Show that if $p$ is prime then $\Bbb Z_p$ is a field

Check my proof please.

Let $\Bbb Z_p:=\Bbb Z/p\Bbb Z$ be the quotient ring modulo $p$. I want to prove that if $p$ is prime then $\Bbb Z_p$ is a field.

Known facts about the quotient rings of the kind $\Bbb Z_n$ (I dont list all the characteristics of a ring, just some ones):

• They are commutative rings with unity.

• $[0]=[n]$ is the identity of addition or zero. Then $[n]+[a]=[a]$ and $[n]\cdot[a]=[n]$ for all $[a]\in\Bbb Z_n$.

• $[1]$ is the multiplicative identity or unity. Then $[1]\cdot[a]=[a]$ for all $[a]\in\Bbb Z_n$.

• The addition is defined as $[a]+[b]=[a+b]$, and the multiplication is defined as $[a]\cdot[b]=[ab]$.

• $[a]=[b]$ means that exists $z\in\Bbb Z$ such that $a=b+nz$.

• $|\Bbb Z_p|=p$.

A ring $\Bbb Z_p$ dont have zero divisors, i.e. doesnt exist $[a],[b]\in\Bbb Z_p$ distinct of $[p]$ such that $[a]\cdot[b]=[p]$. Proof by contradiction: suppose that exists such $[a],[b]$ distinct of $[p]$ that are zero divisors. Then exist some $z_j\in\Bbb Z$ such that:

$$(a+pz_1)(b+pz_2)=pz_3\iff ab+pz_4=pz_3\iff ab=pz_5$$

but this is a contradiction with the assumption that $a,b\neq p$ and $p$ prime, so dont exist divisors of zero on any $\Bbb Z_p$.$\Box$

For any $[a]\in\Bbb Z_p$ distinct of $[p]$ exist $[b]$ such that $[a][b]=[1]$. Because $\Bbb Z_p$ dont have zero divisors and is finite for some $[a]\neq[p]$ we have that if $[b]\neq[c]$ then $[a][b]\neq[a][c]$. Proof by contradiction: if $[a][b]=[a][c]$ in the above conditions then

$$[a][c]=[ac]=[ab]=[a][b]\iff ac+pz_1=ab+pz_2\iff a(c-b)=pz_3$$

for some $z_j\in\Bbb Z$. But $pz_3\in[p]$, then $[a][c-b]=[p]$ but because there are no zero divisors in $\Bbb Z_p$ then or $[a]=[p]$ or $[c-b]=[p]$ what contradicts the above conditions.$\Box$

Because $[a][b]\neq[a][c]$ for $[a]\neq[p]$ and $[b]\neq[c]$ then for any $[a]\neq[p]$ we have that

$$[a]\cdot\Bbb Z_p=\Bbb Z_p\implies \exists [b]\in\Bbb Z_p:[a][b]=[1]$$

In other words: every element of $\Bbb Z_p$ but $[p]$ have a multiplicative inverse as stated above.$\Box$

Because $\Bbb Z_p$ is a commutative ring with $[1]\neq[0]$ and have multiplicative inverse for all their elements but $[p]=[0]$ then $\Bbb Z_p$ is a field.$\Box$

I have looked over your proof. Good job, it is correct. Comments:

• In the proof that inverses exist, you implicitly used the following fact: if $[c-b]=[p]$, then $[b] = [c]$. This is easy to prove, but you could add a sentence of explanation.

• Some of the time, when you introduced $z_j$ in your reasoning, it was unnecessary. For example, if $[ab] = [ac]$, then $[a][b] = [a][c]$, so $[a]([b] - [c]) = [0]$, so either $[a] = [0]$ or $[b] = [c]$. No use of $z_j$ necessary. Anyway, such reasoning is often more commonly written out using the "mod" notation.