If $p(x)\in F[x]$ be a irreducible polynomial over $F$, what are the elements of $(p(x))$?
From my book
$(a) = \{ ra|r \in R\}$
but I don't know what $r$ is going to be in this case.
I see this a lot in other theorems like
If $p(x)$ is a polynomial in $F[x]$ of degree $n\ge 1$, and is irreducible over $F$, there is extension $E$ of $F$ such that $[E:F] = n$ where $p(x)$ has a root.
here on this theorem proof, it begins with
Let $F[x]$ be ring of polynomials in $x$ over $F$ and let $V=(p(x))$ be an ideal of $F[x]$ generated by $p(x)$ ... and goes on and on.
I have difficulty understanding the proof of this theorem from the very beginning.
$r$ is just an element of the ring $R$ - which in your question is the polynomial ring $F[x]$. So in this case (putting $a=p(x)$ and $r = f(x)$ for some $f(x) \in F[x]$), we have:
$$(p(x)) = \left\{f(x)p(x) \mid f(x) \in F[x] \right\}$$
Equivalently, if $g(x) = f(x)p(x)$ for some $f(x) \in F[x]$, what can you say about $g(x)$?