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>=<p(x)>=\{ f(x)p(x)\vert f(x)\in \mathbb{Z}[x]\}.

Note that 2\in <2,x>, hence 2\in <p(x)>.

2=f(x)p(x)

p(x)=2 or -2.

<p(x)>=<2>

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


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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(b).

(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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