Timeline for Solving $z^n=a+bi$ using only radicals of positive real numbers
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 19, 2012 at 0:18 | comment | added | grp | @Hugo: Yes, you're right, that is what I meant to say. | |
| Sep 18, 2012 at 23:33 | comment | added | Hugo Chapdelaine | @grp, you still need to correct slightly your statement, if you take an irreducible cubic polynomial over $Q$ with 3 real roots then the splitting field does not include a root of unity of odd order. I think that what you meant was if $K$ is the splitting field of your polynomial over $F$ and if you embed $K$ in a radical extension, say $M$, then $M$ has to contain a root of unity of odd order. In fact the proof of this statement is straight forward from what I wrote in my second paragraph after Will's answer. | |
| Sep 18, 2012 at 21:24 | comment | added | grp | @Hugo: If you unravel the proof of my comment, you'll find that for the application to cyclotomic polynomials it is no different from the usual Galois theory argument for analyzing constructibility of polygons. So that particular example isn't as compelling an illustration as it may seem to be at first glance. | |
| Sep 18, 2012 at 21:12 | comment | added | Hugo Chapdelaine | I just found this 4 pages document from Brian Conrad which discusses aspects related to my question: see math.stanford.edu/~conrad/210BPage/handouts/radreal.pdf | |
| Sep 18, 2012 at 21:02 | comment | added | Hugo Chapdelaine | Hi @grp, that is quite a nice result. So in particuar using your observation we see that $z^n-1$ is positive solvable iff $\varphi(n)$ is a power of $2$ so iff $n=2^rq_1q_2\ldots q_s$ where the $q_i$'s are distinct Fermat's primes. | |
| Sep 18, 2012 at 20:48 | comment | added | grp | The restriction to degree 3 can be relaxed (and Cardano bypassed): if $f$ is an irreducible polynomial of any odd degree $> 1$ over a field $F$ of characteristic 0 then its splitting field over $F$ contains a root of unity of odd prime order and so cannot be embedded into $\mathbf{R}$. In particular, when $F$ is given as a subfield of $\mathbf{R}$ and the Galois group of $f$ is abelian over $F$ (e.g., cyclic), so the splitting field is generated over $F$ by a single root of $f$, then it is impossible to find any root of $f$ inside a radical tower over $F$ inside $\mathbf{R}$. | |
| Sep 18, 2012 at 17:32 | comment | added | Hugo Chapdelaine | I meant $a$ is a $p$-th power in $K$. | |
| Sep 18, 2012 at 17:31 | comment | added | Hugo Chapdelaine | Note also that the proof that the $3$ real roots cannot be written in terms of radicals of positive quantities follows from the key observation that for a prime number $p$ and an arbitrary field $K$, the polynomial $z^p-a$ with $a\in K$ is either irreducible or $a$ is a $p$-th in $K$. Therefore at the last step of the sequence of fields appearing in the tower one must take a cube root. Since the splitting field is normal and real this would mean that the all 3rd roots of unity are real which is a contradiction. | |
| Sep 18, 2012 at 17:14 | comment | added | Hugo Chapdelaine | Hi Will, indeed I knew about the casus irreducibilis which might one of the key reason (swing factor) for the acceptation by certain mathematicians of complex numbers for solving cubic polynomials. Since as you explained, when an irreducible real cubic polynomial has 3 real roots (so Galois group $\simeq C_3$) then none of the roots can be written in terms of radicals of positive quantities. This means that the cancellation of the imaginary part (which can only takes place in the complex world) is an unavoidable phenomenon when you shoehorn yourself to only allow the use of radicals. | |
| Sep 18, 2012 at 16:49 | history | answered | Will Jagy | CC BY-SA 3.0 |