For which of $m \in \{5, 6\}$ is $\mathbb Z/m\mathbb Z$ a field? I have the following problem that I need to solve:

For which of $m \in \{5, 6\}$ is $\mathbb Z/m\mathbb Z$ a field?

I know how to do this for $m = 5$.
I know that $[1][1] = [1]$, $[2][3] = [1]$, $[4][4] = [1]$.
So every non-zero element of $\mathbb Z/5\mathbb Z$ has an inverse, and so $\mathbb Z/5\mathbb Z$ is a field. 
However, I am not sure how to do this for $m = 6$.
 A: Hint(s): What is $[2][3]$ in $
\mathbb{Z}/6\mathbb{Z}$? 
For a more general result, see if you can prove that $\mathbb{Z}/m\mathbb{Z}$ is a field if and only if $m$ is prime. (a hint for proving this: since $m$ is prime, every nonzero element of $\mathbb{Z}/m\mathbb{Z}$ is relatively prime to $m$!)
A: Note that if $m$ is composite, say $m=p\cdot q$; then $p,q\not \equiv 0 \pmod m$ in $\Bbb Z/m\Bbb Z$ yet $p\cdot q\equiv 0\pmod m$ so $\Bbb Z/m\Bbb Z$ is not even a domain, that is, there are zero divisors!
A: Since $\mathbb{Z}$ is commutative ring with unity and A is ideal of R=$\mathbb{Z}$ then use this result R/A is field if and only if A is maximal ideal.
Here $A=\left\langle 6 \right\rangle$(generated by 6) is not maximal ideal.
I feel still you are not understand 
here $\left\langle 6 \right\rangle \subset \left\langle 2 \right\rangle \subset \mathbb{Z}$ so it is not maximal it implies that$ \mathbb{Z}/ 6\mathbb{Z}$ is not field.  
Additional Ans. of below comments:
For example, Let us consider $\mathbb{Z}/ 4\mathbb{Z}=\{0+4\mathbb{Z},1+4\mathbb{Z},2+4\mathbb{Z},3+4\mathbb{Z}\}$
If possible suppose there is $a+4\mathbb{Z} \in \mathbb{Z}/4\mathbb{Z}$
such that $(2+4\mathbb{Z})(a+4\mathbb{Z})=(1+4\mathbb{Z})$
$(2+4\mathbb{Z})(a+4\mathbb{Z})=2.a+2.4\mathbb{Z}+4\mathbb{Z}.4\mathbb{Z}+a.4\mathbb{Z}$
but all the other elements except '$2.a$' will absorbed in $4\mathbb{Z}$
So we get $2.a+4\mathbb{Z}=(1+4\mathbb{Z})$
it implies that $2.a-1 \in 4\mathbb{Z}$
that is not possible.  
therefore $2+4\mathbb{Z}$ has no inverse in $\mathbb{Z}/4\mathbb{Z}$
