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Let $F \subset K$ be an algebraic field extension, is $K$ a subfield of a splitting field of $S=\{{f_i}\}_{i \in I} $ over $F$? I tried: Let $\{{u_i}\}$ be a $F$-basis of K, $\{{p_i}\}$ be its corresponding minimal polynomials ,can we deduce that $K$ is a subfield of the splitting field of $\{{p_i}\}$ over $F$? I got somewhat confused about the underlying sets. Thanks in advance.

user725757
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    The algebraic closure is the splitting field for $S = \text{every single polynomial in F}$. – David Lui Dec 20 '20 at 07:54
  • Is K embed in the algebraic closure of F? – user725757 Dec 20 '20 at 08:07
  • So take the algebraic closure of K since it’s algebraic closed and algebraic over K hence algebraic of F, it’s also an algebraic closure over F. Then it’s an algebraic closure of F containing K. – user725757 Dec 20 '20 at 08:23
  • Use Zorn's lemma to show that K embeds into the algebraic closure of F https://math.stackexchange.com/questions/2733942/existence-of-extension-of-an-embedding-for-an-algebraic-extension-of-a-field – David Lui Dec 20 '20 at 08:25

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Zorn's lemma (see Existence of extension of an embedding for an algebraic extension of a field) implies that there is an embedding $K \rightarrow \overline{F}$.

Now, $\overline{F}$ is the splitting field for $S = \text{every polynomial in $F[x]$}$ , and $K$ is isomorphic to the image of the embedding in $\overline{F}$.

David Lui
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