# how to interpret theorem about polynomial factorization over modulo ring?

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"?

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Well, in your example, you have $1(1+1) = 1\cdot 0 = 0$, but the condition says $a_n(1 + \sum a_i) \neq 0$. –  Daniel Fischer Jul 27 '13 at 16:01
$x+1$ is itself a linear factor. –  Kunnysan Jul 27 '13 at 16:01
@DanielFischer so the condition isn't satisfied but it has no linear factors, this is why I ask –  AB_ Jul 27 '13 at 16:09
The claim is that that polynomial has no linear factors if and only if $a_n(1 + \sum a_i) \neq 0$. In the case of $X+1$, $a_n(1 + \sum a_i) = 1(1+1) = 0$. So you should expect a linear factor. This linear factor is $X+1$. Where's the confusion? –  Billy Jul 27 '13 at 16:28
(What is a "trivial" linear factor anyway? There are only two linear polynomials - $X$ and $X+1$ - and neither satisfies the condition $a_n(1 + \sum a_i) \neq 0$.) –  Billy Jul 27 '13 at 16:28

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 $a_n(a_1+\dots+a_n)=0$.

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.
$f(X)=0$ does not satisfy the above, as it's not in the form postulated in the hypothesis (there's no $X^n$ with coefficient $1$). Note that it requires that $a_0=1$. –  tomasz Jul 27 '13 at 17:04
Oops, missed that, thanks. But the theorem extends to $f(X)=0$. :) @tomasz –  Thomas Andrews Jul 27 '13 at 17:07