An integral domain \(R\) is said to be an Euclidean domain if there is a function \(N: R \to \{n\in\mathbb{Z} \mid n\geq 0\}\) such that \(N(0)=0\) and for each \(a,b\in R\) with \(b\neq 0\), there exist elements \(q,r\in R\) with \[\begin{align*} a = qb + r, \quad \text{and} \quad r = 0 \, \text{ or } \, N(r) < N(b). \end{align*}\]
Prove:
The ring \(F[[x]]\) of power series over a field \(F\) is an Euclidean domain.
Every Euclidean domain is a PID.
Solution
1. A nonzero series \(f\in F[[x]]\) has a uniquefactorization \(f=x^k u\) with \(u\) a unit and \(u(0)\ne0\),where \(k\) is its least nonzero coefficient index.
Write \(f=\sum_{j\geq0}a_jx^j\) and choose the least \(k\) with \(a_k\ne0\). Then \(f=x^k u\) with \(u=\sum_{j\geq0}a_{k+j}x^j\). For a series \(u=\sum u_jx^j\) with \(u_0\ne0\), define coefficients of its inverse recursively by \[ v_0=u_0^{-1},\qquad v_n=-u_0^{-1}\sum_{i=1}^n u_i v_{n-i}\quad(n\geq1). \] The coefficient of \(x^0\) in \(uv\) is one and every higher coefficient is zero. Thus \(uv=1\). Conversely, a unit must have nonzero constant term by taking constant coefficients in \(uv=1\).
If nonzero series \(f,g\) have least indices \(k,l\), their product has least index \(k+l\), with coefficient \(a_k b_l\ne0\). This also proves that \(F[[x]]\) is an integral domain. The index \(k\) and the shifted series \(u\) are uniquely determined by the coefficients.
2. The least nonzero coefficient index is aEuclidean function on the nonzero series.
Set \(N(0)=0\) and \(N(f)=k\) as in step 1 for \(f\ne0\). Given \(a\in F[[x]]\) and a nonzero \(b=x^k u\), let \(r\) be the polynomial consisting of the terms of \(a\) in degrees less than \(k\). There is a series \(c\) with \(a=r+x^k c\). Set \(q=cu^{-1}\). Commutativity gives \(a=qb+r\). If \(r\ne0\), its least nonzero degree is less than \(k=N(b)\), as required. If \(k=0\), then \(r=0\) and the division is exact. Thus the stated convention, including \(N(0)=0\), is satisfied in all cases.
3. Every Euclidean domain \(R\) is a PID.
Let \(I\) be a nonzero ideal. The nonempty set \(\{N(a):a\in I,\ a\ne0\}\) of nonnegative integers has a least element. Choose \(b\in I\setminus\{0\}\) attaining it. For any \(a\in I\), Euclidean division gives \(a=qb+r\) with \(r=0\) or \(N(r)<N(b)\). The remainder belongs to \(I\) since \(r=a-qb\). The strict inequality contradicts minimality if \(r\ne0\), so \(r=0\) and \(a\in(b)\). Thus \(I\subseteq(b)\); the reverse inclusion follows from \(b\in I\). The zero ideal is generated by zero, so every ideal is principal. Since \(R\) is an integral domain by definition, it is a PID.