$p$ is prime. Show that $\mathbb Z/p\mathbb Z$ is a field 
Let $p$ ∈ $\mathbb N$ be a prime number. Show that $\mathbb Z/p\mathbb Z$ contains no zero-divisors except $0$, hence ($\mathbb Z/p\mathbb Z$)$^∗$ = {$1, 2, . . . , p − 1$}, and $\mathbb Z/p\mathbb Z$ is a field. Where ($\mathbb Z/p\mathbb Z$)$^∗$ is the set of units.

My attempt:
Suppose $ab=0$ for $b\ne 0$. Then $ab\equiv 0\mod p$ and so $p$ divides the product $ab$. Since $p$ is a prime, it follows that $p$ divides a factor. So $p\mid a$ or $p\mid b$. But $b\ne 0$ and so $p\mid a$, i.e., $a\equiv 0\mod p$, i.e., $a=0$ in ${\Bbb Z}_p$.
Now, for the "hence ($\mathbb Z/p\mathbb Z$)$^∗$ = {$1, 2, . . . , p − 1$}" part, I know that $\mathbb Z/p \mathbb Z$ = {$0,1,2,...,p-1$} and I already proved in a previous part of this problem that an element in a ring is either a zero-divisor or a unit. And thus, since $0$ is the zero-divisor, then all the other elements of $\mathbb Z/p \mathbb Z$ are units, therefore ($\mathbb Z/p\mathbb Z$)$^∗$ = {$1, 2, . . . , p − 1$}. Is this correct?
For the "$\mathbb Z/p\mathbb Z$ is field" part, I know that a field is a commutative ring, so all I have to do here is to show that $\mathbb Z/p\mathbb Z$ is commutative? And if so, how can I do that?
 A: ''Now, for the "hence (Z/pZ)∗ = {1,2,...,p−1}" part, I know that Z/pZ = {0,1,2,...,p−1} and I already proved in a previous part of this problem that an element in a ring is either a zero-divisor or a unit. And thus, since 0 is the zero-divisor, then all the other elements of Z/pZ are units, therefore (Z/pZ)∗ = {1,2,...,p−1}. Is this correct?''
In the previous part you proved that the ring contains no zero divisors. By definition, zero is not a zero divisor.
But this does not imply that the ring is a field, i.e., all nonzero divisors are units. For instance, the ring of integers contains no zero divisors, but only $\pm 1$ are units.
To show that ${\Bbb Z}_p$ is a field, it is clear from the construction that it is a commutative quotient ring with elements $0,1,\ldots,p-1$.
Take a nonzero element $a$. Then $\gcd(a,p)=1$ and so by Bezout's theorem,
$ra+sp=1$
for some integers $r,s$. Thus $ra\equiv 1\mod p$, i.e., $ra=1$ in ${\Bbb Z}_p$. Thus $a$ is invertible with inverse $a^{-1}=r$.
A: Note that not all commutative rings are fields; a field is a commutative division ring, which means that there are no zero divisors (as shown in the first part). So the division ring part is done and we are left with commutativity. Now, $\mathbb Z/p\mathbb Z$ is constructed as the integers modulo $p$, so the fact that it is commutative follows simply fro the fact that $\mathbb Z$ is commutative.
A: You could prove that $\mathbb Z /p\mathbb Z$ is commutative with a direct proof. However it’s interesting to understand this more general result: Quotient Ring of Commutative Ring is Commutative and to apply it to your special case noting that $\mathbb Z$ is a commutative ring.
It is true that units are non zero divisors here as we’re considering a finite ring. This is not true in general.
A: This answer only deals with commutativity.
For $n,m\in\mathbb Z$ on base of commutativity of multiplication in $\mathbb Z$ we find by applying definition of multiplication in $\mathbb Z/p\mathbb Z$: $$(n+p\mathbb Z)(m+p\mathbb Z):=nm+p\mathbb Z=mn+p\mathbb Z=(m+p\mathbb Z)(n+p\mathbb Z)$$
A: p is prime implies $p\mathbb{Z}$ is maximal ideal (as well as prime) of $\mathbb{Z} $
Then quotient ring is a field.
