homomorphic mapping from $F_{p^n}$ to $Z_{p^n}$

Is it possible to have a homomorphic mapping from $$F_{p^n}$$ to $${\mathbb Z}_{p^n}$$ that preserves both the add and multiplication operators?

Or if we relax requirement, can we have a homomorphic mapping from the multiplicative group $$F_{p^n}^*$$ to $${\mathbb Z}_{p^n}^*$$ which preserves multiplication?

• You will get a slightly different answer, if you look for a homomorphic mapping in the other direction.
– j.p.
Commented Feb 8, 2022 at 7:09

"Is it possible to have a homomorphic mapping from $$\mathbb F_{p^n}$$ to $$\mathbb Z_{p^n}$$ that preserves both the add and multiplication operators?"
Other than the isomorphism when $$n=1$$, only the very boring mapping that send everything to 0. Consider the multiplicative identity of $$\mathbb F_{p^n}$$. We write this as 1 and consider our putative homomorphism $$\phi$$. We see that by additivity adding $$k$$ copies of 1, for any integer $$k$$ we have $$\phi(1+\cdots+1)=k\phi(1)\mod {p^n}$$ and in particular with $$k=p$$ we see that $$p\phi(1)=\phi(0)=0$$ so that $$\phi(1)=cp^{n-1}$$ for some $$1\le c\le p$$. Moreover, by multiplicativity we have $$\phi(1)=\phi(1\cdot 1)=\phi(1)\phi(1)$$ so that $$\phi(1)=1$$ or $$0$$. We conclude that $$c=p$$ and $$\phi(1)=0$$ (except in the case $$n=1$$). Further, for any $$\alpha\in\mathbb F_{p^n}$$ $$\phi(\alpha)=\phi(1\cdot\alpha)=\phi(1)\phi(\alpha)=0$$.
"Or if we relax requirement, can we have a homomorphic mapping from the multiplicative group $$\mathbb F_{p^n}^\times$$ to $$\mathbb Z_{p^n}^\times$$ which preserves multiplication?"
Only marginally less boring. Note that $$|\mathbb F_{p^n}^\times|=p^n-1$$ and $$|\mathbb Z_{p^n}^\times|=p^n-p^{n-1}$$. The size of the image of any homomorphism has to divide the GCD of these two which is $$p-1$$. We see that the image has to be a subgroup of the $$(p-1)$$th roots of 1 in $$\mathbb Z_{p^n}$$. Now pick any multiplicative generator of $$\mathbb F_{p^n}^\times$$ call this $$\alpha$$. There are precisely $$p-1$$ group homomorphisms depending on which of the $$(p-1)$$th roots 1 is equal to $$\phi(\alpha)$$. The kernel will be the $$\ell$$th powers in $$\mathbb F_{p^n}^\times$$ where $$\ell|(p-1)$$ is the multiplicative order of $$\phi(\alpha)$$ in $$\mathbb Z_{p^n}^\times$$.
• Actually, there is a second possible mapping for the case $n=1$... Commented Feb 5, 2022 at 14:33