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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?

user26857
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KunXian Xia
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6 Answers6

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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.

Marc van Leeuwen
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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.

Sigur
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Ayman Hourieh
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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.

lhf
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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.

user26857
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dust05
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  • This statement is not a bi-implication, 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
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    @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
  • @dust05 Can you please explain why (a,x) cannot be a principal ideal? – Krishan Feb 27 '21 at 18:12
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    K leave it I got it! – Krishan Feb 28 '21 at 14:32
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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 $A$.

3

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.

Manjoy Das
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