Isomorphism: $F = \mathbb{Z}_{5}(\alpha)$, $\alpha^2 +2 =0$, and $F'= \mathbb{Z}_{5}(\beta)$, $\beta ^2 + \beta + 1 = 0$. Let $F = \mathbb{Z}_{5}(\alpha)$, where $\alpha^2 +2 =0$, and let $F'= \mathbb{Z}_{5}(\beta)$, where $\beta ^2 + \beta + 1 = 0$.
Exhibit the isomorphism between $F$ and $F'$. 
Honestly, I don't know how to solve this problem. I tried to do it by finding roots, which are $\alpha = \pm \sqrt{2}i$ and $\beta = \frac{1}{2} \pm \frac{\sqrt{3}i}{2}$
Please help me to solve this provlem. Thank you!

Note that $\alpha^2 = 3$, thus $\alpha^4 = 3^2 = 4$, and so $\alpha^8 = 4^2 = 1$. So the order of $\alpha$ is 8.
Now $(\alpha + 1)^2 = \alpha^2 + 2\alpha + 1 = 3 + 2\alpha + 1 = 2\alpha + 4$, and consequently:
$(\alpha + 1)^3 = (\alpha + 1)(2\alpha + 4) = 2\alpha^2 + 4\alpha + 2\alpha + 4 = 2(3) + 4 + \alpha = \alpha$.
This shows that the order of $\alpha + 1$ must be $24$ (do you see why it cannot be a multiple of $3$ less than $24$? Since we already know the order is not $2$, that leaves only $4$ and $8$- but:
$(\alpha + 1)^8 = (\alpha + 1)^6(\alpha + 1)^2 = \alpha^2(\alpha + 1)^2 = 3(2\alpha + 4) = \alpha + 2 \neq 1$).
So $(\Bbb Z_5(\alpha))^{\ast}$ is generated by $(\alpha + 1)$.
On the other hand we have $\beta^3 = \beta(\beta^2) = \beta(4\beta + 4) = 4\beta^2 + 4\beta = 4(4\beta + 4) + 4\beta$
$= \beta + 1 + 4\beta = 1$.
Since $3 = \dfrac{24}{8} = \dfrac{24}{\gcd(8,24)}$, it seems reasonable that we might hope $(\alpha + 1)^8 = \alpha + 2$ is a root of $x^2 + x + 1$.
I leave it you to show this is indeed the case, and then define the sought-after isomorphism.
