Cardinality of multiplicative group of a field $GF(q)/f(x)$ where $f(x)$ is an irreducible polynomial over GF(q)

Question

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?

Answer 1

$\varphi(q^k)$ is the cardinality of the multiplicative group of units of the ring $\mathbf Z/q^f\mathbf Z$, which has $q^k$ elements too, but is not even an integral domain (it's an artinian local ring, with maximal ideal $(p)/(q^k)$, if $q$ is a power of the prime $p$).

Answer 2

$\varphi(n)$ is the cardinality of $(\mathbb{Z}_n, \cdot)$

However $GF(n)\cong (\mathbb{Z}_n, \cdot)\iff n$ is prime.

Then also the cardinalities match, as $\varphi(n) = n-1$ when $n$ is prime

Answer 3

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 $q^k-1$.