The finite field $GF(256)$ is usually implemented $mod$ 0x11b
to keep the numbers inside that field.
I understand that 0x11b
was picked as it is the smallest irreducible polynomial that is greater than any number of $GF(256)$. This is sufficient for multiplication as any element of the field can be reached by multiplying two other elements from the same field and the result will also be inside that field.
Now I stumbled over powers in finite fields. By trying it myself I noticed that with $g=2$ not every element of $GF(256)$ could be expressed as $g^x$ when using $mod$ 0x11b
. 0x11d
has to be used instead for example. But when using $g=3$ I can use 0x11b
again. And for $g=6$ I can use both. For $g=7$ I can't use either. So the choice of the irreducible polynomial depends on the generator $g$.
Does this mean unless you have tested the choice of irreducible polynomial $p$ throughly, you cannot use it for key exchange as proposed by Diffie-Hellman but you can use it for Elliptic Curves?
Is it even feasible to do both on a finite field?