$R$ 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.
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 $R \cong ℤ/pℤ$.
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 $p$.
