Homework Help: Abstract Algebra: Another Ring Proof

1. Feb 6, 2016

RJLiberator

1. The problem statement, all variables and given/known data
Let R be a ring and suppose r ∈R such that r^2 = 0. Show that (1+r) has a multiplicative inverse in R.

2. Relevant equations
A multiplicative inverse if (1+r)*x = 1 where x is some element in R.

3. The attempt at a solution

We know we have to use two facts.
1. Multiplicative inverse (1+r)*x=1
2. the fact that r^2 = 0.

So I first tried, well, why don't we use (1+r)(1+r) = 0

However, this only got me to 1+r^2+2r = 0 with r^2 =0 we have 1+2r = 0
But that doesn't appear to be what we wanted to show.

Althought, we could subtract -1 from both sides, and subtract r from both sides to get
r = -1-r
r = -(1+r)
But this still isn't what we wanted to show.

If I divide both sides by r, then I get 1 = -1/r (1+r) and this would work, however, I can't assume element r can be divided to get 1. (at least I dont think I can based off the few multiplicative axioms I have for rings/fields).

I have to think that I am on the right path, but any hints to get me closer?

2. Feb 6, 2016

Samy_A

If r²=0, then (-r)²=0.
So if this exercise is correct, what other element of the ring besides 1+r must also have a multiplicative inverse?

3. Feb 6, 2016

RJLiberator

-(1+r) must have an inverse then?

(-(1+r))^2=0
(-(1+r))*\s=1

(-(1+r)) = (-1-r)

(-1-r)*s=1
(1+r)*x=1

(-1-r)(-1-r) = 1+2r+r^2 = 1+2r = 0

Hm

What if we try:
(1+r)(-1-r) = 0
-1-2r-r^2=0
No, that doesn't help there.
Hm...

4. Feb 6, 2016

Samy_A

No, I didn't mean -(1+r).

The exercise says: if (+r)²=0, then 1+r has a multiplicative inverse.
How would you translate this literrally for (-r)²? Set s=-r, write down what the exercise claims about s, and then formulate it again in terms of r.

5. Feb 6, 2016

RJLiberator

Hm, I see what you are saying.
I think what you mean is the following:

If (+r)^2 = 0 then 1+r has a M.I.
So, (-r)^2=0 then 1-r has a M.I.

We want to show that:
(1-r)*x=1
(1+r)*c=1

(1-r)(1-r) = 1-2r=0
(1+r)(1+r) = 1+2r=0

1-2r=1+2r
so r=0
1=1
1 always has a multiplicative inverse in a ring.
(1+r) = 1
Er, is that the correct way of showing it? Somehow one thing just lead to another...

since r = 0
c = 1
x = 1

6. Feb 6, 2016

Samy_A

Yes, that's what I meant.

I'm not sure how you got that.

But you are making this way too complicated.
You now have two ring elements (1+r, 1-r) in search of a multiplicative inverse. Maybe they could help each other?

7. Feb 6, 2016

RJLiberator

I was thinking, since r^2 = 0, just multiply (1+r)^2 = 0 and (1-r)^2 = 0 and then when you factor it out, the r^2 = 0. so that's how I got it :p.

(1+r)(1-r) = 1-r^2 but since r^2 = 0 by assumption,
(1+r)(1-r) = 1
and thus, they are the multiplicative inverses.

Hm, that makes more sense.

8. Feb 6, 2016

Samy_A

But why should (1+r)²=0 be correct?

Correct!

9. Feb 6, 2016

RJLiberator

Ahhh, that makes sense..
I was thinking just set (1+r) = r. But that seems like an 'illegal' move. Could be out of bounds of r.

I appreciate the help here. Simple proof, but hard to see at first (as you could probably tell by my attempts in this thread :p).

10. Feb 6, 2016

Staff: Mentor

This is equivalent to 1 = 0.

Very illegal if you have more than one element to deal with.

It has nothing to do with r.