# Revision history [back]

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  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