Suppose $F$ is a splitting field of an infinite set $T$ of separable polynomials, over $K$. If there is no finite set $S \subseteq T$ such that $F$ is splitting field of $S$ over $K$, is it true that $F$ is infinite dimensional over $K$?
I know that because of separability there must be infinitely many roots for $T$. But how does this lead to infinite dimensional?
This is part of a proof of equivalent definition of algebraic Galois extension. In Hungerford p262, to show (iii) F is splitting field over K of a set T of separable polynomials implies (i) F is algebraic and Galois over K. The proof says it suffices to prove the case where [F:K] is finite, and goes on to say if [F:K] is finite then there are some finite polynomials S in T such that F is splitting field of S over K.
Suppose $F/K$ is the splitting field of a family $T$ of polynomials and not of any finite subset $S\subset T$.
This implies any finite subset $S\subset T$ can be inflated to a bigger finite subset with a bigger splitting field. In particular, this can be done by adjoining a single polynomial to $S$. For, if this wasn't possible, then every other element of $T$ would split in $S$'s splitting field, in which case $T$'s splitting field wouldn't be any bigger than $S$'s, a contradiction.
Thus, there exists a sequence $f_1,f_2,\cdots$ of elements of $T$ for which the splitting fields of $\{f_1,\cdots,f_n\}$ are strictly increasing with $n$, which implies the dimension over $K$ diverges.