Apologies if this is a repeat on the site –– I couldn't find a clear answer. The fundamental theorem of finite abelian groups says that any finite abelian group is isomorphic to the direct product of cyclic groups of a (not necessarily distinct) prime power order. The classic example demonstrating this fact is by taking an abelian group, $G$, of order $4$. Of course, $4=2^2=2 \cdot 2$, so we either have that $G \cong \mathbb{Z}/(4)$ or $G \cong \mathbb{Z}/(2) \times \mathbb{Z}/(2) = V_{4}$, the Klein four-group. Thus, there are two abelian group of order $4$, up to isomorphism.
One way of writing field axioms is as follows:
$\mathbb{F}$ is a field if (i) $(\mathbb{F}, +)$ is an abelian group, (ii) $(\mathbb{F}^{\times}, \cdot)$ is an abelian group, and (iii) the distributive/associative laws hold for both operations. In my textbook, the author constructs a finite field with $4$ elements, $\mathbb{F}_{2^2}$ and claims that $\mathbb{F}_{2^2}$ is not isomorphic to either $V_{4}$ or $\mathbb{Z}/(4)$ without much of an argument. Apologies if I'm missing something obvious, but it seems to me that for $\mathbb{F}_{2^2}$ to be a field, we must have that $(\mathbb{F}_{2^2}, +)$ is an abelian group, implying by the fundamental theorem that it must be isomorphic to one of these four-element groups. I understand why it can't be since both groups have zero divisors, so they aren't even integral domains. Which step of this am I misunderstanding?
Best Answer
It is isomorphic to $V_4$ as an additive group, certainly not as a ring. Indeed, $X^2+X+1$ is an irreducible polynomial of $\mathbf F_2[X]$, and, as a field, we have an isomorphism $$ \mathbf F_4\simeq \mathbf F_2[X]/(X^2+X+1).$$ Denote $\xi$ the congruence class of $X$ modulo $X^2+X+1$. As an $F_2$-vector space, the latter has the basis $\{1, \xi\}$, so it is isomorphic to $F_2\times F_2=V_4$ via the linear map \begin{align} \mathbf F_2[X]/(X^2+X+1)&\longrightarrow F_2\times F_2, \cr a+b\mkern2mu\xi&\longmapsto (a,b). \end{align}