Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to do multiplication in the GF($2^3$) defined by the irreducible minimum binary polynomial $X^3+X^2+1$. I want to multiply $A(x) * B(x)$ where $A(x) = x$ and $B(x) = x^2$. The multiplication is easy, and I get the result: $C'(x) = x^3$ However, I am confused on how to do the modulo in $X^3 + X^2 +1$. I am convinced that it can be done by XOR'ing in long division: $1000 \oplus 1101 = 0101$. Is this the correct result then: $101$?

share|improve this question
up vote 3 down vote accepted

Yes, you are correct.

The modulo operation (in general, not only over polynomials) is defined such that this is always true:

$A \equiv A + B\ \ (\bmod\ B)$

Thus, if $A = X^3$ and $B = X^3+X^2+1$, that implies that $X^3$ is equivalent to the sum of $X^3$ and $X^3+x^2+1$, which is $X^2+1$, which is represented by $101$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.