Showing a polynomial has no roots in $\mathbb{F}_p[x]$ So I'm given the polynomial $$x^5+x^2+1$$ and asked to prove that it is irreducible in $\mathbb{Z}[x]$ by first reducing it modulo 2 and then showing it has no roots in $\mathbb{F}_2$.
After reducing mod 2, the coefficients of the polynomial remain unchanged. How can I then show that $x^5+x^2+1$ has no roots in $\mathbb{F}_2$, meaning it is also irreducible in $\mathbb{Z}[x]$? I've thought about just writing out all elements of $\mathbb{F}_2[x]$ and showing that none of them divide the given polynomial, but I'm not really sure how I'd go about doing that.
 A: You have to prove it has no root in $\mathbf F_2$, so it can't have a linear factor. Therefore the only possibility for the polynomial to be reducible is that it decomposes as the product of a quadratic and a cubic factors (both irreducible).
The only irreducible quadratic polynomial in  $\mathbf F_2[X]$ is $X^2+X+1$, and it is not long to obtain the result of the Euclidean division:
$$X^5+X^2+1=(X^2+X+1)(X^3+X^2)+1.$$
A: If we were to have $f = gh$ in $\Bbb Z[X]$ for $g,h$ polynomials of positive degree, reducing the coefficients modulo $p$ would give $\overline{f} = \overline{g}\overline{h}$. Also $g$ and $h$ have positive degree (they largest degree coefficient of $g$ and $h$ is $\pm 1$, since $f$ was monic).
Hence whenever $f$ is monic being irreducible modulo $p$ implies that $f$ is irreducible in $\Bbb Z[X]$. So far so good, but we have to show now that $x^5+x^2+1$ is irreducbile in $\Bbb F_2[X]$. Note that it has no roots in $\Bbb F_2$ (just plug in $1$ and $0$), so were $f$ to be reducible we should have $q \mid f$ with $\deg q = 2$.
By the previous remark; the polynomial $q$ has to be irreducible, and there's only one such polynomial, namely $q = x^2+x+1$.
Therefore your problem is reduced to showing that $q \not \mid f$ in $\Bbb F_2[X]$. Can you take it from here?
A: Because this polynomial has no roots in $\mathbb{F}_{2}$ we know that it has no linear factors.
There are 4 polynomials in $\mathbb{F}_{2}[x]$ of degree 2 and 8 polynomials in $\mathbb{F}_{2}[x]$ of degree 4 - and some of them have roots in $F_{2}$, so they cannot divide the polynomial that you presented. So you only have to ckeck a small number of possible products of polynomials. Seems a very reasonable way to do this, and it follows your own idea of "writing out all elements of $\mathbb{F}_{2}[x]$ and showing that none of them divide the given polynomial".
