Prove-If n is prime $\mathbb{Z}_n$ is a field. I need to prove that $\mathbb{Z}_n$ is a field if and only if $n$ is prime. And I proved the forward. 
But I am not sure how to prove the backward, 'if n is prime $\mathbb{Z}_n$ is a field.
'
What can I assume from the assumption $n$ is prime? 
and What do I need to show? 
 A: It is enough to show that $\Bbb Z_n$ is an integral domain, because every finite integral domain is a field. Suppose we have $ab = 0 \pmod n$, then $n \mid ab$, so $n\mid a$ or $n \mid b$ since $n$ is prime (Euclid's lemma), so $a$ or $b$ is zero in $\Bbb Z_n$. So $\Bbb Z_n$ is an integral domain. 
To see that every finite integral domain is a field, take powers of a nonzero element $a$. Since there are finitely many elements, we must have repeats, say $a^n= a^m$ for $n > m$. So $a^n-a^m=0$, so $a^{m}(a^{n-m} - 1) = 0$. If $a^m=0$, then $a=0$ by repeatedly applying the integral domain property, so $a^m \neq 0$, since $a$ is nonzero. Thus $a^{n-m}=1$, and $a \cdot a^{n-m-1} = 1$, so $a$ is invertible. Thus, every nonzero element has a multiplicative inverse and every finite integral domain is a field.
A: In fact, this is an 'if and only if' statement. So we will show that.
First, assume that $n$ is prime. We show that it is a field (by showing multiplicative inverses for nonzero elements):
If $n$ is prime and $\overline a\ne0$ (that is, $a$ isn't a multiple of $n$) then $a$ and $n$ are coprime. Then by Bezout's theorem there exist $b,c\in\mathbb Z$ such that
$$
ba+cn=1
$$
but that means that $\overline a\;\cdot\overline b=\overline1$.
Now we show that if $\mathbb{Z}/n\mathbb{Z}$ is a field, then $n$ is prime:
Suppose that this isn't so. So $n$ isn't prime. Then $n=ab$ for $a,b \in \mathbb{Z}$, so $\overline0=\overline a\;\overline b$, where $\overline a\neq \overline0 $ and $\overline b\neq 0$. But then $\mathbb Z/n\mathbb Z$ isn't an integral domain, so it can't be a field!
A: Other than showing trivial axioms, you need to show that every element in $Z/pZ$ has a multiplicative inverse. In other words, given $k$, you need to find $m$ such that $k \cdot m \equiv 1 \pmod p$
This, though, is equivalent to showing that there exists $n$ such that $km = pn + 1$. Because $p$ is a prime, we know that $\gcd(k, p) = 1$. By Euclid's Algorithm/Linear Diophantine Equations, there must exist $(m_0 , n_0)$ such that $km_0 +pn_0 = 1$. The result follows.
I can clarify any of this if you'd like - feel free to comment.
Cheers!
