how to interpret theorem about polynomial factorization over modulo ring?

Question

polynomial $X^n+a_1X^{n-1}+...+a_n \in \Bbb Z_2[X]$ doesn't have linear factors

$\iff a_n(1+\sum a_i) \neq 0$.

How then $f(X)=X+1$ can has no linear factors? Doesn't the condition expands to $a_n(1+\sum a_i) = 1(1+1) =1\cdot2 =2= 0\pmod 2$ ?

shouldn't it be "non-trivial linear factors"?

Answer 1

No, it shouldn't say "non-trivial." If $a_n=0$ then $X$ is a linear factor. If $a_0+\dots+a_n=0$ then $X+1$ is a linear factor.

If it said non-trivial, then the statement would no longer be true.

This theorem is stated badly, because it is of the form $\lnot P\iff\lnot Q$, which is always equivalent to $P\iff Q$.

In this case, you can state the theorem as:

Let $f(X)=X^n+\dots +a_n\in\mathbb Z_2[X]$. Then $f(X)$ has a linear factor if and only if $a_n(a_1+\dots+a_n)=0$.

If you stated instead:

Let $f(X)=X^n+\dots +a_n\in\mathbb Z_2[X]$. Then $f(X)$ has a non-trivial linear factor if and only if $a_n(a_0+\dots+a_n)=0$.

In this statement, $P$ is "$f(X)$ has a non-trivial linear factor" and $Q$ is "$a_n(a_0+\dots+a_n)=0$." With $f(X)=X+1$, $P$ is false and $Q$ is true, so $P\iff Q$ is not true.

So your example is exactly why the theorem doesn't have the added condition.