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Polynomial in a Field of Characteristic p

June 30, 2011 1 comment

I’ll go ahead and it post it here as well:
Hey Ken, I’m trying to check if this proof is correct(it may not be the only way, but its one I thought of)

To Prove: Let F be a field with charF=p. Let q(x)=x^p-x-c where c\in F. Show if q is reducible, then q has a root in F (actually all).

This is actually an iff in the book, but this is the nontrivial direction.

My idea is the following. Let u be a root of q. Let f,g be nonunit irreducible factors of q. Now it is easy to see the set of all roots is the set \{u+i:i=1,\ldots,p\}. Thus, Z(f)=u+I, Z(g)=u+J for some finite sets I,J\subset \mathbb{Z}.

Notice this means \prod_{i\in I}u+i, \prod_{i\in J} u+i are both in F. Let f_1=\prod_{i\in I}u+i and f_2=\prod_{i\in J} u+i. Then u is a root for both f_1-\prod_{i\in I}{x+i} and f_2-\prod_{i\in J}x+i which are in F[x]. But since the degrees of these are |I|,|J|; respectively, the only way this is possible is if |I|=|J|. But we could do this for any irreducible factor so each irreducible factor must have the same degree.

Hence, if q=q_1\cdots q_k with each q_i irreducible, then p=deg(q)=k\cdot n for some n\geq 1. Since p is prime either k or n must be 1. If n=1 then each root is in F. If k=1 then q is irreducible. Since p is reducible the result follows.

After writing this out carefully it feels completely correct (of course I left out some easy details).

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Categories: Algebra questions