Units under Euclidean Valuation

Theorem

Let $\left({D, +, \times}\right)$ be a Euclidean domain whose zero is $0$, and unity is $1$.

Let the valuation function of $D$ be $\nu$.

Then $\forall a, b \in D, a \ne 0, b \ne 0$:

$(1): \quad$ If $a \in D$ is not a unit of $D$ then $\nu \left({b}\right) < \nu \left({a b}\right)$

$(2): \quad a \in D$ is a unit of $D$ iff $\nu \left({a}\right) = \nu \left({1}\right)$

Proof

As $D$ is a Euclidean domain, it is also a principal ideal domain.

$(1): \quad$ If $a \in D$ is not a unit of $D$ then $\nu \left({b}\right) < \nu \left({a b}\right)$:

Consider the principal ideal $U = \left({b}\right)$ of $D$ generated by $b$.

As $\nu$ is a valuation function, we have:

$\forall x \in D, x \ne 0: \nu \left({b}\right) \le \nu \left({x b}\right)$

As $\left({b}\right)$ is a principal ideal:

$\forall c \in \left({b}\right): \exists x \in D: c = x b$

and so:

$\forall c \in \left({b}\right): \nu \left({b}\right) \le \nu \left({c}\right)$

Now suppose $\nu \left({b}\right) = \nu \left({a b}\right)$.

Then from above:

$\forall c \in \left({b}\right): \nu \left({a b}\right) \le \nu \left({c}\right)$

From the argument in Euclidean Domain is Principal Ideal Domain, we have that $U$ is also the ideal generated by $a b$.

Thus $a b$ is a divisor of $b$ itself, that is:

$\exists y \in D: b = a b y$

This leads us to the fact that $a y$ is a unit of $D$ and so $a$ is also a unit of $D$.

So if $a$ is not a unit, then it must be the case that $\nu \left({b}\right) < \nu \left({a b}\right)$.

$\blacksquare$

$(2): \quad a \in D$ is a unit of $D$ iff $\nu \left({a}\right) = \nu \left({1}\right)$:

For any $a \in D$ we have that:

$\nu \left({1}\right) \le \nu \left({1 a}\right) = \nu \left({a}\right)$

by definition of Euclidean valuation.

Let $a$ be a unit of $D$.

Then $\exists b \in D: a b = 1$.

Then:

$\nu \left({a}\right) \le \nu \left({a b}\right) = \nu \left({1}\right)$

and so:

$\nu \left({a}\right) = \nu \left({1}\right)$

Now suppose $\nu \left({a}\right) = \nu \left({1}\right)$.

We can write this as:

$\nu \left({1 a}\right) = \nu \left({1}\right)$

and it follows from $(1)$ that $a$ is a unit of $D$.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.30$: Theorem $58$