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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 for some . Further suppose there is a field E containing F such that we have with . Prove that if is reducible in then . 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: at least in . We have then that the minimal polynomial of , call it divides . Since splits into linear factors over , so must . So what can 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 for some , but I'm not sure how this factorization of in can help me say anything about 's presence in . I guess I could use the binomial theorem to say that the constant coefficient of in is , which means . Can I get from there to , perhaps by using the fact that has characteristic ? 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 , what is ? Is the converse true? You know that divides . So we know that there exists such that . What is the degree of ? What is ? So what can we say about and ? 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 then , and yes the converse is true. But right now I have in with , and I don't understand how we can say anything more specific about and . 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 , we have that , and , so we must have that divides in . Thus, in , we have and where has degree . What now, can we say about ? Continuing, can any factor of not be a power of ? 

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