Discriminant of splitting field Let K number field, $O_K$ be its integer domain. We all know $O_K$ is a free $\mathbb{Z}$-module. If L is a finite (or galois) extension of K, whether $O_L$ is a free $O_K$-module? 
In addition, let $f$ be a irreducible polynomial in $\mathbb{Q}$, α is a root of $f$, $K$ is the splitting field of $f$. $Δ(F)$ denote the discriminant of a number field $F$. Is $Δ(K)$ devides $Δ(\mathbb{Q}(α))^n$ for some integer $n$? 
 A: Answer to your 2nd question:
In the relative situation $K/k$, one defines the discriminant ideal $\Delta (K/k)$ as being the ideal of $O_k$ generated by all the discriminants of all the $k$-bases of $K$ consiting of integral elements. A finer invariant is the different $\mathfrak D(K/k)$, which is an ideal of $O_K$ defined as follows: the elements $x$ of $K$ such that $Tr_{K/k}(x O_K) < O_k$ , where $Tr_{K/k}$ is the trace map of $K/k$ , form a fractional ideal of $K$, and its inverse is $\mathfrak D(K/k)$. The relation between the discriminant and the different is: $\Delta(K/k)$ = $N_{K/k}(\mathfrak D(K/k))$, where $N_{K/k}$ is the norm map in $K/k$. 
Given a tower of number fields $k < K < L$, one has a "transitivity" formula for differents: $\mathfrak D(L/k) = \mathfrak D(L/K).\mathfrak D(K/k)$. Taking norms and using their transitivity, one gets readily: $\Delta(L/k) = N_{K/k}(\Delta(L/K)).\Delta(K/k)^m$, where $m$ is the degree of $L/K$. It remains just to apply this formula to your particular case.
A: In the « absolute situation » of a finite extension of $k/\mathbf Q$, the ring of integers $O_k$ is $\mathbf Z$-free because $\mathbf Z$ is a PID : a finitely generated module without torsion over a PID is free. This is no longer true in the « relative situation » of a finite extension $K/k$ of number fields, where $O_k$ is a Dedekind domain, not a PID in general. Actually, if $R$ is a DD and $M$ is a finitely generated $R$-module, then :
$M$ has no torsion iff $M$ is projective ( = a direct summand of a free $R$-module), iff $M$ is a direct sum of a free $R$-module and a fractional ideal $J$ of $R$. The class of $J$ in the class group $Cl(R)$ is an invariant called the Steinitz class of $M$ .
PS: I don't understand your second question
