I am trying to generalize a problem that I came across previously.
$\mathbf{ Problem:}$ Are the fields $\mathbb{Q}$ and $\mathbb{Q[\sqrt2]}$ isomorphic?
$\mathbf{Generalisation:}$ Let $F$ and $K$ be two fields of characteristic $0$ and such that $F \subset K$. If there exists a polynomial (of finite degree) $p(x) \in F[x]$ with coefficients from copies of $\mathbb{Q}$ such that $F$ does not contain any root of $p(x)$ but $K$ contains every root of it, then $F\ncong K$.
Proof Attempt:
Any field $F$ with $\mathrm{char \ F=0 }$ contains a copy of $\mathbb{Q}$. So, $\mathbb{Q}$ is isomorphic to a subfield of $F$ (thereby $K$).
Now, let us assume that despite of the presence of such a polynomial $p(x)$, $F \cong K$. Let $p(x)=a_0x^n+a_{1}x^{n-1}+...+a_{n-1}x+a_n$. We have $p(\beta)=0$, for some $\beta \in K$ but $\beta \notin F$.
Let $\phi: F \mapsto K$ be such an isomorphism. For some $x \in F$,
$\beta=\phi(x) \implies a_r\beta^{n-r}= \phi(a_rx^{n-r})$ $ \ \ \ \ \ ....(*) $ . [This holds due to the fact that any such isomorphism will fix $\mathbb{Q}$, i.e. $\phi(m)=m$ , $\forall m\in \mathbb{Q}$].
By $(*)$, running $r\in \{1,2,...,n\}$ and adding them gives,
$p(\beta)=0=\phi(p(x)) \implies p(x)=0$. By our assumption, $x \in F$, but $p(x)\neq 0 \ \forall \ x\in F$. A contradiction.
Is the formulation of the general statement correct? Is my "Proof" correct? Kindly verify.
You don't need the full strength of this assumption. All you need is one root in $K$.
Assume $\exists p(x) \in \Bbb Q[x]$ such that $p(x)$ has at least one root $r \in K$ but $p(x)$ has no roots in $F$. Then assume $f:F \to K$ is an isomorphism. Since $f(1)=1$, it follows that $\forall a \in \Bbb Q, f(a)=a,$ so $f(p(x))=p(x)$. But $p(r)=0 \Rightarrow 0=f^{-1}(0)=f^{-1}(p(r))=p(f^{-1}(r)) \Rightarrow f^{-1}(r)$ is a root of $p$ in $F$, contradicting our hypothesis.