0
$\begingroup$

I know that $x^2 + 1$ is irreducible over $\mathbb{R}$, since the $\sqrt{-1}$ is not in the reals.

How can I verify whether or not $x^2 + 1$ is irreducible over $\mathbb{Z_2}$?

So far I have considered that it suffices to show that the polynomial has no reducible factors of degree $1$ or $2$ if it is irreducible.

$\endgroup$
1
  • 1
    $\begingroup$ Since $1=-1$, $X^2+1=(X-1)(X+1)$, so it factors. $\endgroup$
    – Pedro
    Commented Apr 1, 2015 at 0:47

2 Answers 2

2
$\begingroup$

Write $Z_2 = \lbrace 0,1 \rbrace$.Then $0^2 + 1 = 1$ and $1^2 + 1 =2 = 0$ in $Z_2$ i.e. $1$ is a root of $X^2 + 1$.Hence it is reducible in $Z_2$.

$\endgroup$
1
$\begingroup$

Since it is quadratic, the only way it could be reducible would be for it to factor completely. So, to show it's irreducible, you only need to show that it has no roots in $\mathbb{Z}/2$. If you find a root, then it is reducible.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .