Prove that $G$ is a field the multiplicative inverse portion

Question

Let $F$ be a field such that $p(x):=x^2+x+1 \neq 0$ for all $x \in F$.

Let $G = F \times F$, with elements written $(a,b) =a+jb, a,b \in F$ for a formal symbol $j$.

Define addition componentwise to be:

$(a+jb)+(c+jd)=(a+c)+j(b+d)$,

and define multiplication by the rule $j^2+j+1=0$:

$(a+jb)(c+jd)=(ac−bd)+j(ad+bc−bd)$

Question: How can I show the existence of the multiplicative inverse portion for this field $G$?

Thanks in advance.

Answer 1

Hint

$$j(-j-1)=1 \\ (aj+b)(a(-1-j)+b)=a^2-ab+b^2$$

Hint 2

If $b \neq 0$ and $a^2-ab+b^2=0$ then $-ab^{-1}$ is a root of $x^2+x+1$.

Use this idea to prove that $$a^2-ab+b^2=0 \Rightarrow a=0,b=0$$