This is a question related to another question on this site (see below) on the nature of finite extensions of the field of p-adic numbers $\mathbb{Q}_p$ and rings of integers.
Question: Let $\mathbb{Q} \subseteq K$ be a number field and let $\mathfrak{q}_1,\mathfrak{q}_2 \subseteq \mathcal{O}_K$ be primes with $\mathfrak{q}_i \cap \mathbb{Z}=(p)$ for a prime number $p\geq 2$. It follows the extensions $\mathbb{Z}_p \subseteq (\mathcal{O}_K)_{\mathfrak{q}_i}$ are monogenic for $i=1,2$.
Citation: "..what I am looking for is a published proof of examples where for different primes $\mathfrak{q}_1,\mathfrak{q}_2⊆\mathcal{O}_K$, the completions $(\mathcal{O}_K)_{\mathfrak{q}_i}$ are non isomorphic and have different degrees over $\mathbb{Z}_p$ - can you give an explicit reference?"
I am asking for an explicit reference to the litteratore on the above topic. I formulate this as a question since part of the discussion on the below link was deleted and I do not think people will read the comments and respond.
Proof that all extensions of $\mathbb{Q}_p$ are of the form $\mathbb{Q}_p[\sqrt[n]{a}]$
For a very elementary example, let $f(x)=x^3+10x^2-5x+10$. $f$ is Eisenstein for the prime $5$, so is irreducible over $\mathbb{Z}$, and let $K$ be the number field generated by one root $\alpha$ of $f$.
Now, $K/\mathbb{Q}$ is unramified over $3$, as $f$ splits with simple roots in $\overline{\mathbb{F}_3}$. Assume that the conductor of $\mathbb{Z}[\alpha]$ is coprime to $3$: then, by Neukirch’s Algebraic Number Theory, Proposition 8.3 in Chapter I, since $f(x)=(x^2+1)(x+1)$ mod $3$, with $x^2+1$ and $x+1$ irreducible mod $3$, it follows that there are two prime ideals $\mathfrak{q}$ and $\mathfrak{r}$ in $O_K$ above $3$ and that the completed field extensions are unramified of degrees $1$ and $2$ over $\mathbb{Q}_3$.
Now, it remains to see that the conductor of $\mathbb{Z}[\alpha]$ is coprime to $3$. But we see in the same book, in the proof of Theorem 2.5 of Chapter III, that the conductor $\mathfrak{f}$ of $\mathbb{Z}[\alpha]$ contains $f’(\alpha)$. Since $f$ is separable and separable mod $3$, $(f’(\alpha),3)=O_K$, $\mathfrak{f}$ is indeed coprime to $3$ and we are done.