I'm trying to understand the McEliece cryptosystem and I'm looking to this paper http://www.mif.vu.lt/~skersys/vsd/crypto_on_codes/goppamceliece.pdf
On page 26 they are calculating syndrome and somehow they got $1 + a^{10}z$ from $ (a^8+ a^4+ a^{10}) + (a^7+ a^{11} + a)z$
Could somebody please explain me how is that possible?
At the start of the section on page 25 there's a reference to Section 2.6.2. In Section 2.6.2, it's stated that $a$ (actually $\alpha$) is a primitive element satisfying $a^4+a+1 = 0$. So $a^4 = -a-1 = a+1$ (the final equality is because $-1 = 1 \mod{2}$).
Substitute $a^4 = a+1$ into $(a^8+a^4+a^{10})+(a^7+a^{11}+a)z$ and reduce. You'll find that it's equal to $1+(a^2+a+1)z$. Note that $a^{10}=a^2+a+1$.