Let \(O\) be the ring of integers of \(\mathbf{Q}(\sqrt2)\). Show that \(O\) is a Euclidean domain with respect to the absolute field norm \[ \delta(a+b\sqrt2)=\left|a^2-2b^2\right|. \]
Solution
1. The ring of integers is \(O=\mathbb Z[\sqrt2]\).
For \(a,b\in\mathbb Z\), the element \(a+b\sqrt2\) satisfies the monic polynomial \[ T^2-2aT+(a^2-2b^2)\in\mathbb Z[T], \] so it is an algebraic integer.
Conversely, write an algebraic integer in \(\mathbb Q(\sqrt2)\) as \(\alpha=a+b\sqrt2\), with \(a,b\in\mathbb Q\). If \(b=0\), the rational root theorem applied to a monic integer polynomial for \(\alpha\) gives \(a\in\mathbb Z\).
Suppose \(b\ne0\). Then \(\alpha\) has degree \(2\), with minimal polynomial \(T^2-2aT+(a^2-2b^2)\). A monic factor over \(\mathbb Q\) of a monic integer polynomial has integer coefficients, by Gauss’s lemma [@DF04]. Therefore \[ 2a=m\in\mathbb Z,\qquad a^2-2b^2=n\in\mathbb Z, \qquad 8b^2=m^2-4n\in\mathbb Z. \] Write \(b=u/v\) in lowest terms with \(v>0\). The last integrality condition gives \(v^2\mid8u^2\), hence \(v^2\mid8\) and \(v\in\{1,2\}\). Thus \(b=k/2\) for an integer \(k\), and \[ m^2-2k^2=4n. \] If \(k\) were odd, this would give \(m^2\equiv2\pmod4\), impossible because squares modulo \(4\) are \(0\) and \(1\). Hence \(k\) is even; then the same equality makes \(m\) even. Consequently \(a=m/2\) and \(b=k/2\) are both integers. This proves the asserted equality of rings.
2. The absolute norm is multiplicative and is a positive integeron every nonzero element of \(O\).
Conjugation \(\tau(a+b\sqrt2)=a-b\sqrt2\) is an automorphism of the field. Thus \[ N(\alpha)=\alpha\tau(\alpha),\qquad N(\alpha\beta)=N(\alpha)N(\beta), \] and absolute values give multiplicativity of \(\delta=|N|\). Step 1 makes \(N\) integer-valued on \(O\). If \(\alpha\ne0\), then both \(\alpha\) and \(\tau(\alpha)\) are nonzero field elements, so \(\delta(\alpha)\) is a positive integer.
3. Euclidean division holds for \(\delta\).
Let \(\alpha,\beta\in O\) with \(\beta\ne0\), and write \[ \frac{\alpha}{\beta}=u+v\sqrt2,\qquad u,v\in\mathbb Q. \] Choose integers \(m,n\) such that \(|u-m|\leq1/2\) and \(|v-n|\leq1/2\), and put \(q=m+n\sqrt2\in O\) and \(r=\alpha-\beta q\in O\). If \(r=0\), the division has zero remainder. Otherwise step 2 gives \[ \begin{aligned} \delta(r) &=\delta(\beta)\left|(u-m)^2-2(v-n)^2\right|\\ &\leq\delta(\beta)\bigl((u-m)^2+2(v-n)^2\bigr)\\ &\leq\frac34\delta(\beta)<\delta(\beta). \end{aligned} \] Thus \(\alpha=\beta q+r\) with zero remainder or strictly smaller positive integer size. Since \(O\) is a subring of a field, it is an integral domain, and this proves that it is Euclidean. Under the definition also requiring \(\delta(\alpha)\leq\delta(\alpha\beta)\) for nonzero \(\alpha,\beta\), that condition follows from multiplicativity and \(\delta(\beta)\geq1\).