What does it mean to be an irreducible polynomial over a field? (need clarification on the definition) My understanding is to find out if a polynomial is reducible, we can use substitution. For example, to see if $x^2+1$ is reducible over $F_2$, we can substitute $0$ and $1$ for $x$, and see that neither of them result in $0$ mod $2$.
But we know $x$ is irreducible because it's of degree $1$ so you can't reduce it any further. Yet when you substitute $x$ for $0$, you get $0$. Does substituting $1$ have to yield $0$ as well? 
So generally, when I say $x$ "is a root" in a field, given some polynomial over $x$, do I want to check that ALL substitutions for $x$ with the elements in the field give me $0$? or if there exists such an element such that the polynomial gives $0$?
 A: A polynomial $f$ is irreducible over a field $\mathbb F$ if and only if it is an irreducible element of the ring $\Bbb F[x]$, id est, if and only if there are no $g,h\in\Bbb F[x]\setminus\{0\}$ such that $\deg g<\deg f$, $\deg h<\deg f$ and $gh=f$.
For instance, $x^2+1$ is not irreducible over $\Bbb F_2$ because $(x+1)\cdot(x+1)=x^2+1$ in $\Bbb F_2[x]$, while $x$ (and any polynomial of degree $1$) is irreducible over $\Bbb F_2$ because the product of two polynomials of degree strictly smaller than $1$ has degree $0$.
A polynomial has a root in $\Bbb F$ if and only if there is $\xi\in\Bbb F$ such that $f(\xi)=0$.
Due to the fact that, if $\Bbb F$ is a field, $x-\xi$ divides $f$ in $\Bbb F[x]$ if and only if $f(\xi)=0$ (sometimes it is referred to as Ruffini's theorem, I think), the following holds:

Let $\Bbb F$ be a field and let $f\in\Bbb F[x]$ with $1<\deg f\le 3$. Then, $f$ is irreducible over $\Bbb F$ if and only if it does not have roots in $\Bbb F$.

This "only if" part of the above does not extend to $\deg f>3$. For instance, in $\Bbb R[x]$, $$x^4+1=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$$ but $\xi^4+1\ne 0$ for all $\xi\in\Bbb R$.
