# How to show this is a Hom.

1. Feb 27, 2014

### Bachelier

Define $\Omega: \mathbb{Z_{p^n}} \rightarrow \mathbb{Z_{p^n}}$ where $p$ is prime

with $\ \ \ \ \ \ \Omega(x) = x^{p}$

I am trying to prove this is a $Hom$ under addition.

any ideas?

2. Feb 27, 2014

### jgens

Essentially just apply the binomial theorem. Since the ring has characteristic p all the undesired terms vanish.

3. Feb 27, 2014

### Bachelier

The ring $K$ has order $p^n$ which makes it $\cong \mathbb{Z_{p^n}}$

Last edited: Feb 27, 2014
4. Feb 27, 2014

### Bachelier

You did notice that the ring I am working with has $p^n$ elements. It is an extension field of $\mathbb{Z_{p^n}}$.

Last edited: Feb 27, 2014
5. Feb 27, 2014

### jgens

Oh whoops you are right! Unless I am mistaken it turns out the result is actually false! Take p = n = 2 and consider the map Ω:Z4Z4 given by Ω(x) = x2. Then Ω(1+1) = Ω(2) = 22 = 0 but Ω(1)+Ω(1) = 12+12 = 2.

6. Feb 27, 2014

### Bachelier

yes, even in the general case $\Omega(1+1) = 2^p \neq 2 = \Omega(1) + \Omega(1) \ \forall n, p \geq 2$