Q:

On showing that $\mathbb Z[\sqrt{-2}] / (5) \simeq \mathbb F_{25}$.

I want to show that $\mathbb Z[\sqrt{-2}] / (5) \simeq \mathbb F_{25} = Z_5[x] / (x^2+x+1)$.
my first attempt:
First I see that all elements in the ideal $(5)$ are given by $\{ (a+b \sqrt{-2}) 5 | a,b \in Z \}$ so a generic element has the form $5a+ 5 \sqrt{-2} b $.
Now I want to show that there are 25 equivalence classes in $\mathbb Z[\sqrt{-2}] / (5) $ where two elements $x,y$ are equivalent $\iff (y_1 +y_2 \sqrt{-2}) -(x_1+x_2 \sqrt{-2}) \in    (5) $.
So I obtain the condition that $y_1 - x_1$ must be of the form $5a$ and $y_2 -x_2$ must be of the form $5b$ Thus
$$y_1 + y_2 -x_1 -x_2 = 5(a+b)$$ so I must have that $y_1 + y_2 -x_1 -x_2$ is a multiple of 5 for the system to have solution this seems to lead me to the solution that $\mathbb Z[\sqrt{-2}] / (5) \simeq \mathbb Z_{5}$ that is not correct.
my second attempt:
$$\mathbb Z[\sqrt{-2}]
\cong
\frac{\mathbb Z[x]}{(x^2+2)}
\implies
\frac{\mathbb Z[\sqrt{-2}]}{(5)}
\cong
\frac{\mathbb Z[x]}{(x^2+2,5)}
\cong
\frac{\mathbb Z}{(3,5)}$$
Where the last isomorphism is given by the map $f(x) \rightarrow f(1)$ this still seems incorrect.
Could someone please explain to me the flaws in my reasoning and try to rectify them?

A:

Everything is easier if you know that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain with respect to the usual norm function (essentally the same proof as for $\mathbb{Z}[\sqrt{-1}]$).
We can now show that $5$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$. If $5=xy$, then $5^2=|x|^2|y|^2$. A proper factorization would have $|x|^2=5$, but
$$
|a+b\sqrt{-2}|^2=a+2b^2=5
$$
is satisfied by no pair of integers $a$ and $b$.
Therefore $\mathbb{Z}[\sqrt{-2}]/(5)$ is a field and it obviously has characteristic $5$. Since it is generated over $\mathbb{F}_5$ by an element whose square is $-2=3$, it is a proper quadratic extension of $\mathbb{F}_5$, because $3$ is not a square in it; hence it is $\mathbb{F}_{25}$.
About your first attempt, you are correct up to a certain point but jump to a wrong conclusion: you can arbitrarily choose $a$ and $b$ among $\{0,1,2,3,4\}$, which makes for $25$ elements. You indeed prove that $x_1\equiv x_2\pmod{5}$ and $y_1\equiv y_2\pmod{5}$.
In the second attempt, you also do a wrong step: the correct isomorphism is
$$
\mathbb{Z}[x]/(x^2+2,5)\cong(\mathbb{Z}/5\mathbb{Z})[x]/(x^2+2)
$$
which is essentially the same as the first attempt, but requires a (short) proof.

