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I was studying field theory when a question came to my mind.

Let K be a field, F be the set of non-constant polynomials with coefficients in K, L be a splitting field for F over K (i.e. an extension of K over which every polynomial of F splits completely). My question is the following: is L algebraically closed?

In other terms, it is known that every polynomial in K splits completely over L, but is it true for all the polynomials in L? If this was true, the algebraic closure could be characterised in terms of splitting fields, which is the occasion I popped up into this question.

Amanda Wealth
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  • Yes. See here and here. – Arturo Magidin Aug 04 '24 at 17:07
  • For example, it suffices to show that every polynomial with coefficients in $L$ divides a polynomial with coefficients in $K$. – Qiaochu Yuan Aug 04 '24 at 17:42
  • @QiaochuYuan And how could that be proved? – Amanda Wealth Aug 05 '24 at 07:59
  • Hmm, maybe that isn't the cleanest way to do it. The argument I had in mind requires separability so you can use Galois theory (if the coefficients of a polynomial with coefficients in $L$ generate a Galois extension of $K$ then you can just act on it by the Galois group and multiply the results together, and you can consider the splitting field of the minimal polynomials of the coefficients inside $L$), but then a different argument is needed in the inseparable case. Maybe better to use the other arguments. – Qiaochu Yuan Aug 05 '24 at 09:43

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