I think you are slightly confused.
Ok, so let's just agree that there exists a field $k$ such that all the roots of $p$ lie in $k$ (see here). Ok, so what we can then do is analyze individual roots $\alpha,\beta,\gamma,\cdots,$ of $p$. To do this we can consider fields such as $F(\alpha)$ which is the smallest field containing both $F$ and $\alpha$ (i.e. the intersection of all subextensions of $k/F$ which contain $F$). What is definitively true is that $F(\alpha)\cong F[x]/(p)$. Now, this was true for ANY root $\alpha$ of $p$ and so you may conclude that, in particular, $F(\alpha)\cong F(\beta)$ for any two roots $\alpha,\beta\in k$ of $p$.
That said, while the $k$ I said contains all the roots of $p$ any extension containing A root of $p$ need not contain all the roots of $p$. For example, consider the irreducible polynomial $x^3-2\in\mathbb{Q}[x]$. Then, the complex numbers $\mathbb{C}$ are a field containing all the roots $x^3-2$--they are $\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$ where $\omega$ is a primitive third root of unity. Note then that while $\mathbb{Q}(\sqrt[3]{2})$ contains A ROOT of $p(x)$ it definitively does not contain ALL the roots of $p(x)$ since $\omega$ is complex and all the elements of $\mathbb{Q}(\sqrt[3]{2})$ are real (if this last fact isn't obvious consider that $\mathbb{R}$ is a field containing $\mathbb{Q}$ and $\sqrt[3]{2}$).
I hope this helps.
EDIT: Just like the fields $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\sqrt[3]{2}\omega)$ and $\mathbb{Q}(\sqrt[3]{2}\omega^2)$ it seems you now have three isomorphic responses. Thank god that fields permit no more roots.