field isomorphism extension theorem and galois transitive action

Question

Here is a version of isomorphism extension theorem: Given $\phi: F \to F'$ field isomorphism, let $K$ be the splitting field of some $f \in F[x]$ and $K'$ be the splitting field of $\phi(f)$, we have

1. there exists an isomorphism $\psi: K \to K'$ whose restriction to $F$ is $\phi$
2. Furthermore, we can choose a pair of roots $(a, a')$ from irreducible factor $(g, \phi(g))$ of $(f,\phi(f))$ such that $\psi(a) = a'$

If the theorem I stated is correct, then can't we inductively apply it (i.e.

1. start with $A=F=F'$ and $D=K=K'$ and pick an irreducible factor and a pair of roots $(a,a')$
2. Now restrict the isomorphism from step 1 to get an isomorphism from $F=A[a], F'=A[a']$, pick another irreducible factor and a pair of roots $(b,b')$.
3. $F=A[a,b], F'=A[a',b']$ and so on...)

and conclude that

1. We can always pick pairs of roots from EACH irreducible factor of $f$ and there will be some automorphism in $Gal(K/F)$ that map each pair?
2. similarly, call those pairs $[a_1,..,a_n]$ and $[a_1',...,a_n']$. If I am given $n$ isomorphisms from $F[a_i] \to F[a_i']$, then I have an automorphism that extends all of them?

Answer 1

The key point to take home is that irreducible polynomials can become reducible in a larger extension. This might seem obvious once said out loud, but it's still important to take note of it.

So 1 fails because I can take $f(x) = (x^2+1)(x^2+4)$ and $F = F' = \Bbb Q$, because if I specify the pair $(i,i)$ for $x^2+1$, then the pair $(2i,-2i)$ for $x^2+4$ is not realizable.

And 2 fails because once you consume the pair $(a_1, a_1')$ and go to the larger field, then $a_2$ and $a_2'$ might no longer belong to the same irreducible factor. Essentially the same counter-example works, and if you want a counter-example with $f$ irreducible then I suggest you think about it as an exercise.