I am asked to prove that $L=\bigcup_{n=1}^\infty\mathbb{Q}(\sqrt[n]2)$ is an algebraic field extension over $\mathbb{Q}$. So far I have:

Let $\beta\in L$, then by definition of union there exists a $N\in\mathbb{Z}_{\geq1}$ such that $\beta\in\mathbb{Q}(\sqrt[N]{2})$. $\sqrt[N]{2}$ is algebraic in $\mathbb{Q}$ (take $f=X^N-2$), so $\beta$ is algebraic as well. Since $\beta$ was chosen arbitrarily, every element of $L$ is algebraic over $\mathbb{Q}$. This makes $L$ an algebraic field extension.

All examples I've seen so far are finite extensions, and $L$ is obviously not finite. I'm wondering if this proof suffices, or if I'm missing something important.

  • 1
    $\begingroup$ For my money, this proof suffices, as long as you understand that the union actually is a field. Is the sum (product) of any two elements an element of the field? What if one is in $K_6$ and the other is in $K_7$, where $K_n=\mathbb Q(2^{1/n})$ ? $\endgroup$ – Lubin Apr 1 '14 at 23:09
  • $\begingroup$ Well, by the same reasoning, every $\beta\in K_N$ has to have a $\beta^{-1}$ also in $K_N$, since $K_N$ itself is a field. This means that all elements of $L$ have an inverse. $\endgroup$ – Kaj Apr 1 '14 at 23:13
  • $\begingroup$ I would say it is not trivial that the union $L$ is again a field; it is instructive to verify all the axioms. $\endgroup$ – Servaes Apr 1 '14 at 23:30

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