When will $F[x]/\langle p(x)\rangle$ strictly contain $F$? Let $F$ be a field, $p(x)\in F[x]$ with $\deg p(x)\geq 1$. I know that when viewing $F$ as its isomorphic copy in $F[x]/\langle p(x)\rangle$, then $F\subseteq F/\langle p(x)\rangle$. 
However, when will $F[x]/\langle p(x)\rangle$ strictly contain $F$, i.e., $F\subsetneq F[x]/\langle p(x)\rangle$ (ps: view $F$ as its isomorphic copy in $F[x]/\langle p(x)\rangle$)? I have been headache about these materials in the early steps, so I think I should ask for the help.
What I have tried is that: I know there's a theorem that $p(x)$ is reducible over $F$ iff $F[x]/\langle p(x)\rangle$ is a field. So when $p(x)$ is reducible, then $F[x]/\langle p(x)\rangle$ is not a field, then $F\subsetneq F[x]/\langle p(x)\rangle$ in this case. Am I correct? And how if $p(x)$ is reducible?
 A: $F[x] / \langle p(x) \rangle$ is an $F$-vector space; your question boils down to asking whether the dimension of this vector space is strictly greater than one.
It's not too hard to show that $\dim\left( F[x] / \langle p(x) \rangle \right) = \deg(p)$; e.g. by writing down a basis.
A: There is always an injective map $F \mapsto F[x]/(p(x))$, unless $(p(x))$ is the unit ideal. 
To see this, suppose $p(x)$ is not a unit. Then $F[x]/(p(x)) \neq 0$, so the composition $F \to F[x] \to F[x]/(p(x))$ is nonzero (it sends $1$ to $1 \neq 0$). But the kernel of this map is an ideal of $F$, which means it's $(0)$. In other words, the map is injective. 
$\textbf{EDIT}$: If we want this map to only be injective, we have to require that $deg(p(x)) > 1$
A: In general, if you gave an integral domain (or in your case a field) $F$ and an ireducible polynomial $p(x)$ then $F[x]/<p(x)>\cong A$ where $A$ is a field. The elemet of $A$ are the equivalence classes of the division remainders with $p(x)$. So in $A$ $[a(x)]=[r(x)]$ where $a(x)=b(x)p(x)+r(x)$.This means that if whatever we do, if we $\deg(p(x))\geq2$ then $A$ will contain $F$ as long with other elements (say for example the class $[x+p]$). 
Now if $\deg(p(x))\leq1$ then $A$ will contain at best $F$ (depending if $p(x)$ is irreducible or not) so we conclude that for $A$ to be contain $F$ $\deg(p(x))\geq 2$
