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February 22nd, 2011, 06:12 PM   #1
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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...
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February 23rd, 2011, 08:08 PM   #2
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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?
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February 23rd, 2011, 09:44 PM   #3
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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.
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February 23rd, 2011, 10:42 PM   #4
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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.
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February 24th, 2011, 07:52 AM   #5
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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.
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February 24th, 2011, 01:13 PM   #6
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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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