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I've recently been looking at the relationships between divisibility of polynomials and their degrees. This following idea has eluded me for a while.

Suppose $p$ and $q$ are polynomials in $F[X]$ for $F$ a field, with $p$ irreducible. If $d$ is another irreducible polynomial that divides $p(q(X))$, then how can you conclude that the degree of $p$ divides the degree of $d$?

I've whittled it down as much as I can. If $p$ has degree $n$ and $q$ has degree $m$ then $p(q(X))$ has degree $nm$, and if $p(q(X))=d(X)f(X)$, where $\deg(d)=a$ and $\deg(f)=b$, then $nm=a+b$. Then it's enough to show $n$ divides either $a$ or $b$, but I'm not sure how to see that, or why irreducibility of $p$ and $d$ are a factor. I thought maybe if I assume neither is divisible by $n$, I could find a contradiction, but I did not see anything immediate.

Thanks.

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up vote 4 down vote accepted

Hint (assuming that you have done some field theory): Let $\alpha$ be a zero of $d(x)$ in some extension field of $F$. Because $d(x)$ is irreducible, we know that $\deg d(x)=[F(\alpha):F]$. Show that $p(x)$ is a minimal polynomial of $q(\alpha)\in F(\alpha)$, and...

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Jyrki, Here's a +1 for you to prevent deletion again. @Clara: please don't delete your questions after you got an answer. Jyrki spent some time thinking about your problem and giving you an indication of how to solve it. So it is somewhat impolite to delete the records of that. –  t.b. Nov 17 '11 at 18:19
    
@t.b. Clara is obviously a new user, so probably just a misunderstanding. –  Jyrki Lahtonen Nov 17 '11 at 18:54
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@Clara: Just before you deleted the question, I was about to suggest that you should write up your argument in detail, and post it as an answer! That way you would get more feedback about your solution, also upvotes, and comments from several users. Some of them more knowledgable than me. That would have helped you even more. Also IMHO your question was very much ok, and nothing to be ashamed of. You even solved it yourself, and deserve a bit of credit for that! What to do, is, of course, strictly your call :-) –  Jyrki Lahtonen Nov 17 '11 at 18:58
    
Oh, I'm sorry. I thought since it looked trivial in comparison to most other questions I should just get rid of it. Thanks for thinking about it! –  Clara Nov 19 '11 at 9:47

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