Exercise 4.4.1

Proof.

(a)

In Example 4.2.4, we have seen that the Schönemann-Eisenstein Criterion implies that, for all $n\ge 2$, and $p$ prime, $$f=x^n+\mathit{px}+p$$

is irreducible over $\mathbb{Q}$. Let $\alpha$ a root of $f$ in $ℂ$. Since $f$ is irreducible over $\mathbb{Q}$, the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $f$, and

$$[\mathbb{Q}(\alpha ):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n.$$

As $[\mathbb{Q}(\alpha ):\mathbb{Q}]<\infty$, every element of $\mathbb{Q}(\alpha )$ is algebraic, so

$$\mathbb{Q}\subset \mathbb{Q}(\alpha )\subset \bar{\mathbb{Q}}.$$

$L=\mathbb{Q}(\alpha )$ is so an answer to the question.

(b)

Suppose on the contrary that $[\bar{\mathbb{Q}}:\mathbb{Q}]<\infty$. The tower theorem gives then

$$[\bar{\mathbb{Q}}:\mathbb{Q}]=[\bar{\mathbb{Q}}:\mathbb{Q}(\alpha )]\times [\mathbb{Q}(\alpha ):\mathbb{Q}]\ge [\mathbb{Q}(\alpha ):\mathbb{Q}]\ge n.$$

Then for all integer $n\ge 2$, $[\bar{\mathbb{Q}}:\mathbb{Q}]\ge n$, thus $[\bar{\mathbb{Q}}:\mathbb{Q}]=\infty$, which is a contradiction.

Conclusion : $[\bar{\mathbb{Q}}:\mathbb{Q}]=\infty$.

$\bar{\mathbb{Q}}$ is an algebraic extension of $\mathbb{Q}$, with infinite dimension.