Finding a subfield of $\mathbb R$ such that $e^2$ is of degree five over that subfield.

Question

My work so far is as follows: \begin{align*} x & = e^2 \\ x^5 & = e^{10} \\ x^5 - e^{10} & = 0. \end{align*} Thus I have a fifth-degree polynomial in $\mathbb R[x]$ with $e^2$ as a zero. But $x^5 - e^{10}$ is reducible over $\mathbb R.$ Thus, I want to find a field $F$ such that $x^5-e^{10}$ is in $F[x]$ and is irreducible over $F.$ Starting with $\mathbb Q$, over which $e$ is transcendental, I formed the extension field $F = \mathbb Q(e^{10}).$ Then $x^5-e^{10}\in F[x],$ and $x^5 - e^{10}$ has no zeros over $F.$ Thus, $x^5-e^{10}$ is not divisible by any linear factors in $F[x]$. However, I haven't yet proven that it is irreducible over $F.$ Could it be the product of an irreducible quadratic and an irreducible cubic in $F[x]?$ I'm not sure how to proceed.

Answer 1

The number $\alpha=e^{10}$ is transcendental over $\mathbb{Q}$. Consider $F=\mathbb{Q}(\alpha)$. Then $e^2$ is a root of $X^5-\alpha$.

The polynomial $X^5-\alpha$ is irreducible over $\mathbb{Q}[\alpha]$, which is a PID, by Eisenstein's criterion, so it's irreducible also over the field of fractions.

Therefore the degree of $e^2$ over $\mathbb{Q}(\alpha)$ is $5$.