Is $\mathbb{Z}[x]$ a principal ideal domain?

Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a principal ideal domain?

Solution 1:

If $\Bbb Z[X]$ were a principal ideal domain, then its quotient by the ideal generated by$~X$, an element that is obviously irreducible, would have to be a field. But it is clear that $\Bbb Z[X]/(X)\cong\Bbb Z$ which is not a field.

Solution 2:

Hint: Consider the ideal $(2, x)$. Show that it's not principal.

Suppose $(2, x) = (p(x))$ for some polynomial $p(x) \in \mathbb Z[x]$. Since $2 \in (p(x))$, then $2 = p(x) q(x)$ for some polynomial $q(x)\in \mathbb Z[x]$. Since $\mathbb Z$ is an integral domain, we have $\operatorname{degree} p(x)q(x) = \operatorname{degree}p(x) + \operatorname{degree}q(x)$. Thus, both $p(x)$ and $q(x)$ must be constant. The only possible options for $p(x)$ are $\{\pm 1, \pm 2\}$. Each possibility gives a contradiction. I'll let you show this.

Solution 3:

Here is a general result:

If $D$ is a domain, then $D[X]$ is a PID iff $D$ is a field.

Solution 4:

No. Consider the ideal $(2, x)$. In general, $F[x]$ is PID if and only if $F$ is a field. Integer set is not a field.

Solution 5:

As Ayman pointed out, one can consider $I=(2,x)\triangleleft\mathbb Z[x]$.

For contradiction, suppose $I=(p(x))$ for some $p(x)\in\mathbb Z[x]$. Consider that $2x+2=2(x+1)\in I$. Then $2(x+1)=p(x)q(x)$ for some $q(x)\in\mathbb Z[x]$. By hypothesis, we must have $p(x)=2$.

But now observe that $x+2\in A$. However, there is no $h(x)\in\mathbb Z[x]$ such that $x+2=2h(x)$. So $p(x)$ cannot generate $A$.