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Is it known whether the Fundamental Theorem of Algebra is a theorem or non-theorem of the first-order theory of the complex field (i.e. $\mathbb{C}$ together with $+,\times,0,1$)?

Every proof I've seen uses some topological properties of the plane, but I was wondering if anyone had answered the question of their necessity from this viewpoint.

EDIT: I see now that the way I have worded this question makes it either unanswerable, or trivial (after all, the FTA is a true theorem of the complex field, hence, in the theory). What I think I was originally curious about is the following, motivated by the question of whether there exists a purely algebraic proof of the FTA, can each sentence in the following schema, for each $n$,

$$\forall x_0\forall x_1\cdots\forall x_n\exists z (x_nz^n+\cdots+x_1z+x_0=0)$$

be given first-order proofs in some reasonable extension of the first-order axiomatization of fields, which would somehow characterize $\mathbb{C}$, however, there are obvious obstacles to the second part of this statement (via Loewenheim-Skolem Theorems). I will have to rethink my curiosity. Thanks.

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How about starting with a statement of the FTA in the language of +,×,0,1 ... ??? Such a statement, if it exists, would belong to the theory, at least in the way I understand that word. –  GEdgar Jan 7 '12 at 1:06
@ismythe: What is the first-order theory of the complex field? As you know, no first-order theory with infinite models can be absolutely categorical... –  Zhen Lin Jan 7 '12 at 1:16
You are right, there is an issue with my wording, as the full statement of the FTA requires a natural number quantifier. The next best thing would be to consider, for each $n$, the sentence (taking $z,x_0,x_1,\ldots$ as my list of variables in the usual language of fields): $\forall x_0\forall x_1\cdots\forall x_n \exists z (x_nz^n+x_{n-1}z^{n-1}+\cdots+x_1z+x_0 = 0)$ –  Iian Smythe Jan 7 '12 at 1:18
Also, see my edit. –  Iian Smythe Jan 7 '12 at 1:40

2 Answers 2

First off, the Fundamental Theorem of Algebra isn't even a statement in the stated language. However, you can express it as a schema -- roughly speaking, as the collection of statements "every degree-$n$ polynomial has a root" for each positive integer $n$.

So, assuming you mean this theorem schema, the precise question you asked can be answered trivially: "yes", by the very definition of what "theory of the complex field" means -- it is the first-order theory whose theorems are those statements true about the complex field.

The question I think you mean to ask might look something like writing down an axiomatic presentation of some theory, and asking if you could prove the fundamental theorem (schema) of algebra for that theory.

As it turns out, I expect the most common axiomatization I would see for a theory like that would be:

  • The field axioms
  • One axiom for each arithmetic fact about complex numbers
  • The fundamental theorem schema of algebra

The fundamental theorem schema of algebra is a theorem of this theory, but the proof is trivial.

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Yes, that is what my original intention was, but I see now that I cannot really formulate the question I had in mind without making it trivial. –  Iian Smythe Jan 7 '12 at 1:40
The fundamentals of logic are sort of weird. If you don't know them, the subject seems deep and mysterious. But then when you take your first real look at them, it seems trivial and devoid of content. But once you use it to reorganize the way you think about things, things seem deep and interesting again. :) –  Hurkyl Jan 7 '12 at 2:00

When we write down the axioms for an algebraically closed field of characteristic $0$, we ordinarily include a specific infinite list $(A_n)$ of axioms, one for each $n \ge 1$. For any positive integer $n$, axiom $A_n$ asserts that every polynomial of degree exactly $n$ has a root. It is not difficult to write down, for any $n$, a sentence $A_n$ that does the job.

So every instance of the Fundamental Theorem of Algebra is a theorem of the first-order theory of algebraically closed fields of characteristic $0$, for the simple reason that it is one of the usual axioms of the theory.

The usual formulation of the Fundamental Theorem is not a sentence in our first-order language. We cannot in our language quantify over degree.

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