MHB Unit of Ring Question - Cbarker1

[cbarker1]

Dear Everyone,

I am having trouble with an exercise problem. Here is the problem: Dummit and Foote Ed.2 pg 231: "Let $R$ be a ring with 1. Prove that if $u$ is a unit in R then so is $-u$."

My Attempt:

Suppose $u$ is a unit in $R$. Then, from Prop 1 (4) (if $R$ has an identity, then the identity is unique and $-1a=-a$), let $a=u$. Then $-u$ is in $R$. QED

What did I do wrong and/or correct? Thanks,
Cbarker1

 

[caffeinemachine]

Dear Everyone,

I am having trouble with an exercise problem. Here is the problem: Dummit and Foote Ed.2 pg 231: "Let $R$ be a ring with 1. Prove that if $u$ is a unit in R then so is $-u$."

My Attempt:

Suppose $u$ is a unit in $R$. Then, from Prop 1 (4) (if $R$ has an identity, then the identity is unique and $-1a=-a$), let $a=u$. Then $-u$ is in $R$. QED

What did I do wrong and/or correct? Thanks,
Cbarker1

What you have shown is that $-u$ is in the ring. That is clear since ring is closed under multiplication. What you need to show is that $-u$ has a multiplicative inverse in $R$, that is, you need to show the existence of an element $a\in R$ such that $(-u)\cdot a=1$.

 

[Olinguito]

Hi Cbarker1.

You have to show that there is some element $v\in R$ such that $v(-u)=(-u)v=1$.

Now if $u$ is a unit in $R$ then there is some element $w\in R$ such that $wu=uw=1$. I’ll let you carry on from there. (Hint: use the fact that for any $x,y\in R$, $(-x)y=x(-y)=-xy$ and $(-x)(-y)=xy$.)