Abstract Algebra: Another Ring Proof

  • #1
RJLiberator
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Homework Statement


Let R be a ring and suppose r ∈R such that r^2 = 0. Show that (1+r) has a multiplicative inverse in R.

Homework Equations


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

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?
 

Answers and Replies

  • #2
Samy_A
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Homework Statement


Let R be a ring and suppose r ∈R such that r^2 = 0. Show that (1+r) has a multiplicative inverse in R.

Homework Equations


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

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?
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?
 
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  • #3
RJLiberator
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-(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
Samy_A
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-(1+r) must have an inverse then?
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.
 
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  • #5
RJLiberator
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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
Samy_A
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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.
Yes, that's what I meant.

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
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?
 
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  • #7
RJLiberator
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I'm not sure how you got that.

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.

You now have two ring elements (1+r, 1-r) in search of a multiplicative inverse. Maybe they could help each other?

(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
Samy_A
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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.
But why should (1+r)²=0 be correct?

(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.
Correct!
 
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  • #9
RJLiberator
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But why should (1+r)²=0 be correct?

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
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Ahhh, that makes sense..
I was thinking just set (1+r) = r.
This is equivalent to 1 = 0.

But that seems like an 'illegal' move.
Very illegal if you have more than one element to deal with.

Could be out of bounds of r.
It has nothing to do with r.
 
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