$\mathbb{Z} _{29}$ is a field. True or False. 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. 
 A: Yes it is a field; more generally $\mathbf{Z}/n\mathbf{Z}$ is a field iff $n$ is a prime number. But no your argument is not sufficient, but it is if you argue that it is an integral domain (to see this notice that $(n)$ is a prime ideal iff $n$ is 0 or a prime number; in case $n=0$ this is $\mathbf{Z}$ which is not a field). It is certainly not the case that every commutative ring is a field, but a field is by definition a commutative ring.
To make the argument explicit for your convenience: First show that every finite integral domain $A$ is a field (hint: for $a\neq 0$ in $A$ consider the mapping $A\rightarrow A$ defined by $x\mapsto ax$; show that it is injective and hence bijective since $A$ is finite). Then show that $\mathbf{Z}/n\mathbf{Z}$ (a finite ring for $n$ at least $1$) is an integral domain iff $n=0$ or $n$ a prime (use the classification of prime ideals of $\mathbf{Z}$).
A: 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. 
