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.
Does an algebraically closed field contain all algebraic field extensions
abstract-algebrafield-theoryseparable-extension
Best 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