Splitting fields construction My question is about the standrad construcion which shows that every polynomial has a splitting field.
The construction is presented in wikipedia here.
https://en.wikipedia.org/wiki/Splitting_field
In the construction we take the base field F and add to it roots of the polynomial f(x).
My question is - in the end of this proccess do we get a splitting field (which is minimal in a sense) or we can get a field which splitts f(x) but is not minimal ?
The reason I'm assking is because my prof' proved that every two splitting fields of p(x) over F is isomorphic with reliance on the fact that every to algebraic closures are isomorphic.
He didn't prove that fact so I was wondering if it's possible to prove the theorm without using the algebraic closures.
(A proof that algebraic closures are isomorphic will help too)
Thanks to all the helpers.
 A: The construction is minimal. This is due to the fact that for a field extension $E/F$ and elements $α, β ∈ E$,
$$(F(α))(β) = F(α,β).$$
So let $K$ be a field, $f ∈ K[X]$ and $K_0, K_1, …, K_r$ a sequence of fields such that $f$ completely splits in $L = K_r$ and for every $i = 1, …, r$ $K_i = K_{i-1}(α_i)$ for some zero $α_i$ of $f$, then by induction
$$L = K(α_1,…,α_r)$$
for some zeros $α_1, …, α_r$ of $f$. But since all zeros of $f$ are in $L$, $L$ is certainly generated by them and the construction is indeed minimal.
This can indeed be used to construct a field homomorphism $L → L'$ for any other splitting field $L'$ of $f$ by taking the inclusion $σ \colon K → L'$ and extending it at every step of the construction with the universal property of polynomial rings and factor rings to field morphisms $σ_i \colon K_i → L'$ for $i = 1, …, r$. Since the final field morphism $τ$ is injective and sends zeroes of $f$ to zeroes of $f$, it has to surjectively hit all zeroes of $f$, so it’s surjective overall, thus it’s a field isomorphism $L → L'$.
