# Silly question...?!

• Dec 20th 2010, 04:57 AM
Primality
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 $A$?
• Dec 20th 2010, 05:10 AM
Swlabr
Quote:

Originally Posted by Primality
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!)
• Dec 20th 2010, 05:17 AM
Primality
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.
• Dec 20th 2010, 05:18 AM
TheArtofSymmetry
Quote:

Originally Posted by Primality
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.
• Dec 20th 2010, 05:22 AM
Swlabr
Quote:

Originally Posted by Primality
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.
• Dec 20th 2010, 02:50 PM
Primality
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?
• Dec 20th 2010, 11:42 PM
Swlabr
Quote:

Originally Posted by Primality
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).
• Dec 21st 2010, 04:39 AM
Primality
I agree, but what if this is not the case?