Proving the Falsehood of the Homomorphism Property for Ω:Zp^n→Zp^n

[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?

 

[jgens]

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

 

[Bachelier]

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

 

[Bachelier]

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

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

 

[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:Z₄→Z₄ given by Ω(x) = x². Then Ω(1+1) = Ω(2) = 2² = 0 but Ω(1)+Ω(1) = 1²+1² = 2.

 

[Bachelier]

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:Z₄→Z₄ given by Ω(x) = x². Then Ω(1+1) = Ω(2) = 2² = 0 but Ω(1)+Ω(1) = 1²+1² = 2.

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