# Splitting field over K of an infinite set of polynomial

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 $$x^2+1$$.

• 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 $\{f_n\}$. – Arturo Magidin Mar 24 at 6:00
• 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? – Dene Mar 24 at 11:51
• If the extensions is finitely generated and algebraic, then it is finite, and then you can take a single polynomial. – Arturo Magidin Mar 24 at 17:12
• 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. – Dene Mar 24 at 19:55
• 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. – Arturo Magidin Mar 24 at 21:19

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.