Set-theoretic equality of splitting fields within a fixed algebraic closure

Question

Let $F$ be a field and let $f(x)\in F[x]$ be a polynomial. Recall the following two facts:

(1) algebraic closures are unique up to isomorphism

(2) splitting fields are unique up to isomorphism

Fix an algebraic closure $\overline F$ of $F$. Is it true that any two (necessarily isomorphic) splitting fields $E,E'\subseteq \overline F$ for $f$ are equal $as$ $sets$ in addition to being isomorphic?

My intuition tells me that this should be the case: fixing an algebraic closure allows us to fix roots of $f$ within this particular ambient field, and since any splitting field is the smallest field containing these roots, any two such fields must actually be equal as sets, in addition to being isomorphic. Does this sound about right? Is this totally trivial?

Answer 1

Yes, this is correct. The only splitting field for $f$ over $F$ in $\overline{F}$ is the subfield of $\overline{F}$ generated by $F$ and all the roots of $f$ in $F$.

(To be clear, for this to be true, by "splitting field for $f$ over $F$ in $\overline{F}$" we must mean a splitting field for $f$ over $F$ which is a subfield of $\overline{F}$ (so the field operations are the same) and whose copy of $F$ is the same as the copy of $F$ inside $\overline{F}$.)

Answer 2

I don't think they are equal as sets. Remember that in our way to get splitting field for $\;f(x)\in F[x]\;$ , we first get (assuming $\;f\;$ is irreducible, otherwise we take one of its irreducible factors) the quotient ring (field) $\;K:=F[x]/\langle f(x)\rangle\;$ .

Here, we already have no more $\;F\;$ but an isomorphic copy of $\;F\;$ within $\;K\;$ , and there could be several ways to obtain that copy...