I am doing some self-study in the area of Cryptography. I am using the Third Edition of the book "Cryptography Theory and Practice" by Douglas R. Stinson. Based upon the information on page 105, in the field $F_{2^8}$, I believe the following two polynomials are multiplicative inverses.
$x^6 + x^4 + x + 1$
$x^7 + x^6 + x^3 + x$
If they are and I multiply them together, I should get the irreducible polynomial for the field right? Here is what I get when I multiply the two polynomials together in $F_{2^8}$. \begin{eqnarray*} (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& (x^6)(x^7 + x^6 + x^3 + x) + \\ && (x^4)(x^7 + x^6 + x^3 + x) \\ &+& (x + 1)(x^7 + x^6 + x^3 + x) \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& (x^{13} + x^{12} + x^9 + x^7) + \\ && (x^{11} + x^{10} + x^7 + x^5) \\ &+& (x + 1)(x^7 + x^6 + x^3 + x) \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& (x^{13} + x^{12} + x^9 + x^7) + \\ && (x^{11} + x^{10} + x^7 + x^5) + \\ && (x^8 + x^7 + x^4 + x^2) + (x^7 + x^6 + x^3 + x) \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{13} + x^{12} + x^{11} + x^9 + x^{10} + x^5 + \\ && x^8 + x^7 + x^4 + x^2 + x^7 + x^6 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{13} + x^{12} + x^{11} + \\ && x^{10} + x^9 + x^8 + x^6 + x^5 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{5} + x^{4} + x^{3} + x^{2} + x^9 + x^8 + x^6 + x^5 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{4} + x^{3} + x^{2} + x^9 + x^8 + x^6 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{4} + x^{3} + x^{2} + x + x^8 + x^6 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^8 + x^6 + x^4 + x^2 \\ \end{eqnarray*} This is not the irreducible polynomial. What am I missing?
Thanks
Bob
