# Splitting field of a set of polynomials

Given $X\subseteq F[x]$ where $F$ is a field, how to prove that there exist a splitting field of $X$ over $F$?

In the case that $X$ is finite, I think the answer can be solved using Kronecker's Theorem, which states that every for every polynomial $f(x)\in F[x]$ there exist an extension field $E$ of $F$ containing a root of $f(x)$. Letting $f(x)$ be the product of all polynomial in $X$, one can proof by induction over $\deg f(x)$ that there is an extension field $K$ of $F$ containing a root of $f(x)$ and, therefore, an extension field $E$ of $F$ such that $f(x)$ splits in linear factors over $E[x]$; from this the result.

But what if $X$ is not finite? In general, how to proceed?

• The field generated by the roots of the elements of $X$ in an algebraic closure of $F$. – Bernard May 3 '16 at 21:43
• if $X$ is infinite, in particular when $X = F[x]$, you'll clearly need to define the whole algebraic closure. I don't see why there would be much more to say except maybe that "Using Zorn's lemma, it can be shown that every field has an algebraic closure" en.wikipedia.org/wiki/Algebraic_closure – reuns May 3 '16 at 21:52
• Can't you give a proof that has any subset $X \subset F[x]$ has a splitting field and then consider the specific case $X = F[x]$ to obtain the algebraic closure? – Ethan Alwaise May 3 '16 at 22:01
• Could we Zorn's lemma this without assuming the existence of an Algebraic Closure? – Ravi May 3 '16 at 23:04
• I concur with several of the comments above, especially @Jake’s. If you feel free to use the existence of an algebraic closure of $F$, then Bernard’s solution does the trick. If you do not feel so free, it’s a different problem entirely, and as far as I can see, you do need to Zornify in one way or another: either prove the existence of an algebraic closure, or prove that there’s a splitting field of your given $X$. – Lubin May 4 '16 at 0:56