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Let $F$ be a field and let $K$ be an extension of $F$. Show that if $\alpha\in K$ is algebraic over $F$, $F[\alpha]=\{p(\alpha)\mid p(x)\in F[x]\}$ is the smallest subfield of $K$ containing $F$ and $\alpha$.

user72870
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  • That's only true if $\alpha$ is algebraic over $F$. – Daniel Fischer Dec 31 '13 at 15:09
  • Thanks, I've edited :) Any idea for the proof? – user72870 Dec 31 '13 at 15:12
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    Show $F[\alpha]$ is a subfield, and show any subfield containing $F$ and $\alpha$ must contain $F[\alpha]$. – Pedro Dec 31 '13 at 15:14
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    @user72870 The interesting part is division, for this, consider the minimal polynomial $g(x)$ for $\alpha$ over $F$, note that it must be irreducible... given $p(x)\in F[x]$ not having $\alpha$ as a root, can you find $f(x),h(x)\in F[x]$ such that $f(x) p(x) + h(x) g(x) = 1$ ? what does that imply? – Pablo Rotondo Dec 31 '13 at 15:15
  • For division, it is enough to show that for every non-zero $\beta$ that is algebraic over $F$, $1/\beta \in F[\beta]$. Applying this to an arbitrary (non-zero) element $\beta = p(\alpha) \in F[\alpha]$ shows that $F[\alpha]$ is a field. – Magdiragdag Dec 31 '13 at 15:58

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You have to show that (1) $F[\alpha]$ is a field, and (2) that whenever $K$ is a field with $F \subseteq K$ and $\alpha \in K$, then $F[\alpha] \subseteq K$.

Once you have shown (1), then (2) is not too hard: Assume that $K$ is a field with $F \subseteq K$ and $\alpha \in K$. Then surely all elements of the form $$a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n$$ with $a_i \in F$ and $n \in \mathbb{N}$ must be elements of $K$, i.e. $F[\alpha] \subseteq K$.

Thus it remains to show (1). The proof relies on the fact that $\alpha$ is algebraic over $F$. Let $p(x)$ denote the minimal polynomial of $\alpha$ over $F$. Then in fact the ring $F[x]/(p(x))$ will be isomorphic to $F[\alpha]$ (as rings). If you can show this, and that $F[x]/(p(x))$ is a field, then you're done.

Ulrik
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