Proving $ord_p(ζ_p-1)=1/(p-1)$
$begingroup$
After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.
My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.
I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.
abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
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add a comment |
$begingroup$
After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.
My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.
I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.
abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
$endgroup$
3
$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19
1
$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20
$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32
$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52
add a comment |
$begingroup$
After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.
My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.
I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.
abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
$endgroup$
After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.
My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.
I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.
abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity
abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
asked Dec 7 '18 at 22:13
Pascal's WagerPascal's Wager
371315
371315
3
$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19
1
$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20
$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32
$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52
add a comment |
3
$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19
1
$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20
$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32
$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52
3
3
$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19
$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19
1
1
$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20
$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20
$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32
$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32
$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52
$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52
add a comment |
1 Answer
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$begingroup$
$|a|_p = p^{-v_p(a)}$
Since $zeta_p^p = 1$ then $|zeta_p|_p=1$
$sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$
$prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.
$1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$
whence (for $p nmid n$)
$|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$
letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$
Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
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1 Answer
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$begingroup$
$|a|_p = p^{-v_p(a)}$
Since $zeta_p^p = 1$ then $|zeta_p|_p=1$
$sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$
$prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.
$1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$
whence (for $p nmid n$)
$|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$
letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$
Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
$endgroup$
add a comment |
$begingroup$
$|a|_p = p^{-v_p(a)}$
Since $zeta_p^p = 1$ then $|zeta_p|_p=1$
$sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$
$prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.
$1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$
whence (for $p nmid n$)
$|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$
letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$
Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
$endgroup$
add a comment |
$begingroup$
$|a|_p = p^{-v_p(a)}$
Since $zeta_p^p = 1$ then $|zeta_p|_p=1$
$sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$
$prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.
$1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$
whence (for $p nmid n$)
$|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$
letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$
Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
$endgroup$
$|a|_p = p^{-v_p(a)}$
Since $zeta_p^p = 1$ then $|zeta_p|_p=1$
$sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$
$prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.
$1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$
whence (for $p nmid n$)
$|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$
letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$
Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
answered Dec 7 '18 at 22:46
reunsreuns
21.2k21351
21.2k21351
add a comment |
add a comment |
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3
$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19
1
$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20
$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32
$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52