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

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 $$(p^n-1)$$ ?

• Commented Feb 8 at 6:08

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.