# On Showing that $\mathbb{Z}[x]$ is not a Principal Ideal Domain

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?

• If $(x,7)=(a)$, then $a$ would be a common divisor of $a$ and of $7$ in $\mathbb{Z}[x]$. There aren't many of those, and you can check none of them yield $(x,7)$, because prime ideals can't be the whole thing. But this just requires you to know that $(x,7)\neq\mathbb{Z}[x]$, so showing it is prime seems like overkill to me. Commented May 12, 2022 at 19:52
• I know you're not taking this approach, but it's worth having in your pocket: the Krull dimension of $\mathbb Z[x]$ is $2$, but the Krull dimension of any PID is $1$. Commented May 12, 2022 at 19:55
• @ArturoMagidin - Do you mean that $a$ would be a common divisor of $x$ and of 7? Either way, I'm not sure how you can come to that conclusion Commented May 12, 2022 at 20:06
• @Oderus: If $(x,7)=(a)$, then $x\in (a)$, so $x=ap$ for some $p\in\mathbb{Z}[x]$. So $a$ divides $x$. Symmetrically, since $7\in (x,7)=(a)$, you would have $a|7$. Any element that lies in a principal ideal $(a)$ must be a multiple of the generator, so the generator is a divisor of every element in the ideal. Commented May 12, 2022 at 20:07
• @ArturoMagidin I think I understand now - under this reasoning, $a|7$ so either $a = 1$ or $a = 7$, but it cannot be that $a=1$ since otherwise $(a)$ would be the entire ring, hence $a = 7$. But $7\nmid x$, hence the contradiction. Commented May 12, 2022 at 20:10

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.