TITLE: $\mathbb{Z} _{29}$ is a field. True or False.
QUESTION [5 upvotes]: My answer was True and this is my argument:
Since $\mathbb{Z}_{n}$ has got $2$ operations plus the other properties of a ring, I figured that it is indeed a ring. On the other hand, since $\mathbb{Z}_{n}$ is abelian (commutative) and has the multiplicative identity $1$, It can be concluded that it is a $Commutative$ Ring with an $identity$, hence implies that it is a Field. 
I've also read a theorem which states that: "Every finite integral domain is a field." 
Is my argument acceptable and can I also use the above theorem as an argument?
Your help would be really appreciated.

REPLY [2 votes]: A commutative ring with identity is not necessarily a field. You should check your definitions. For instance, $\Bbb Z_6$ has $2\cdot 3=0$, hence isn't even an integral domain. 
You absolutely can use your theorem. Note that $\Bbb Z_{29}$ has no zero divisors (why?), and hence it is an integral domain, and hence by your theorem is a field. 
However, one need not appeal to such advanced machinery. Bezout's theorem says that if $gcd(a,n)=1$ then there's a pair of integers $(p,q)$ such that $pa+qn=1$. Picking $n=29$ this shows that every nonzero element of $\Bbb Z_{29}$ is invertible (why?). Since $\Bbb Z_{29}$ is a nonzero commutative ring, it is a field.