Prove $F[x]/\langle p(cx)\rangle\cong F[x]/\langle p(x)\rangle$

Question

This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise E5.

Recall the definition of $F(a)$. It is a field such that $F \subseteq F(a)$, $a \in F(a)$, and any field containing $F$ and $a$ contains $F(a)$. Let $F \subseteq K, c \in F$, and $a \in K$. Let $p(x)$ be irreducible, and let $a$ be a root of $p(cx)$.

Prove $F[x]/\langle p(cx)\rangle \cong F(a)$ and $F[x]/ \langle p(x)\rangle \cong F(ca)$.

Conclude that $F[x]/\langle p(cx)\rangle\cong F[x]/\langle p(x)\rangle$.

[Edited]

1. $p(x)$ is irreducible implies $p(cx)$ is irreducible, for otherwise let $y=cx\implies p(y)$ is reducible over $F$; a contradiction. Note $c\ne 0$ for otherwise it doesn't make sense for $a$ to be a root of $p(cx)$.
2. $a$ is a root of $p(cx)\implies F[x]/\langle p(cx)\rangle\cong F(a)$.
3. $a$ is a root of $p(cx)\implies ca$ is a root of $p(x) \implies F[x]/\langle p(x)\rangle\cong F(ca)$.
4. By Exercise E1, $F(a)=F(ca)$. Hence, $F[x]/\langle p(cx)\rangle\cong F(a) = F(ca)\cong F[x]/\langle p(x)\rangle$.

Correct?

Answer 1

It's mostly correct. The argument in 1. isn't quite right because if $p(cx)$ is reducible that means that $p(cx) = q(x)r(x)$ not necessarily $q(cx)r(cx)$ so simply substituting $y = cx$ doesn't quite work. You can fix this by defining $\bar q(x) = q(x/c)$ and $\bar r(x) = r(x/c)$ then $p(cx) = \bar q(cx) \bar r(cx)$.

And then the rest of the argument is exactly right.