Field with $p^2$ elements where $p \equiv 3 \pmod 4$ is a prime number

Question

Let $p$ be a prime number of the form $4n+3$.

Is $GF(p^2)$, the field with $p^2$ elements, isomorphic to $GF(p)(i)$ (constructed from $GF(p)$, the field with $p$ elements, the same way that the complex numbers are constructed from the real numbers)?

Answer 1

Yes, because the polynomial $X^2+1$ is irreducible modulo $p$, so $GF(p)[X]/(X^2+1)$ is a field with $p^2$ elements, and a finite field with a given number of elements is unique, up to isomorphism.

Answer 2

Every two finite fields of the same order are isomorphic, so the only thing to check is whether $X^2 + 1$ is irreducible modulo $p$. However,$X^2 + 1$ is reducible in $\mathbb F_p$ if and only if $-1$ is a square modulo $p$, which is the case if and only if $4 \mid p - 1$.

So $X^2 + 1$ is indeed irreducible if and only if $p \equiv 3 \mod 4$.

Every two finite fields of the same order $n$ are isomorphic because they have the same prime field, so both are the splitting fields of $X^{n-1} - 1$ over their common prime field.