I give up. I'm new in the fields world, and
I'm trying to give a sufficient and necessary condition for $\mathbb{Z}_{p}[\sqrt{k}]=\{a+b\sqrt{k}:a,b\in \mathbb{Z}_{p}\}$ to be a field ($p$ is a prime and $k$ is a positive integer).
I claim that the condition is that $p$ doesn't divide $k$, but I don't know if this is true, I've tried to prove this but I've got stuck in the "sufficient" part.
It would be great if some of you can help me, thanks!.
If $F$ is a field, then $F[\sqrt{a}]$ is always a field.
Case I: $a$ is a square in $F$, $b^2 = a$. Then $F[\sqrt{a}] = F[b] = F$.
Case II: $a$ is not a square in $F$. Then the polynomial $X^2 - a$ is irreducible, so $F[\sqrt{a}] \cong F[X]/(X^2-a)$ is a field.
More generally, if $R$ is a ring with no zero divisors that contains $F$, and it is finite-dimensional as a vector space over $F$, then it is a field.
The issue raised in the comments is that we might define $F[\sqrt{a}]$ to be $F[X]/(X^2-a)$, rather than the smallest subring of $\overline{F}$ containing $F$ and $\sqrt{a}$ (this appears to be the definition you intend). But when $a$ is a nonzero square, this is isomorphic to $F\times F$, which is not a field, and when $a=0$, it is isomorphic to $F[\epsilon]/\epsilon^2$ (there is a very minor exception when $\operatorname{char}K = 2$, in which case nonzero squares also yield $F[\epsilon]/\epsilon^2$).