Let $f(x) = x^6 + x + 1$ and define the field $F = \mathbb{Z}_2[x]/f(x)$
Compute the following in this field:
1. $(x^5 + x + 1)(x^3 + x^2 +1)$
I start by multiplying (in $\mathbb{Z}_2[x]$):
$(x^5 + x + 1)(x^3 + x^2 +1)$ = $(x^8 + x^7 + x^5 +x^4 + x^2 + x + 1)$
Then dividing the result with $f(x)$:
$(x^2 + x)$ and the remainder $(x^5 + x^4 + x^3 + x^2 + 1)$
Is this the right approach for solving this problem? Do I understand it correct that I want the result of my multiplication mod $f(x)$? Can I think of it as a simple modulus calculation:
$11$ mod $7 = a$
$7*1 + 4$ mod $7 = 4$
In my case I have
$(x^6 + x +1)*(x^2 + x) + \mbox{remainder}$ mod $(x^6 +x + 1) = \mbox{remainder}$
So my answer to the question would be the remainder, $(x^5 + x^4 + x^3 + x^2 + 1)$?
2. $(x + 1)^{-1}$
I read (wiki) that the inverse to $(x + 1)$ could be found by using the extended euclidean alg. for $a = (x^6 + x +1)$, $b = (x+1)$ but I don't really get it since $a$ is irreducible.
Any hints in the right direction would be appreciated!