## Is Z[x] a Principal Ideal Domain?

In the previous post, we showed that a Euclidean domain is a Principal Ideal Domain (PID).

Consider the Polynomial Ring $\mathbb{Z}[x]$. We can show that it is not a PID and hence also not a Euclidean domain.

Proof: Consider the ideal $<2,x>=\{ 2f(x)+xg(x)\vert f(x), g(x) \in \mathbb{Z} [x]\}$.

Suppose to the contrary $<2,x>==\{ f(x)p(x)\vert f(x)\in \mathbb{Z}[x]\}$.

Note that $2\in <2,x>$, hence $2\in $.

2=f(x)p(x)

p(x)=2 or -2.

<p(x)>=<2>

However, $x\in <2,x>$ but $x\notin <2>$. (contradiction!)

## Proof that a Euclidean Domain is a PID (Principal Ideal Domain)

Previously, we defined what is a Euclidean Domain and what is a PID. Now, we will prove that in fact a Euclidean Domain is always a PID (Principal Ideal Domain). This proof will be elaborated, it can be shortened if necessary.

Proof:

Let R be a Euclidean domain.

Let I be a nonzero ideal of R. (If I is a zero ideal, then I=(0) )

Choose $b\in I, b\neq 0$ such that $d(b)=\min \{ d(i): i\in I\}$, where d is the Euclidean function. By the well-ordering principle, every non-empty set of positive integers contains a least element, hence b exists.

Let $a\in I$ be any element in I. $\exists q,r \in R$ such that $a=bq+r$, with either r=0, or d(r)<d(b). (This is the property of Euclidean domain.)

We can’t have d(r)<d(b) as that will contradict minimality of d(b). Thus, r=0, and a=bq. Hence every element in the ideal is a multiple of b, i.e. I=(b). Thus R is a PID (Principal Ideal Domain).

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## Definition of Euclidean Domain and Principal Ideal Domain (PID)

A Euclidean domain is an integral domain $R$ with a function $d:R\setminus \{0\}\to \mathbb{N}$ satisfying the following:

(1) $d(a)\leq d(ab)$ for all nonzero $a,b$ in $R$.

(2) for all $a,b \in R$, $b\neq 0$, $\exists q, r, \in R$ such that $a=bq+r$, with either $r=0$ or $d(r).

(d is known as the Euclidean function)

On the other hand, a Principal ideal domain (PID) is an integral domain in which every ideal is principal (can be generated by a single element).

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