Let $(A,+,* )$ be a ring with $9$ elements. Show that the following statements are equivalent
- For every $x \in A\setminus\{0\}$ there exists $a \in \{-1,0,1\}$ and $b\in\{-1,1\}$ such that $x^2+ax+b=0$
- $(A,+,* )$ is a field
I just started taking algebra and I don't really know how to deal with this problem. Can somebody please help me?
I'm assuming $A$ to be commutative and $1\in A$.
$1 \implies 2$ : $\forall x\neq 0$, $$x^2 +ax + b= 0 \iff x(x+a) = -b$$ Since $b = \pm 1$, either $x+a$ or $-(x+a)$ is the inverse of $x$. Thus $A$ is a field.
$2 \implies 1$ : Assume there exists $x$ s.t. for all $(a,b) \in \{-1,0,1\}\times \{-1,1\}$, $x^2+ax+b \neq 0$. Then $x \notin \{-1, 0, 1\}$ and thus $x^{-1} \notin \{-1, 0, 1\}$.
But since $x^2+ax = x(x+a) \neq \pm 1$ for all $a\in \{-1,0,1\}$, we have $x^{-1} \notin \left \{ \pm(x-1), \pm x, \pm (x+1) \right \}$.
Because $x\notin \{-1,0,1\}$, this last set has cardinal $6$ and doesn't intersect $\{-1,0,1\}$. But this leaves no more possibility for $x^{-1}$, because $A$ has cardinal $9$ : thus $A$ is not a field, a contradiction.