Notes - Rings and Modules HT24, Euclidean domains

For all $a, b \in R$ with $b \ne 0$, $\exists q, r \in R$ such that $a = qb + r$ and either $r = 0$ or $N(r) < N(b)$.

Overall idea: Ideally, $\frac a b$ could be done exactly and we would have a Gaussian integer as a result. But this is not always the case. So we pick the closest Gaussian integer to the result of the division, and show that the bit left (the remainder) has a small modulus.

Want to show that for all $a, b \in R$ with $b \ne 0$, $\exists q, r \in R$ such that $a = bq + r$ and either $r = 0$ or $N(r) < N(b)$.

Suppose $a, b \in \mathbb Z[i]$ and $b \ne 0$. Then $a/b \in \mathbb C$, so $a/b = u + iv$ where $u, v \in \mathbb Q$. Let $s, t$ be the closest integers to $u$ and $v$, so that $ \vert s - u \vert , \vert t - v \vert \le 1/2$.

Define $q = s + it$ and $r = qb - a$ so that $a = qb + r$. Then

\[\begin{aligned} N(r) &= |qb - a|^2 \\\\ &=\left|q - \frac a b\right|^2 |b|^2 \\\\ &= \left|(s + it) - (u + iv)\right|^2 |b|^2 \\\\ &= |(s - u) + i(v - t)|^2 |b|^2\\\\ &= |s - u|^2 + |v - t|^2 |b|^2\\\\ &\le \left(\frac 1 4 + \frac 1 4 \right)|b|^2 \\\\ &= \frac 1 2 |b|^2\\\\ &= \frac 1 2 N(b) \end{aligned}\]

So $r$ has the required properties.

(A similar argument also works for more general $\mathbb Z[\omega]$ where $\omega$ is the third root of unity. You can use the same norm but get a slightly different bound on the furthest a point in $\mathbb C$ can be from a lattice point in $\mathbb Z[\omega]$).

Let $I$ be a nonzero ideal. Take $d \in I$ such that $N(d)$ is minimal. Then $m = qd + r$ where $r = 0$ or $N(r) < N(d)$. But $r = m - q d \in I$, so the minimality of $N(d)$ implies $r = 0$ and $m = qd$. Then $I \subseteq Rd$, and since $d \in I$, $Rd \in I$, so $Rd \subseteq I$ so $I = Rd$ as required.

Suppose $3 = (a + bi)(c + di)$. Taking norms and using the fact the norm is multiplicative, this would imply

\[9 = (a^2 + b^2) (c^2 + d^2)\]

Which is impossible for integers $a, b, c, d$.

All Euclidean domains are in particular PIDs:

Let $I$ be a nonzero ideal. Take $d \in I$ such that $N(d)$ is minimal. Then $m = qd + r$ where $r = 0$ or $N(r) < N(d)$. But $r = m - q d \in I$, so the minimality of $N(d)$ implies $r = 0$ and $m = qd$. Then $I \subseteq Rd$, and since $d \in I$, $Rd \in I$, so $Rd \subseteq I$ so $I = Rd$ as required.

But $\langle x, y \rangle$ is an ideal in $\mathbb F[x, y]$ that is not principal, so it can’t be a Euclidean domain:

Suppose $\langle x, y \rangle = \langle p \rangle$ for some $p$. Then $x = i p$ and $y = jp$ for some $i, j \in \mathbb F[x]$. So it follows that the degree of $p$ is less than or equal to $1$. It can’t be $0$, since then $p$ is a constant and since we are in a field, this is a unit, but $\langle x, y \rangle \ne \mathbb F[x, y]$. So it would have to be degree $1$. But then $p = ax + by + c$ for some $a, b, c \in \mathbb F$. But then $x = i(ax + by + c)$ means $b = 0$ and likewise $y = j(ax + c)$ means $a = 0$, which would imply $a$ were a constant. Again, a contradiction. So no such $p$ exists.