Finite fields and arithmetic

Question

For every prime number $p$ and every positive integer $k$, there is a field with exactly $p^k$ elements. When $k=1$, it's just the integers$\bmod p^k$, and when $k>1$, it's not. So if I want the elements of the field to be the congruence classes$\bmod p^k$, then the addition and multiplication will differ from adding and multiplying$\bmod p^k$. Is there any way to define addition and multiplication on those classes that makes them a field and interacts in some elegant way with the usual arithmetic operations?

Answer 1

Actually, for $k>1$ (and even for $k=1$ but in this case you'll have the same result) you could construct a field $K$ of charateristic $p$ with $p^k$ elements. As $\mathbf{Z} / p^k$ has also $p^k$ elements, it is in bijection via a map $\varphi : \mathbf{Z} / p^k \rightarrow K$ with $K$. (Note that $\varphi$ cannot be a morphism of ring if $k>1$ because $p$ is a zero divisor in $\mathbf{Z} / p^k$ but not in $K$). Now for $x,y\in \mathbf{Z} / p^k$ define $x +^\varphi y := \varphi^{-1} (\varphi(x) + \varphi(y))$ and $x {\times}^{\varphi} y := \varphi^{-1} (\varphi(x) \times \varphi(y))$. If $0^{\varphi}$ and $1^{\varphi}$ correspond to $0$ and $1$ via $\varphi$ then $\left( \mathbf{Z} / p^k, +^\varphi, \times^\varphi, 0^\varphi, 1^\varphi \right)$ will be a field, isomorphic to $K$.

Now your/the question is, could we find a $\varphi$ such that the additive law $+^\varphi$ is equal to the canonical additive law $+$ (the one coming from $\mathbf{Z}$) on $\mathbf{Z} / p^k$ ? The answer is no. Because, as in $\mathbf{Z}$, the additive law determines uniquely the multiplicative law. Indeed $$k \times l = \underbrace{l + \ldots + l}_{\textrm{$k$ times}} = \underbrace{1 + \ldots + 1}_{\textrm{$k l$ times}}$$