Given two fields, constructing a field containing both In the text by Dummit and Foote, the authors introduce the idea of an algebraic closure, and also prove that one always exists for every field $F$.  They then comment that we can "...speak sensibly of the composite of any collection of algebraic extensions by viewing them as subfields of an algebraic closure."
But I'm a bit confused by this.  Suppose we have two fields $K_1$ and $K_2$.  Then we have the two algebraic closures $\overline{K_1}$ and $\overline{K_2}$, but it doesn't necessarily follow that either one of these algebraic closures contains both $K_1$ and $K_2$...
So I have two questions: 


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*How do we construct the big field that contains both $K_1$ and $K_2$?

*The quoted comment said that $K_1$ and $K_2$ must be algebraic extensions...why is this?
 A: The first thing to note here is that $K_1$ and $K_2$ must be algebraic extensions of the same base field.  There's no way we can find a field $L$ such that both $\mathbb{F}_9$ and $\mathbb{Q}(\sqrt{2})$ are subfields of $L$.
So let's assume that $K_1$ and $K_2$ are both algebraic extensions of some base field $F$.  The general idea here is that $K_1 = F(S_1)$ and $K_2 = F(S_2)$, where $S_1$ and $S_2$ are (possibly countably infinite) collections of roots of polynomials in $F[x]$.  Because the algebraic closure of $F$ is constructed by adjoining all the roots of every possible polynomial in $F[x]$ to $F$, we must have $F(S_1) \subset \overline{F}$, and likewise for $F(S_2)$.  In many cases, we do not have to go all the way to the algebraic closure; notice that both $F(S_1)$ and $F(S_2)$ will be contained in $F(S_1 \cup S_2)$.  For a concrete example, $\mathbb{Q}( \sqrt{2})$ and $\mathbb{Q}( \sqrt{3})$ are both contained in $\mathbb{Q}( \sqrt{2}, \sqrt{3})$.
Technically, it seems that one can find a larger field containing both $K_1$ and $K_2$ as long as these fields have the same characteristic -- i.e. both have $\mathbb{Q}$ or both have $\mathbb{Z}_p$ (for the same $p$) as a subfield.  Even if $K_1$ and $K_2$ are not strictly algebraic extensions, they will nevertheless be generated by (i.e. have a basis of) some collection of algebraic elements together with some collection of transcendental elements.  One can find a field containing both $K_1$ and $K_2$ by considering the field generated by the union of all of the algebraic / transcendental generators for both fields.  

Recall that we can think of extensions of a field $F$ as a vector space over $F$.  Because of this, we are able to make analogies with vector spaces we're comfier working with to help grasp the concept:
Suppose we have two vector spaces $S_1$ and $S_2$, which are subspaces of $\mathbb{R}^{10}$.  Let's denote the "standard basis vectors" as $\mathbf{e}_k$, which have all zeros except for a $1$ in the $k^\text{th}$ slot.  Let $S_1$ be the subspace of $\mathbb{R}^{10}$ with the basis $\mathbf{e}_1, \mathbf{e}_5, \text{ and } \mathbf{e}_9$, and let $S_2$ be the subspace with basis $\mathbf{e}_2$ and $\mathbf{e}_3$.  That is to say:
$$S_1 = \text{Span} \Big( \mathbf{e}_1, \mathbf{e}_5, \mathbf{e}_9 \Big)$$
$$S_2 = \text{Span} \Big( \mathbf{e}_2, \mathbf{e}_3 \Big)$$
One can ask: "What is the smallest subspace of $\mathbb{R}^{10}$ that contains both $S_1$ and $S_2$?".  It isn't too hard to convince ourselves that the answer will be the subspace spanned by $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3, \mathbf{e}_5, \text{ and } \mathbf{e}_9$.  In other words, the subspace requested has as its basis elements the union of the basis elements from the two original subspaces $S_1$ and $S_2$.
A: After some thought, I don't think my question is too complicated.  The first thing to note (as was pointed out in the comments section and by Kaj Hansen) is that the fields $K_1$ and $K_2$ must both be extensions of the same base field $F$.  
Now to answer the two questions I originally asked:


*

*How do we construct the big field that contains both $K_1$ and $K_2$?
Since $K_1 / F$ is algebraic over $F$, we must have $K_1 \subseteq \overline{F}$, where $\overline{F}$ is one choice of algebraic closure for $F$ (there are many choices, but they are all isomorphic).  If we stick with this same choice of algebraic closure, then $K_2 \subseteq \overline{F}$.  So $\overline{F}$ is the "big field".


*The quoted comment said that $K_1$ and $K_2$ must be algebraic extensions...why is this?
I think it is still possible to construct a big field extension of $F$ containing $K_1$ and $K_2$ if $K_1$ and $K_2$ are not both algebraic over $F$.  But it is more straightforward if $K_1$ and $K_2$ are algebraic over $F$, since we can use $\overline{F}$ as our big field.   
