Let $f(X)\in \mathbb Q[X]$. Consider the family $\mathcal F(f):=\{\mathbb K|_\mathbb Q : f(X)$ is reducible in $\mathbb K[X]\}$.
Then does $\mathcal F (f)$ always have a minimal element (of course it is non-empty) ? If it does, then is any two such minimal elements are isomorphic as fields ?
Assuming a splitting field $F$ for $f$ has been fixed and working only with subfields of $F$, then a minimal extension such that $f$ is reducible exists, because the set is finite: the extension $F/\mathbb{Q}$ is Galois, and so it admits a finite number of intermediate fields.
As far as the isomorphism is concerned, the answer is no.
Consider $f(x)=x^4+1$, which is irreducible over $\mathbb{Q}$. It is reducible over $\mathbb{Q}[i]$ as $(x^2+i)(x^2-i)$; it is also reducible over $\mathbb{Q}[\sqrt{2}]$ as $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$. Both extensions have degree $2$, so they're surely minimal.