# Primitive n-th roots

1. Jan 20, 2012

### autre

1. The problem statement, all variables and given/known data

Show that primitive n-th roots of unity have the form $e^{i2\pi k/n}$ for $k\in\mathbb{Z},n\in\mathbb{N}$, $k$ and $n$ coprime.

The attempt at a solution

So the n-th roots of unity $z$ have the property $z^{n}=1$. I have previously shown that $(e^{2\pi ik/n})^{n}=e^{2\pi ik}=(e^{2\pi i})^{k}=1^{k}=1$. However, I'm not sure where to start in proving that primitive n-th roots of unity have that property. Any ideas on where I could get started?

2. Jan 20, 2012

### Dick

The problem isn't to show that if z is a primitive n-th root of 1 that z^n=1. You know that. A primitive n-th root of 1 is also an n-th root of 1. What's the definition of a primitive root?

3. Jan 21, 2012

### autre

A primitive n-th root has the smallest such n that z^n = 1. So if k and n aren't coprime then they would have a common factor except 1, because if they did, you could have a smaller n s.t. z^n = 1.

4. Jan 21, 2012

### Dick

Well, yes, that's the really sloppy explanation.

5. Jan 21, 2012

### autre

We never covered it in class so I've had to make do with various online sources and have had some difficulty coming up with a formal definition to help me with the proof (though I basically understand what's going on). Do you have a moreformal definition that I could use as a starting point? Thanks!

6. Jan 21, 2012

### Dick

z is a primitive nth root of unity if $z^n=1$ and $z^k \ne 1$ for $0<k<n$.

7. Jan 21, 2012

### autre

How's this for starters?

Let $z\in\mathbb{C}$ be a primitive n-th root of unity. Then, for $n\in\mathbb{N}$ and $k\in\mathbb{Z}$ s.t. $0<k<n$ ,$z^{n}=1$ and $z^{k}\neq1$. Suppose $n=jk$ for some j$\in\mathbb{Z}$. Then, since z is a root of unity, $z=e^{2k\pi i/n}$ and $z^{k}=z^{n/j}=e^{(2\pi ik/n)^{n/j}}=e^{(2\pi ik)/j}=(e^{2\pi i})^{k/j}=1^{k/j}=1$. But $z^{k}\neq1$.

I know I've only proven (or tried to) that $n$ is not a multiple of $k$. Any ideas on what I'm missing?

8. Jan 21, 2012

### Dick

If n and k are not coprime then there is an s>0 such that n=sn' and k=sk'.

9. Jan 21, 2012

### autre

Let $z\in\mathbb{C}$ be a primitive n-th root of unity. Then, for $n\in\mathbb{N}$ and $k\in\mathbb{Z}$ s.t. $0<k<n$ ,$z^{n}=1$ and $z^{k}\neq1$. Suppose k and n are not coprime. Then, $\exists s\in\mathbb{N}$ s.t. $n=sn'$ and $k=sk'$ for some $k'\in\mathbb{Z},n'\in\mathbb{N}$. Then, since $z$ is a root of unity, $z=e^{2k\pi i/n}$ and $z^{k}=z^{nk'/n'}=e^{(2\pi ik/n)^{nk'/n'}}=e^{(2\pi ik)k'/n'}=(e^{2\pi i})^{kk'/n'}=1^{kk'/n'}=1$. But $z^{k}\neq1$. Contradiction, so k, n coprime.

10. Jan 21, 2012

### Dick

That 'proof' should be making you feel at least a little uncomfortable. Pick n=6 and k=2. You are claiming z^2=1? That's not true. Tell me what's wrong with it.

11. Jan 21, 2012

### autre

Do I need to exclude 2 somehow? It seems like 2 is the only prime number where the proof would fail.

12. Jan 21, 2012

### Dick

Pick n=15 and k=3. z^3 still isn't 1. Put the numbers in and trace through your steps in the proof. Figure out where something fishy happens.

13. Jan 21, 2012

### autre

Ah, the problem seems to be that I should have $s\neq k$.

14. Jan 21, 2012

### Dick

You are just guessing. Look at this. $e^\frac{2 i \pi}{3}={(e^{2 i \pi})}^\frac{1}{3}=1^\frac{1}{3}=1$. True or false?

15. Jan 21, 2012

### autre

$e^\frac{2 i \pi}{3}\neq{(e^{2 i \pi})}^\frac{1}{3}$. That was some sloppy algebra on my part -- $x^{2/3}=(x^{2})^{1/3}$ iff $x>0$.

16. Jan 21, 2012

### Dick

You have to be careful. Fractional powers of complex numbers aren't well defined. It's only 'safe' to say (a^b)^c=a^(b*c) if c is an integer. You are throwing fractional powers around in your 'proofs' in a bad way.

17. Jan 21, 2012

### autre

Okay, so going back to $k=sk'$ and $n=sn'$, can I use the fact that for s>1, $n/s \leq n$ but $z^{n/s}=e^{2\pi i(k/s)}=1$? Or am I still using fractional powers improperly?

18. Jan 21, 2012

### Dick

No. That's fine. n/s=n' is an integer. So your are allowed to bring that into the power of the exponent. And k/s=k' is an integer. So, sure, you get 1.