An integral domain \(A\) is said to be an Euclidean domain if there is a function \(\delta\colon A\to\Z_+\) such that if \(a,b\in A\) are nonzero then there exists \(q,r\in A\) such that
Also the division algorithm for polynomials in \(F[X]\) shows that \(F[X]\) is an Euclidean domain if we define \(\delta\colon F[X]\to\Z_+\) by \(f(X)\mapsto 2^{\deg(f)}\text{.}\)
Consider nonzero elements \(x,y\in\Z[i]\text{.}\) If \(\delta(a)\lt\delta(b)\) then we take \(q=0\in\Z[i]\) and \(r=a\in\Z[i]\) to obtain (5.3.1). So we assume that \(\delta(a)\geq\delta(b)\text{.}\) Let \(b=m_1+n_2i\text{.}\) Since \(0\neq b\) its inverse \(c=\tfrac{m_1}{m_1^2+n_1^2}-\tfrac{n_1}{m_1^2+n_1^2}i\) exists in \(\C\text{.}\) Therefore, \(ac=k+\ell i\in\C\) with \(k,\ell\in\Q\text{.}\) Choose \(r,s\in\Z\) such that \(|k-r|\leq\tfrac{1}{2}\) and \(|\ell-s|\leq\tfrac{1}{2}\text{.}\) Put \(M=k-r\) and \(N=\ell-s\text{.}\) Thus \(ac=(M+r)+(N+s)i\in\C\text{.}\) In other words, \(a=b(r+si)+b(M+Ni)\text{.}\) Note that \(M,N\in\Q\text{.}\) However, \(b(M+Ni)=a-b(r+si)\in\Z[i]\text{.}\) Following the definition of \(\delta\) it is easy to see that \(\delta\left(b(M+Ni)\right)\lt\delta(b)\text{.}\)
Theorem5.3.5.
Every Euclidean domain is a principal ideal domain.
Let \(A\) be an Euclidean domain with a map \(\delta\colon A\to\Z_+\text{.}\) Since the zero ideal is principal we show that every nonzero ideal in \(A\) is principal. Suppose that \(\mfa\) is a nonzero ideal. Choose \(0\neq a\in\mfa\) with \(\delta(a)\leq\delta(x)\) for every nonzero element \(x\in\mfa\text{.}\) Thus for any \(x\in\mfa\) there exists \(q_x,r_x\in A\) with \(\delta(r_x)\lt\delta(a)\) and \(x=aq_x+r_x\text{.}\) By the choice of \(a\) we thus have \(r_x=0\text{.}\) This shows that \(\mfa=(a)\text{,}\) i.e., \(\mfa\) is a principal ideal.
