- Thread starter
- #1

- Thread starter James
- Start date

- Thread starter
- #1

- Moderator
- #2

- Feb 7, 2012

- 2,793

I assume that x_2 = 1_R should read $x^2 = 1_R$.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

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.

- Feb 15, 2012

- 1,967

- 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.

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:

- Feb 15, 2012

- 1,967

that is, i believe, what Opalg was hinting at.

- Thread starter
- #8

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?that is, i believe, what Opalg was hinting at.

- Feb 15, 2012

- 1,967

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.