There are no elliptic curves over $mathbb{F}_8$ with $7$ or $11$ points
$begingroup$
This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).
One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.
However, the question advises to use the previous part, which states:
Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.
For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.
It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.
Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?
elliptic-curves
edited Dec 29 '18 at 6:16
Did
248k23224463
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
$endgroup$
add a comment |
$begingroup$
This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).
One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.
However, the question advises to use the previous part, which states:
Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.
For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.
It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.
Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?
elliptic-curves
edited Dec 29 '18 at 6:16
Did
248k23224463
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
$endgroup$
$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41
$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55
$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31
add a comment |
$begingroup$
This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).
One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.
However, the question advises to use the previous part, which states:
Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.
For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.
It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.
Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?
elliptic-curves
edited Dec 29 '18 at 6:16
Did
248k23224463
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
$endgroup$
This is taken from The Arithmetic of Elliptic Curves by Silverman on page 154, Q5.10(f).
One way of directly solving this problem is to work out on sage all 8^5 possibilities of elliptic curves and show that no such curve with the required number of points exist. This has been done and in that sense, the problem has been solved.
However, the question advises to use the previous part, which states:
Let $p^i$ be the largest power of $p$ such that $p^{2i}|q$. Then $tr(phi)=0mod p iff tr(phi)=0mod p^i$.
For the case of part (f), this does not seem to be at all useful as $p=2$ and $q=8$, which implies that $i=1$ and so the previous part does not yield any new information.
It might be the case that there is a typo but by testing for several cases of $q$, this was found to not be the case.
Is there anything being misunderstood, especially with regards to the application of part (e) to part (f)?
elliptic-curves
elliptic-curves
edited Dec 29 '18 at 6:16
Did
248k23224463
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
edited Dec 29 '18 at 6:16
Did
248k23224463
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
edited Dec 29 '18 at 6:16
Did
248k23224463
edited Dec 29 '18 at 6:16
Did
248k23224463
edited Dec 29 '18 at 6:16
Did
248k23224463
248k23224463
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
asked May 1 '14 at 21:34
Haikal YeoHaikal Yeo
1,027823
1,027823
$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41
$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55
$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31
add a comment |
$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41
$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55
$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31
$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41
$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41
$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55
$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55
$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31
$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31
add a comment |
1 Answer
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$begingroup$
You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
$$
rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
$$
(Here, $lceil k/2rceil$ is the ceiling function.)
In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.
answered Dec 29 '18 at 5:10
GispsanGispsan
363
$endgroup$
add a comment |
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$begingroup$
You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
$$
rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
$$
(Here, $lceil k/2rceil$ is the ceiling function.)
In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.
answered Dec 29 '18 at 5:10
GispsanGispsan
363
$endgroup$
add a comment |
$begingroup$
You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
$$
rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
$$
(Here, $lceil k/2rceil$ is the ceiling function.)
In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.
answered Dec 29 '18 at 5:10
GispsanGispsan
363
$endgroup$
add a comment |
$begingroup$
You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
$$
rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
$$
(Here, $lceil k/2rceil$ is the ceiling function.)
In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.
answered Dec 29 '18 at 5:10
GispsanGispsan
363
$endgroup$
You cannot apply part (e) as written, but in fact you can prove in the same way something slightly stronger which does the trick: if $q=p^k$, then
$$
rm{tr}(phi)equiv 0 mod p iff rm{tr}(phi)equiv 0 mod p^{lceil{k/2}rceil}.
$$
(Here, $lceil k/2rceil$ is the ceiling function.)
In this particular case, you get that $rm{tr}(phi)$ must be a multiple of 4.
answered Dec 29 '18 at 5:10
GispsanGispsan
363
answered Dec 29 '18 at 5:10
GispsanGispsan
363
answered Dec 29 '18 at 5:10
GispsanGispsan
363
answered Dec 29 '18 at 5:10
GispsanGispsan
363
363
add a comment |
add a comment |
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$begingroup$
I think you are right. I notice that if you replace 8 with 16 then one may use part (e.) to prove this result (leaving the other numbers the same).
$endgroup$
– sdf
Jul 9 '14 at 3:41
$begingroup$
I have tried out all possible examples and the case of $q=8$ has not failed.
$endgroup$
– Haikal Yeo
Jul 9 '14 at 11:55
$begingroup$
what I meant was in reference to your question about using part (e.) in that I think possibly Silverman meant to put $mathbb{F}_{16}$ instead so that one can use the previous part to deduce part (f.).
$endgroup$
– sdf
Jul 9 '14 at 14:31