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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2 Responses to Proof that a Euclidean Domain is a PID (Principal Ideal Domain)

  1. Pingback: Is Z[x] a Principal Ideal Domain? | Singapore Maths Tuition

  2. Pingback: Z[Sqrt(-2)] is a Principal Ideal Domain Proof | Singapore Maths Tuition

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