Number of elements in a ring with identity.

Click For Summary

Discussion Overview

The discussion revolves around determining the number of elements in a ring with identity, specifically under the condition that for all non-zero elements \( x \) in the ring \( R \), \( x^2 = 1_R \). Participants explore various properties of the ring, including its structure and implications for its cardinality.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that \( x^2 = 1_R \) should be interpreted as \( x^2 = 1_R \) for all non-zero \( x \) in \( R \), leading to a consideration of the element \( x + 1_R \) and its implications.
  • Another participant conjectures that the cardinality of \( R \) could be 2 or 3 and proposes that \( U(R) = R - \{0\} \), indicating that \( R \) might be a division ring.
  • It is noted that if \( R \) is an integral domain, then certain properties must hold, such as \( xy = 0 \) implying \( y = 0 \) if \( x \neq 0 \), and that \( R \) is commutative.
  • A participant argues that \( R \) is a field, as every element satisfies \( x^2 = 1 \) for \( x \) in \( R^* \), and mentions that \( R \) must be finite with at most 2 roots for the polynomial \( x^2 - 1 \).
  • Another participant raises a concern about the reasoning regarding \( (a + 1)a = a + 1 \) and the implications of additive inverses, questioning whether \( a + 1 \) could be zero.
  • One participant points out that division is not generally defined in arbitrary rings and provides an example with integers mod 3 to illustrate that \( |R| \) could be 3 without contradiction.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the properties of the ring \( R \) and its potential cardinality. There is no consensus on the exact number of elements in \( R \), as different interpretations and examples are presented.

Contextual Notes

Some limitations include the dependence on definitions of elements and operations in rings, as well as the unresolved nature of certain mathematical steps regarding the properties of \( R \).

AkilMAI
Messages
73
Reaction score
0
1_R=identity in the ring R.
/=...not equal
Having some issues with this any help will be great:
Let R be a ring with identity, such that
x_2 = 1_R for all 0_R /= x ,where x belongs to R. How many elements are in R?
Thanks
 
Physics news on Phys.org
James said:
1_R=identity in the ring R.
/=...not equal
Having some issues with this any help will be great:
Let R be a ring with identity, such that
x_2 = 1_R for all 0_R /= x ,where x belongs to R. How many elements are in R?
Thanks
I assume that x_2 = 1_R should read $x^2 = 1_R$.

Suppose that $x\ne0_R$. Start by looking at the element $x+1_R$. There are two possibilities: either $x+1_R = 0_R$ (in which case $x=-1_R$), or $x+1_R \ne 0_R$, in which case $(x+1_R)^2 = 1_R.$ See what you can deduce from that last equation.
 
i conjecture |R| = 2 or 3. i will go further, it appears that U(R) = R- {0}, so we have a division ring. and by a theorem of wedderburn...
 
Take $$ a\in R $$ s.t. $$ a\not= 0 $$ then $$ (a+1).a= ...$$ using distributivity and $$ a.(a+1)= ...$$ so from the uniqueness of the identity element in R ..
 
Last edited:
there's a problem with your reasoning, hmmm16. we have no guarantee that (a+1)a = a+1 implies a = 1 because a+1 might be 0.

some things that are true:

R is an integral domain: suppose xy = 0 and x ≠ 0. then y = 1y = xxy = x0 = 0.

R is commutative: suppose x,y ≠ 0. then xy ≠ 0, so $(xy)^2 = 1 = x^2y^2$. that is:

xyxy = xxyy
xyxyy = xxyyy
xyx = xxy
xxyx = xxxy
yx = xy

R is a field: since $x^2 = 1$ for all x in R*, U(R) = R*, since $x^{-1} = x$ for x in R*.

R is finite: since R is a field, and since every element of R* is a root of the polynomial $x^2 - 1$, there can be at most 2 such roots.
 
Last edited:
yeah but we just treat this case first and use the uniqueness of additive inverses right?
 
that is, i believe, what Opalg was hinting at.
 
Deveno said:
that is, i believe, what Opalg was hinting at.
if (x+1_r)^2=1_r =>x=-1/2...but x^2=1 so this is a contradiction...therefore x=-1=>|R|=2(because of the powers of x).Correct?
 
no, you can't just write: "x = 1/2". for one thing, division isn't defined for arbitrary rings, only for fields. for another, in a general ring, "2" might not exist. in the field with just 2 elements F = {0,1}, also known as the integers mod 2, 2 DOES NOT EXIST. more generally, in boolean rings, A+A = 0, so you can't even have "2 of something".that is, you have no reason to suppose that the multiplicative inverse of x+x exists.

as a matter of fact, if R = the integers mod 3, we have:

(1)(1) = 1
(2)(2) = 4 = 1 (mod 3),

and it is easy to verify that in this case (1+1)^2 = 2^2 = 1, so there is no contradiction, and |R| = 3.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 24 ·
Replies
24
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
20
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K