I'm working on a question that asks to consider the ring $R = \mathbb{Z}[x]$. I need to show that $(x)$ is a prime ideal, show that $(x,7)$ is a prime ideal and use these facts to conclude that $R$ is not a principal ideal domain.
I believe I've gotten to the point where I have shown that both $(x)$ and $(x,7)$ are prime ideals, but is using these facts to conclude that $\mathbb{Z}[x]$ is not a principal ideal domain somewhat trivial? From what I understand, in a principal ideal domain all ideals are principal, i.e., all of the form $(a)$ for some element $a \in \mathbb{Z}[x]$.
Since I have shown that $(x,7)$ is a prime ideal, does this in fact mean that $\mathbb{Z}[x]$ is not a PID? Or do I need to go further and prove that somehow the prime ideal $(x,7)$ is not equal to some other principal ideal $(a)$? If so, how could I go about showing that $(x,7)$ is in fact not principal?
You know that $(x)$ and $(x,7)$ are prime ideals. Since $7\notin(x)$ we have $(x)\subsetneq(x,7)$. Now suppose that $R$ is a PID. There is $p\in R$, a prime element, such that $(x,7)=(p)$. From $(x)\subsetneq(p)$ we get $p\mid x$ and since $x$ is prime, hence irreducible, $p$ is associated to $x$, so $(x)=(p)$, a contradiction.