Proof the set $\mathbb{Z}_p$ is a field

Question

How do we prove that for a given number $n>1$ and defining $\mathbb{Z_n}= \{0, 1, 2, . . . , n − 1\}$, for any prime $p$, the set $\mathbb{Z_p}$ with the addition mod p and multiplication mod p, and congruence mod p, is a field.

How do I go about this proof? I have no idea where to begin except I know that:

• F is closed under addition and multiplication, i.e. $\forall a, b \in$ F, we have $a + b$ and $a × b$ are also both in F.
• commutitivity: $a + b = b + a$ and $a × b = b × a$
• associativity: $a + (b + c) = (a + b) + c$ and $a × (b × c) = (a × b) × c$
• unique neutral element property: there are neutral elements 0 and 1 for addition and multiplication.
• Unique inverse element for each given element: and element a of F has unique additive inverse and a unique multiplicative inverse.
• distributivity of multiplication over addition: $a × (b + c) = a × b + a × c$.

Answer 1

It is very easy to show the first four propertis and the last one, actually it heredates the abelian group structure from $\mathbb{Z}$ (I don't know if you know about groups and quotient groups) but is easy to do the calculations the only hard part is to proof that there exists a multiplicative inverse for every non zero element, that's what I will show here. (If you need help with the other I can help too)

Lets take a non zero element of $\mathbb{Z}_p$ let's say it is $a$ then $p$ prime says $gcd(a,p)=1$ then, there exists $r,s \in \mathbb{Z}$ such that $ra+sp=1$ then you have $sp=1-ra$ which implies $p|1-ra$ and then $ra\equiv 1 (mod$ $p)$ and r is the multiplicative inverse of $a$ and we have showed the fifth property and if you can prove the other properties then you have that $\mathbb{Z}_p$ is a field.

Answer 2

Probably the most non-trivial step is showing the existence and uniqueness of the multiplicative inverse. For this, we apply Bezout's lemma: given $0 \le a < p$, there will be - because $a$ and $p$ must be co-prime - integers $m$ and $n$ such that $am + pn = 1$. Modulo $p$, this will say that $am = 1$, and then we may write $m = pk + m'$, where $0 \le m' < p$, to obtain $am' = 1$. So $a$ will have the multiplicative inverse $m'$.

If we also have $ak = 1$ for some other $k \in \mathbb{Z}_p$, then $a(k-m') = 0$, which would mean that either $a$ is zero, or else $k - m'$ is zero modulo $p$, in which case we must have $m' = k$.