In Integer and modular addition of Cyclic group:
The integer and modular addition operations, used to define the cyclic groups, are both the addition operations of commutative rings, also denoted Z and Z/n. If p is a prime, then Z/p is a finite field, and is usually instead written as Fp or GF(p). Every field with p elements is isomorphic to this one.
Per my understanding, the Z/p should be a finite field no matter p is prime or not, is it right?
It'll be finite, but it won't be a field. A field requires that every element has a multiplicative inverse.
Take for example $Z/6$. In this ring, $2 \cdot 3 = 6 = 0$, but neither $2$ nor $3$ is zero. An elementary consequence of this is that neither of them can have multiplicative inverses.
No; if $p$ is not prime, then $\Bbb{Z}_p$ is not a field. If $p$ is not prime, then let $ab=p$, where $a\neq 1,p$. Then in $\Bbb{Z}_p$, we have $ab = 0$. If $\Bbb{Z}_p$ were a field, then there would exist $c$ such that $ac = 1$. However, we would then have $b = cab = c\cdot 0 = 0$, which is false (if $b=0$ mod $p$ then $a=1$).
An important sometimes overlooked addition to the provided answers:
Although there exist finite fields on $p^n$, it is not the case that $\mathbb{Z}/p^n\mathbb{Z}$ is a field for $n>1$. $p^{k-1}$ and $p$ are both non-zero elements of $\mathbb{Z}/p^n\mathbb{Z}$, but their product is obviously $p^n=0$. Constructions of finite fields for $n>1$ can be tricky, although there is a simple example for $n=2$ and $p=4k+3$ given by $$(\mathbb{Z}[x]/(x^2+1))/(p)=\mathbb{Z}[i]/(p))\cong \mathbb{F}_{p^2}$$ The reason this fails to work for other values of $p$ is that $i$ is the solution to $x^2+1=0$, but this equation already has solutions when $p=4k+1$ and when $p=2$ which makes the reduction trivial.
