= Multiplicative self-maps of a finite field
{title2=$f(xy)=f(x)f(y)$}
For a <finite field> with $q$ elements, exactly $q+1$ <functions> from the field to itself preserve multiplication, without requiring additivity or preservation of the identity. They are the constant-zero map, the constant-one map, and the $q-1$ power maps $x\mapsto x^k$ for $1\le k\le q-1$.
To prove completeness, $f(1)^2=f(1)$ forces $f(1)\in\{0,1\}$ because a field has no <zero divisors>. The zero choice forces the constant-zero map. For $f(1)=1$, nonzero elements must map to nonzero elements, and the restriction is a <group homomorphism> of the cyclic <multiplicative group of a finite field>. Its generator has $q-1$ possible images, giving the $q-1$ power maps on nonzero elements. Setting $f(0)=0$ extends each. The only further possibility is $f(0)=1$, which forces every value to be one. For $q=5$, this gives six maps. These are generally not <field homomorphisms>, because preservation of addition was never required.
Back to article page
