Extension Galois field, multiplication of elements

Question

Consider the extension field $GF(2^4)$ and the primitive polynomial $P(x)=x^4+x^3+1$. How would I find the result of multiplication of elements $1011$ and $1100$?

My work:

$1011$ corresponds to $\{1x^3 + 0x^2 + 1x^1 + 1x^0\} = x^3 + x + 1$

$1100$ corresponds to $\{1x^3 + 1x^2 + 0x^1 + 0x^0\} = x^3 + x^2$

Multiplying: $(x^3 + x + 1 ) * ( x^3 + x^2 )\\ = x^6 + x^ 5 + x^4 + 2x^3 + x^2\\ = x^2(x^3+1) + x(x^3+1) + (x^3+ 1) + 2x^3 + x^2\\ = x^5 + x^2 + x^4 + x + x^3 + 1 + 2x^3 + x^2 \\ = x(x^3 + 1) +x^2 + (x^3 + 1) + x + x^3 + 1 + 2x^3 + x^2 \\ = x^4 + x + x^2 + x^3 + 1 + x + x^3 + 1 + 2x^3 + x^2 \\ = (x^3 +1) + x^2 + x^3 + 1 + x + x^3 + 1 + 2x^3 + x^2 \\ = 5x^3 + 2x^2 + x + 3 = x^3 + x + 3 \\ $

Binary representation = 1011

But, I'm not sure if this is correct.

Answer 1

You could do this in binary, using long hand (carryless - xor instead of add) multiplication and (borrowless - xor instead of subtract) division:

Multiply

    1011
    1100
    ----
    0000
   0000
  1011
 1011
 -------
 1110100  == x^6 + x^5 + x^4 + x^2

Divide

            101
      ---------
11001 | 1110100
        11001
        -----
         01000
         00000
         -----
          10000
          11001
          -----
           1001   == x^3 + 1

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Or following the method in the original question, noting that x^i + x^i == 0:

$(x^3 + x + 1 ) * ( x^3 + x^2 )\\ = (x^6 + x^4 + x^3) + (x^5 + x^3 + x^2) \\ = x^6 + x^5 + x^4 + x^2 \\ = x^2(x^3 + 1) + x(x^3 + 1) + (x^3 + 1) + x^2 \\ = (x^5 + x^2) + (x^4 + x) + x^3 + 1 + x^2 \\ = x^5 + x^4 + x^3 + x + 1 \\ = x(x^3 + 1) + (x^3 + 1) + x^3 + x + 1 \\ = (x^4 + x) + (x^3 + 1) + x^3 + x + 1 \\ = x^4 \\ = x^3 + 1 $