{0,1,2,.....n-1} is a field⇔p is prime

Note that since this is an “if and only if” proof,
 we need to show both directions:
“if” (a.k.a. “⇐”):
 Let p be prime.

1. the addition is closed inherit from integers
2. Commutative inherit from integers
3. Associativity inherit fro integers
4. ℤ_p contains additive identity(0)
5.  Each element has additive inverse 
6.  Each element has an additive inverse: for each x ∈ ℤ_p take −x (mod p). Then by the properties of modular arithmetic

                   x + (−x) ≡ 0 mod p

1. We get closure under multiplication from the integers since for x, y ∈ ℤ_p xy mod p is in ℤ_p
2. _{Multiplication is commutative since its commutative in the integers.}
3. _{Multiplication is associative since it is in the integers.}
4.  _{We have a multiplicative identity, 1 which acts the same as it does in the integers.}
5. _{We need multiplicative inverses for each element.}

            Let x ∈ Zp.         

     Note that we cannot just say that the inverse is 1 x since this does not look like an element of Zp. Since p is prime, it is relatively prime to every number, including x. Therefore by the hint we know that ∃ a, b ∈ Z such that
           ax + bp = 1
Modding out by p gives us:
         ax ≡ 1 mod p
Therefore a is the multiplicative inverse of x. (D) The distributive law is inherited from the integers since modular arithmetic behaves well. So this is a field!


Now we need the other direction: “only if” a.k.a. “⇒” We do this by proving that if p is not prime then Zp is not a field. (This is the contrapositive of the statement, if Zp is a field then p is prime.)


Since p is not a prime we can say

                     P=n.m ∈ℤ and ≠0 or1

Now 

     If ℤ_p is a field there exist m^-1 such that m.m^-1=1

Then

    n.1=n.(m.m^-1)=(n.m).m^-1=p.m^-1=0 (Beacuse P is there as factor using mod(p) arthmatci answer will be 

0)

But we assumed n≠0 hence cotradiction 

hence assumption is false Contrapostive statment is True

hence if ℤ_p is field then P is Prime