Cardinality of multiplicative group of a field $GF(q)/f(x)$ where $f(x)$ is an irreducible polynomial over GF(q) If you have an irreducible polynomial $f(x)$ over $GF(q)$, then the following should be true: 
$GF(q)/f(x) \cong GF(q^k)$, where $k = deg(f(x))$.
So the multiplicative group should have $\varphi(q^k)$ elements, where $\varphi$ is the totient.
On the other hand, since $f(x)$ is irreducible, doesn't that mean that there are no divisors of $f(x)$ in $GF(q)/f(x)$, so all of its elements should be part of the multiplicative group, save for $0$. So, arguing that way, the group should have $q^k-1$ elements.
It seems to me that I have a misunderstanding of the problem at hand, but I cannot figure out where my misconception lies. Can anyone point out what I'm doing wrong?

For what reason should the multiplicative group of $\operatorname{GF}(q^k)$ have $\varphi(q^k)$ elements? In general it is not true that $\operatorname{GF}(q^k)\cong\Bbb{Z}/q^k\Bbb{Z}$. In fact, if $k>1$ then $\Bbb{Z}/q^k\Bbb{Z}$ is not a even field because $q\neq0$ and $q^{k-1}\neq0$ in $\Bbb{Z}/q^k\Bbb{Z}$ but their product is zero.
Your second line of reasoning is correct; because $\operatorname{GF}(q^k)$ is a field, every element other than zero is a unit, so the cardinality of its unit group is $q^k-1$.
