Cyclotomic integers are based the equation x

^{n}= 1 (they get their name from this equation which is known as the cyclotomic equation). The solutions to this equation are known as the Roots of Unity. The solutions to n=1 and n=2 are well known (1, -1). What is less well known is that as n increases, there are complex solutions that also solve this equation (these are also known as DeMoivre Numbers).

For some, it may seem strange that complex numbers emerge out of x

^{3}= 1. The insight to why this is so comes from considering Euler's Identity:

e

^{iπ}=-1

In my next blog, I will go into the details behind this equation. Still, using this equation, we can see that three solutions to x

^{3}= 1 are:

1, e

^{(2iπ)/3}, e

^{(4iπ)/3}

Since (e

^{(2iπ)/3})

^{3}= e

^{2iπ}= (-1)

^{2}= 1.

The solution then for any value of n is e

^{(2iπ)/n}and this solution is by convention represented by the Greek symbol zeta ζ. The set of integers then is Z[ζ] where each integer can be constructed from the rational integers a,b using a + bζ. This construction follows the same pattern as what Euler did for Z[√-3] (see here) and what Gauss did for Z[i] (see here).

For any value n, there are n different roots of unity which can be generated by e

^{(2iπk)/n}where k is any integer 0, 1, 2, ..., up to n-1. For this reason, one speaks about the value ζ (equal to

**e**

^{2iπ/n}) as a primitive root of unity.

One might reasonably ask if all cyclotomic integers created in this way are characterized by unique factorization. In using quadratic integers, Euler made a mistake by assuming that Z[√-3] has unique factorization which it does not. Gabriel Lame assumed that all cyclotomic integers where characterized by unique factorization when he presented his proof of Fermat's Last Theorem. This assumption turns out to be the major flaw in Lame's proof.

This then becomes the major question which Kummer sought to resolve. Under what circumstances are cyclotomic integers characterized by unique factorization? The answer to this question led Kummer to his important partial proof of Fermat's Last Theorem. He showed that Fermat's Last Theorem holds for a certain set or primes which he called "regular primes." I talk more about the properties of cyclotomic integers here.

References

- Roots of Unity, Wikipedia
- Cyclotomic Equation, MathWorld

## 8 comments:

You have a minor error in your:

"Still, using this equation, we can see that the three solutions to x^3=-1 are:

1, -1, exp(2iπ)/3".

Of course, 1 is

nota solution to x^3=-1, because 1^3=1. The third solution is exp((4iπ)/3), as can be seen from what you write below.HTH,

Hi Veky,

Thanks very much for noticing that. That was a typing mistake. I just fixed it.

Cheers,

-Larry

Question about generalizing the definition of cyclotomic integers from the definition of gaussian integers. (I studied algebra, years ago - it's been a while.)

Let zeta be a primitive nth root of unity, whose minimial polynomial has degree m. (Side question. Refresh my memory - is m equal to phi(n), where phi a.k.a. euler-phi is the number of integers from 2 to n that are relatively prime to n?)

Do you want the set of cyclotomic integers to be the set of integers of the form a_0 + a_1 * zeta? Or rather do you want it to be the set of integers of the form

a_0 + a_1 * zeta + a_2 * zeta^2 + ... + a_{m-1} * zeta^{m-1} ?

Here, the a_i are integers.

Hi N. B. Horvath,

Thanks for your question.

Yes, you are correct. If p is a prime number, then all cyclotomic integers based on the pth root of unity can be reduced to an equation of order p-1.

One minor correction. The Euler Phi function is the number of positive integers coprime with a given number. But you are correct that Euler-phi(p) is p-1.

I believe that I answer your questions in more detail here

-Larry

This could be a silly enquiry, mind you, but I am deeply perplexed with this:

three solutions to x3 = 1 are:

1, -1, e(2iπ)/3

How can -1 be a solution to x³=1? Ain't the cube of -1 = -1??!!!

Hi Veefessional,

You are right. 1^3 cannot = -1. That is a mistake. I will fix it.

-Larry

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