# AES. Algorithm of multipication in finite field [duplicate]

I multiply 2 numbers using their polynomial representation and addition via XOR. Do I need to find the remainder of the division by $$x ^ 8 + x ^ 4 + x ^ 3 + x + 1$$, using the operator "%" or by division, using the rules of finite fields?

• Welcome to Cryptography. Keep a while we find you a good duplicate. In short $x ^ 8 = x ^ 4 + x ^ 3 + x + 1$ use this identity. That is very common in Finite Field applications that eliminates the polynomial division/remainder. – kelalaka May 6 '20 at 16:59
• – kelalaka May 6 '20 at 17:06
• Does this answer your question? How is multiplication in field $GF(2^{8})$ done? – hardyrama May 6 '20 at 18:14