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 February 22nd, 2011, 07:12 PM #1 Newbie   Joined: May 2010 Posts: 12 Thanks: 0 Root of a reducible polynomial Hey guys, just took an exam it completely kicked my ass. This was one of the problems that I could just not make any headway on: Suppose F is a field of characteristic p > 0 and let $f(x)=x^p-a \in F[x]$ for some $a \not= 0$. Further suppose there is a field E containing F such that we have $\alpha \in E$ with $f(\alpha)=0$. Prove that if $f(x)$ is reducible in $F[x]$ then $\alpha \in F$. Can someone explain how this proof goes? I don't think I'll be able to sleep tonight until I find out...
 February 23rd, 2011, 09:08 PM #2 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Root of a reducible polynomial Claim: $x^p - a= x^p - \alpha ^ p = (x - \alpha)^p$ at least in $E$x$$. We have then that the minimal polynomial of $\alpha$, call it $m_{\alpha}(x) \in F$x$$ divides $x^p-a$. Since $x^p-a$ splits into linear factors over $E$, so must $m_{\alpha}$. So what can $m_{\alpha}$ be?
 February 23rd, 2011, 10:44 PM #3 Newbie   Joined: May 2010 Posts: 12 Thanks: 0 Re: Root of a reducible polynomial That is about as far as I got. I think that $m_\alpha= (x-\alpha)^n$ for some $n < p$, but I'm not sure how this factorization of $m_\alpha$ in $E[x]$ can help me say anything about $\alpha$'s presence in $F$. I guess I could use the binomial theorem to say that the constant coefficient of $m_\alpha$ in $F[x]$ is $\alpha^n$, which means $\alpha^n \in F$. Can I get from there to $\alpha\in F$, perhaps by using the fact that $F$ has characteristic $p$? I can't think of how to do that.
 February 23rd, 2011, 11:42 PM #4 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Root of a reducible polynomial In the end, you're worrying a little too much about complicated things. Try to keep it simple. If $\alpha \in F$, what is $m_{\alpha}$? Is the converse true? You know that $m_{\alpha}$ divides $x^p - a$. So we know that there exists $g(x) \in F$x$$ such that $x^p - a= m_{\alpha} \cdot g$. What is the degree of $g$? What is $g(\alpha)$? So what can we say about $m_{\alpha}$ and $g$? See where this takes you.
 February 24th, 2011, 08:52 AM #5 Newbie   Joined: May 2010 Posts: 12 Thanks: 0 Re: Root of a reducible polynomial If $\alpha \in F$ then $m_\alpha=x-\alpha$, and yes the converse is true. But right now I have $f(x)=m_\alpha (x)g(x)=(x-\alpha)^n(x-\alpha)^m$ in $E[x]$ with $m+n=p$, and I don't understand how we can say anything more specific about $m$ and $n$. Clearly I'm missing something obvious here.
 February 24th, 2011, 02:13 PM #6 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Root of a reducible polynomial Since $g= (x - \alpha)^{p-n}$, we have that $g(\alpha)= 0$, and $g \in F$x$$, so we must have that $m_{\alpha}$ divides $g$ in $F$x$$. Thus, in $F$x$$, we have $g= m_{\alpha} \cdot g#39;$ and $x^p - a= (m_{\alpha})^2 \cdot g#39;$ where $g'$ has degree $p - 2n$. What now, can we say about $g'$? Continuing, can any factor of $x^p - a$ not be a power of $m_{\alpha}$?

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