# Field Extensions as $F$ adjoin some element

Let $F$ be a field and $E$ an extension of $F$. Is it always possible to write $E=F(\alpha_1,\alpha_2,\ldots)$?

If $E$ is a finite extension then I think it is possible to write $E=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$. My reason is that if we take $\alpha\in E$ then as $[E:F]<\infty$ for some $n$ we must have $\alpha^n\in\text{Span}\{\alpha, \ldots,\alpha^{n-1}\}$. Meaning that $\alpha$ satisfies an (irreducible) polynomial in $F[x]$. If we keep doing this for each element in $E$ then we get $E=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$. Is this correct?

What about the case when $E$ is not a finite extension?

Thanks

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Is $E$ an algebraic extension of $F$? –  lhf Jun 11 '12 at 12:08
Does it matter whether it is algebraic or not? For example: if it is algebraic, then can I write it as $F(\alpha_1,\ldots)$. And not when it is not algebraic. –  Sebastian Jun 11 '12 at 12:35
You can have transcendental extensions. –  Eugene Jun 11 '12 at 12:53
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## 3 Answers

An example is $\mathbb R$ as extension of $\mathbb Q$.

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+1 Presumably you proffer this as a counterexample, i.e. a field extension that cannot be generated by adjoining countably many elements? –  Jyrki Lahtonen Jun 11 '12 at 12:18
@JyrkiLahtonen : exactly. –  Mohamed Jun 11 '12 at 12:22
@JyrkiLahtonen $\Bbb{R}$ is not an algebraic extension of $\Bbb{Q}$, if we require that our $E/F$ is algebraic but not finite can we say that $E$ is $F$ adjoined a countable number of elements? –  fpqc Jun 12 '12 at 5:01
@BenjaminLim: I think that, if we select $F=\mathbb{R}(x)$ with $x$ transcendental over the reals, i.e. $F$ is the field of rational functions with real coefficients, then the extension field $E=F(\{\sqrt{x-a}\mid a\in \mathbb{R}\})$ "probably" cannot be gotten by adjoining a countable number of elements. Yet $E/F$ is algebraic. I don't have a proof though. –  Jyrki Lahtonen Jun 12 '12 at 6:26
@JyrkiLahtonen Hmmmm perhaps I should ask on MO? –  fpqc Jun 12 '12 at 6:27
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When $E$ is a finite extension of $F$ we can always do this. For suppose that $[E:F] = n$. Then this means that we can choose $\alpha_1 \in E \setminus F$. Now consider $F(\alpha_1)$ as a subspace of $E$. As a subspace of a finite dimensional vector space it is finite dimensional so that $F(\alpha_1)/F$ is a finite extension. Now if $[F(\alpha_1):F] = n$, then we are done because $E = F(\alpha_1)$. Otherwise if it is less than $n$, we repeat this procedure again and find $\alpha_2 \in E \setminus F(\alpha_1)$ and look at $[F(\alpha_1,\alpha_2):F] = [\big(F(\alpha_1)\big)(\alpha_2):F]$. Eventually we will stop because $\dim_F E$ is finite, so that

$$E = F(\alpha_1,\ldots, \alpha_n)$$ for some $\alpha_1,\ldots \alpha_n \in E$.

Edit: Qiaochu has posted an example here on MO to show that it is not true that every algebraic extension is obtained by adjoining a countable number of elements.

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Yes I agree Ben. What about the case when $E/F$ is algebraic but not finite? Can we write $E=F(\alpha_1,\alpha_2,\ldots)$ for some countable set of element $\{\alpha_1,\alpha_2,\ldots\}$? –  Sebastian Jun 12 '12 at 3:10
@Sebastian Hmmm I am not so sure about that. Would you like to meet tomorrow at the library to discuss things? –  fpqc Jun 12 '12 at 5:09
@Sebastian Please see my edit. I have crossposted this on MO. –  fpqc Jun 12 '12 at 6:40
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If $E/F$ is a finite extension, then $E$ is a finite dimensional vector space over $F$ with some basis $\{\alpha_1, \ldots, \alpha_n\}$ and thus $E = F(\alpha_1, \ldots, \alpha_n)$.

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