I'm trying to parse McEliece's paper on Factorization of Polynomials over Finite Fields, particularly the algorithm rather than the proof of why it works, and although I understand part of it, something escapes me.
On p. 865 the paper says that to factor the squarefree $\bar{f} = x^7+x^5+x^4+x+1$ (represented in binary as 10110011) in $GF(2)[x]$, we compute $T_1(x) = \sum\limits_{k=0}^{N-1} x^{2^k} \bmod \bar{f} = x^6 + x^2 + x + 1$ (1000111) and then because of Theorem 1, we can write $\bar{f}$ as
$$\bar{f}=\gcd(\bar{f},T_1(x)) \cdot \gcd(\bar{f},g(x))$$
where $g(x) = x^5 + x + 1$ (100011) just appears in the paper and I'm not sure where it comes from.
Where did this polynomial come from?
Theorem 1 implies that $$\overline{f}=\prod_{a\in GF(2)}\gcd(\overline{f},T_1(x)-a).$$ With $a=0$ you get the factor $\gcd(\overline{f};T_1(x))$. The other factor is then surely $\gcd(\overline{f},T_1(x)-1)$. The polynomial $g(x)$ that you are asking about comes from the observation $$ T_1(x)-1=x g(x). $$ The trivial factor $x$ of $T_1(x)-1$ can be left out, because $x\nmid \overline{f}$.