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Let $\mathcal{A}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Prove that $[\mathcal{A}:\mathbb{Q}] = \infty$.

I can show this using $[\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q}] = n$ for all $n \in \mathbb{N}$ and $[\mathcal{A}:\mathbb{Q}] = [\mathcal{A}:\mathbb{Q}(\sqrt[n]{2})][\mathbb{Q}(\sqrt[n]{2}):\mathbb{Q}]$. Does anyone know an alternative proof? It's just curiosity.

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Lucas
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2 Answers2

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You don't need to say "the algebraic closure of $\Bbb{Q}$ in $\Bbb{C}$": (up to isomorphism) the algebraic closure of a field $k$ depends on $k$ alone and not on any embedding of $k$ in a larger field.

One alternative proof that $[{\cal A}:\Bbb{Q}] = \infty$ would be to argue that $$\Bbb{Q}[\sqrt{2}] \subseteq \Bbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \Bbb{Q}[\sqrt{2},\sqrt{3}, \sqrt{5}] \subseteq \ldots$$ is a strictly increasing sequence of $\Bbb{Q}$-vector spaces.

Rob Arthan
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    This is not quite easy to prove. See https://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of – lhf Apr 05 '18 at 23:26
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    @inf: Sure. I didn't say it was an easy alternative. It is an obvious alternative to try. And if you do the work, it works! – Rob Arthan Apr 07 '18 at 22:15
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If $p$ is prime, then $[\mathbb{Q}(\zeta_p):\mathbb{Q}] = p-1$, where $\zeta_p$ is a primitive $p$-th root of unity.

Therefore, $[\mathcal{A}:\mathbb{Q}] \ge p-1 \to \infty$ since there are infinitely many primes.

lhf
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