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?
6 Answers
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.
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why would it's quotient by
be a Field? – Krishan Feb 23 '21 at 17:57 
1@Krishan The quotient of a PID $R$ by its ideal generated by a nonzero element $a$ is a field iff $a$ is irreducible in $R$. – Marc van Leeuwen Feb 24 '21 at 06:05
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.
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Here is a general result:
If $D$ is a domain, then $D[X]$ is a PID iff $D$ is a field.
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2This was already mentioned in the answer by dust05, and I think we only have to require the obvious condition that $\;D\;$ has to be a commutative unitary ring. – DonAntonio Sep 21 '13 at 11:34


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.

This statement is not a biimplication, the theorem only says that if F is a field, then F[x] is a PID. Kindly check once. – Krishan Feb 23 '21 at 18:04

@Krishan, which direction fails? What counterexample? Perhaps you could ask a separate question. – lhf Feb 23 '21 at 19:59

1@Krishan suppose $F$ is a ring which is not a field. Take an element $a/in F $ which has no multiplicative inverse. Consider an ideal $(a,x)$ in $F[x]$. This cannot be principal ideal. So when polynomial ring is PID, then the coefficient ring is a field. – dust05 Feb 24 '21 at 01:40


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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 $A$.
Let $S$ be an ideal generated by $2$ and $x$ in $\Bbb{Z}[x]$.
Then $S=2f(x)+xg(x):f(x),g(x)\in \Bbb{Z}[x]$
Now let $S$ be a principal domain generated by $h(x)$.
Then $2 \in S\implies 2\in h(x)\implies 2=h(x)h_1(x)\;\text {for some}\;\; h_1(x)\in\Bbb{Z}[x] \tag {1}$
and $x \in S\implies x\in h(x)\implies x=h(x)h_2(x)$ for some $h_2(x)\in\Bbb{Z}[x]$
This gives $2h_2(x)=xh_1(x)$. This expression shows that the coefficients of $h_1(x)$ are even.
So let $h_1(x)=2p(x)$ for some $p(x)\in \Bbb{Z}[x]$
So $(1)$ gives $2=2h(x)p(x)\implies h(x)p(x)=1\implies 1\in\langle h(x)\rangle=S$
Since $1\in S\implies 2q(x)+xr(x)=1\tag{2}$ for some $q(x), r(x)\in \Bbb{Z}[x]$
Let $q(x)=a_0+a_1x+a_2x^2+\cdots$ and $r(x)=b_0+b_1x+b_2x^2+\cdots$
So $(2)$ gives $2a_0=1$ which is an impossibility.
So $S$ is not a principal ideal and consequently $\Bbb{Z}[x]$ is not a PID.
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