# Fundamental Theorem of Algebra, via algebra

I know there are already a couple of questions on this on the site, but I haven't seen an answer to this particular form...

We know, from the Fundamental Theorem of Algebra, that the complex algebraic numbers contain a unique maximal ordered subfield, namely the real algebraic numbers, and the complex algebraic numbers are obtained by adjoining a square root of $$-1$$ to the real algebraic numbers.

This is a purely algebraic statement (I think!) and one could reasonably ask for a purely algebraic proof. Is there such a proof? Or is there a barrier to such a proof (such as a model of ZF where the statement is false)?

I'm aware of the standard (purely algebraic, or at least I'm happy to call it so) Zorn's Lemma proof that the complex algebraic numbers exist, since any field $$K$$ has an algebraic closure and this is unique up to isomorphism. However the 'up to isomorphism' here includes a great many isomorphisms of the complex algebraic numbers that fix the rationals. I think a more or less equivalent question to what I am asking is: from this construction, is there a good algebraic way to pick out the privileged automorphism $$i\to-i$$ and put an order on the subfield this automorphism fixes?

There do exist proofs of the usual Fundamental Theorem of Algebra which use a $$`$$minimal' amount of analysis, such as https://arxiv.org/abs/1504.05609 (due to Piotr Błaszczyk) which doesn't really use more analysis than required to define the reals. However even this is more than (I think!) I am asking for.

• It isn't a purely algebraic statement since you can't define the complex numbers in a purely algebraic fashion. The complex numbers are much bigger than an algebraic closure of the rationals. Commented Dec 18, 2019 at 15:28
• I agree with the other comment. FTA is by far more of an analytic statement than an algebraic one because $\mathbb C$ is an analytic object more so than an algebraic one. However, I am interested in your other question, which I would phrase as asking to show algebraically that an algebraically closed field in characteristic 0 has an order 2 automorphism. Commented Dec 18, 2019 at 15:35
• Constantin-Nicolae - I don't see this; for example I can happily show $\sqrt{2}$ isn't rational, and adjoin an element $x$ such that $x^2-2=0$, to obtain a bigger field than $\mathbb{Q}$, without having to construct the reals. If you give me anything isomorphic to a subfield of the reals, then I know (but I only know via analysis) I can put an order on that field; if you give me a field containing a solution to $x^2+1=0$ then I can (trivially and algebraically) prove there is no order on the field. Commented Dec 18, 2019 at 16:24
• We have the following statement: Every polynomial over a field $K$ has a root in $K(\sqrt{-1})$. Then FTA says the statement is true for $K=\mathbb R$. Obviously, it's not true for every field $K$, we need some properties. A minimal set of properties would be: (1) $X^4-a^2$ has a root in $K$ for every $a\in K$, (2) Every polynomial in $K[X]$ of odd degree has a root in $K$, (3) $K$ is not finite. (Note that (1) simply states that either $a$ or $-a$ has a square root in $K$.) Commented Dec 18, 2019 at 17:19
• One relevant (negative) result is that there is no first-order definition of the real algebraic numbers in the structure $(\mathbb{C}^{alg},+,\cdot)$.
– user44143
Commented Dec 18, 2019 at 17:50

Despite the extensive discussion in the comments, I'm not sure I completely understand the question, but I think the following fact may settle the question in the negative.

Work of Läuchli implies that there is a model of ZF in which $$\mathbb Q$$ has an algebraic closure with no real-closed subfield. For more details, see for example Wilfrid Hodges, "Läuchli's algebraic closure of $$Q$$," Math. Proc. Cambridge Philos. Soc. 79 (1976), 289–297.

I think your desired conclusion requires every algebraic closure of $$\mathbb Q$$ to have a real-closed subfield. If you agree, then the above result shows that you at least need some weak form of AC, which arguably is "not algebraic." But this depends on what you mean by "algebraic"; maybe a weak form of choice doesn't automatically disqualify a proof from being "algebraic"?

• Some questions. The statement: $\forall P\in \mathbb{Q}[X], \exists a\in \mathbb{Q}[i] \text{ such that } |P(a)|\le 1$ 1) is it "purely algebraic"? 2) is it true in Läuchli's model? 3) is it then provable in a "purely algebraic" way? (No AC, or whatever "purely algebraic" may mean...) That to me would amount as close as can be to an algebraic proof of the Fundamental Theorem. Commented Dec 19, 2019 at 9:50
• @YaakovBaruch, what to call algebraic is a matter of personal opinion, but I personally would not use the phrase "purely algebraic" for $|P(a)|\le 1$.
– user44143
Commented Dec 21, 2019 at 5:05
• @MattF. : Why not? Suppose we write it this way: $\exists (p+qi) \in {\mathbb Q}[i]$ such that $p^2 + q^2 \le 1$. The only objection I can think of is that the order relation on the integers is "not algebraic." The OP seems to think that Zorn's lemma is okay, so why not the order relation on the integers? It seems a bit of a stretch to call the order relation on the integers "analytic". Commented Dec 21, 2019 at 16:48
• @TimothyChow, I like the clarity of first-order logic. So I would call order a part of the algebra of $\mathbb{R}$ (because $a\ge b \leftrightarrow \exists c (a=b+c^2)$) or $\mathbb{Q}$ or $\mathbb{Z}$ (because of the four-squares definition). But I would not call norms or conjugates part of the algebra of $\mathbb{C}=\mathbb{R}[i]$ or $\mathbb{Q}[i]$ or $\mathbb{Z}[i]$, where they are not definable from the ring operations.
– user44143
Commented Dec 21, 2019 at 18:59
• Update: Norms actually can be defined algebraically for $\mathbb{Z}[i]$ and $\mathbb{Q}[i]$. This follows from Julia Robinson's 1958 result (in "The Undecidability of Algebraic Rings and Fields") that $\mathbb{Z}$ can be defined in $\mathbb{Z}[i]$, and $\mathbb{Z}[i]$ can be defined in $\mathbb{Q}[i]$. The first definition is even short: $n$ is an integer iff (quantifying over $\mathbb{Z}[i]$) $$\exists f,g,h(\tau(0,f,g,h)\wedge\forall a(\tau(a,f,g,h)\implies a=n\vee\tau(a+1,f,g,h)))$$ $$\text{where }\tau(a,f,g,h)\text{ abbreviates }(f\neq 0)\wedge (a+1|f)\wedge (a+2|f)\wedge (1+ag|h)$$
– user44143
Commented Dec 23, 2019 at 6:46