Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
The situation is even slightly better; two finite fields of the same cardinal are in fact equal inside a fixed algebraic closure. –  vgty6h7uij Jul 26 '12 at 9:34
1  
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). –  Geoff Robinson Jul 26 '12 at 11:13
add comment

1 Answer

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.

share|improve this answer
    
I fixed my question. I agree, it was unclear as stated. –  Sam Lane Jul 26 '12 at 9:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.