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I am trying to show that $ [\bar{\mathbb{Q}}:\mathbb{Q}]=\infty$ where $\bar{\mathbb{Q}}$ denotes the algebraic closure of $\mathbb{Q}$. I decided to approach the problem as follows:

For every $n \in \mathbb{N}$ there exists an irreducible polynomial of degree greater than or equal to $n$. For instance, consider the Eisenstein polynomials: $x^n-p$ where $n \in \mathbb{N}$ and $p$ is prime. Considering the roots of these polynomials, we know that they are linearly independent over $\mathbb{Q}$. Hence, $[\bar{\mathbb{Q}}:\mathbb{Q}] \geq n$ as desired.

Does this proof work? I believe the same proof holds for cyclotomic polynomials.

Any and all suggestions are appreciated!

KangHoon You
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    That proof should work fine, with a few changes. The roots are NOT necessarily linearly independent over $\mathbb{Q}$ - take for example $x^{2} - 2$, where $+\sqrt{2}$ and $-\sqrt{2}$ are clearly not linearly dependent over $\mathbb{Q}$. I'm not sure what linear independence has to do with anything though. As you noted, $x^{n}-p$ is irreducible of degree $n$ by Eisenstein, so letting $\eta$ be a root, we see that $\mathbb{Q} \subset \mathbb{Q}(\eta) \subset \bar{\mathbb{Q}}$. Thus, since $[\mathbb{Q}(\eta):\mathbb{Q}]=n$, we establish $[\bar{\mathbb{Q}}:\mathbb{Q}] \geq n$ for all $n$. – Alex Wertheim Feb 05 '15 at 15:10
  • As an aside, the $n$th cyclotomic polynomial is irreducible of degree $\phi(n)$, where $\phi$ is the Euler totient function, so if you can prove that $\phi(n)$ is unbounded as $n$ increases, that works fine. I don't think $\phi(n)$ assumes all integer values though, so I think your first approach is probably better. – Alex Wertheim Feb 05 '15 at 15:12
  • @AlexWertheim Indeed, all odd values $k>1$ are not assumed by $\phi(n)$, and also many even values are not assumed, see this question. – Dietrich Burde Oct 18 '18 at 18:58

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We can also show $[\overline{\Bbb{Q}}:\Bbb{Q}]=\infty$ by using cyclotomic polynomials $\Phi_n(x)$ of degree $\phi(n)$, because $\phi(n)$ gets arbitrarily large for $n\mapsto \infty$. In fact, we have $\phi(n) \geq \frac{\sqrt{n}}{\sqrt{2}}$ for all $n\ge 1$. There is even a much better lower bound, namely $$ \phi(n) > \frac{n}{e^\gamma \log \log n + \frac{3}{\log \log n}} $$ for $n>2.$

Reference: Is the Euler phi function bounded below?

Dietrich Burde
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