Finding zeta function of an elliptic curve
$begingroup$
Let $p equiv 3 pmod 4$ be a prime, and $E/F_{p^r}$ be
the elliptic curve given by $y^2 = x^3 − x$. Find the zeta-function of
$E/F_p$ and use it to determine $|E(F_{p^r} )|$ for all r>0.
number-theory finite-fields elliptic-curves zeta-functions
$endgroup$
add a comment |
$begingroup$
Let $p equiv 3 pmod 4$ be a prime, and $E/F_{p^r}$ be
the elliptic curve given by $y^2 = x^3 − x$. Find the zeta-function of
$E/F_p$ and use it to determine $|E(F_{p^r} )|$ for all r>0.
number-theory finite-fields elliptic-curves zeta-functions
$endgroup$
1
$begingroup$
Hint: Look at the polynomial $m(x)=x^3-x.$ It is odd, right? Because $pequiv3pmod4$ exactly one of $m(a)$ and $m(-a)$ is a quadratic residue (obviously $aneq0,pm1$).
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 21:56
1
$begingroup$
Partially duplicate of this older question.
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– Jyrki Lahtonen
Nov 20 '15 at 22:01
$begingroup$
Is this an assignment*homework question? What have you tried so far?
$endgroup$
– Klangen
Dec 17 '18 at 10:09
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@Klangen See my edit for a simpler way of typesetting a congruence. Also, the question is 3 years old, and the asker hasn't been heard from since then. It is safe to assume that the eventual homework is overdue. If you are inclined to answer the question, feel free to do so! The question is a bit lacking in context though.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 13:04
add a comment |
$begingroup$
Let $p equiv 3 pmod 4$ be a prime, and $E/F_{p^r}$ be
the elliptic curve given by $y^2 = x^3 − x$. Find the zeta-function of
$E/F_p$ and use it to determine $|E(F_{p^r} )|$ for all r>0.
number-theory finite-fields elliptic-curves zeta-functions
$endgroup$
Let $p equiv 3 pmod 4$ be a prime, and $E/F_{p^r}$ be
the elliptic curve given by $y^2 = x^3 − x$. Find the zeta-function of
$E/F_p$ and use it to determine $|E(F_{p^r} )|$ for all r>0.
number-theory finite-fields elliptic-curves zeta-functions
number-theory finite-fields elliptic-curves zeta-functions
edited Dec 18 '18 at 13:01
Jyrki Lahtonen
110k13171386
110k13171386
asked Nov 20 '15 at 21:49
mathnerd007mathnerd007
452
452
1
$begingroup$
Hint: Look at the polynomial $m(x)=x^3-x.$ It is odd, right? Because $pequiv3pmod4$ exactly one of $m(a)$ and $m(-a)$ is a quadratic residue (obviously $aneq0,pm1$).
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 21:56
1
$begingroup$
Partially duplicate of this older question.
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 22:01
$begingroup$
Is this an assignment*homework question? What have you tried so far?
$endgroup$
– Klangen
Dec 17 '18 at 10:09
$begingroup$
@Klangen See my edit for a simpler way of typesetting a congruence. Also, the question is 3 years old, and the asker hasn't been heard from since then. It is safe to assume that the eventual homework is overdue. If you are inclined to answer the question, feel free to do so! The question is a bit lacking in context though.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 13:04
add a comment |
1
$begingroup$
Hint: Look at the polynomial $m(x)=x^3-x.$ It is odd, right? Because $pequiv3pmod4$ exactly one of $m(a)$ and $m(-a)$ is a quadratic residue (obviously $aneq0,pm1$).
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 21:56
1
$begingroup$
Partially duplicate of this older question.
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 22:01
$begingroup$
Is this an assignment*homework question? What have you tried so far?
$endgroup$
– Klangen
Dec 17 '18 at 10:09
$begingroup$
@Klangen See my edit for a simpler way of typesetting a congruence. Also, the question is 3 years old, and the asker hasn't been heard from since then. It is safe to assume that the eventual homework is overdue. If you are inclined to answer the question, feel free to do so! The question is a bit lacking in context though.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 13:04
1
1
$begingroup$
Hint: Look at the polynomial $m(x)=x^3-x.$ It is odd, right? Because $pequiv3pmod4$ exactly one of $m(a)$ and $m(-a)$ is a quadratic residue (obviously $aneq0,pm1$).
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 21:56
$begingroup$
Hint: Look at the polynomial $m(x)=x^3-x.$ It is odd, right? Because $pequiv3pmod4$ exactly one of $m(a)$ and $m(-a)$ is a quadratic residue (obviously $aneq0,pm1$).
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 21:56
1
1
$begingroup$
Partially duplicate of this older question.
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 22:01
$begingroup$
Partially duplicate of this older question.
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 22:01
$begingroup$
Is this an assignment*homework question? What have you tried so far?
$endgroup$
– Klangen
Dec 17 '18 at 10:09
$begingroup$
Is this an assignment*homework question? What have you tried so far?
$endgroup$
– Klangen
Dec 17 '18 at 10:09
$begingroup$
@Klangen See my edit for a simpler way of typesetting a congruence. Also, the question is 3 years old, and the asker hasn't been heard from since then. It is safe to assume that the eventual homework is overdue. If you are inclined to answer the question, feel free to do so! The question is a bit lacking in context though.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 13:04
$begingroup$
@Klangen See my edit for a simpler way of typesetting a congruence. Also, the question is 3 years old, and the asker hasn't been heard from since then. It is safe to assume that the eventual homework is overdue. If you are inclined to answer the question, feel free to do so! The question is a bit lacking in context though.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 13:04
add a comment |
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$begingroup$
Hint: Look at the polynomial $m(x)=x^3-x.$ It is odd, right? Because $pequiv3pmod4$ exactly one of $m(a)$ and $m(-a)$ is a quadratic residue (obviously $aneq0,pm1$).
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 21:56
1
$begingroup$
Partially duplicate of this older question.
$endgroup$
– Jyrki Lahtonen
Nov 20 '15 at 22:01
$begingroup$
Is this an assignment*homework question? What have you tried so far?
$endgroup$
– Klangen
Dec 17 '18 at 10:09
$begingroup$
@Klangen See my edit for a simpler way of typesetting a congruence. Also, the question is 3 years old, and the asker hasn't been heard from since then. It is safe to assume that the eventual homework is overdue. If you are inclined to answer the question, feel free to do so! The question is a bit lacking in context though.
$endgroup$
– Jyrki Lahtonen
Dec 18 '18 at 13:04