Splits completely of a prime ideal

Question

Suppose that $K$ is a number field and $\mathfrak p$ is a prime ideal non-zero.

In general always exists a finite extension of $L$ of $K$ such that $\mathfrak p$ is ramified, for example $L=K(\sqrt f)$ with $f$ is a generator of maximal ideal of $\mathcal O_{K,\mathfrak p}$.

My question is about the existence of a finite extension $L$ of $K$ such that the factorization of $\mathfrak p$ is

$$\mathfrak p=\mathfrak P\mathfrak Q\prod J,$$ with $\mathfrak P$ and $\mathfrak Q$ are two distinct prime ideals and J is a ideal of $\mathcal O_L$ i.e. such that $\mathfrak p$ we have two distinct prime ideals above.

Or more generally if exist be a finite extension $L$ of $K$ such that $$\mathfrak p=\mathfrak P_1\cdots\mathfrak P_r$$ with $r>1$ i.e. $\mathfrak p$ splits completely.

Answer 1

There always is a monic irreducible polynomial $f\in{\mathcal O}_K[t]$ that splits into distinct linear factors mod $\mathfrak p$. Then adjoining the root of such a polynomial gives an extension in which $\mathfrak p$ splits completely. To see that there is such a polynomial start with a (monic) not necessarily irreducible polynomial that splits completely into distinct linear factors mod $\mathfrak p$ and then, using the Chinese Remainder Theorem, add elements $\equiv0\mod\mathfrak p$ to the coefficients so that it becomes Eisenstein at some other prime so that it will be irreducible (maybe there is an easier way to do this?).

Edit: This only gives polynomials of degree $\leq N_{K/\Bbb Q}(\mathfrak p)$. You can get polynomials of arbitrary degree with the desired property by modifying this idea slightly. Take any monic polynomial $f\in{\mathcal O}_K[t]$ that splits completely into distinct linear factors over ${\mathcal O}_K$. Then modify its coefficients so that the resulting polynomial is still $\mathfrak p$-adically close to $f$ and will therefore still split into distinct linear factors over $K_{\mathfrak p}$, but is Eisenstein at some other prime and therefore irreducible. Then adjoining a root of that polynomial gives an extension in which $\mathfrak p$ splits completely.