Irreducible Polynomials in Finite Fields I'm reading through some notes online concerning finite fields, and attempting to come up with a proof that all finite fields of the same size are isomorphic. But I'm getting stuck at a certain point, and I was wondering of you might have any hints.
If $F$ is any field with $p^d$ elements and $m(x)$ has coefficients all from $\mathbb{F}_p$ and is an irreducible polynomial of degree $d$ over $\mathbb{F}_p$, then $m(x)$ has roots in $F$. Is this immediately obvious? It seems to be stated so but I can't see why.
 A: This is false as stated (ie without some clarification) for all $d>1$. If $m(x)$ is irreducible over $\mathbb{F}_{p^d}$, then it cannot be factored into a product of two polynomials of positive degree. Thus it cannot have any roots in $F$ unless the polynomial is linear, which contradicts the assumption that $d>1$.  This has been fixed in an edit.
On the other hand, if you're working over $\mathbb{F}_p$ and $m(x)$ has coefficients all from $\mathbb{F}_p$ and is irreducible over $\mathbb{F}_p$, then the statement is true. Unfortunately, I cannot think of a reason right now why this is true without appealing to the result you are trying to prove.
A different avenue of proof may be the following, if you're familiar with a theorem about splitting fields:
Theorem: Any two splitting fields of a fixed polynomial over a ground field $F$ are isomorphic.
Proof: Exercise.
If you can show that every field of characteristic $p$ contains a subfield isomorphic to $\mathbb{F}_p$ and find a polynomial over $\mathbb{F}_p$ which has exactly $p^n$ distinct roots and find a clever use for the above theorem, you will have a proof.
