Polynomials to split in order to get an algebraic closure

Question

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

Following the topic in this question, let's consider a field K and an extension L/K such that L is algebraically closed. It is known that the set of elements of L algebraic over K is an algebraic closure of K. Thought in terms of polynomials, it can be said that a splitting field F (in particular, in L) over K of the set of non-constant polynomials in K is an algebraic closure of K.

My question is basically the following: how can one reduce the set of polynomial and still get an algebraic closure? i.e., which subsets of non-constant K-polynomials are such that the K-extension generated by their roots in L is an algebraic closure of K (that is, their roots over L generate the same extension F)?

Since a splitting field of polynomials over K is always an algebraic extension of K, the latter question is equivalent to asking what families of polynomials give (in the above sense) an algebraically closed extension. One possibility is to consider the set of irreducible polynomials of K, but this choice does not appear to be minimal in general (e.g., over the real numbers, a single polynomial can be enough, if well-chosen). I'm looking for other possibilities or related propositions, both in the general case and in notable circumstances.

Answer 1

In general I don't think it's possible to say anything more specific than "all monic irreducible polynomials." In special cases, of course, more can be said.

For example, over a finite field $\mathbb{F}_p$ we know that every element in an algebraic extension of $\mathbb{F}_p$ has finite order under the Frobenius map, so satisfies $x^{p^n} = x$ for some $n$. This means that instead of considering all polynomials we can just consider the polynomials $x^{p^n - 1} - 1$; that is, we can adjoin all $p^n - 1$-th roots of unity, and this already constructs the algebraic closure $\overline{\mathbb{F}_p}$. This has the benefit of at least being a specific and explicit class of polynomials, but note that it's not hard to prove that the irreducible factors of $x^{p^n} - x$ are exactly the monic irreducible polynomials of degree $d \mid n$, so we actually have not avoided considering every such polynomial, they've just been wrapped up in a convenient package so we don't have to find them individually. This description still has redundancies, which we could actually reduce further by considering the cyclotomic polynomials $\Phi_{p^n-1}(x)$, which are now relatively prime; I think this is as good as it gets.

But already over, say, $\mathbb{Q}$ I don't think it's possible to say anything meaningfully more specific than "all monic irreducible polynomials." Of course this is not minimal and contains redundancies but I don't really think much is gained by trying to get rid of the redundancies; to my mind this actually makes the construction of the algebraic closure seem more complicated than it is. The first construction I saw was a complicated inductive construction that involved repeatedly adjoining roots, but it can actually be done in a single step by considering the ring obtained by formally splitting all monic irreducible polynomials, then quotienting by a maximal ideal; I learned this from Keith Conrad. There is also a really nice argument for both existence and uniqueness using the compactness theorem; I learned this from Joel David Hamkins.