tag:blogger.com,1999:blog-9004452969563620551.post4675502850342345455..comments2017-09-25T12:48:32.845+05:30Comments on Paramanand's Math Notes: Gauss and Regular Polygons: Cyclotomic Polynomialstesternoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-9004452969563620551.post-57839041111626326962014-06-25T13:07:58.444+05:302014-06-25T13:07:58.444+05:30@Kazbich-the-bandit,
Let's now see your second...@Kazbich-the-bandit,<br />Let's now see your second issue. Suppose $d_{1}, d_{2}$ are two distinct divisors of $n$. Suppose $k, l$ are two integers such that $k$ is coprime to $d_{1}$ and $l$ is coprime to $d_{2}$. Then $(\cos(2\pi k/d_{1}) + i\sin(2\pi k/d_{1}))$ is a primitive $d_{1}^\text{th}$ root and similarly $(\cos(2\pi l/d_{2}) + i\sin(2\pi l/d_{2}))$ is a primitive $d_{2}^\text{th}$ root. If they are same then $k/d_{1} = l/d_{2}$. This is not possible unless $d_{1} = d_{2}$ and $k = l$ because the ratios $k/d_{1}$ and $l/d_{2}$ are in their lowest forms.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-46408021774329582022014-06-25T12:00:20.418+05:302014-06-25T12:00:20.418+05:30@Kazbich-the-bandit,
First we deal with "repl...@Kazbich-the-bandit,<br />First we deal with "replacing $k/n$ with $c/d$" thing. Clearly this means that $k/n = c/d$ and $c, d$ are coprime to each other. We have thus reduced $k/n$ into lowest form $c/d$ and $c, d$ have no common factors. In your example $n = 15, k = 4$ so that $k/n = 4/15$ is already in lowest form and then $c = 4, d = 15$. Thus note that given $k, n$ the values $c, d$ are not arbitrary like you try to take $d = 5$.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-28069349065295504182014-06-25T02:20:33.127+05:302014-06-25T02:20:33.127+05:30I am also finding difficulty understanding the div...I am also finding difficulty understanding the divisors issue. Every number is a divisor of itself. If any nth root of unity is also a primitive dth root of some divisor of n, then how can it be true that d and n as distinct divisors of n do not share primitive roots. I assume that by any 'nth root' you also include the primitve nth roots. I wondered if by distinct divisor you meant coprime but that seemed improbable because you would have specified it if it were the case.Kazbich-the-bandithttps://www.blogger.com/profile/18247629194967955510noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-10303324716503004842014-06-25T01:13:47.352+05:302014-06-25T01:13:47.352+05:30we can replace k/n with a fraction c/d in lowest t...we can replace k/n with a fraction c/d in lowest terms so that d∣n, it follows that any nth root of unity is a primitive dth root of unity for some divisor d of n and vice-versa<br /><br />I found this part hard to accept. I can accept the vice-versa part. Clearly, any dth root of unity is also an nth root of unity for some divisor d of n. But it is not true that any nth root of unity is a primitive dth root of unity. Take any nth root of unity, say for n= 15, k= 4. and let d be 5. Clearly, (Z^4)^5 is not equal to 1. I guess I must have misunderstood what you mean by replace the fraction. Would greatly appreciate an explanationKazbich-the-bandithttps://www.blogger.com/profile/18247629194967955510noreply@blogger.com