Solvable number fields

For a number field $K$ take

• $\mathfrak{p}_2,\mathfrak{p}_3,\mathfrak{p}_5$ some prime ideals above $2,3,5$,

• take $h_2\in O_K/\mathfrak{p}_2[x]$ monic irreducible of degree $4$,

  take $h_3\in O_K/\mathfrak{p}_3[x]$ monic irreducible of degree $3$,

  take $h_5\in O_K/\mathfrak{p}_5[x]$ monic irreducible of degree $5$,

• Take $f\in O_K[x]_{monic}$ of degree $5$ such that $$f\equiv x h_2\bmod \mathfrak{p}_2,f\equiv x^2 h_3\bmod \mathfrak{p}_3,f\equiv h_5\bmod \mathfrak{p}_5$$

Let $L$ be the splitting field of $f$.

$O_L$ will contain a prime $\mathfrak{P}_2$ above $\mathfrak{p}_2$ and $f=xh_2\bmod \mathfrak{P}_2$ will split completely, so $$4\ | \ [O_L:\mathfrak{P}_2:O_K/\mathfrak{p}_2] \ \implies 4 \ | \ [L:K]$$

And so on, you'll get that $4\cdot 3\cdot 5$ divides $[L:K]$.

This implies that $Gal(L/K)$ is either $S_5$ or $A_5$, unsolvable.

Number fields satisfy Hilbert's irreducibility theorem, which implies that there are infinitely many extensions with Galois group $S_n$ (the symmetric group, which is non-solvable for $n \geq 5$) for any $n$. Thus, there are infinitely many non-solvable quintic polynomials over any number field.