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Among the many techniques available at our disposal to prove FTA, is there any purely algebraic proof of the theorem?

That seems reasonably unexpected, because somehow or the other we are depending on the topological nature of $R$, and Wikipedia supports the claim in these statements: "In spite of its name, there is no purely algebraic proof of the theorem, since any proof must use the completeness of the reals (or some other equivalent formulation of completeness), which is not an algebraic concept."

Is it a proven fact that no pure algebraic proof is possible?

And so if I assume we are relying on the topological properties of $R$ and $C$ to prove the theorem, then given any arbitrary field how can one test whether it is algebraically closed or not?

Again in Wikipedia I found this result stated, "The classic example is the theory of algebraically closed fields of a given characteristic. Categoricity does not say that all algebraically closed fields of characteristic 0 as large as the complex numbers C are the same as C; it only asserts that they are isomorphic as fields to C. It follows that although the completed p-adic closures Cp are all isomorphic as fields to C, they may (and in fact do) have completely different topological and analytic properties." so i now want to rephrase my question as, given any field with different topological properties than C and which is in no simple way isomorphic to $C$ how can we generalize the proof of FTA to check whether FTA is valid on those fields?

And what about characteristic p fields? i can see an example that algebraic closure of $F_p((t))$ is an example of infinite, characteristic p field that is by construction closed, but if we had in our arsenal other ways to describe the field and suppress the fact that it is algebraic closure of another field then how can one prove that it is algebraically closed?

I don't understand anything about categoricity and such things, and I was only interested in the result taken, and I am sorry if it is a repost.

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That depends on what you mean by "pure algebraic proof." There is a proof in which the only topological input needed is the intermediate value theorem, and only to prove that polynomials of odd degree always have roots (see…). Is that pure enough? – Qiaochu Yuan Jul 3 '12 at 4:27
In general, to prove a field is algebraically closed, you simply... have to prove it. There's no shortcut. You just do it. – Qiaochu Yuan Jul 3 '12 at 4:28
Sometimes the fact that any two algebraically closed fields of characteristic $0$ are elementarily equivalent enables us to prove a result for all algebraicallly closed fields of charaacteristic $0$ by using topological properties of the complex numbers. However, that does not seem to be relevant here. – André Nicolas Jul 3 '12 at 5:09
I know of a professor who objects to the name "fundamental theorem of algebra" on the grounds that it is a misnomer and will therefore confuse people. – Michael Hardy Jul 3 '12 at 5:49
It also depends on what we take "algebra" to mean. Back when algebra meant the art of finding unknowns, the FTA was indeed a fundamental theorem of algebra. It is modern/abstract algebra that it is not a theorem of. – Henning Makholm Jul 3 '12 at 11:47

No, there is no purely algebraic proof of FTA. So, as someone already noted, FTA is a misnomer.

I think the following proof is one of the most algebraic ones, though it's not purely algebraic.

Assumptions We assume the following facts.

(1) Every polynomial of odd degree in $\mathbb{R}[X]$ has a root in $\mathbb{R}$.

(2) Every polynomial of degree 2 in $\mathbb{C}[X]$ has a root in $\mathbb{C}$.

Note: (1) can be proved by the intermediate value theorem. (2) can be proved by the fact that every polynomial of degree 2 in $\mathbb{R}[X]$ has a root in $\mathbb{C}$.

Notation We denote by $|G|$ the order of a finite group $G$.

Lemma Let $K$ be a field. Suppose every polynomial of odd degree in $K[X]$ has a root in K. Let $L/K$ be a finite Galois extension. Then the Galois group $G$ of $L/K$ is a 2-group.

Proof: We can assume that $L \neq K$. By the theorem of primitive element, there exists $\alpha$ such that $L = K(\alpha)$. By the assumption, the degree of the minimal polynomial of $\alpha$ is even. Hence $|G|$ is even. Let $|G| = 2^r m$, where $m$ is odd.

Let $P$ be a Sylow 2-subgroup of $G$. Let $M$ be the fixed subfield by $P$. Since $(M : K) = m$ is odd, $m = 1$ by the similar reason as above. QED

The fundamental theorem of algebra The field of complex numbers $\mathbb{C}$ is algebraically closed.

Proof: Let $f(X)$ in $\mathbb{R}[X]$ be non-constant. It suffices to prove that $f(X)$ splits in $\mathbb{C}$. Let $L/\mathbb{C}$ be a splitting field of $f(X)$. Since $L/\mathbb{R}$ is a splitting field of $(X^2 + 1)f(X)$, $L/\mathbb{R}$ is Galois. Let $G$ be the Galois group of $L/\mathbb{R}$. Let $H$ be the Galois group of $L/\mathbb{C}$.

By the assumption (1) and the lemma, $G$ is a 2-group. Hence $H$ is also a 2-group. Suppose $|H| > 1$. Since $H$ is solvable, $H$ has a nomal subgroup $N$ such that $(H : N) = 2$. Let $F$ be the fixed subfield by $N$. Since $(F : C) = 2$, this is a contradiction by the assumption (2). Hence $H = 1$. It means $L = \mathbb{C}$. QED

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Thanks for the proof, so are the two analysis assumptions or there equivalent (or advanced) forms absolutely necessary for any proof of the theorem? What I mean to ask is whether it is a proven fact that you have to make some basic assumptions outside the realms of algebra or algebraic constructions to prove the theorem??? – Aranya Lahiri Jul 4 '12 at 4:11
If you want to prove it formally, you have to define exactly what an algebraic proof is. Is that what you want? – Makoto Kato Jul 4 '12 at 6:32
Okay, to define everything formally I will run into logical difficulties I suppose?? But from an intuitive point of view my question is what essential properties of R are we using here ,and how much of can that be generalized and(or) relaxed? And instead of mean value theorem I think you mean intermediate value theorem right?? so intermediate value theorem can be generalized to other spaces as well, can we use that fact to expand the proof further? – Aranya Lahiri Jul 4 '12 at 7:32
My proof is valid on any field F which satisfies the following conditions: Let F be a field of characteristic 0. Let K/F be a finite Galois extension. (1) Every polynomial of odd degree in F[X] has a root in F. (2) Every polynomial of degree 2 in K[X] has a root in K. Then K is algebraically closed. – Makoto Kato Jul 4 '12 at 11:57
This proof works for all real closed fields; in fact, characterizes them. – lhf Jul 4 '12 at 12:33

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