Field $\mathbb F_2$, Reducible and irreducible polynomials In a lecture on ring theory, we stated that if $K$ is a field, then $K[X]$ is factorial and then looked at examples of polynomials in $K[X]$ (for instance for $K=\mathbb C$). Then we chose $K=\mathbb F_2$. We wrote down: $$(X-a)(X-b) = \begin{cases}X^2, \text{if} \ a=0=b, \\ X^2+X, \text{if} \ a=1, b = 0 \ \text{and vice versa}, \\ X^2+1, \text{if}\ a = 1 = b.\end{cases} $$

We continued by stating that $X^2+X+1$ would be irreducible over $F_2$, since this polynomial doesn't appear in the above list. But if $a=1$ and $b=0$, then the factor $(X-0)$ is a unit, or I am mistaken? Therefore - at least for me - the above list does not necessarily state all the reducible polynomials with degree $2$ ... 

Clarification would really be appreciated! 

Kind regards, 
MathIsFun
$X = X-0$ is not a unit.  if it were, there would be an $f(X) \in \Bbb{F}_2[X]$ such that $X f(X) = 1$, but this product always produces a polynomial one degree higher than that of $f$.

For a general field of coefficients, one should take care of units, which in $F[x]$ are the nonzero elements of $F$, again by arguing through degrees.  However when the field is $\Bbb{F}_2$, there is only one unit and it is adequately represented in the list of possible factorizations you have written.
$X = X-0$ is not a unit.  if it were, there would be an $f(X) \in \Bbb{F}_2[X]$ such that $X f(X) = 1$, but this product always produces a polynomial one degree higher than that of $f$.

For a general field of coefficients, one should take care of units, which in $F[x]$ are the nonzero elements of $F$, again by arguing through degrees.  However when the field is $\Bbb{F}_2$, there is only one unit and it is adequately represented in the list of possible factorizations you have written.