Timeline for Fundamental Theorem of Algebra, via algebra
Current License: CC BY-SA 4.0
30 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2020 at 14:31 | comment | added | David Lampert | @MattF. $d$ is an odd number. | |
| Jan 7, 2020 at 18:11 | comment | added | Timothy Chow | For some further comments, see the Foundations of Mathematics mailing list, specifically the thread about Artin-Schreier for Q. | |
| Jan 5, 2020 at 22:08 | comment | added | Timothy Chow | @user36212 : I'm feeling more confident that you can construct something that has all the properties you want, without AC and without "analysis." Even without AC, one can prove that all countable algebraic closures of $\mathbb Q$ are isomorphic. If we construct an algebraic closure by lexicographically ordering $\mathbb Q$ and then lexicographically ordering polynomials over $\mathbb Q$, adjoining roots one by one, then the algebraic closure is countable by construction, so no funny business can occur. | |
| Jan 5, 2020 at 17:47 | comment | added | David Lampert | Maybe this statement can be proved in arithmetic (by repeated interval bisection)? $\forall N,d \in \mathbb{N}, a_1,...,a_d \in \mathbb{Z}, d$ odd $\implies \exists m,n,t_1,..,t_4 \in \mathbb{Z}$ such that $t_1^2+t_2^2+t_3^2+t_4^2+N^2(m^d+a_1nm^{d-1}+...+a_dn^d)^2 = n^{2d}$ (meaning $|(f(m/n)| \leq 1/N$). | |
| Jan 5, 2020 at 12:54 | comment | added | user36212 | Thanks! I can believe Artin-Schreier might be choice free; it certainly looks like algebra rather than analysis. But doesn‘t it leave the possibility that the constructed closure doesn‘t have any finite index real closed subfield? (Actually this must be what happens in the Läuchli model you referenced, so one needs to see somehow that the construction of a closure doesn‘t accidentally pick this closure) | |
| Jan 4, 2020 at 19:18 | comment | added | Timothy Chow | I think one just needs to walk through the proof of the Artin-Schreier theorem and check that in the specific case of $\mathbb Q$, one doesn't need Zorn. | |
| Jan 4, 2020 at 19:10 | comment | added | Timothy Chow | @user36212 : This math.SE question explains how to construct an algebraic closure of $\mathbb Q$ without Zorn's lemma. The main point is that $\mathbb N$ is already given to us with a well ordering, so we can well order $\mathbb Q$ and all polynomials with rational coefficients without having to ask Zorn to do it for us. The part I'm not immediately sure about is proving the existence of the real closed subfield you want. I suspect this is straightforward but there are details to check. | |
| Jan 4, 2020 at 11:45 | comment | added | user36212 | I agree that use of choice is probably not really what I want to exclude with 'algebraic'; I think I really would like to avoid having to construct the reals or otherwise do something morally equivalent along the way. I certainly agree that as far as the students are concerned I won't ever need to know this - but I would still like to understand it better. | |
| Jan 4, 2020 at 11:42 | comment | added | user36212 | This is a good point. OK, I guess I would like to know the existence of an algebraic closure of $\mathbb{Q}$, which is countable, and which contains an ordered subfield such that adjoining a square root of $-1$ we get the entire field. Your answer says I can't hope that all such algebraic closures have this structure without using some choice, so at least I'd better be paying attention to the construction... I don't immediately see how to construct an algebraic closure without any choice; don't I at some point need choice to decide which polynomial to adjoin a root of next? | |
| Jan 3, 2020 at 21:39 | comment | added | Timothy Chow | @user36212 : It might be worth stating more precisely what you mean by "go from the rationals directly to a countable algebraic closure with the structure we see in the Argand plane." This is a somewhat different question from what I thought you were asking. It is provable in a very logically weak system that there exists an algebraic closure of $\mathbb Q$, because you can just construct it directly from polynomial equations. It's only uniqueness that requires some form of choice. Proving that this specific algebraic closure has the properties you want may not require "analysis". | |
| Jan 3, 2020 at 19:33 | comment | added | user36212 | Of course, no student ever asked me anything close to this kind of thing; realistically, they're probably happy enough if they think they more or less understand how the standard proof works. That doesn't stop me from wanting to know what answer I might give such a hypothetical student. | |
| Jan 3, 2020 at 19:27 | comment | added | user36212 | Or to put it another way: can we go from the rationals directly to a countable algebraic closure with the structure we see in the Argand plane, namely there are two ordered axes, without having to go via something uncountable? Of course what the students would like is a proof that just uses a bunch of high-school algebraic manipulations as with the quadratic formula, which I can happily tell them doesn't exist and go read Abel if they want to know more. But that's not really ruling out a more serious 'algebraic' proof. | |
| Jan 3, 2020 at 19:18 | comment | added | user36212 | Sure; and that's how I stated it too. But then I started thinking about how I would answer questions along this line... Like most such courses we build up to this via integers / rationals / (sloppily) reals, and make fairly heavy use of the order along the way, and then introduce the complex numbers as what you get when you have the reals and want to solve $x^2+1=0$, and then magic occurs that in fact this turns out to be algebraically closed. Now the magic has really already occurred when we construct the reals here, so (at least I think) it's natural to ask whether we need that. | |
| Jan 3, 2020 at 16:13 | comment | added | Timothy Chow | @user36212 : In a 1st-year undergrad context, surely FTA just means "$\mathbb C$ is algebraically closed" rather than the convoluted "purely algebraic" statements we are struggling to devise? If so, then it would seem that the trivial answer suffices: analysis is needed even to state FTA. | |
| Jan 3, 2020 at 12:19 | comment | added | user36212 | These students are for sure not going to be happy with Zorn's lemma (not least because they don't know anything like axiomatic set theory) and hence would not likely want to know about different models of ZF, but at least I am happy that I could answer the hypothetical question now... | |
| Jan 3, 2020 at 12:14 | comment | added | user36212 | @Timothy Chow - my interest here is that I'm happy to teach (first year undergrad) students some standard analytic proof of FTA, but I've never been quite sure what I should answer (beyond 'well, I don't know how to avoid it') if a student asked why we need analysis to prove this; not that one ever did. So your answer would let me say: even if we are generous enough to assume there is an algebraic closure of the rationals, we can't prove that this algebraic closure looks like the complex numbers without doing something that (these) students will identify as not like algebra they know. | |
| Dec 23, 2019 at 14:37 | comment | added | user44143 | @YaakovBaruch, I’d be interested to see what that statement and that proof would look like! It probably goes easier as a statement about homogeneous polynomials in two variables with coefficients in $\mathbb{Z}[i]$. | |
| Dec 23, 2019 at 13:40 | comment | added | Yaakov Baruch | @MattF. This result seems quite amazing. Is there any remaining obstruction to a potentially purely algebraic proof of the above? (Non-constant rational polynomials on $\mathbb{Q}[i]$ taking values arbitrarily close to 0.) | |
| Dec 23, 2019 at 6:46 | comment | added | user44143 | 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 $\mathbb{Z}[i]$) $$\exists f,g,h(\tau(0,f,g,h)\wedge\forall a(\tau(a,f,g,h)\implies a=n\vee\tau(a+1,f,g,h)))$$ $$\text{where }\tau(a,f,g,h)\text{ abbreviates }(f\neq 0)\wedge (a+1|f)\wedge (a+2|f)\wedge (1+ag|h)$$ | |
| Dec 22, 2019 at 3:57 | comment | added | Timothy Chow | @MattF. : Okay, I think I see what you mean. The question then seems to be a (near-)duplicate of this question on math.stackexchange. In particular, it remains unclear what "the fundamental theorem of algebra" even means in such a framework, let alone whether it can be proved. | |
| Dec 21, 2019 at 21:03 | comment | added | Yaakov Baruch | @TimothyChow: norms even allow to do away with order without existence predicates, since $|a|\le 1$ is equivalent to $|a|+|1-|a||=1$. So I now agree with Matt F. that norms lead into a non-algebraic landscape. | |
| Dec 21, 2019 at 18:59 | comment | added | user44143 | @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. | |
| Dec 21, 2019 at 18:08 | comment | added | Yaakov Baruch | @MattF. I'm inclined to agree with you. I was myself thinking that the order relation is the spoiler, and I cannot think of a statement that avoids using it while still managing to capture the spirit of the Fund. Thm. | |
| Dec 21, 2019 at 16:48 | comment | added | Timothy Chow | @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". | |
| Dec 21, 2019 at 5:05 | comment | added | user44143 | @YaakovBaruch, what to call algebraic is a matter of personal opinion, but I personally would not use the phrase "purely algebraic" for $|P(a)|\le 1$. | |
| Dec 19, 2019 at 9:50 | comment | added | Yaakov Baruch | 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. | |
| Dec 19, 2019 at 2:46 | comment | added | Timothy Chow | @user36212 : Although you've accepted my answer, I feel that I should mention a couple of caveats. Without AC, one cannot even prove that all algebraic closures of $\mathbb Q$ are isomorphic. Conversely, if we allow enough choice to prove uniqueness, then it sounds like you still feel that everything is "algebraic." So maybe AC is not really getting at your question? What's wrong with this: construct a specific algebraic closure, pick out $i\to-i$, and then carry it over to any other algebraic closure using AC? This doesn't use any "analysis" does it? | |
| Dec 18, 2019 at 23:31 | comment | added | user36212 | Thanks! I think that's exactly what I was looking for. As you say, it would still be interesting to know how much of AC one needs to kill models of this type; and it would be interesting to know how 'analytic' the proof has to get (which, I think is an I know-it-when-I-see-it question). | |
| Dec 18, 2019 at 23:26 | vote | accept | user36212 | ||
| Dec 18, 2019 at 22:47 | history | answered | Timothy Chow | CC BY-SA 4.0 |