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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?

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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. Thank you in advance.

elliptic-curves cryptocurrency

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edited 2 hours ago

kelalaka

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asked 8 hours ago

inherithandleinherithandle

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add a comment | 

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. Thank you in advance.

elliptic-curves cryptocurrency

share|improve this question

edited 2 hours ago

kelalaka

8,74532351

asked 8 hours ago

inherithandleinherithandle

1061

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. Thank you in advance.

elliptic-curves cryptocurrency

share|improve this question

edited 2 hours ago

kelalaka

8,74532351

asked 8 hours ago

inherithandleinherithandle

1061

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 2 hours ago

kelalaka

8,74532351

asked 8 hours ago

inherithandleinherithandle

1061

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 2 hours ago

kelalaka

8,74532351

asked 8 hours ago

inherithandleinherithandle

1061

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago

kelalaka

8,74532351

edited 2 hours ago

kelalaka

8,74532351

asked 8 hours ago

inherithandleinherithandle

1061

New contributor

inherithandle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago

inherithandleinherithandle

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$begingroup$

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 $y^2=x^3-7$. With characteristic $p$, when you add a number $p$ times you will get the identity.

$$kP= (k bmod p) P = P+P+cdots+P.$$

Therefore, when you add the $P$ itself, again and again, you will reach the identity.

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

share|improve this answer

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

answered 6 hours ago

kelalakakelalaka

8,74532351

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

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 $y^2=x^3-7$. With characteristic $p$, when you add a number $p$ times you will get the identity.

$$kP= (k bmod p) P = P+P+cdots+P.$$

Therefore, when you add the $P$ itself, again and again, you will reach the identity.

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

share|improve this answer

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

answered 6 hours ago

kelalakakelalaka

8,74532351

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $y^2=x^3-7$. With characteristic $p$, when you add a number $p$ times you will get the identity.

$$kP= (k bmod p) P = P+P+cdots+P.$$

Therefore, when you add the $P$ itself, again and again, you will reach the identity.

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

share|improve this answer

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

answered 6 hours ago

kelalakakelalaka

8,74532351

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour 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 $y^2=x^3-7$. With characteristic $p$, when you add a number $p$ times you will get the identity.

$$kP= (k bmod p) P = P+P+cdots+P.$$

Therefore, when you add the $P$ itself, again and again, you will reach the identity.

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

share|improve this answer

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

answered 6 hours ago

kelalakakelalaka

8,74532351

$endgroup$

share|improve this answer

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

answered 6 hours ago

kelalakakelalaka

8,74532351

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

edited 1 hour ago

Squeamish Ossifrage

22.2k132100

answered 6 hours ago

kelalakakelalaka

8,74532351

answered 6 hours ago

kelalakakelalaka

8,74532351