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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 http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra). Is that pure enough?2012-07-03
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    In general, to prove a field is algebraically closed, you simply... have to prove it. There's no shortcut. You just do it.2012-07-03
  • 0
    *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.2012-07-03
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    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.2012-07-03
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    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.2012-07-03
  • 0
    @ Qiaochu Yuan , i know i am asking something vague, but can i have a brief worked out example where we are proving algebraic closedness of some field, or maybe highlight some general techniques for special cases....2012-07-03
  • 6
    How do you even define $\mathbb{C}$ without analysis? If you can't, then you can't even state the theorem, much less prove it.2012-12-16
  • 0
    [This](http://mathoverflow.net/a/25045) sounds interesting for the context.2015-12-29
  • 0
    I studied 4 different proof of this theorem: one made with complex analysis, one with elementary calculus, one with Galois theory (the one below) and finally, one purely algebraic, which made use of symmetric polynomials. But now I don't have no reference, unfortunately2016-01-18
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    ***Completeness*** is obviously not required for the proof because $\overline{\mathbb Q}$ satisfies FTA as well.2018-04-30

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