Results 1 to 8 of 8

Math Help - Silly question...?!

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    4

    Silly question...?!

    Say we have a finite ring A with x \in A and x^k = x for some k \geq 4. Is it true that x^{k-1}=1 in A?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Primality View Post
    Say we have a finite ring A with x \in A and x^k = x for some k \geq 4. Is it true that x^{k-1}=1 in A?
    No. Take x=0.

    Do you perhaps mean non-zero element? Then certainly the result holds if there are no zero divisors, as

    x^k=x \Rightarrow x^k-x=0 \Rightarrow x(x^{k-1}-1)=0.

    However, it is not true in general. Take a ring with a non-zero and non-unital idempotent ( x^2=x). Then clearly this will be a counterexample.

    There exists a ring of order 4 with this property. Can you find it?

    (Also, no question is ever silly in maths!)
    Last edited by Swlabr; December 20th 2010 at 06:21 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    4
    Yes you're quite right, I should have said non-zero element. And yes, you're right that the result holds if there are no zero divisors (in this case we have a finite integral domain, hence a field), but I'm really interested in the case where there are zero divisors.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2010
    Posts
    95
    Quote Originally Posted by Primality View Post
    Say we have a finite ring A with x \in A and x^k = x for some k \geq 4. Is it true that x^{k-1}=1 in A?
    You assumed that a finite ring A has a multiplicative unit. Not all finite rings have a multiplicative unit.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Primality View Post
    Yes you're quite right, I should have said non-zero element. And yes, you're right that the result holds if there are no zero divisors (in this case we have a finite integral domain, hence a field), but I'm really interested in the case where there are zero divisors.
    I've edited my above post.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2010
    Posts
    4
    Thanks very much for the help. I guess the answer to my question as stated is "no" although I'm not sure the replies above show this completely but am sure they can be easily adapted. (I don't think I've assumed that A has a multiplicative unit...if you can prove there is a ring with x^k=x doesn't have a multiplicative unit then this just shows that the answer is "no" and the slight problem with taking ring with a non-zero and non-unital idempotent is that I'd specifically said k \geq 4.)

    I'll state the problem I'm really interested in:

    Suppose f(x) \in \mathbb{Z}[x] and assume we know that f(x) = ((f(x))^k in the ring \frac{\mathbb{Z}[x]}{(x^r-1,n)} where r,n,k \in \mathbb{Z} and k \geq 4. Suppose we also have that (k-1) \mid (b-c) for integers b,c. Then show that (f(x))^b = (f(x))^c in \frac{\mathbb{Z}[x]}{(x^r-1,n)}.

    I thought I must be missing something obvious (which led to the original question), but perhaps not...Can anyone help?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Primality View Post
    Thanks very much for the help. I guess the answer to my question as stated is "no" although I'm not sure the replies above show this completely but am sure they can be easily adapted. (I don't think I've assumed that A has a multiplicative unit...if you can prove there is a ring with x^k=x doesn't have a multiplicative unit then this just shows that the answer is "no" and the slight problem with taking ring with a non-zero and non-unital idempotent is that I'd specifically said k \geq 4.)

    I'll state the problem I'm really interested in:

    Suppose f(x) \in \mathbb{Z}[x] and assume we know that f(x) = ((f(x))^k in the ring \frac{\mathbb{Z}[x]}{(x^r-1,n)} where r,n,k \in \mathbb{Z} and k \geq 4. Suppose we also have that (k-1) \mid (b-c) for integers b,c. Then show that (f(x))^b = (f(x))^c in \frac{\mathbb{Z}[x]}{(x^r-1,n)}.

    I thought I must be missing something obvious (which led to the original question), but perhaps not...Can anyone help?
    The result you wanted to hold will hold here if (x^r-1, n) is a prime ideal (as then your quotient ring will be an integral domain).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2010
    Posts
    4
    I agree, but what if this is not the case?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Silly question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: February 12th 2011, 05:05 AM
  2. Silly question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: September 28th 2010, 02:33 AM
  3. silly question - does cos^2 x = (cos x)^2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 14th 2009, 05:33 PM
  4. A silly question
    Posted in the Calculus Forum
    Replies: 12
    Last Post: June 28th 2007, 07:06 AM
  5. Silly question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 9th 2006, 12:25 PM

/mathhelpforum @mathhelpforum