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the following codes for 2^3 field. for the same field how to check the curve parameter A,B are the element of the field

asked 2019-08-05 02:28:17 -0600

santoshi gravatar image

updated 2019-08-07 07:31:46 -0600

Iguananaut gravatar image

.

p=(2^3) 
F1.<x>=GF(2)[] 
F.<a>=GF(2^3,'a',modulus=x^3+x+1);F.modulus();
for i,x in enumerate(F):  print("{} {}".format(i, x))

A=a^2+a;A 
B=a+1

E = EllipticCurve(F, (1,A,0,0,B));E
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Changing the numbers doesn't make your question any clearer. Please clarify with words what you want to know / what you are asking for.

rburing gravatar imagerburing ( 2019-08-05 08:33:21 -0600 )edit
1

what is the sage command to check A and B belongs to the field F.

santoshi gravatar imagesantoshi ( 2019-08-08 02:10:29 -0600 )edit

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answered 2019-08-08 05:37:01 -0600

rburing gravatar image

updated 2019-08-08 05:38:24 -0600

To check that A and B belong to the field F:

sage: A in F
True
sage: B in F
True
sage: A in F and B in F
True

Elements of F are polynomials in the generator a with coefficients in GF(3). Here a satisfies the relation a^3 + a + 1 == 0 due to the choice of modulus in the definition of F. So, to get all elements of F it suffices to take polynomials of degree less than the degree of the modulus, i.e. of degree less than 3. (This is because any higher degree polynomial in a can be simplified to one of degree < 3, using the relation a^3 + a + 1 == 0.) This explains why you only get (all) polynomials of degree less than 3 in a when you enumerate(F).

Since A and B are defined as polynomials in a, they automatically belong to the field F. You can even define, say, C = a^5, and it will also belong to F. Using the relation that a satisfies you can see that C == a^2 + a + 1.

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Asked: 2019-08-05 02:28:17 -0600

Seen: 44 times

Last updated: Aug 08