I know $\frac{\mathbb{F}_5[x]}{(x^2+x+1)} $ and $ \frac{\mathbb{F}_5[x]}{(x^2 -2)} $ is isomorphic because they are both 2-degree extension of $ \mathbb{F}_5 $ .
But I cannot contract explicit isomorphism between them.
Could you show me isomorphism map between $\frac{\mathbb{F}_5[x]}{(x^2+x+1)}$ and $ \frac{\mathbb{F}_5[x]}{(x^2 - 2)} $?
I tried to show $$ f\colon\mathbb{F}_5[x]\to\mathbb{F}_5[x]/(x^2+x＋1)$$ is isomorphism, which sends $x$ to $g(x)＋c＋(x^2＋x-1)$. I couldn't find $g(x)$, and show $f$ is ring hom and bijection（surjection is clear）. I couldn't find $g(x)$ because constant term doesn't vanish. Thank you for your help.
As requested, I outline the isomorphism. If we fix $y:=2x+1\in\mathbb F_5[x]/\left(x^2+x+1\right),$ then we see that $$y^2=4x^2+4x+1=2.$$ In particular, we can construct a map $$\mathbb F_5[y]\to\frac{\mathbb F_5[x]}{(x^2+x+1)}$$ by sending $y\mapsto2x+1.$ It's not difficult to check that this is surjective, for $3y+2\mapsto x.$ It remains to show that our kernel is $\left(y^2-2\right).$ The kernel requires $p(y)\in\mathbb F_5[y]$ to have $$p(2x+1)\equiv0\pmod{x^2+x+1}.$$ However, this is equivalent to $$p(2x+1)\equiv0\pmod{(2x+1)^2-2},$$ so indeed $p(y)\in\left(y^2-2\right).$ This finishes the construction of our isomorphism.
At a high level, we were looking for an automorphism of $\mathbb F_{5^2},$ so it makes sense to look for linear maps to do the trick. This is what motivates the $y:=2x+1$ substitution.