Understanding the field of order $p^2$

Question

So I understand that for any prime $p$, there exists a field of order $p^2$. What's confusing me, is that a field is also an abelian group, and the only abelian groups of order $p^2$ up to isomorphisms are $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p \times \mathbb{Z}_p$, neither of which are fields. Can anyone please help me wrap my head around how its possible for a field of order $p^2$ to both be an abelian group of order $p^2$, and simultaneously not be isomorphic to any abelian group of order $p^2$ that exists?

Answer 1

Let $p$ be a prime. Then the field with $p^2$ elements has the underlying additive group $\mathbb{Z}_p\times \mathbb{Z}_p$.

You state that $\mathbb{Z}_p\times \mathbb{Z}_p$ is "not a field," and indeed it is not when you view it with the standard multiplication entrywise in each component. But the field with $p^2$ has a different multiplication structure which allows you to turn the additive group $\mathbb{Z}_p\times \mathbb{Z}_p$ into a commutative ring where every nonzero element is invertible with respect to this multiplication.