Field generated by element contains all other roots of minimum polynomial? I just want to check my understanding of this point of field theory:
Let $F$ be a field, and $c$ algebraic over $F$ with minimum polynomial $p(x)$. Then we know that $F(c)$ (the field generated by $F$ and $c$) is isomorphic to $F[x]/\langle p(x)\rangle$.
Does this imply that if $d$ is any other root of $p(x)$, then $d\in F(c)$?
 A: No, it does not. What you want is called the splitting field, which is roughly the smallest field containing your base field and all roots of your polynomial. A simple example is $F = \mathbb{Q}$ and $p(x) = x^3 + x + 1$. This polynomial has a unique real (irrational) root since it's always increasing. Your quotient can be thought of as (embedded in) the reals, so neither of its complex roots came along for the ride when creating $\mathbb{Q}[x]/\langle p(x)\rangle \cong \mathbb{Q}(\alpha) \subset \mathbb{R}$ ($\alpha$ being the real root). (There are significant technical details being swept under the rug here, but this is the intuition.)
The usual construction of the splitting field essentially just iterates the quotient operation you described repeatedly. If the minimal polynomial factors in the larger field, just apply the quotient operation to one of the (non-linear) irreducible factors. Eventually it'll finish as your original polynomial will factor more and more each time. Indeed, this suggests a proof of the fact that the splitting field of $p(x)$ over $F$ will have degree dividing $n!$, where $n$ is the degree of $p$.
