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Can you tell me why doing scalar multiplication of a point on a Elliptic curve over a finite field gets to a point at infinity?

What is the relationship between p (prime), n (order) and h (cofactor) of an elliptic curve?What is the point at infinity on secp256k1 and how to calculate it?Modulus for elliptic curve point multiplicationGraphically representing points on Elliptic Curve over finite fieldElliptic curve group over a prime finite field $F_p$Scalar Multiplication for Elliptic CurveUsage of parameter “b” of an elliptic curve over GF(p)Elliptic curve scalar point multiplicationElliptic curve point multiplication — who is wrong?Understanding elliptic curve point addition over a finite fieldPoint-at-infinity in the scalar multiplicationelliptic curve infinity point implementation returns exception

2

1

$begingroup$

I am reading Programming Bitcoin. The author said:

Another property of scalar multiplication is that at a certain multiple, we get to the point at infinity (remember, the point at infinity is the additive identity or $0$). If we imagine a point $G$ and scalar-multiply until we get the point at infinity.

He doesn't explain why. So I don't understand why. I would like you to give me a plain explanation, without a serious mathematical proof, if that could be possible.

elliptic-curves cryptocurrency

share|improve this question

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

asked 13 hours ago

inherithandleinherithandle

1111

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

1

$begingroup$

I am reading Programming Bitcoin. The author said:

Another property of scalar multiplication is that at a certain multiple, we get to the point at infinity (remember, the point at infinity is the additive identity or $0$). If we imagine a point $G$ and scalar-multiply until we get the point at infinity.

He doesn't explain why. So I don't understand why. I would like you to give me a plain explanation, without a serious mathematical proof, if that could be possible.

elliptic-curves cryptocurrency

share|improve this question

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

asked 13 hours ago

inherithandleinherithandle

1111

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I am reading Programming Bitcoin. The author said:

Another property of scalar multiplication is that at a certain multiple, we get to the point at infinity (remember, the point at infinity is the additive identity or $0$). If we imagine a point $G$ and scalar-multiply until we get the point at infinity.

He doesn't explain why. So I don't understand why. I would like you to give me a plain explanation, without a serious mathematical proof, if that could be possible.

elliptic-curves cryptocurrency

share|improve this question

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

asked 13 hours ago

inherithandleinherithandle

1111

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

asked 13 hours ago

inherithandleinherithandle

1111

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

asked 13 hours ago

inherithandleinherithandle

1111

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

edited 4 hours ago

Maarten Bodewes♦

55.7k679196

asked 13 hours ago

inherithandleinherithandle

1111

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 13 hours ago

inherithandleinherithandle

1111

1 Answer
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3

$begingroup$

The points on the Elliptic Curves are forming an additive group with the identity $mathcalO$, the point at infinity.

The scalar multiplication $k P$ this actually means adding $P$, $k$-times itself

$$kP=underbraceP+P+cdots+P_text$k$ times.$$

Bitcoin uses Secp256k1 which has characteristic $p$ and it is defined over the prime field $mathbbZ_p$ with the curve equation $y^2=x^3-7$.

Point addition in $mathbbZ_p$ has an interesting property since the number of elements is finite if you add a point $P$ itself many times eventually you will get the identity $mathcalO$.

$$underbraceP+P+cdots+P_text$t$ times = mathcalO$$

The smallest $t$ will be the order of the subgroup generated by the $P$. For security, we want this order huge.

Note 1: a point $P$ may not generate the whole group but it generates a cyclic subgroup.

Note 2: As pointed by SqueamishOssifrage, The Smart showed that if the order of the curve and order of the base field are same then the discrete logarithm on this curves runs in linear time.

share|improve this answer

edited 4 hours ago

answered 12 hours ago

kelalakakelalaka

8,75532351

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The order of the scalar ring is not the characteristic or order of the coordinate field. The orders are related, but are not the same except in cases that are trivially breakable as Nigel Smart showed.
  $endgroup$
  – Squeamish Ossifrage
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SqueamishOssifrage thanks and for the links.
  $endgroup$
  – kelalaka
  4 hours ago
  
  

  
  
  
  

add a comment | 

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3

$begingroup$

The points on the Elliptic Curves are forming an additive group with the identity $mathcalO$, the point at infinity.

The scalar multiplication $k P$ this actually means adding $P$, $k$-times itself

$$kP=underbraceP+P+cdots+P_text$k$ times.$$

Bitcoin uses Secp256k1 which has characteristic $p$ and it is defined over the prime field $mathbbZ_p$ with the curve equation $y^2=x^3-7$.

Point addition in $mathbbZ_p$ has an interesting property since the number of elements is finite if you add a point $P$ itself many times eventually you will get the identity $mathcalO$.

$$underbraceP+P+cdots+P_text$t$ times = mathcalO$$

The smallest $t$ will be the order of the subgroup generated by the $P$. For security, we want this order huge.

Note 1: a point $P$ may not generate the whole group but it generates a cyclic subgroup.

Note 2: As pointed by SqueamishOssifrage, The Smart showed that if the order of the curve and order of the base field are same then the discrete logarithm on this curves runs in linear time.

share|improve this answer

edited 4 hours ago

answered 12 hours ago

kelalakakelalaka

8,75532351

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The order of the scalar ring is not the characteristic or order of the coordinate field. The orders are related, but are not the same except in cases that are trivially breakable as Nigel Smart showed.
  $endgroup$
  – Squeamish Ossifrage
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SqueamishOssifrage thanks and for the links.
  $endgroup$
  – kelalaka
  4 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The points on the Elliptic Curves are forming an additive group with the identity $mathcalO$, the point at infinity.

The scalar multiplication $k P$ this actually means adding $P$, $k$-times itself

$$kP=underbraceP+P+cdots+P_text$k$ times.$$

Bitcoin uses Secp256k1 which has characteristic $p$ and it is defined over the prime field $mathbbZ_p$ with the curve equation $y^2=x^3-7$.

Point addition in $mathbbZ_p$ has an interesting property since the number of elements is finite if you add a point $P$ itself many times eventually you will get the identity $mathcalO$.

$$underbraceP+P+cdots+P_text$t$ times = mathcalO$$

The smallest $t$ will be the order of the subgroup generated by the $P$. For security, we want this order huge.

Note 1: a point $P$ may not generate the whole group but it generates a cyclic subgroup.

Note 2: As pointed by SqueamishOssifrage, The Smart showed that if the order of the curve and order of the base field are same then the discrete logarithm on this curves runs in linear time.

share|improve this answer

edited 4 hours ago

answered 12 hours ago

kelalakakelalaka

8,75532351

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The order of the scalar ring is not the characteristic or order of the coordinate field. The orders are related, but are not the same except in cases that are trivially breakable as Nigel Smart showed.
  $endgroup$
  – Squeamish Ossifrage
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SqueamishOssifrage thanks and for the links.
  $endgroup$
  – kelalaka
  4 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The points on the Elliptic Curves are forming an additive group with the identity $mathcalO$, the point at infinity.

The scalar multiplication $k P$ this actually means adding $P$, $k$-times itself

$$kP=underbraceP+P+cdots+P_text$k$ times.$$

Bitcoin uses Secp256k1 which has characteristic $p$ and it is defined over the prime field $mathbbZ_p$ with the curve equation $y^2=x^3-7$.

Point addition in $mathbbZ_p$ has an interesting property since the number of elements is finite if you add a point $P$ itself many times eventually you will get the identity $mathcalO$.

$$underbraceP+P+cdots+P_text$t$ times = mathcalO$$

The smallest $t$ will be the order of the subgroup generated by the $P$. For security, we want this order huge.

Note 1: a point $P$ may not generate the whole group but it generates a cyclic subgroup.

Note 2: As pointed by SqueamishOssifrage, The Smart showed that if the order of the curve and order of the base field are same then the discrete logarithm on this curves runs in linear time.

share|improve this answer

edited 4 hours ago

answered 12 hours ago

kelalakakelalaka

8,75532351

$endgroup$

share|improve this answer

edited 4 hours ago

answered 12 hours ago

kelalakakelalaka

8,75532351

answered 12 hours ago

kelalakakelalaka

8,75532351

answered 12 hours ago

kelalakakelalaka

8,75532351