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I proved this simple thing, but using some simple field theory. I want to know whether I can prove it with simpler tools. The proof is not difficult, it uses only a little field theory, like the idea of extensions of finite degree. But can I prove it only using the definitions?

Another related question: If I know that the element $u$ satisfies some polynomial equation $p(x)=0$, can I find the polynomial for $1/u$ only knowing $p(x)$?

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    Hint: laimonylop2011-07-21
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    If you stated what you have proved, it would be easier to suggest easier proofs. Voting to close as not a real question.2011-07-21
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    @André Nicolas: I put your hint into a well-known search engine before I thought to reverse it. It is amazing how many hits there are with varying numbers of capitals.2011-07-21
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    @André Nicolas Nice hint. I couldn't understand it at first but now it all makes sense =P2011-07-21
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    There is a real question nestled in the post.2011-07-21
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    @Adrian: no, it makes esnes.2011-07-21
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    I wish Andre's hint was an answer just so that I could upvote it. I like it.2011-07-21
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    Me too. LOLs in the morning2011-07-21
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    @Daniel: You can assume constant term is $\ne 0$. If $p(x)$ has degree $n$, divide through by $x^n$, and let $y=1/x$. Then $q(y)$ works. Example: If $p(x)=x^3 +2x^2+13x+4$, then $q(y)=4y^3+13y^2+2y+1$. Please accept some more good answers!2011-07-21

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