Welcome to our community

Be a part of something great, join today!

Number of elements in a ring with identity.

James

New member
Jan 30, 2012
15
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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
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.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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...
 

hmmm16

Member
Feb 25, 2012
31
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:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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:

hmmm16

Member
Feb 25, 2012
31
yeah but we just treat this case first and use the uniqueness of additive inverses right?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
that is, i believe, what Opalg was hinting at.
 

James

New member
Jan 30, 2012
15
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?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.