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When one says that $K$ is a splitting field of a family of polynomials over $F$ is it assumed that the family of polynomials is always finite? If infinite families are allowed then I wonder if the splitting field would be algebraic?

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AFAIK infinite families are allowed. The splitting field of $\{p_i\mid i \in I\}$ is then generated by the collection $\{z_{i,k} \mid i \in I, k \}$ where $z_{i,k}$ denote the zeros of $p_i$. As all $z_{i,k}$ are algebraic (there are zeros of the polynomials $p_i$), the splitting field is. But of course it may be an infinite extension. –  martini Sep 18 '12 at 18:37

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When I see "splitting field," I do not assume a finite family of polynomials. Splitting fields tend to come up when one talks about Galois Theory, in which we like splitting fields a lot to identify when we have Galois groups.

For example, let's take the field $\mathbb{Q}$ and consider its algebraic closure $\overline{\mathbb{Q}}$. It's clearly normal, it's Galois over $\mathbb{Q}$, and 'clearly' the 'splitting field' of all polynomials with coefficients in $\mathbb{Q}$. But it's not at all finite dimensional.

To play devil's advocate, perhaps someone would explicitly define a splitting field as only for a finite family of polynomials, and they would call what I mentioned above instead a 'normal extension' (whereas you might say that splitting field and normal are the same). They must have some name for those extensions that allow infinite Galois groups, so there is some word.

So, in short, I would say that splitting field does not require only a finite family of polynomials.

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If $K$ is a field, then a splitting field over $K$ for a set $\{f_i:i\in I\}$ of polynomials $f\in K[X]$ is an extension $F$ of $K$ such that (a) each $f_i$ splits completely in $F[X]$ and (b) $F=K(\{a\in F:a\text{ is a root of some }f_i\})$. So $F$ must have all the roots of the $f_i$ and it must be generated over $K$ by these roots. There's no need to assume the set of polynomials is finite, although, if it is, it is clear that $[F:K]$ is finite because the conditions show that $F$ can be written as a tower of successive simple extensions of $K$ (in other words, order the finite set of roots of your finite set of polynomials in $F$, then adjoin the first one to $K$, then adjoin the second to the extension you just made, etc.). Splitting fields over $K$ are always algebraic over $K$. This follows from the fact that if $F/K$ is an extension of fields and $F^\prime$ is the set of elements of $F$ which are algebraic over $K$, then $F^\prime$ is a field.

Splitting fields are unique up to generally non-unique isomorphism. This follows from the result in field theory that you can extend an isomorphism of fields to an isomorphism of splitting fields. More precisely, if $\sigma:K^\prime\cong K$ is an isomorphism of fields, and $S$ is a set of polynomials in $K^\prime[X]$ (finite or infinite), with $\sigma(S)=\{\sigma(f):f\in S\}$ the corresponding set in $K[X]$, then $\sigma$ can be extended to an isomorphism $F^\prime\cong F$ where $F^\prime$ is a splitting field over $K^\prime$ for $S$ and $F$ a splitting field for $\sigma(S)$ over $K$. An algebraic closure $F$ of $K$ is the same thing as a splitting field over $K$ for the set of all irreducible polynomials in $K$, so in particular, thinking of algebraic closures as splitting fields allows one to prove that they are unique (up to non-unique isomorphism over the base field). So you want to allow infinite sets of polynomials!

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