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 Fp((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.

**Answer**

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 R[X] has a root in R.

(2) Every polynomial of degree 2 in C[X] has a root in 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 R[X] has a root in 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≠K.

By the theorem of primitive element, there exists α such that L=K(α).

By the assumption, the degree of the minimal polynomial of α is even.

Hence |G| is even.

Let |G|=2rm, 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 C is algebraically closed.

Proof:

Let f(X) in R[X] be non-constant.

It suffices to prove that f(X) splits in C.

Let L/C be a splitting field of f(X).

Since L/R is a splitting field of (X2+1)f(X), L/R is Galois.

Let G be the Galois group of L/R.

Let H be the Galois group of L/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=C.

**QED**

**Attribution***Source : Link , Question Author : Aranya Lahiri , Answer Author : Makoto Kato*