Revision history [back]
 | 1 | initial version |
Let us start with some finite field $F$ of characteristic $2$ with $q = 2^r$ elements, and some irreducible polynomial $g$ of (small) degree $s$ over this field.
Then $L=F[Y]/g(Y)$ is a field with $q^s=2^{rs}§ elements, so the corresponding Frobenius morphism is the identity isomorphism of $L$.
Now let us fix $P=P(y)=P(Y)\mod g(Y)$. Let $$ Q=P^{\displaystyle 2^{rs-1}}\ . $$ Then $$ Q^2=\left(P^{\displaystyle 2^{rs-1}}\right)^2 =P^{\displaystyle 2^{rs}}=P\ . $$
Example:
F.<a> = GF(2^5)
R.<Y> = PolynomialRing(F)
g = Y^3 + Y + 1
print "Is g = %s irreducible? %s" % ( g, g.is_irreducible() )
S.<y> = R.quotient(g)
P = a*y
Q = P^(2^(3*5-1))
print "P = %s" % P
print "Q = %s" % Q
print "Is Q^2 == P? %s" % bool( Q^2 == P )
This gives:
Is g = Y^3 + Y + 1 irreducible? True
Q = (a^4 + a^3 + a + 1)*y^2 + (a^4 + a^3 + a + 1)*y
Is Q^2 == P? True
| | 2 | No.2 Revision |
Let us start with some finite field $F$ of characteristic $2$ with $q = 2^r$ elements, and some irreducible polynomial $g$ of (small) degree $s$ over this field.
Then $L=F[Y]/g(Y)$ is a field with $q^s=2^{rs}§ $q^s=2^{rs}$ elements, so the corresponding Frobenius
morphism is the identity isomorphism of $L$.
Now let us fix $P=P(y)=P(Y)\mod g(Y)$. Let $$ Q=P^{\displaystyle 2^{rs-1}}\ . $$ Then $$ Q^2=\left(P^{\displaystyle 2^{rs-1}}\right)^2 =P^{\displaystyle 2^{rs}}=P\ . $$
Example:
F.<a> = GF(2^5)
R.<Y> = PolynomialRing(F)
g = Y^3 + Y + 1
print "Is g = %s irreducible? %s" % ( g, g.is_irreducible() )
S.<y> = R.quotient(g)
P = a*y
Q = P^(2^(3*5-1))
print "P = %s" % P
print "Q = %s" % Q
print "Is Q^2 == P? %s" % bool( Q^2 == P )
This gives:
Is g = Y^3 + Y + 1 irreducible? True
Q = (a^4 + a^3 + a + 1)*y^2 + (a^4 + a^3 + a + 1)*y
Is Q^2 == P? True
