How to define the multiplicative group on the additive group of a finite field?

Question

For a finite field $\mathbb{F}_{p^n}$ with characteristic $p$ , we can with the Fundamental Theorem of Finitely Generated Abelian Groups and the elementary divisor decomposition to show that the additive group of $\mathbb{F}_{p^n}$ is isomorphic to $(\mathbb{Z}/p\mathbb{Z})^{n,\oplus}$ , the direct sum of $n$-copies of $\mathbb{Z}/p\mathbb{Z}$ . And then how can we define the multiplication on $(\mathbb{Z}/p\mathbb{Z})^{n,\oplus}$ to make the multiplicative group be isomorphic to $Z_{(p^n-1)}$ ? A natural idea is use the Hadamard product: $$ \boldsymbol{a}\circ\boldsymbol{b}=(a_1,\cdots,a_n)\circ(b_1,\cdots,b_n)=(a_1b_1,\dots,a_nb_n)\ , $$ and the multiplicative identity of such multiplication is $(1_1,\cdots,1_n)$ but there exist many zero divisors like $(0_1,1_2,\cdots,1_n)$ . So how to define the multiplication on $(\mathbb{Z}/p\mathbb{Z})^{n,\oplus}$ to construct a cyclic group of order $(p^n-1)$ ?

Answer 1

One way to do is to take an irreducible polynomial $f \in (\Bbb Z/p\Bbb Z)[x]$ of degree $n$ and then identify the additive groups of $(\Bbb Z/p\Bbb Z)^{n,\oplus}$ and $(\Bbb Z/p\Bbb Z)[x]/(f)$ via $(a_1,a_2, \ldots,a_n) \mapsto a_1+a_2x+a_3x^2+\ldots+a_nx^{n-1}+(f)$. You can transfer the multiplication in the quotient ring $(\Bbb Z/p\Bbb Z)[x]/(f)$ via this isomorphism. This depends on a choice of $f$, but I don't think you can get an explicit formula without making any such choices.