$x^2+ax+b=0 \overset ? \impliedby (A,+,\cdot)$ is a field Let $(A,+,\cdot)$ be a ring with $|A|=9$. Prove that the next statements are equivalent:
a) For all $x\in A-\{0\}$ there exist $a \in \{-1,0,1\}$ and $b \in \{-1,1\}$ s.t.   $x^2+ax+b=0$.
b) $(A,+,\cdot)$ is a field

*

*a) $\implies$ b)
If  $a \in \{-1,0,1\}$ then
\begin{aligned}
ax&=xa \\
x(x+a)&=(x+a)x=b \in\{-1,1\}.
\end{aligned}
So $x$ is invertible and $(A,+, \cdot)$ is a field.
I don't know how to show that b) $\implies$ a). Can somebody help me, please?
 A: The proof of (a)$\implies$(b) is not complete as you have not shown that multiplication is commutative. This follows from Wedderburn's theorem that every finite division ring is commutative, but we can avoid using the sledgehammer.
You know that $F=\{0,1,-1\}$ is a subfield of $A$ and is certainly contained in the center. Hence $A$ is a two-dimensional vector space over $F$, so it has a basis $\{1,x\}$ and every element is of the form $a+bx$, for $a,b\in F$. Hence any two elements commute. (Note that for this you don't even need that $A$ is a division ring, so this might go before the proof that every element of $A\setminus\{0\}$ is invertible.)
More generally, if $A$ is a ring of cardinality $p^2$, with $p$ a prime, then $A$ is commutative.
For the converse, you know that $A$ has characteristic $3$, so it is a two-dimensional field extension of $F=\{0,1,-1\}$. For the nonzero elements of $F$ there is no problem in finding the suitable $a$ and $b$. If $x\in A\setminus F$, it has a degree $2$ minimal polynomial over $F$.
