Talk:Root of unity/Archive 1

Note that, contrary to first appearances, not all coefficients of all cyclotomic polynomials are 1, −1, or 0; the first polynomial where this occurs is Φ105, since 105=3×5×7 is the first product of three odd primes.

It's not immediately clear to me why this property of 105 implies it's the first cyclotomic polynomial with coefficient not equal to 1, -1, or 0. Is this really easy to see, or something I've just forgotten? If it's pretty easy to show, then a quick justification would be good, if it's not very easy to show, I don't think the fact should be mentioned, it leaves the reader dangling.

The case $n=0$ is included in the definition, and it is stated that there are $n$ solutions to $z^{n}=1$. Actually the equation $z^{0}=1$ has infinitely many solutions, rather than zero. I'll exclude $n=0$ . Bo Jacoby 14:03, 20 September 2005 (UTC)

unitary matrix

"These different roots of unity can be arranged to form the elements of a unitary matrix, and are thus orthogonal to each other. "

Are you talking about the n n'th roots of unity, or about the primitive roots of unity ? Please show me how {+i,-i} is arranged into a unitary matrix. Bo Jacoby 13:34, 23 September 2005 (UTC)

That is a standard treatment given in most books on number theory.

A detailed exposition of the orthogonality relationship is given in the article character group.

Admittedly, the exposition could be made simpler, with explicit examples for n=2,3,4,5. linas 16:10, 23 September 2005 (UTC)

The characters of a group representation are orthogonal, but the characters of the cyclic group representation are all the roots of unity, not just the primitive ones. I agree with Bo that this must have been a reference to the DFT matrix, and not just to the primitive roots. If something involving just the primitive roots is also possible, please be more explicit. —Steven G. Johnson 18:53, 26 September 2005 (UTC)

I never said anything about the primitive roots. Either this article or the article on the character group can/should be expanded to talk about representations of the cyclic group. I have no idea of what a "DFT matrix" is. Oh, I see , I just followed the link, Discrete Fourier transform and indeed, it does seem to describe the same marix that is given in character group. Not to flaunt my ignorance, I had no idea that DFT had anything to do with his topic; it is not a connection that group theory or number theory people ever bother to mention. linas 23:44, 26 September 2005 (UTC)

1. I made the example of the unitary matrix for n=4 as Linas requested, but Stevenj has removed it.

2. My confusion regarding primitive roots making a unitary matrix is because the text was "these different roots of unity" instead of "the different roots of unity". That has been corrected and the correction is not yet removed by Stevenj.

3. I am pleased that Linas has learned from my insight into the connection between roots of unity and fourier transform. I would like to simplify the articles on fourier transforms using this insight, but I expect that it will be removed by Stevenj. Bo Jacoby 09:26, 27 September 2005 (UTC)

The article still links to discrete Fourier transform, which already has liberal mentions of the roots of unity. This connection is well known, and I agree that it should be linked to from this page. Linking to DFT and discussing connections between concepts has nothing to do with adopting nonstandard notations. —Steven G. Johnson 23:15, 27 September 2005 (UTC)

Notation

I'd be happier is the notation $1^{\frac {k}{m}}$ was introduced before it was used. That is a highly non-standard notation; I've never seen anything like that before. I assume its historical in some way? Is there a reference for this notation? linas 16:10, 23 September 2005 (UTC)

Hi Linas. You are right of cause, the definition must precede the use. I'll fix it. I have used $\ 1^{x}$ instead of $\ e^{2\pi ix}$ for many years as a simple ink-saving notation, but frankly I do not remember from where it came. Perhaps I have invented it myself. I tend to invent things now and then. If $\ e^{2\pi i}=1$, then why not write $\ 1$ instead of $\ e^{2\pi i}$  ? It is really no big deal. It could simplify the confusing notation for fourier transforms. The alternative interpretation, $\ 1^{x}=1$ , is uninteresting. Confusion is unlikely to occur because nobody will write $\ 1^{x}$ meaning $\ 1$. See exponentiation. Bo Jacoby 10:14, 26 September 2005 (UTC)

I've removed this notation. I agree that it is nonstandard. (I tend to think it is also misleading, because it gives the impression that there is a unique choice of primitive root $1^{\frac {1}{m}}$, when in fact this is wholly arbitrary.) Wikipedia should not be in the business of inventing major new notations from whole cloth. —Steven G. Johnson 18:51, 26 September 2005 (UTC)

$1^{\frac {1}{m}}$ is not an arbitrary primitive root, but is equal to $e^{\frac {2\pi i}{m}}$ which is well defined. To the readers most of the material in Wikipedia is new, and of cause they don't mind. By deleting material just because is new to you, you prevent Wikipedia from exceeding your level. Think about it. Get wiser. Bo Jacoby 09:07, 27 September 2005 (UTC)

Hey, please, be nice. Personal notations are not good here, and that is a solid matter of policy. Charles Matthews 10:52, 27 September 2005 (UTC)

I'm sorry. But what is the policy of Wikipedia: that everybody can contribute, or that a censor removes anything that he hasn't seen before simply because he hasn't seen it before ? Bo Jacoby 06:39, 28 September 2005 (UTC)

The policy in cases of content disputes is that matters are resolved by discussion, in an atmosphere of mutual respect. The policy in the matter of innovation is that WP is not a place to innovate. Charles Matthews 07:35, 28 September 2005 (UTC)

Properties

It can be proved in any number of ways

1. Is is really important to stress that there are more ways to prove it? I think not.
2. Nor do I think that it is true that it can be proved in essentially more than one way. Please provide an alternative proof. Bo Jacoby 10:09, 27 September 2005 (UTC)

Proof #2: the roots of unity are eigenvectors of the discretized Laplacian with periodic boundary conditions, which is Hermitian and therefore has orthogonal eigenvectors. Proof #3: multiply the sum S of the roots of unity by one of the roots of unity z; by the rearrangement theorem you get the same sum in a different order; thus, Sz = S and therefore S = 0. Proof #4: the group theoretic proof already mentioned, based on representation theory. See also Charles' proofs below. The existence of the various proofs of the orthogonality of the roots of unity is indeed important, because it means that this orthogonality appears in different ways in many branches of mathematics. —Steven G. Johnson 23:16, 27 September 2005 (UTC)

The proofs: geometric by barycentre of regular n-gon, algebraic by coefficient of x^{n − 1}, complex number by ζS = S — I think these are different ways. By the way, the point is that we may include material from books, but don't have to. Charles Matthews 15:00, 27 September 2005 (UTC)

Thank you very much for these alternative proofs. They clarify the claim that there are several proofs. Without that clarification the claim is confusing to the reader. Can the set of proofs of a mathematical statement be counted ? From the point of view of Gödel a countable infinity of proofs can be trivially generated from a single proof, but that is cheating. That makes the claim that there are several proofs meaningless, because it applies for any provable statement. So that is not what is meant. The 'different' proofs must be somehow logically independent. Are the above proofs logically independent ? Can the set of logically independent proofs of a provable mathematical statement be counted ? I don't know the answer. Does anybody ? I do not think that these thoughts are relevant to the subject of summing the roots of unity. So I still don't think that the number of ways in which it can be proved should be mentioned in the article. Bo Jacoby 09:32, 29 September 2005 (UTC)

Well, you can cut it out. 'Logical independence' of proofs isn't a useful idea, I think. Most people recognise the idea that there can be proofs from different starting points. You are wrong about it being irrelevant. WP's articles aim for completeness, so certainly mentioning how one topic is seen in different fields is relevant. Charles Matthews 13:09, 29 September 2005 (UTC)

What is the definition of $\omega _{n}$

"The primitive root $e^{2\pi i/n}$ (or its conjugate) is often denoted by $\omega _{n}$, especially in the context of discrete Fourier transforms."

1. What is the definition of $\omega _{n}$ ? $e^{2\pi i/n}$ or $e^{-2\pi i/n}$ ?

It could be either—there is no universal definition, as far as I can tell. Hence the or in parentheses. —Steven G. Johnson

1. Do you want to teach people bad mathematics just because it is often found in the books, or do you want to present quality mathematics ? Bo Jacoby 10:09, 27 September 2005 (UTC)

Wikipedia does not present original research. If you think that the mathematics in standard textbooks is bad, then you should publish elsewhere and convince people. Wikipedia is not the place for promulgation of new standards. —Steven G. Johnson

WP is a reference, not a textbook. Readers might be using WP to look up something forgotten, or to get a more complete idea of a subject that is already familiar.

Thus, among other things, an article should discuss alternate notations that are in common use, so that when the reader compares their textbook to the WP article, they find them to be consistent. The few readers who might be trying to learn something completely new should be directed to textbook references.

If the 1^{k/n} notation is indeed in "common use", then we should indeed give it. However, you've given zero evidence of this. —Steven G. Johnson 23:11, 27 September 2005 (UTC)

As to Steven's edits, perhaps they were too drastic. Please try to find a middle ground. linas 13:30, 27 September 2005 (UTC)

Notation, again

Bo,

Please stop using the $1^{k \over n}$ notation until such time as you can provide a book reference for this notation. After the discussion above, I am surprised that you put it back in again. linas 20:33, 30 September 2005 (UTC)

Well, Linas, I didn't find the discussion decisive. You requested a compromise, and I considered "the notation is convenient though nonstandard" a compromise. Wasn't PAR too in favour for the notatation? Bo Jacoby 06:32, 3 October 2005 (UTC)

It's not a standard notation. It doesn't belong on WP, under policy. Charles Matthews 07:00, 3 October 2005 (UTC)

n=0

Now please explain to me why the case n=0 is included when in fact it is not defined? My change of $\mathbb {N} _{0}$ to $\mathbb {N}$ was revoked. Bo Jacoby 06:32, 3 October 2005 (UTC)

Bo, no one has any problem with minor clarifications like this. The problem is that you mix them in with large-scale changes of notation in the article, which are hard to undo except by reversion...then your littler changes get lost in the shuffle. It also doesn't help that you react with moral outrage to every perceived slight. —Steven G. Johnson 05:09, 6 October 2005 (UTC)

Thank you. I think your last edit changed $\mathbb {N}$ into $\mathbb {N} _{0}$ . It should have been the other way round. $\mathbb {N}$ excludes 0 and is correct. $\mathbb {N} _{0}$ includes 0 and is incorrect.

Hermitian matrix

Steven,

You added the text

The roots of unity also appear as the eigenvectors of a Hermitian matrix (a discretized one-dimensional Laplacian with periodic boundaries), from which the orthogonality properties also follow.

Can you give an explicit form for this matrix? I am left to guess what it might be. Is it the tridiagonal matrix with -1 2 -1 along the diagonal? Different readers will no doubt make other guesses. linas 20:33, 30 September 2005 (UTC)

Almost... -1 2 -1 would be -∇², whereas +∇² would be 1 -2 1 (except for the boundary elements which have to wrap around). Not that it matters; an overall sign flip only affects the eigenvalues and not the eigenvectors, and a negative Laplacian is actually more natural in the physical context of a wave equation. (I guess a more precise description would be the center-difference discretized Laplacian.) It's discussed in this article by Strang:

The Discrete Cosine Transform

I'll add a reference to the article. —Steven G. Johnson 01:23, 1 October 2005 (UTC)

FYI, the discretized Laplacian is not, of course, the only Hermitian matrix that has the roots of unity as eigenvectors. Any discretized Hermitian operator with discrete translational symmetry under cyclic shifts will do. This follows, once again, from representation theory, because the eigenvectors of a Hermitian operator can always be chosen to transform as one of the representations of its symmetry group, which in this case is a cyclic group...as usual, everything is connected. For the same reason, it actually doesn't matter what discretization of the Laplacian you use, as long as it is symmetric (in order to get a Hermitian matrix) and periodic. —Steven G. Johnson 01:34, 1 October 2005 (UTC)

orthogonality

Steven. The $n$^th roots of unity can be arranged to form an $n\times n$ matrix whose $(j,k)$^th entry is

$U_{j,k}=n^{-{\frac {1}{2}}}\omega _{n}^{jk}$

is a lie. The elements of the matrix are not the roots of unity. I keep correcting it and you keep repeating it. Please argue in the discussion forum before you insist on removing my edits. Bo Jacoby 08:55, 5 October 2005 (UTC)

Um, the article doesn't state that the entries of the matrix are roots of unity. It clearly gives the sqrt(n) factor, and the matrix is indeed based on an arrangement of the roots of unity. If you want to be pedantic about the word "arranged" (even though the equation makes things perfectly clear, whereas English is more flexible than you seem to think), fine, I replaced it with "used". Throwing around accusations of "lies" only makes you look foolish. —Steven G. Johnson 17:07, 5 October 2005 (UTC)

By the way, your edit was reverted not because of the minor changes in the text, but because of an unhelpful wholesale change of notation. This was clearly explained in my edit comments. —Steven G. Johnson

The reader has bigger difficulties than the author in understanding the text. So the author should lend ear to the reader's comments. As a reader I changed several formulations which were simply wrong, such as the case n=0. But you reverted everything - not merely what you disagreed on, but everything. Do you realize know how many times you have done that? Please stop it. It is a backwards step instead of a forward step. I edited for a reason. A word that confuses rather than clarifies should be removed. The word arrange is explained in the Longman Dictionary of Contemporary English: to put into a correct, pleasing or desired order. The examples given are: To arrange flowers in a vase, and: The books are arranged on the shelves in alphabetical order. Please don't argue that it can ever mean divide by the square root of the number of items, and then copy them to fit into a square matrix. Any reader is pedantic about words until he gives up reading. Please don't argue ad hominem that I am specially pedantic and therefore should be taken less seriously. I would not be this foolishly indiplomatic if I had seen diplomacy work. But to your surprise I wish to say that I do enjoy working with you in improving WP. It is fun and it is worth while. Let's be friends. Bo Jacoby 09:33, 6 October 2005 (UTC)

Bo, I apologize for any ad hominem comments. On your part, please avoid charged language like "lies" etcetera. As I mentioned above, the reason for the reverts were because you mix your small corrections in with wholesale changes of notation that I could not agree with, and the only practical way to undo such a large change is a revert. I've tried to scan through the diff to see if you made other substantive changes that I should try to keep, but it is easy to miss little changes of wording. —Steven G. Johnson 15:40, 6 October 2005 (UTC)

(PS. If you arrange flowers in a vase, that doesn't mean that you haven't put anything else into the vase. That doesn't mean I mind changing the wording, but I do think you overreacted. —Steven G. Johnson)

I agree that my charged language make me look like a fool and I am not happy about it. I apologize. Bo Jacoby 12:26, 7 October 2005 (UTC)

Wikipedia:Lamest edit wars ever

Not quite there, but certainly this page is heading in the direction ... Charles Matthews 20:07, 5 October 2005 (UTC)

Geometrical Representation of Roots

It was mentioned that "It can be shown that they are located on the unit circle of the complex plane and that in that plane they form the vertices of a n-sided regular polygon with one vertex on 1.". Could somebody please show it? Thanks a lot.

If x is an n th root of unity, then xⁿ=1. Take the complex conjugate of that equation to see that the conjugate x* is a root of unity: (x*)ⁿ=1. Multiply these two equations to see that the positive real number xx*=|x|² is a root of unity. So xx*=1, because 1 is the only positive real root of unity. So |x|=1. That means that x is on the unit circle. This was the first part of the assertion. Now consider the regular n-sided polygon with center in 0 and one of the vertices in 1. Consider the first vertex A in clockwise direction from 1. The triangle (0;1;A) is similar to triangle (0;A;A²) according to Complex_number#Geometric_interpretation_of_the_operations_on_complex_numbers. So A² is the 2'nd vertex on the polygon. And (by induction) A^k is the kth vertex on the polygon. The numbers A, A², ... , Aⁿ=1. are n'th roots of unity if A is. This was the second part of the assertion. Bo Jacoby 15:29, 30 November 2005 (UTC)

I suspect "show it" may have been a request for an illustration, although the proof might be good to add too. Doops | talk 18:31, 30 November 2005 (UTC)

Cyclotomic polynomials, incorrect statement

Applying Möbius inversion to the formula gives

$\Phi _{n}(z)=\prod _{d\,\mid n}(z^{n/d}-1)^{\mu (d)},$

where μ is the Möbius function.

The Möbius inversion is about sums, not about products. And the cyclotomic polynomial is irreducible and thus cannot be written as a product of polynomials in a non-trivial way. I will remove the sentence from the article. Bo Jacoby 11:26, 21 June 2006 (UTC)

I am afraid that change is not valid: the expression given, and the reason (Mobius inversion) are both correct - see page 206 of "Algebraic Number Theory" by Frohlich and Taylor. I will re-instate it. Madmath789 11:54, 21 June 2006 (UTC)

Example. n = 4. The controversial formula gives:

Φ₄(z) = (z¹−1)^μ(4)(z²−1)^μ(2)(z⁴−1)^μ(1) = (z¹−1)⁰(z²−1)⁻¹(z⁴−1)¹.

UPS !

Φ₄(z) = z²+1

You are right and I was wrong. Thank you, and apology. Could you please clarify the connection to the Möbius inversion which is about sums? Bo Jacoby 12:44, 21 June 2006 (UTC)

No problem - what I have done is added a very short "generalisations" section to the Möbius inversion article, which states the multiplicative version. Madmath789 12:53, 21 June 2006 (UTC)

Incidentally, care is also needed with the statement "the cyclotomic polynomial is irreducible", as for example (5th cyclotomic polynomial:

$x^{4}+x^{3}+x^{2}+x+1=(x^{2}+{\frac {-1+{\sqrt {5}}}{2}}x+1)(x^{2}+{\frac {-1-{\sqrt {5}}}{2}}x+1).$ So $\Phi _{5}(x)$ is reducible over $\mathbb {Q} ({\sqrt {5}}){\mbox{ but irreducible over }}\mathbb {Q} .$

${\mbox{If }}F(n)=\prod _{d|n}f(d),{\mbox{ then }}f(n)=\prod _{d|n}F(n/d)^{\mu (n/d)}.$ should be ${\mbox{If }}F(n)=\prod _{d|n}f(d),{\mbox{ then }}f(n)=\prod _{d|n}F(n/d)^{\mu (d)}.$ Agree? Bo Jacoby 15:13, 21 June 2006 (UTC)

Yes - either

${\mbox{If }}F(n)=\prod _{d|n}f(d),{\mbox{ then }}f(n)=\prod _{d|n}F(n/d)^{\mu (d)},$ or ${\mbox{If }}F(n)=\prod _{d|n}f(d),{\mbox{ then }}f(n)=\prod _{d|n}F(d)^{\mu (n/d)},$ but not the version I stated - I will correct it, thanks. Madmath789 15:29, 21 June 2006 (UTC)

There is a small mistake in the calculation - a typo I'm guessing - of the 6th cyclotomic polynomial. The term (z-1) in the product should be to the 1st power not to the -1 power as the mobius function maps 6 to 1 not -1. This would lead to a prdouct giving a polynomial of degree 2 as desired. 71.194.89.197 01:27, 20 March 2007 (UTC)

Indeed, you are right. Fixed now. -- EJ 12:58, 20 March 2007 (UTC)

So I checked this page and there is still a -1 power right next to the (z-1) term in the calculation. I thought I would be proactive nad look to see if I could edit it myself but it appears that it is edited correctly (as you say) but the display shows something different, what gives? 71.194.89.197 03:44, 21 March 2007 (UTC)

Convergent Points

I think a section mentioning that $z^{\infty }=1$ for all points on the unit circle. But it should also mention those points which are not finite roots of unity. That is, the theoretical "infiniteth" roots of unity, whos infinite exponentation converge to one, but no finite exponentation there of then actually reaches one. It can be shown that these points distance along the unit circle (Not lateral distance. Distance on the circle) is pi divided by some irrational number iff it is a primitive infiniteth root. (Because the distance will for nth roots of unit for finite n is pi, the circumference of the unit circle, divided by n, times k, for all natural k less than n.) But I don't think infinitethis the right word to use in article. Anyone with better terminology, ideas, or objections to this going in the article? -- He Who Is^{[ Talk ]} 12:20, 23 June 2006 (UTC)

No, there should be no infinity in the article. It is purely algebraic.

By the way, the only solution to

$\lim _{n\rightarrow \infty }z^{n}=1$

is

$\ z=1$

so there is no nontrivial infinite'th root of unity.

The number

$\ e^{2\pi i/x}$

where x is irrational, might be called an irrational root of unity. Perhaps that is what you are talking about. Bo Jacoby 08:06, 26 June 2006 (UTC)

Well the kth nth root of unity (Let's assume the nth roots of unity are ordered by the value of k in the following formula for the moment) we know equals $e^{2\pi ik/n}$ for all natural k less than n. The distance from 1 to this number along the unit circle will be irrational for all kth nth root of unity where k and n are integers, because pi is irrational.but what about if ${\frac {k}{n}}$ is irrational? Is it still a root of unity? For some "irrational root of unity" z, as you put it, no finite power could bring it to one. So I would say that it only stands to reason that it in fact can be shown that:

$\lim _{n\to \infty }z^{n}=1$

Though I may be wrong. I haven't really given this much thought. -- He Who Is^{[ Talk ]} 17:13, 26 June 2006 (UTC)

If $z=e^{2\pi ia},$ with a irrational, then the sequence $z^{n}$ cannot converge to 1, and in fact, (if I remember correctly) the values $\{z^{n}\}$ are actually dense in the unit circle. Madmath789 21:05, 26 June 2006 (UTC)

So then only kπ for rational k represent the distance on the unit circle from 1 to a root of unity? Or did I not understand corrrectly? -- He Who Is^{[ Talk ]} 21:36, 26 June 2006 (UTC)

Yes, that is how it is - rational k gives an nth root of unity and irrational k do not. Madmath789 21:54, 26 June 2006 (UTC)

Notation

A common notation for $e^{2\pi i/n}$ I've noticed in many books is $\zeta _{n}$ or just $\zeta$. Should this be mentioned here or is it rarer notation than I think? I daren't edit this page for fear of starting a lame edit war :) Moxmalin 22:44, 1 April 2007 (UTC)

Edit wars are made by disrespectful editors who do not follow the rules: Help:Reverting#When_to_revert, not by those who in good faith improve the articles. Be bold and edit. Bo Jacoby 09:44, 2 April 2007 (UTC).

Cyclotomic polynomials

Quote:

While in general polynomials of degree higher than 4 cannot be solved by radicals, cyclotomic polynomials can. This is because their Galois group is abelian, and hence solvable. Thus, every root of unity has an expression in radical form. Algorithms exist for calculating such expressions.

Trivially every root of unity has an expression in radical form: ${\sqrt[{n}]{1}}$. What is the nontrivial content of the quote? Bo Jacoby 20:04, 9 April 2007 (UTC).

The only possibility I can think of is that the quote was trying to say that the real and imaginary parts of roots of unity can be expressed by real radical expressions. However, even if that is true (which I do not know), the first two sentences still make no (nontrivial) sense, the quote needs a reformulation and a solid reference. I'm cutting it out for the moment. -- EJ 12:14, 10 April 2007 (UTC)

Aha, I'm talking nonsense. Let p be an irreducible integer polynomial of degree 3 with 3 distinct real roots, whose discriminant has a rational square root (e.g., $x^{3}-3x-1$). Then roots of p cannot be expressed using real radicals ("casus irreducibilis" [1][2]). OTOH, its splitting field is an Abelian extension of $\mathbb {Q}$ (being of degree 3), hence it is contained in some $\mathbb {Q} (\zeta _{n})$ by the Kronecker-Weber theorem. Thus, $\Re \zeta _{n}$ and $\Im \zeta _{n}$ cannot be expressed using real radicals either. -- EJ 10:21, 11 April 2007 (UTC)

I've been away for a long time, so I haven't responded. But well, basically, all that I wanted to convey is the content of [3]. I'm not really sure what the best way of putting it is - the issue is that the fact that the 3rd root of unity is 1/2(1 + i * root 3) (and not some unwritedownable transcendental value, say) is not a fluke, and that for any given n, it is possible to construct some finite combination of integers, elementary operations, i, and the real kth root to give the root of unity exactly. I don't really understand what you mean by it being impossible, since well, we have examples of it being true, and an algorithm to find such radicals. (The article attributes the proof to Gauss)--Fangz 20:35, 20 June 2007 (UTC)

By being impossible I mean that it is literally impossible: there exists an n such that the real and imaginary part of the primitive nth root of unity cannot be expressed using rational operations and real roots. If you do not understand the proof of this fact above, I can explain it in more detail, just ask. Notice that this does not contradict the Weber-Keckeisen paper: the radical expression they construct do involve roots of nonreal complex numbers, as can be seen from the examples.

The problem is that it is totally unclear what the paper is trying to accomplish. They do not give their working definition of "radical expression", and the standard definition as used in algebra textbooks allows to adjoin any radical of a previously constructed number: then roots of unity are trivially expressible as, well, roots of unity.

I strongly believe that no information is better than false or incomprehensible information. The second paragraph you included is blatantly false. The first one was originally either trivial or it had unclear and unexplained meaning, and after your modification, it is also false. I will thus remove it again.

I do not have access to Disquisitiones Arithmeticae; it would be most wonderful if somebody who can looked it up to find the definition of "radical expression" it really uses, instead of trying to guess it, which just proved not to work. -- EJ 10:03, 21 June 2007 (UTC)

pedantry

Yes, it's pedantic -- I know; but as far as I know the convential definition requires unities to coincide (i.e. considers selection of unity to be an operation). In any case, the interesting case is when the unity is stable. --VKokielov 18:39, 4 June 2007 (UTC)

My addition

Because the roots of unity are algebraic, and because they are given by the trigonometric functions sine and cosine, it follows that the sine (or cosine) of any angle which is a rational fraction of the revolution 2π must be computable by a finite number of additions, multiplications, and square roots.

Even if I'm right, I don't want to stick it in out of context. --VKokielov 17:12, 5 June 2007 (UTC)

Square roots!.. I'm thinking of powers of two. If the root isn't a power of two, then these needn't be square roots at all. --VKokielov 17:39, 5 June 2007 (UTC)

I think that is true, but the fact that the numbers are algebraic is not enough - since not all algebraic integers are expressible by a finite number of additions, multiplications, and roots. (See Abel–Ruffini theorem). Fortunately, however, in the case of roots of unity, they can, (see referenced article above), but again, this is a property beyond that of just being algebraic.--Fangz 20:49, 20 June 2007 (UTC)