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Suppose $F$ is a finite splitting field over $K$ of $X=\lbrace f_i(x)\rbrace_{i\in I}$, some infinite set. Is there necessarily a finite set $Y\subseteq X$ such that $F$ is a finite splitting field of $Y$?

I'm curious if there is a way to generalize: $F$ is a splitting field over $K$ of a finite set $\lbrace f_1,…,f_n\rbrace$ of polynomials in $K[x]$ if and only if $F$ is a splitting field over $K$ of the single polynomial $f=f_1f_2⋯f_n$.

The example I have in mind is $\mathbb{C}$ over $\mathbb{R}$. Clearly $\mathbb{C}$ is the splitting field for $X=\lbrace ax^2+bx+c\mid a,b,c\in \mathbb{R}, b^2-4ac<0 \rbrace$ and but also for just $x^2+1$.

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  • $\begingroup$ No, in general it is not possible to find a finite subset. If you could, as you note, the finite subset could then be replaced by a single polynomial, and so the extension would necessarily be a finite extension. However, you can find infinite extensions, e.g., of $\mathbb{Q}$: the extension given by the splitting field of the set of polynomials $f_n(x) = x^n-1$ for all $n\geq 1$ is not given as the splitting field of any finite set of polynomials, let alone a subset of $\{f_n\}$. $\endgroup$ – Arturo Magidin Mar 24 at 6:00
  • $\begingroup$ That was another example I was thinking about. Although $\mathbb{Q}(\sqrt{2},\sqrt{3},...)$ is an infinite extension. If we assumed our restriction was finite then could we, for some reason, take some finite subset? $\endgroup$ – Dene Mar 24 at 11:51
  • $\begingroup$ If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial. $\endgroup$ – Arturo Magidin Mar 24 at 17:12
  • $\begingroup$ Yes, this is actually what I'm trying to prove. This issue has arised because the question is worded in such a way that our only assumption is that the extension is finite and normal. Then I am to prove that it is the splitting field of a single polynomial. So I'm curious if the set of polynomials can always be finite. $\endgroup$ – Dene Mar 24 at 19:55
  • $\begingroup$ No, “a set of polynomials can always be finite” is not correct, as already noted. But a finite extension is necessarily finitely generated, and you only need one polynomial for each generator, and then you can take the product of all of these polynomials. $\endgroup$ – Arturo Magidin Mar 24 at 21:19
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It is a classic result in Galois theory that a finite extension $F/K$ is the splitting field of a single polynomial (i.e. generated by all the roots of a single polynomial) if and only if $F/K$ is normal, i.e. every polynomial in $K$ with a root in $F$ splits completely. See this post, for instance.

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  • $\begingroup$ Actually, this is the question I was being asked. But was worded in such a way that we don't necessarily have a finite set of polynomials. When our set was finite I had already produced a proof. If the extension is finite and the set of polynomials is infinite then as of right now my proof doesn't apply. $\endgroup$ – Dene Mar 24 at 11:59

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