Integral Domains
Table of Contents
1 Euclidean Domains
Definition Let \( R \) be an integral domain, then \( R \) is a Euclidean domain if there exists some function \( d: R \backslash \{0\} \to \mathbb{Z}^{\ge 0} \) which satisfies:
- If \( a \in R, \ b \ne 0 \in R \) then \( \exists q, r \in R \) such that \( a = qb + r \) and either \( r = 0 \) or \( d(r) < d(b) \)
- For all \( a, b \in R \ne 0 \) we have \( d(a) \le d(ab) \)
So Euclidean Domains generalise the Division Theorem to (commutative) rings. We will see for an integral domain, having the property that it is a Euclidean domain is a fairly strong condition.
1.1 Examples
- The Gaussian integers \( \mathbb{Z}[i] \) are a Euclidean domain. This can be seen by taking \( d(a + bi) = |a + bi| = a^2 + b^2 \).
- Any field \( F \) is a Euclidean domain since we can take \( d(0) = 0 \) and \( d(x) = 1 \) for \( x \ne 0 \). To see this satisfies (1), given some \( a, b \in R \) we can choose \( r = 0, \ q = ab^{-1} \) or \( r = a \) and \( q \) arbitrary if \( b = 0 \), and have \( a = qb + r \) in either case as required. (2) is trivially satisfied.
- If \( K \) is a field, then the polynomial ring \( K[X] \) is a Euclidean domain, with \( d(P(x)) = deg(P(x)) \).
2 Principle Ideal Domains
Definition Let \( R \) be an integral domain, then \( R \) is a PID if for all ideals \( I \) in \( R \exists a \in R \) such that \( I = \langle a \rangle \).
Theorem All Euclidean domains are Principle Ideal Domains.
Proof Let \( R \) be a Euclidean domain with Euclidean function \( d \), and let \( I \ne {0} \) be some ideal of \( R \). To prove the Theorem we show this must be principle.
Let \( b \) be any nonzero element in \( R \) such that \( d(b) \) is minimal (there always exists such a \( b \) since the codomain of \( d \) is the nonnegative integers). We claim:
\[ \left< b \right> = I \]
Let \( s \in R \) and suppose for a contradiction that \( s \not \in \left< b \right> \). By (2) we can write:
\[ s = bq + r \]
For some \( q, r \in R \) such that \( d(r) < d(b) \). Now rearranging the above equation we get:
\[ r = s - bq \]
Which implies \( r \in I \) since ideals are closed under addition. But then since \( d(b) \) is minimal in \( I \) we must have \( d(b) \le d(r) \), a contradiction. Hence we must conclude that \( s \in \left< b \right> \) and thus \( I \) is principle.