Theorem 17.5 Gallian: A polynomial generates a maximal ideal if and only if it is irreducible.

Question

Let $F$ be a field and $p\in F[x]$ a polynomial. Then $\langle p(x)\rangle$ is maximal if and only if $p(x)$ is irreducible.

The proof goes as follow:

Suppose first that $\langle p(x)\rangle$ is a maximal ideal $\operatorname{in} F[x] .$ Clearly, $p(x)$ is neither the zero polynomial nor a unit in $F[x],$ because neither $\{0\}$ nor $F[x]$ is a maximal ideal in $F[x] .$ If $p(x)=g(x) h(x)$ is a factorization of $p(x)$ over $F,$ then $\langle p(x)\rangle \subset\langle g(x)\rangle \subset F[x] .$ Thus, $\langle p(x)\rangle=\langle g(x)\rangle$ or $F[x]=\langle g(x)\rangle$.

I don't understand from the factorization of $p(x)$, we can deduce the relationship between $\langle p(x)\rangle,\langle g(x)\rangle$ and $F[x]$

Answer 1

If $p=gh$ then $p\in\langle g\rangle$, and so $\langle p\rangle\subseteq\langle g\rangle$, and clearly $\langle g\rangle\subseteq F[x]$.