If $k>0$ is a positive integer and $p$ is any prime, when is $\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in\mathbb Z_p\}$ a field. 
If $k>0$ is a positive integer and $p$ is any prime, when is $\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in\mathbb Z_p\}$ a field? Find necessary and sufficient condition.

Attempt: Since we know that a finite integral domain is a field and since $\mathbb Z_p[\sqrt{k}]$ is finite, it suffices to find the condition when $\mathbb Z_p[\sqrt{k}] =\{a + b\sqrt k~|~a,b \in \mathbb Z_p\}$ forms an integral domain.
The given set forms a commutative ring with unity. Hence, the only condition that needs to be satisfied is the absence of any zero divisors.
$(a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k) = (a_1a_2+b_1b_2k)+(a_1b_2+a_2b_1)\sqrt k$
Case $1$: When $k$ is not a perfect square
$(a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k)=0 \implies (a_1a_2+b_1b_2k) \bmod p=0~~ \& ~~ (a_1b_2+a_2b_1) \bmod p =0$
Case $2$: When $k$ is a perfect square $ = u^2$
$(a_1+b_1 \sqrt k)(a_2+b_2 \sqrt k) = 0 \implies (a_1a_2+b_1b_2 u^2 + (a_1b_2+a_2b_1)u ) \bmod p=0 $
Both the above cases can have many cases and seems a bit complicated. Am I missing out on something? The book which I am reading (Gallian) hasn't introduced Quadratic Residues as of yet. Thank you for the help.
 A: Let $\alpha=a+b\sqrt{k}$ be a zero-divisor then there exists a non-zero $\beta=c+d\sqrt{k}$ such that $\alpha \beta \equiv 0 \pmod{p}$. It is equivalent to $\alpha \bar{\alpha} \beta \bar{\beta} \equiv 0 \pmod{p}$ (where $\bar{\alpha}$ is the conjugate of $\alpha$). This reduces to
$$(a^2-kb^2)(c^2-kd^2) \equiv 0 \pmod{p}.$$
Using the prime property, we get $p$ should divide at least one of them. Thus $\mathbb{Z}_{n}[\sqrt{k}]$ will NOT be a field if $x^2 \equiv k \pmod{p}$ has a solution. 
I know you have said that you have not studied quadratic residues but the simplest criterion for your question eventually relates to that. So if $k$ has no square root modulo $p$ then you will have a field otherwise not.
A: If $k$ is a perfect square than $ a+ b \sqrt{k} $ is already in $\mathbb Z _ p $ and you're result should follow.  
A: It is easy task to prove that R= Z_p[√k] is a commutative ring with unity.Recall that a finite integral domain is a field. As R is finite commutative ring with unity , only one condition left R to be an integral domain is that R has no zero divisors. Now , two cases may arise:
CASE 1 : k is a perfect square.
                 In this case , R reduces to Z_p and  consequently R is an integral domain as p is prime and hence R is a field , using the fact that a finite integral domain is a field.
CASE 2 : k is not a perfect square.
                Let a + b√k be a nonzero element of R. Then a and b can't be simultaneously zero. Consider the product
                 ( a + b√k) (x +y√k)=0 in Z_p .............(1).
If we can prove that the only possibility for x and y is zero , a+b√k is not a left zero divisor as well as not a right zero divisor ( as R is commutative) and consequently a+b√k is not a zero divisor and hence R becomes an integral domain and R will be a field.
    Now , from (1), 
     (ax+kby)+(bx+ay)√k =0 in Z_p
  =>        ax+kby =0 in Z_p
   and     bx+ay =0 in Z_p.
Thus we have a homogeneous system of linear equations in x and y  seeking for a unique zero solution in Z_p. Now , by Cramer's rule , the above homogeneous system will have a unique zero solution in Z_p iff  its coefficient determinant not = 0 in Z_p 
i.e., 
    a^2 - kb^2 not=0 in Z_p ..................................(2).
Now, two subcases may arise of CASE 2:
Subcase 1:  k is a multiple of p (as well as not a                             perfect square).
Let k=pm, m€N. Then k=0 in Z_p and from (2), 
      a^2 not=0 in Z_p => a not=0 in Z_p => the non zero-divisor elements in R are of the form a+b√k ,where a€Z_p* and b€Z_p. This suggests that there are some nonzero elements in R which are zero divisors namely of the form b√k, b€Z_p*. For instance , take p=7 and k=14 and see that 
                       2√14. 3√14 =0 in Z_7[√14].
In this subcase , R is not an integral domain.
Subcase 2: Let k is not a multiple of p as well as not a perfect square.
 Then by division algorithm, there exist integers q and r such that
                   k = q p  + r ; 0< r  c^2 not= r in Z_p , where c = ab^(-1) € Z_p
 => c^2 not= qp + r in Z_p
 => c^2 not= k in Z_p.
 Hence combining all the cases, we conclude that
Z_p[√k] is filed iff either k is a perfect square or         
      x^2 not= k in Z_p , x€Z_p*.

