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.

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    $\begingroup$ The situation is even slightly better; two finite fields of the same cardinal are in fact equal inside a fixed algebraic closure. $\endgroup$ – vgty6h7uij Jul 26 '12 at 9:34
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    $\begingroup$ Hint: Set $q = p^{d}.$ Then every element $f \in F$ satisfies $f^{q} = f$ (consider non-zero $f$ and use a little elementary group theory). $\endgroup$ – Geoff Robinson Jul 26 '12 at 11:13

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.

  • $\begingroup$ I fixed my question. I agree, it was unclear as stated. $\endgroup$ – Sam Lane Jul 26 '12 at 9:47

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