Prove that $A$ is a field if and only if a certain condition is true

Question

Let $(A, +, \cdot)$ be a ring with $8$ elements. Prove that $A$ is a field if and only if for any $x \in A - \{0\}$ there exist $a, b \in \{0, 1\}$ such that $x^3 + ax^2 + bx + 1 = 0$.

I haven't managed to do anything meaningful yet.

Thank you!

Answer 1

For one direction, choose any $\alpha\in A-\{0\}$ and locate $a, b\in \{0,1\}$ such that $\alpha$ is a root of $x^3+ax^2+bx+1$. Then $1=(-\alpha^2-a\alpha-b)\alpha=\alpha(-\alpha^2-a\alpha-b)$, so $(-\alpha^2-a\alpha-b)$ is an inverse of $\alpha$. This shows that $A$ is a division ring. An $8$-element division ring (or any finite division ring) is a field.

Before proceeding with the other direction, let me point out that I just used a moderately deep fact unnecessarily: I used Wedderburn's Little Theorem that a finite division ring is a field. I could have avoided this by noting that if $A$ is an $8$-element division ring, then the set $A-\{0\}$ under multiplication is a $7$-element group. Any $7$-element group is cyclic, hence commutative, so $A$ is commutative. This shows that $A$ is a field.

For the other direction, let $A=\mathbb F_8$ be an 8-element field. It is a degree $3$ extension of its prime field, $\mathbb F_2$, hence every element of $\mathbb F_8$ has degree $3$ or degree $1$ over the prime field. Choose $\alpha\in\mathbb F_8-\{0\}$ and let $m(x)$ be the (monic, irreducible) minimal polynomial of $\alpha$ over the ground field $\mathbb F_2$. Since $\alpha\neq 0$ and $m(x)$ is irreducible, $x$ is not a factor of $m(x)$. This forces $m(x)$ to be a monic polynomial with constant term $1$. If the degree of $m(x)$ is $3$, then it must have the form $x^3+ax^2+bx+1$ for some $a, b\in \{0,1\}$. If the degree of $m(x)$ is $1$, then $\alpha$ must be $1$, which is a root of $(x+1)^3=x^3+x^2+x+1$, which has the desired form.