A question about an exercise in Marcus book “Number Fields” The exercise is the number 27 in chapter 3.
“Let $\alpha^5=5(\alpha+1), K=\mathbb{Q}(\alpha)$ and let $p\neq3$ a prime of $\mathbb{Z}.$ Show that the prime decomposition of $p\mathcal{O}_K$ can be determined by factoring $x^5-5x-5 \mod{p}.$
I know Kummer theorem. A simple calculation shows that $disc(\alpha)=5^5\cdot3^2\cdot41.$
Since $disc(\alpha)=ind(\alpha)^2\cdot disc(K),$ clearly $41\nmid ind(\alpha).$ So the only prime that can create some problems is $5.$ How can I prove that $5\nmid ind(\alpha)?$
 A: I looked for another approach to this problem as the accepted solution was a little too "bare hands" for my liking.
Letting $S = \mathbb{A} \cap K$ we have that $\alpha^5 = 5(\alpha+1)$; a previous exercise has you show that in $K$, $\alpha +1$ is a unit.  Therefore $5S = (\alpha)^5$ and so 5 is totally ramified over $\mathbb{Q}$ ($e = 5, f = 1, r = 1$).   Another exercise has you show that for any prime, $p^{k}$ divides the discriminant ($k = n - \sum f_i = n - 1$ here).  Therefore for an integral basis of $K$, $5^4$ must divide $\text{disc}(S)$.
Coming back to the problem statement: $\text{disc}(\alpha) = (d_1 d_2 d_3 d_4)^2\text{disc}(S)$ where the $d_i$ are denominators of the integral basis.  $d_1 d_2 d_3 d_4$ is also the order of $R[\alpha]$ in $S$.  $\text{disc}(\alpha) = 5^5 \cdot 3^2 \cdot 41$ so the factor of $5^4$ must be in $\text{disc}(S)$, so $d_1 d_2 d_3 d_4 \in \{1, 3\}$.  Therefore for any prime $p \neq 3$, $p \not\mid |S:R[\alpha]|$ and we can apply the Dedekind approach to determine its factorization in $S$.
Postscript: you can see with Sage that $d_1 d_2 d_3 d_4$ is actually $3$:
sage: K.<a> = QQ.extension(x^5 - 5*x - 5); K.integral_basis()
[2/3*a^4 + 2/3*a^3 + 2/3*a^2 + 2/3*a + 1/3, a, a^2, a^3, a^4]

A: The polynomial is Eisenstein for $p = 5$. Thus $5$ is completely ramified in the extension, and you're done.
