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Theorem Let $p(x)$ be an irreducible polynomial in $F[x]$ and let $u$ be a root of $p(x)$ in an extension $E$ of $F$. Then if the degree of $p(x)$ is $n$, the set $(1,u,\ldots,u^{n-1})$ forms a basis of $F(u)$ over $F$.

What I did:

Note that $F(u)$ is by definition the smallest subfield of $E$ generated by $F$ and $u$. I proved that $F(u) = F[u]$ and I tried to prove the theorem, I found weird because since $F[u]$ is by definition the set $b_0+b_1u+\cdots+b_mu^m\in E$ such that $b_0+b_1x+\cdots+b_mx^m \in F[x]$ the basis should be $(1,u,\ldots,u^{n-1},u^n)$ with $n$ elements. where am I wrong? I need help.

Thanks

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up vote 2 down vote accepted

By definition, $F[u]$ is the set of all $f(u)$ with $f\in F[X]$. Note that $f$ may have arbitrarily high degree. But modulo $p(X)$, you can transport everything to degree $<n$. Thus at least $(1, u, \ldots, u^{n-1})$ is a generating system of $F[u]$. Remains to show that they are linearly independent. But what would a linear dependence imply?

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what is modulo $p(X)$? –  user42912 Oct 21 '12 at 21:55
    
@user42912 One way of looking at this is that you can reduce everything to the remainder you get after dividing by $p(x)$ - which gives you degree less than $\deg(p(x))$ –  Mark Bennet Oct 21 '12 at 22:12
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@user42912: If $f$ is a polynomial, division with remainder lets you write $f(X)=q(X)\cdot p(X)+r(X)$ with $\deg r<\deg p$. Then $p(u)=0$ implies $f(u)=r(u)$ and $r(u)$ is a linear combination of $1,\ldots, u^{n-1}$. –  Hagen von Eitzen Oct 22 '12 at 5:55
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