### Archive

Archive for the ‘Algebra questions’ Category

## 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).