Fundamental Theorem of Algebra for fields other than $\Bbb{C}$, or how much does the Fundamental Theorem of Algebra depend on topology and analysis? When proving the Fundamental Theorem of Algebra, we need to appeal to analytic and/or topological properties of $\Bbb{C}$ and $\Bbb{C}[z]$. Is this going to be necessary in general, and if so, to what extent?
That is, suppose a field $K$ is given, and we desire to show that $K$ is algebraically closed. Is there any amount of purely algebraic data (save that $K = F^{alg}$ for some field $F$) that will allow us to say that $K$ is algebraically closed? Of course, the phrase "purely algebraic data" isn't well-defined, but I loosely mean information about $K$ in relation to fields $E_i$ of which $K$ is an extension, such as the degrees of the extensions $K/E_i$, if $K = E_i(\alpha)$ for some $\alpha\in K$, their Galois groups $\operatorname{Gal}\left(K/E_i\right)$, what $\operatorname{char}K$ is, and so on (where this information isn't precisely the information that $K$ is constructed as the algebraic closure of some field).
If this isn't possible, at what point does it become necessary to appeal to the topology and analysis of $K$ and $K[x]$, and how does this "point" depend on $K$? I realize that exactly how and where analysis and topology come into play for different $K$ will depend on the nature of the $K$ with which one is working, but I would be interested in knowing what properties of $K$ have the greatest effect on the necessity of analysis and topology in a proof that $K$ is algebraically closed. Of course, the more general these properties, the better.
A note: this question deals with this idea to some extent, although the discussion there is more focused on $\Bbb{C}$, and I would like to consider a more general setting: the conditions required to show a certain type of field is algebraically closed, and how those conditions differ for different types of fields.
 A: this is borrowed from the comments in the link to the question you posted:
Let $F$ be a field of characteristic $0$, and $K / F$ be a finite Galois extension. Suppose every polynomial of odd degree in $F[x]$ has a root in $F$, and every polynomial of degree $2$ in $K[x]$ has a root in $K$. Then $K$ is algebraically closed.
Also, it turns out that algebraically closures are rarely finite extensions. In fact, if $F$ is a field and $C / F$ a finite extension that is algebraically closed, then $C=F(i)$ where $i^{2}=-1$, and $F$ has characteristic $0$. Also for every nonzero $a \in F$, either $a$ or $-a$ is a square, and finite sums of nonzero squares are nonzero squares. The source for this is Keith Conrad's notes:
http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf
A: There’s one other way of getting an algebraically closed field: start with $\mathbb F_p=\mathbb Z/p\mathbb Z$, and adjoin an $m$-th root of unity for every natural number $m$ relatively prime to $p$. In other words, adjoin all possible roots of unity to $\mathbb F_p$. That’s the construction, and now, of course, there’s the exercise of showing that the result is indeed algebraically closed.
A: As is very common to note, the Fundamental Theorem of Algebra is not a theorem of algebra, but of the analysis of $\mathbb R$ and $\mathbb C$. It is the result that "by adjoining a solution to the equation $x^2+1=0$ and closing under field operations, once obtains precisely all solutions to all polynomials with coefficients in this new field". 
To address your second paragraph, I don't think there is a lot of hope here. There is no information in the extension $K:K$, and I don't see how information about the extensions over proper subfields will capture $K$ being algebraically closed. I don't have a good argument though or a particular convincing counter example. 
To address your third paragraph, there is immediately a big problem here. If $K$ is an arbitrary field, then there isn't necessarily any topology on $K$, so for general fields the question doesn't even jump-start. 
So, you need to have some topology present on a field $F$, and then ask if the topological properties of $F$ are strong enough to guarantee that adjoining a single solution to $x^2+1$ will result in an algebraically closed field. This seems horribly too general, but again, I don't have any clear argument or convincing examples. 
So, my answer doesn't actually answer anything, but rather tries to clarifies the situation, hoping others will contribute more insightful answers.  
