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.
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 $L → L'$.