$\forall p \in \mathbb P\backslash\{2\}$ there exists a field that has $p^2$ elements

Question

Prove that for each odd prime $p$, there is a field with $p^2$ elements.

I know that i have to show that there is an irreducible quadratic polynomial over $\mathbb Z_p$ to which end i can show that there has to exist $a \in \mathbb Z_p$ which is not the square of another element in $\mathbb Z_p$. But i am completely stuck from here on out.

I really appreciate any kind of help, thank you!

Answer 1

The squaring map $\phi: \Bbb Z_p \to \Bbb Z_p$ defined by $\phi(a)=a^2$ is not injective as when $p > 2$ we always have $1 \neq -1$ and $\phi(1)=\phi(-1)$. In fact any polynomial $x^2 -a $ has at most two roots (in any field), so $\phi$ is at most 2 to 1. (As $0$ is the only field element with just one square root ($p$ is odd) there are exactly $\frac{p-1}{2} + 1$ squares in $\Bbb Z_p$.)

For a finite set $F$ any map $f: F \to F$ is injective iff its is surjective iff it is bijective. So $\phi$ is not sutjective and some $a \notin \phi[\Bbb Z_p]$ exists. Then $x^2-a$ is irreducible (a degree $2$ polynomial over a field is irreducible iff it has no roots) and its splitting field is of size $p^2$.

Answer 2

The squaring map $a\mapsto a^2$ in $\Bbb Z_p$ is quite easily seen to be non-injective (we have $1^2 = (p-1)^2$, and $p\geq 3$ means $1\neq p-1$). Since $\Bbb Z_p$ is finite the squaring map can therefore not be surjective.

Answer 3

Actually you can prove it without the restriction $p\neq 2$. There are $p^2$ normalized polynomials of degree 2. A quadratic polynomial is reducible iff it's a product of two linear one. Therefore we have $p + \frac{p(p-1)}{2}$ reducible polynomials ( $p$ squares, $\frac{p(p-1)}{2}$ products of two different linear polynomials) and $ 0 < p^2-\frac{p(p+1)}{2}=\frac{p(p-1)}{2}$ irreducible polynomials. So we always have an irreducible $f\in\mathbb{F}_p$ such that $\mathbb{F}_p[X]/(f)$ is a field with $p^2$ elements.