A Question about Irreducible Polynomials 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) &amp;=&amp;
    (x^6)(x^7 + x^6 + x^3 + x) + \\
    &amp;&amp; (x^4)(x^7 + x^6 + x^3 + x) \\
    &amp;+&amp; (x + 1)(x^7 + x^6 + x^3 + x) \\
(x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &amp;=&amp;
    (x^{13} + x^{12} + x^9 + x^7) + \\
    &amp;&amp; (x^{11} + x^{10} + x^7 + x^5) \\
   &amp;+&amp; (x + 1)(x^7 + x^6 + x^3 + x) \\
(x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &amp;=&amp;
    (x^{13} + x^{12} + x^9 + x^7) + \\
    &amp;&amp; (x^{11} + x^{10} + x^7 + x^5) + \\
    &amp;&amp; (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) &amp;=&amp;
    x^{13} + x^{12} + x^{11} + x^9 + x^{10} + x^5 + \\
    &amp;&amp; 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) &amp;=&amp; x^{13} + x^{12} + x^{11} + \\
    &amp;&amp; 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) &amp;=&amp;
    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) &amp;=&amp;
    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) &amp;=&amp;
    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) &amp;=&amp; x^8 + x^6 + x^4 + x^2 \\
\end{eqnarray*}
This is not the irreducible polynomial. What am I missing?

Thanks

Bob
In the page $104$ you can see:
$$F_{2^8}=\frac{Z_2[x]}{x^8+x^4+x^3+x+1}.$$
this means that irreducible polynomial that generate the $F_{2^8}$ is $x^8+x^4+x^3+x+1$. so, in multiplications, you should note that $x^8=-x^4-x^3-x-1=x^4+x^3+x+1$. Therefore $x^{8+i}=x^8\cdot x^i=x^{4+i}+x^{3+i}+x^{1+i}+x^i$.