Why is $X^2+1$ reducible in $F_2$?

Question

One of my modules contains parts of ring theory which I have not done in over a year. So I appreciate this question may seem very basic as I am giving myself a crash course in ring theory by studying examples.

Why is $X^2+1$ reducible in $F_2$? And why is $X^2-X+1$ irreducible in $F_2$?

Answer 1

It is reducible since it has a root in the field...Note : don't use the converse that if a polynomial doesn't have a root in the field means it is irreducible.This is not the case always.But is true only if you have a polynomial of degree 2 or 3.

Answer 2

$$\forall\;\text{ prime}\;\;p\;,\;\;(a+b)^p=a^p+b^p\pmod p\implies (x^2+1)=(x+1)^2\pmod2$$

As for $\;x^2-x+x^2+x+1\pmod 2\;$ : check this quadratic has not roots in $\;\Bbb F_2\;$ and is thus irreducible (because a polynomial of degree less than four is irredicible over a field iff it has no roots in that field)

Answer 3

$X^2 + 1$ is reducible because it is the product of smaller degree polynomials: $(x+1)(x+1) = x^2 + 1$.

Now, if $X^2 -X + 1 $ was reducible, then it would be the product of two linear factors. That means that the polynomial would have a root in $F_2$. But, you can check (easily) that the polynomial does not have any roots.