root of irreducible polynomial I have just started studying finite fields and I'm confused by the language around irreducible polynomial and find the following definition confusing:
"If $f$ is irreducible in $\mathbb{F}_{q}[x]$ of degree $m$ then $f$ has a root a in $\mathbb{F}_{q^{m}}$ "
This seems to contradict itself because $f$ is supposed to be irreducible and therefore shouldn't have a root. I suppose I must be missing something obvious. Can someone help explain.
 A: When we say a polynomial is irreducible over $\mathbb{F}_{q}[x]$, what we mean is that it does not factor (and therefore has no roots) in $\mathbb{F}_{q}$.  It will factor over an extension field, basically there is a larger field containing $\mathbb{F}_{q}$ over which the polynomial will factor.  One interesting result is that all fields of order $q^{m}$ are isomorphic, so any irreducible polynomial in $\mathbb{F}_{q}[x]$ will factor completely in an extension field of order $q^{m}$.
One way to think of this, there are irreducible polynomials in $\mathbb{Q}[x]$ such as $x^{2}-3$, $x^{2}+1$, $x^{4}+x^{2}+x+1$, but all of these polynomials will factor completely over $\mathbb{C}$. However, the behavior is a bit different here from the finite field situation, because there are many different nonisomorphic degree 2 extensions of $\mathbb{Q}$. For example, $x^{2}-3$ factors over the degree 2 extension $\mathbb{Q}[\sqrt{3}]$, but $x^{2}+1$ does not.
A: If $f$ is irreducible of degree m, $F_q[x]/f$ is a field and a $m$-dimensional vector space,thus has $q^m$ elements, the image of $x$ in $F_q[x]/f$ is a root of $f$ in $F_q[x]/f$.
