5
$\begingroup$

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

  • 0
    You can have transcendental extensions.2012-06-11

3 Answers 3

5

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

  • 0
    @BenjaminLim: I think that my "counterexample" should be seen as an analogue of adjoining square roots of distinct primes to the rationals - to get all those square roots you need them all. The proofs for those might carry over to this case, where we have an "uncountable set of primes of the form $x-a$". The reason the standrad proofs might work is that each individual element of the extension field only involves a finite subset of those square roots, so...2012-06-12
2

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.

  • 0
    @Sebastian Please see my edit. I have crossposted this on MO.2012-06-12
0

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