Use of frobenius map of an elliptic curve












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I was reading about elliptic curves from https://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf. Page No. 44 defines Frobenius map. It defines the frobenius map as $f(x,y)=(x^p,y^p) mod p$. Isn't it just an identity map by fermat's little theorem? What's the use of this map in elliptic curves?






group-theory elliptic-curves frobenius-groups







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  • 1




    Fermat's little theorem says that $x^p = x$ in the finite field $mathbb{F}_p = mathbb{Z}/p$. However, this is false for any nontrivial field extension of $mathbb{F}_p$. In this case, we are working in the algebraic closure of $mathbb{F}_p$.
    – Slade
    Dec 1 at 17:33












  • @Slade I am new to this area. May I know what is a field extension and algebraic closure of $F_p$?
    – satya
    Dec 1 at 17:37










  • I'd read Silverman to find out more about the Frobenius map.
    – Wuestenfux
    Dec 1 at 18:03






  • 2




    Let $E : y^2 = x^3+x$ then $E(mathbb{F}_{5^k}) = { (x,y) in mathbb{F}_{5^k}, y^2= x^3+x} cup {O }$, there is a group law on it and the full curve is $bigcup_k E(mathbb{F}_{5^k}) = E(overline{mathbb{F}_{5}})$. You are supposed to know how to construct those finite fields with $5^k$ elements (and how the Frobenius act on them) and get some intuitions on the roots of "the polynomial map defining $P mapsto n P, P in E(overline{mathbb{F}_{5}})$"
    – reuns
    Dec 1 at 18:15


















0














I was reading about elliptic curves from https://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf. Page No. 44 defines Frobenius map. It defines the frobenius map as $f(x,y)=(x^p,y^p) mod p$. Isn't it just an identity map by fermat's little theorem? What's the use of this map in elliptic curves?






group-theory elliptic-curves frobenius-groups







share|cite|improve this question


















  • 1




    Fermat's little theorem says that $x^p = x$ in the finite field $mathbb{F}_p = mathbb{Z}/p$. However, this is false for any nontrivial field extension of $mathbb{F}_p$. In this case, we are working in the algebraic closure of $mathbb{F}_p$.
    – Slade
    Dec 1 at 17:33












  • @Slade I am new to this area. May I know what is a field extension and algebraic closure of $F_p$?
    – satya
    Dec 1 at 17:37










  • I'd read Silverman to find out more about the Frobenius map.
    – Wuestenfux
    Dec 1 at 18:03






  • 2




    Let $E : y^2 = x^3+x$ then $E(mathbb{F}_{5^k}) = { (x,y) in mathbb{F}_{5^k}, y^2= x^3+x} cup {O }$, there is a group law on it and the full curve is $bigcup_k E(mathbb{F}_{5^k}) = E(overline{mathbb{F}_{5}})$. You are supposed to know how to construct those finite fields with $5^k$ elements (and how the Frobenius act on them) and get some intuitions on the roots of "the polynomial map defining $P mapsto n P, P in E(overline{mathbb{F}_{5}})$"
    – reuns
    Dec 1 at 18:15
















0












0








0







I was reading about elliptic curves from https://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf. Page No. 44 defines Frobenius map. It defines the frobenius map as $f(x,y)=(x^p,y^p) mod p$. Isn't it just an identity map by fermat's little theorem? What's the use of this map in elliptic curves?






group-theory elliptic-curves frobenius-groups







share|cite|improve this question













I was reading about elliptic curves from https://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf. Page No. 44 defines Frobenius map. It defines the frobenius map as $f(x,y)=(x^p,y^p) mod p$. Isn't it just an identity map by fermat's little theorem? What's the use of this map in elliptic curves?






group-theory elliptic-curves frobenius-groups




group-theory elliptic-curves frobenius-groups






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 at 17:24









satya

857




857








  • 1




    Fermat's little theorem says that $x^p = x$ in the finite field $mathbb{F}_p = mathbb{Z}/p$. However, this is false for any nontrivial field extension of $mathbb{F}_p$. In this case, we are working in the algebraic closure of $mathbb{F}_p$.
    – Slade
    Dec 1 at 17:33












  • @Slade I am new to this area. May I know what is a field extension and algebraic closure of $F_p$?
    – satya
    Dec 1 at 17:37










  • I'd read Silverman to find out more about the Frobenius map.
    – Wuestenfux
    Dec 1 at 18:03






  • 2




    Let $E : y^2 = x^3+x$ then $E(mathbb{F}_{5^k}) = { (x,y) in mathbb{F}_{5^k}, y^2= x^3+x} cup {O }$, there is a group law on it and the full curve is $bigcup_k E(mathbb{F}_{5^k}) = E(overline{mathbb{F}_{5}})$. You are supposed to know how to construct those finite fields with $5^k$ elements (and how the Frobenius act on them) and get some intuitions on the roots of "the polynomial map defining $P mapsto n P, P in E(overline{mathbb{F}_{5}})$"
    – reuns
    Dec 1 at 18:15
















  • 1




    Fermat's little theorem says that $x^p = x$ in the finite field $mathbb{F}_p = mathbb{Z}/p$. However, this is false for any nontrivial field extension of $mathbb{F}_p$. In this case, we are working in the algebraic closure of $mathbb{F}_p$.
    – Slade
    Dec 1 at 17:33












  • @Slade I am new to this area. May I know what is a field extension and algebraic closure of $F_p$?
    – satya
    Dec 1 at 17:37










  • I'd read Silverman to find out more about the Frobenius map.
    – Wuestenfux
    Dec 1 at 18:03






  • 2




    Let $E : y^2 = x^3+x$ then $E(mathbb{F}_{5^k}) = { (x,y) in mathbb{F}_{5^k}, y^2= x^3+x} cup {O }$, there is a group law on it and the full curve is $bigcup_k E(mathbb{F}_{5^k}) = E(overline{mathbb{F}_{5}})$. You are supposed to know how to construct those finite fields with $5^k$ elements (and how the Frobenius act on them) and get some intuitions on the roots of "the polynomial map defining $P mapsto n P, P in E(overline{mathbb{F}_{5}})$"
    – reuns
    Dec 1 at 18:15










1




1




Fermat's little theorem says that $x^p = x$ in the finite field $mathbb{F}_p = mathbb{Z}/p$. However, this is false for any nontrivial field extension of $mathbb{F}_p$. In this case, we are working in the algebraic closure of $mathbb{F}_p$.
– Slade
Dec 1 at 17:33






Fermat's little theorem says that $x^p = x$ in the finite field $mathbb{F}_p = mathbb{Z}/p$. However, this is false for any nontrivial field extension of $mathbb{F}_p$. In this case, we are working in the algebraic closure of $mathbb{F}_p$.
– Slade
Dec 1 at 17:33














@Slade I am new to this area. May I know what is a field extension and algebraic closure of $F_p$?
– satya
Dec 1 at 17:37




@Slade I am new to this area. May I know what is a field extension and algebraic closure of $F_p$?
– satya
Dec 1 at 17:37












I'd read Silverman to find out more about the Frobenius map.
– Wuestenfux
Dec 1 at 18:03




I'd read Silverman to find out more about the Frobenius map.
– Wuestenfux
Dec 1 at 18:03




2




2




Let $E : y^2 = x^3+x$ then $E(mathbb{F}_{5^k}) = { (x,y) in mathbb{F}_{5^k}, y^2= x^3+x} cup {O }$, there is a group law on it and the full curve is $bigcup_k E(mathbb{F}_{5^k}) = E(overline{mathbb{F}_{5}})$. You are supposed to know how to construct those finite fields with $5^k$ elements (and how the Frobenius act on them) and get some intuitions on the roots of "the polynomial map defining $P mapsto n P, P in E(overline{mathbb{F}_{5}})$"
– reuns
Dec 1 at 18:15






Let $E : y^2 = x^3+x$ then $E(mathbb{F}_{5^k}) = { (x,y) in mathbb{F}_{5^k}, y^2= x^3+x} cup {O }$, there is a group law on it and the full curve is $bigcup_k E(mathbb{F}_{5^k}) = E(overline{mathbb{F}_{5}})$. You are supposed to know how to construct those finite fields with $5^k$ elements (and how the Frobenius act on them) and get some intuitions on the roots of "the polynomial map defining $P mapsto n P, P in E(overline{mathbb{F}_{5}})$"
– reuns
Dec 1 at 18:15












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In the notation of the notes you link to $E(K)$ stands for the points $(x,y)in K^2$ that satisfy the equation $y^2=x^3 + Ax + b$ that defines the elliptic curve together with the point at infinity. Here $K$ can be any field in which the coefficients $A,B$ are contained. On $E(mathbb{F}_p)$ the Frobenius map $(x,y)mapsto (x^p,y^p)$ is indeed the identity. But the Frobenius map is defined on $E(overline{mathbb{F}_p})$.



So to understand it you first need to understand what $overline{mathbb{F}_p}$ is. It is an algebraic closure of $mathbb{F}_p$, i.e. an algebraic extension of $mathbb{F}_p$ in which every non-constant polynomial has a root. This is an infinite field that contains $mathbb{F}_p$.



In fact $mathbb{F}_p$ is the subset of elements in $overline{mathbb{F}_p}$ that are fixed by the map $xmapsto x^p$. Hence $E(mathbb{F}_p)$ is the subset of elements of $E(overline{mathbb{F}_p})$ that are fixed by the Frobenius map.






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    In the notation of the notes you link to $E(K)$ stands for the points $(x,y)in K^2$ that satisfy the equation $y^2=x^3 + Ax + b$ that defines the elliptic curve together with the point at infinity. Here $K$ can be any field in which the coefficients $A,B$ are contained. On $E(mathbb{F}_p)$ the Frobenius map $(x,y)mapsto (x^p,y^p)$ is indeed the identity. But the Frobenius map is defined on $E(overline{mathbb{F}_p})$.



    So to understand it you first need to understand what $overline{mathbb{F}_p}$ is. It is an algebraic closure of $mathbb{F}_p$, i.e. an algebraic extension of $mathbb{F}_p$ in which every non-constant polynomial has a root. This is an infinite field that contains $mathbb{F}_p$.



    In fact $mathbb{F}_p$ is the subset of elements in $overline{mathbb{F}_p}$ that are fixed by the map $xmapsto x^p$. Hence $E(mathbb{F}_p)$ is the subset of elements of $E(overline{mathbb{F}_p})$ that are fixed by the Frobenius map.






    share|cite|improve this answer


























      2














      In the notation of the notes you link to $E(K)$ stands for the points $(x,y)in K^2$ that satisfy the equation $y^2=x^3 + Ax + b$ that defines the elliptic curve together with the point at infinity. Here $K$ can be any field in which the coefficients $A,B$ are contained. On $E(mathbb{F}_p)$ the Frobenius map $(x,y)mapsto (x^p,y^p)$ is indeed the identity. But the Frobenius map is defined on $E(overline{mathbb{F}_p})$.



      So to understand it you first need to understand what $overline{mathbb{F}_p}$ is. It is an algebraic closure of $mathbb{F}_p$, i.e. an algebraic extension of $mathbb{F}_p$ in which every non-constant polynomial has a root. This is an infinite field that contains $mathbb{F}_p$.



      In fact $mathbb{F}_p$ is the subset of elements in $overline{mathbb{F}_p}$ that are fixed by the map $xmapsto x^p$. Hence $E(mathbb{F}_p)$ is the subset of elements of $E(overline{mathbb{F}_p})$ that are fixed by the Frobenius map.






      share|cite|improve this answer
























        2












        2








        2






        In the notation of the notes you link to $E(K)$ stands for the points $(x,y)in K^2$ that satisfy the equation $y^2=x^3 + Ax + b$ that defines the elliptic curve together with the point at infinity. Here $K$ can be any field in which the coefficients $A,B$ are contained. On $E(mathbb{F}_p)$ the Frobenius map $(x,y)mapsto (x^p,y^p)$ is indeed the identity. But the Frobenius map is defined on $E(overline{mathbb{F}_p})$.



        So to understand it you first need to understand what $overline{mathbb{F}_p}$ is. It is an algebraic closure of $mathbb{F}_p$, i.e. an algebraic extension of $mathbb{F}_p$ in which every non-constant polynomial has a root. This is an infinite field that contains $mathbb{F}_p$.



        In fact $mathbb{F}_p$ is the subset of elements in $overline{mathbb{F}_p}$ that are fixed by the map $xmapsto x^p$. Hence $E(mathbb{F}_p)$ is the subset of elements of $E(overline{mathbb{F}_p})$ that are fixed by the Frobenius map.






        share|cite|improve this answer












        In the notation of the notes you link to $E(K)$ stands for the points $(x,y)in K^2$ that satisfy the equation $y^2=x^3 + Ax + b$ that defines the elliptic curve together with the point at infinity. Here $K$ can be any field in which the coefficients $A,B$ are contained. On $E(mathbb{F}_p)$ the Frobenius map $(x,y)mapsto (x^p,y^p)$ is indeed the identity. But the Frobenius map is defined on $E(overline{mathbb{F}_p})$.



        So to understand it you first need to understand what $overline{mathbb{F}_p}$ is. It is an algebraic closure of $mathbb{F}_p$, i.e. an algebraic extension of $mathbb{F}_p$ in which every non-constant polynomial has a root. This is an infinite field that contains $mathbb{F}_p$.



        In fact $mathbb{F}_p$ is the subset of elements in $overline{mathbb{F}_p}$ that are fixed by the map $xmapsto x^p$. Hence $E(mathbb{F}_p)$ is the subset of elements of $E(overline{mathbb{F}_p})$ that are fixed by the Frobenius map.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Dec 1 at 18:12









        MichalisN

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