$R$ is a ring with p elements where p is a prime and has no zero divisors. Prove that $R$ is a field and is isomorphic to $\mathbb{Z}_p$. 
$R$ is a ring with p elements where p is a prime and has no zero divisors. Prove that $R$ is a field and is isomorphic to $\mathbb{Z}_p$.

I can prove that $R$ is a division ring and I know that by Wedderburn's theorem it must be a field. However, this problem is on the part of the text before the theorem is introduced so I'm wondering if there is another way to prove the commutativity of multiplication.
 A: You only need half of your assumptions if you assume that rings have a unit.
If $R$ is a ring with unit $1$, then there’s a map $ℤ → R$ given by $1 ↦ 1$ uniquely extendend to a ring homomorphism, whose kernel must be $nℤ$ for some $n ∈ ℕ₀$. Can’t be $n = 1$, though, because otherwise $1$ in $ℤ$ (which is mapped to $1 ≠ 0$ in $R$) would be in the kernel (unless of course $R = 0$).
By the first isomorphism theorem, $ℤ/nℤ$ must be isomorphic to some subring of $R$ whose cardinality must be, by Lagrange’s theorem (when considering the subring as an additive subgroup), a divisor of $\# R = p$ (if you assume $R$ to be finite of cardinality $p$). But $\# ℤ/nℤ ≠ 1$, since $n ≠ 1$.
So $n = p$ and the subring is $R$ itself and $R \cong ℤ/pℤ$.
A: You can directly determine the addition and multiplication tables. The additive group is a group of order $p$, thus is isomorphic to the additive group of $\mathbb{Z}_p$. So we can label the elements by $n=\underbrace{1+1+...+1}_{n\text{ times}}$, where $n<p$. Then using distributivity we determine the multiplication: $$m\times n=\underbrace{\underbrace{(1\times 1+...+1\times 1)}_{n\text{ times}}+\underbrace{(1\times 1+...+1\times 1)}_{n\text{ times}}+...+\underbrace{(1\times 1+...+1\times 1)}_{n\text{ times}}}_{m\text{ times}}$$ This is just $mn\pmod p$, if the ring is unital with unit $1$. 
In general the multiplication is thus determined by $1\times 1$, and is commutative (the subring generated by a single element is commutative, and this entire ring is generated by $1$.) 
You get another ring if $1\times 1=0$, but this has rather a lot of zero divisors! If $1\times 1$ were some other $k$, then I claim the ring is still unital, and isomorphic to $\mathbb{Z}_p$: $n\times m$ is, as calculated above, given by $knm\pmod{p}$, and in particular the multiplicative inverse of $k\pmod{p}$ will give an identity, since $k^{-1}\times m=m\pmod{p}$ and similarly on the other side. So we get an isomorphism $\mathbb{Z}_p$ to our new ring $R$ by sending $m\mapsto mk^{-1}$, which preserves addition by definition and multiplication because it sends $1\times 1$ in $\mathbb{Z}_p$ to $k^{-1}\times k^{-1}$ in $R$, and which has inverse $mk\mapsto m$.
A note that $\mathbb{Z}_p$ is somewhat risky notation: I might suggest using $\mathbb{Z}/p$ or $\mathbb{Z}/p\mathbb{Z}$ to avoid confusion with two other important rings, namely the $p$-adic integers and the integers localized at $p$.
