Addition and multiplication calculation of GF (2^8)

        When preparing for the cryptography exam , see AES Yes GF（2^8） The above calculation , At first, the calculation method in the book , I don't understand much , I also read the textbook of mathematical basis of information security , To sum up, in GF（2^8） Addition and multiplication on fields . We use reduced polynomials （ It cannot be decomposed into the product of two polynomials ）X^8+X^4+X^3+X+1 For example .

The following is the process of my handwriting （ Font ignore ）

2. Is to use the long division method to find the modulus of polynomials

3. The remainder after long division is polynomial modulus X^8+X^4+X^3+X+1 Result .