In my lecture, my professor wrote within a proof: $q_1$ in the principal ideal domain $A$ is irreducible, therefore, the ideal $q_1A$ is a maximal ideal. I don't understand how that's true.
I tried proving it by supposing that there exists a principal ideal $(p)$ generated by p, such that $ q_1A \subsetneq (p) \subsetneq A$ and to show that if it's the case then $q_1$ is not irreductible, therefore, $q_1 A$ is a maximal ideal.
If $ q_1A \subsetneq (p) \subsetneq A$ then there exists $c$ invertible such that:$ q_1 = cp \implies p=c^{-1}q \implies q_1 = c(c^{-1}p)$
if I show that $c$ and $ c^{-1}p$ are both invertible then I've proved that $q_1$ is not irreducible.
I know that $c$ is invertible but what about $ c^{-1}p$, in order to prove that it is invertiple, $p$ would have to be invertible but I have no information on p other than $ q_1A \subsetneq (p) \subsetneq A$.
Am I heading in the right direction and If so how do I finish the proof?
You are going in the right direction. Suppose there is $p \in A$ s.t. $(q_1)\subsetneq (p)$ since $q_1 \in (p)$ you have that there is a $c \in A$ (not necessary invertible) s.t. $q_1=cp$. If $c$ is not invertible then $q_1$ is not irreducible and you are done. If $c$ is invertible you have $p=c^{-1}q_1$ which means that $p \in (q_1)$ but then you have that $(p)=(q_1)$ and you are done.