Timeline for Fundamental Theorem of Algebra, via algebra
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2019 at 10:02 | comment | added | Yaakov Baruch | I posted a comment to Timothy Chow's answer that I think is also relevant to the discussion here. | |
| Dec 18, 2019 at 23:26 | vote | accept | user36212 | ||
| Dec 18, 2019 at 22:47 | answer | added | Timothy Chow | timeline score: 10 | |
| Dec 18, 2019 at 22:08 | comment | added | user44143 | @user36212, The reference is just to quantifier elimination: the only first-order definable subsets of $\mathbb{C}^{alg}$ are Boolean combinations of sets of the form $\{z: p(z)=0\}$, which are all either finite or co-finite, and therefore can’t be $\mathbb{R}^{alg}$. Meanwhile, it’s true that “$S$ admits an order” is not a first-order condition, but “the squares in $S$ define an order on $S$” is first-order. | |
| Dec 18, 2019 at 20:27 | comment | added | Wojowu | @user36212 No, what you describe is not possible. It follows from Artin-Schreier Theorem as stated e.g. here (Theorem 3.1). For the latter question, consider the field $\mathbb Q(i,\sqrt{-2})$. It has an automorphism which takes $i$ to $-i$ but fixes $\sqrt{-2}$. We can extend this automorphism to $\overline{\mathbb Q}$. | |
| Dec 18, 2019 at 19:58 | comment | added | user36212 | @Wojowu - and is it possible that there is an automorphism of the algebraics which swaps the roots of $x^2+1=0$, the fixed subfield doesn't have an order, and also adding $i$ back to the fixed subfield gives us the entire algebraics? (Actually I'd be very interested to see an example with only the first two of these three properties, do you have a reference?) | |
| Dec 18, 2019 at 19:16 | comment | added | user36212 | @Matt F. - that sounds interesting, do you have a reference? I guess ‘admitting an order’ is not a first order condition, but still. | |
| Dec 18, 2019 at 19:14 | comment | added | user36212 | @Alexandre Eremenko - what I’d like to say is a definition is: the subfield admitting an order and such that adding $i$ we get the algebraic closure of $\mathbb{Q}$. If this isn’t a definition (I.e. there are multiple such subfields) then you’ve answered my question in the negative. Or at least this is the main thing I’d like to know; I might come up with something more to ask if your examples suggest I should just add another algebraic condition to the above ‘definition’. | |
| Dec 18, 2019 at 18:46 | comment | added | Constantin-Nicolae Beli | @Wojowu My mistake. I thought you meant (3) follows from (1), but you actually said it follows from (2), which is right. | |
| Dec 18, 2019 at 18:33 | comment | added | Constantin-Nicolae Beli | @Wojowu You are right, (3) is superfluous, but it doesn't always follow from (1). If $K={\mathbb F}_q$, where $q$ is a prime power, $q\equiv 3\pmod 4$, then $-1$ is not a square in $K$ so for every $a\in K^*$ exactly one of $\pm a$ is a square. If $q\equiv 1$ then for half of the values of $a\in K^*$ both $\pm a$ are squares and for the other half neither of them is, so (1) fails. In all cases though, (3) follows from (2). If $P=X^q-X+1$, with odd degree $q$, then for every $a\in K$ we have $P(a)=1$ so $P$ has no roots in $K$ so (2) fails. | |
| Dec 18, 2019 at 18:33 | comment | added | Constantin-Nicolae Beli | @Wojowu I forgot to mention that $F$ is not of characteristic $2$. In characteristic $2$ we have equations of the type $X^2+X+a=0$, which have roots in the so called Artin-Schreier extensions. | |
| Dec 18, 2019 at 17:50 | comment | added | user44143 | One relevant (negative) result is that there is no first-order definition of the real algebraic numbers in the structure $(\mathbb{C}^{alg},+,\cdot)$. | |
| Dec 18, 2019 at 17:50 | comment | added | Wojowu | @Constantin-NicolaeBeli I believe your (3) follows from (2). | |
| Dec 18, 2019 at 17:19 | comment | added | Constantin-Nicolae Beli | 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 $K$.) | |
| Dec 18, 2019 at 17:07 | comment | added | LSpice | A purely algebraic definition of $\mathbb C$ is "an algebraically closed, characteristic-0 field of cardinality the continuum." (Inside here we can pick out the algebraic numbers as the algebraic closure of the embedded copy of $\mathbb Q$.) This is also a definition of $\overline{\mathbb Q_\ell}$. If we could pick out $\mathbb R$ canonically inside $\mathbb C$, what would we be picking out inside $\overline{\mathbb Q_\ell}$? (Just thinking out loud.) | |
| Dec 18, 2019 at 16:43 | comment | added | Alexandre Eremenko | To make an algebraic statement with algebric proof you first have to DEFINE "real algebraic numbers" without using complex numbers. | |
| Dec 18, 2019 at 16:33 | comment | added | Wojowu | @user36212 Actually, it need not be true that an automorphism of a field swapping the two roots of $x^2+1$ has totally orderable fixed field. This of course happens for the complex conjugation, but needn't happen to other such automorphisms. | |
| Dec 18, 2019 at 16:30 | comment | added | user36212 | @Wojowu - good point. I'm tempted to say this should follow from the fact that the algebraic closure of $\mathbb{Q}$ is the same as the field of $\mathbb{Q}$-algebraic numbers. But either this is so obviously true I don't see why, or there is a gap... In any case, I also don't see (maybe due to my stupidity) how to put an order on the subfield this automorphism fixes without invoking some analysis. This is what I would really like to be able to do, because for me a fundamental characteristic of $\mathbb{R}$ and its subfields is that there is an order. | |
| Dec 18, 2019 at 16:24 | comment | added | user36212 | 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. | |
| Dec 18, 2019 at 16:17 | comment | added | Wojowu | @user36212 This only guarantees that there is an automorphism which swaps $i$ with $-i$, but not that this automorphism has order $2$ (i.e. $\sigma(\sigma(x))=x$ for all $x$ in the field). As some food for thought, one can show that there cannot be any automorphism of finite order larger than $2$, so however you want to show that an order $2$ automorphism exists, you have to somehow make significant use of the number $2$. | |
| Dec 18, 2019 at 16:13 | comment | added | user36212 | @Wojowu - I'm not exactly an expert, nor did I think too much - but isn't it more or less obvious that there is such an automorphism, precisely because $x^2+1=0$ has two solutions which are necessarily conjugate and not in the base field? But this applies equally well to $x^2-2=0$. The point is that the first automorphism fixes a subfield which we happen to know is ordered, whereas (I think - actually one should maybe check this properly!) the second does not. Again, though, the only way I know how to order the first subfield goes via constructing $\mathbb{C}$. | |
| Dec 18, 2019 at 16:10 | comment | added | Constantin-Nicolae Beli | Completely without analysis you can't. Otherwise you could prove that every equation with rational coefficients has solutions in ${\mathbb Q}(i)$. The very definition of $\mathbb R$ involves some analysis. The one "most algebraic" proof I know uses two facts that require some analysis: (1) If $a>0$ then the equation $X^2-a=0$ has a real root. (2) Every polynomial of odd degree has a real root. If we write the degree of $P\in{\mathbb R}[X]$ as $n=2^km$, with $m$ odd, then the proof is done by induction on $k$. | |
| Dec 18, 2019 at 16:07 | comment | added | user36212 | I'm well aware that there are many more complex numbers than complex algebraic numbers - but that's the entire point of this question. The linked arxiv proof already shows that 'all' you need to prove FTA in the usual form is the Dedekind description of the reals, which is basically the same as constructing $\mathbb{C}$. But I don't see that talking about 'complex algebraic numbers' means you need to talk about the complex numbers as a whole. I agree it's slightly sloppy language, but, if you prefer, take it as simply meaning the algebraic closure of $\mathbb{Q}$. | |
| Dec 18, 2019 at 15:35 | comment | added | Wojowu | 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. | |
| Dec 18, 2019 at 15:28 | comment | added | Simon Wadsley | 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. | |
| Dec 18, 2019 at 14:51 | history | asked | user36212 | CC BY-SA 4.0 |