Sharpening a lemma on the uniqueness of splitting fields

Question

I am familiar with the proof of uniqueness of splitting fields that employs this lemma (Dummit & Foote, Abstract Algebra, Theorem 13.1.8):

Let $\varphi : F \to F'$ be an isomorphism of fields. Let $p(x) \in F[x]$ be an irreducible polynomial and let $p'(x) \in F'[x]$ be the irreducible polynomial obtained by applying the map $\varphi$ to the coefficients of $p(x)$. Let $\alpha$ be a root of $p(x)$ (in some extension of $F$) and let $\beta$ be a root of $p'(x)$ (in some extension of $F'$). Then there is an isomorphism $$\begin{aligned} \sigma : F(\alpha) &\to F'(\beta) \\ \alpha &\mapsto \beta \end{aligned}$$ mapping $\alpha$ to $\beta$ and extending $\varphi$, i.e., such that $\sigma$ restricted to $F$ is the isomorphism $\varphi$.

I would like to know if it is ever unnecessary to demand the existence of $\beta$ in this lemma. More precisely, for which $K$ and $K'$ is the following true?

Let $\varphi : F \to F'$ be an isomorphism of fields. Let $p(x) \in F[x]$ be an irreducible polynomial and let $p'(x) \in F'[x]$ be the irreducible polynomial obtained by applying the map $\varphi$ to the coefficients of $p(x)$. Let $K/F$ and $K'/F'$ be field extensions. If $p(x)$ has a root $\alpha \in K$, then $p'(x)$ has a root in $K'$.

In the application to splitting fields, $K'$ is (contains) a splitting field for $p'(x)$, so $p'(x)$ automatically has a root in $K'$. But I am also interested if it is true in the following situations:

1. $K = K'$
2. $K \cong K'$ as fields
3. There exists a field isomorphism $\widetilde{\varphi} : K \to K'$ which restricts to $\varphi$ on $F$

Answer 1

1. This is not sufficient. Let $F=\Bbb Q(s), F'=\Bbb Q(t)$ where $s,t$ are algebraically independent transcendental symbols and $K=K'=\Bbb Q(\sqrt{s},t)$. Then $F,F'$ are subfields of $K=K'$. Let $\varphi:F\to F'$ be defined by $s\mapsto t$ and consider the polynomial $p(x)=x^2-s$. Clearly $p(x)$ has a root $\alpha=\sqrt{s}\in K$ but $p'(x)=x^2-t$ does not have a root in $K'$.
2. See 1.
3. This is sufficient. If $\alpha\in K$ is a root of $p'(x)$ then $\widetilde{\varphi}(\alpha)$ is a root of $p'(x)$ in $K'$ (just apply $\widetilde{\varphi}$ to the equation $p(\alpha)=0$).