is -v.