I am trying to clarify a statement regarding the homomorphic image of a maximal ideal
Let $\varphi: F\rightarrow F'$ be an isomorphism of fields. The map $\varphi$ induces a ring homomorphism (also denoted $\varphi$) $\varphi:F[x]\rightarrow F'[x]$ defined by applying $\varphi$ to the coefficients of a polynomial in $F[x]$. Let $p(x)\in F[x]$ be an irreducible polynmial and let $p'(x)\in F'[x]$ be the polynomial obtained by applying the map $\varphi$ to the coefficients of $p(x)$. The isomorphism maps the ideal $(p(x))$ to the ideal $(p'(x))$.
Is it true that $(p'(x))$ is also maximal? I don't think the homomorphic image of maximal ideal is necessarily maximal.
EDIT: $\varphi$ in fact induces a ring isomorphism, which is why this statement is true.
The statement is true in greater generality. Suppose $\varphi\colon A\to B$ is a surjective ring homomorphism. Then, for every maximal ideal $I$ such that $\ker\varphi\subseteq I$, the image $\varphi(I)$ is a maximal ideal of $B$.
Indeed, the homomorphism theorems provide an isomorphism $\bar{\varphi}\colon A/I\to B/\varphi(I)$ (here just $\ker\varphi\subseteq I$ is needed, not maximality). Then $\varphi(I)$ is maximal if and only if $I$ is maximal.
In your case $\varphi\colon F[x]\to F'[x]$ is an isomorphism, so the condition $\ker\varphi\subseteq I$ is trivially satisfied.
As a side note, if $\ker\varphi\not\subseteq I$, then $\varphi(I)=B$.