# Does an algebraically closed field contain all algebraic field extensions?

Let $$K$$ be a field and $$Ω$$ an algebraically closed field that contains $$K$$. Is it the case that for every algebraic extension $$L/K$$, that $$L ⊆ Ω$$? Intuitively this makes sense because colloquially, $$Ω$$ "is the largest algebraic extension of $$K$$". I cannot, however, formally exclude yet that neither $$L ⊆ Ω$$ nor $$Ω ⊆ L$$ holds. I suspect there must be a quick and simple argument that shows that one of the inclusions (or rather, embeddings) must hold.

• Are you sure you want $L\subseteq Q$? If so, then the answer is no, an algebraically closed extension does not contain all algebraic extensions. It does so up to isomorphism though: every element in an algebraic extension is a root of a polynomial over the ground. That polynomial splits in the algebraic closure, so the algebraic closure contains an isomorphic copy of the element you started with. This can be made precise, and requires a bit of work. Commented Apr 19, 2020 at 22:06
• Ah yes. Well that's the reason I switched to "embeddings" actually. I'll try working this out! Commented Apr 19, 2020 at 22:08
• Ok different question. I have shown that $L$ itself has a separable closure $L^{\text{sep}}$, and that each pair of separable extensions of $K$ is $K$-isomorphic. It remains to show that $L^{\text{sep}}$ is itself a separable extension of $K$, which is not obvious to me. Commented Apr 19, 2020 at 22:54

## 1 Answer

The statement in your question is incorrect. You cannot define the algebraic closure to be the union of all algebraic extensions. The basic problem is that you can add any element in when you extend a field. For example when we go from the reals to the complex numbers we add in an element $$i$$ according to the rule that $$i^2 = -1$$. We could equally have used in the place of $$i$$ any thing you could dream of so that if an algebraic closure was to contain every algebraic extension then as a set it would have to contain everything which is not possible. The standard proof of this uses Zorn's lemma to get around this problem

• I see, so must we take some sort of 'fancy union' that nicely 'glues together' the overlapping parts? Commented Apr 19, 2020 at 22:11
• Yeah you just have to make precise the idea of adding in elements that solve equations you cannot currently solve until you can solve everything. I'm not the best person to explain how you actually prove algebraic closures exist because I've never studdied the proof myself Commented Apr 19, 2020 at 22:26