I have the following theorem:
Let $k^a$ be the algebraic closure of $k$ and take $K_i$ to be a splitting field of $f_i$ in $k^a$. Then the compositum of the $K_i's$ is a splitting field for $\{f_i\}_{i\in I}$
My question is why do we need here the splitting field in $k^a$, because we have defined a normal splitting field to be an extension of $f$ such that $f$ splits into linear factors in it and we have never mentionned an algebraic closure. So I don't see why they use it here, is then $K_i\subset k^a$?
Thanks for your help
You did not tell us what book or other reference you are reading. More precisely, is the author actually using an algebraic closure for any other purpose than to see there is a splitting field?
You do not need algebraic closures to show a polynomial has a splitting field or to do Galois theory for finite extensions of fields. That you can construct a splitting field of a polynomial inside an algebraic closure does not mean you must use them. Just because an author mentions algebraic closures to get a splitting field doesn't mean you have to pay any attention to it.
However, I think I know what is going on. You are referring to an indexed set of polynomials $f_i$. If that is a finite set, then you could let $f = \prod f_i$, which is another polynomial, and a splitting field of $f$ will be a splitting field for all the $f_i$'s together. But what if $\{f_i\}$ is an infinite set of polynomials? How do you know there is a splitting field for all of them? For example, what is a splitting field over $\mathbf Q$ of all the polynomials $x^3 - nx - n$ where $n$ runs over the integers? To show there is a splitting field for an arbitrary set of polynomials, possibly being an infinite set, then it is understandable why an author might like to appeal to the algebraic closure and find the splitting field in there. That could be technically simpler than trying to build up an infinite abstract field extension from scratch that plays the role of a splitting field for all $f_i$.