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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.

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Stahl
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    There are, frankly, very few fields that are algebraically closed other than by definition. Most algebraically closed fields are introduced as being algebraically closed. Exceptions are the complex numbers, for which you need analysis, and the union of the Laurent series fields ${\mathbf C}((X^{1/n}))$ over all $n \geq 1$, for which knowledge of ramification is used. Also the algebraic numbers are algebraically closed, but this is almost "by definition" if you grant the transitivity of algebraicity for field extensions, so it has a lot less meat to it than proofs for the other two fields. – KCd Apr 02 '13 at 03:12
  • I don't think there is a clean-cut answer to what properties have "the greatest effect" on proving the field is alg. closed if it isn't so by definition. – KCd Apr 02 '13 at 03:14
  • @KCd What do you mean when you say you need the Laurent series fields $\Bbb C((X^{1 / n}))$ to show $\Bbb C$ is algebraically closed? Or am I misunderstanding you? – Dylan Yott Apr 02 '13 at 03:25
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    @Dylan, KCd is posing (1) the complex numbers and (2) the union of Laurent fields, as two distinct examples of where algebraic closure arise naturally (not by definition, roughly) - he is not posing (2) as a requirement for seeing item (1). (This is seen by observing the comma structure in the sentence.) – anon Apr 02 '13 at 03:32
  • Okay, that makes a lot more sense. So what is the bold $C$? Is it the complex numbers? – Dylan Yott Apr 02 '13 at 03:51
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    Dylan, yes, ${\mathbf C}$ is the complex numbers. I despise blackboard bold in LaTeX. – KCd Apr 02 '13 at 04:40
  • $\mathbb{R}$ is defined by its order and Dedekind-completeness properties -- so it should be no surprise that if we are to prove something about $\mathbb{R}$ which doesn't follow from just the field and order axioms (other than the Dedekind one), and so of its derived (non-ordered but topologically related) field $\mathbb{C}$, we will have to invoke the Dedekind-completeness property at some point. And if you invoke Dedekind-completeness, you're doing analysis almost by definition. – The_Sympathizer Jun 21 '14 at 10:47

3 Answers3

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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

Dylan Yott
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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.

Lubin
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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.

Ittay Weiss
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    I disagree with your first sentence. There are purely analytic proofs of the FTA (every holomorphic map $\mathbb{P}^1\to\mathbb{P}^1$ is surj.) but the most illuminating proofs of the FTA are those that combine the alg. and analytic. For example, while the creation of $\mathbb{R}$ can be motivated almost entirely with an analytic flare (the completion of a metric space) the creation of $\mathbb{C}$ is, in the majority of instances, motivated by algebraic terms--wanting to solve equations. So, I don't think we can say the FTA is a theorem about either analysis or algebra, but one about both. – Alex Youcis Apr 02 '13 at 03:34
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    @AlexYoucis I see your point. I agree with the slightly schizophrenic nature of the theorem. While the sophisticated proofs all tell you something nice and illuminating about the theorem, I find the straightforward proof (basically, a maximum modulus analysis without any of the deep theorems (since when the function is a polynomial you don't need anything deep)) is most illuminating. I learned several proofs of the FDA, and several years later came across the simple proof. I now consider the sophisticated proofs as using a rocket to kill a fly. – Ittay Weiss Apr 02 '13 at 03:47