Existence of a field with $p^2$ elements Let $F$=$\mathbb{Z}/p\mathbb{Z}$ where $p \in \mathbb{Z}$ is prime
I need to show there is a field with $p^2$ elements.
I am not sure where to start with this.
 A: If $p=2$, consider $\mathbb{F}_p[x]/{(x^2+x+1)}$. If $p$ is odd, there is $b\in\mathbb{F}_p$ such that $b^2-1$ is a quadratic non-residue - if all the quadratic non-residues were followed by a quadratic non-residue, $0$ would be a quadratic non-residue, contradiction. Hence $x^2-2bx+1$ is an irreducible quadratic polynomial over $\mathbb{F}_p$, and $\mathbb{F}_p[x]/(x^2-2bx+1)$ is a field with $p^2$ elements.
A: Another approach, a little more advanced than the basic one in Jack's answer: take the polynomial $\;x^{p^2}-x\in\Bbb F_p[x]\;$ and let $\;K\;$ be its splitting field over $\;\Bbb F_p\;$.
Now just prove that $\;\left|K\right|=\left|\Bbb F_{p^2}\right|=p^2\;$ and we're done by the uniqueness of finite fields.
A: Hint :
suppose $p\neq 2$ and consider $\eta : \mathbb{F}_p^*\rightarrow \mathbb{F}_p^*$ defined as $\eta(x)=x^2$


*

*Prove that $\eta$ is not injective (This does not behave properly in $p=2$ case)

*Thus, $\eta$ is not surjective (Why ??)

*So, you can find some $a\in \mathbb{F}_p^*$ that is not a square. (Why ??)

*Consider $\mathbb{F}_p[\sqrt{a}]=\{m+n\sqrt{a}  : m,n \in \mathbb{F}_p\}$ (Is this a field?)

*What is the cardinality of $\mathbb{F}_p[\sqrt{a}]$?


It is your turn to construct field of order $4$ (you may even get some answers in this same site).
Another way of looking at the general case is...
You can surely say there does exists an irreducible polynomial in $\mathbb{F}_p[x]$ of degree $2$
Just check no of polynomials of degree $2$ in $\mathbb{F}_p[x]$ and look for all possible combinations like $(x-a)(x-b)$ (be careful in counting)
so, you would know that there does exist some irreducible polynomial $f(x)\in \mathbb{F}_p[x]$ and consider $\mathbb{F}_p[x]/(f(x))$
