# Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm

## Statement

Suppose $R$ is a commutative unital ring and $N$ is a Euclidean norm on $R$ -- in particular, $R$ is a Eulidean ring. We can define a new Euclidean norm $\tilde{N}$ on $R$ as follows (for $a \ne 0$):

$\tilde{N}(a) = \min \{ N(ax) \mid x \in R, ax \ne 0 \}$.

In other words, it is the minimum of the norms of nonzero elements in the principal ideal generated by $a$.

$\tilde{N}$ is a multiplicatively monotone norm and further, $\tilde{N}(a) \le N(a)$ for any $a \ne 0$. Thus, $\tilde{N}$ is a smaller multiplicatively monotone Euclidean norm on $R$.

## Proof

Given: A ring $R$ with a Euclidean norm $N$. Define, for $a \ne 0$, $\tilde{N}(a) = \min \{ N(ax) \mid x \in R, ax \ne 0 \}$.

To prove: $\tilde{N}$ is Euclidean, $\tilde{N}(ab) \ge \max \{ \tilde{N}(a), \tilde{N}(b) \}$ for $ab \ne 0$, and $\tilde{N}(a) \le N(a)$ for $a \ne 0$.

Proof:

1. Proof that $\tilde{N}(a) \le N(a)$ for all $a \ne 0$: This is direct since $\tilde{N}(a)$ is the minimum over a collection of numbers that includes $N(a)$.
2. Proof that $\tilde{N}(ab) \ge \max \{ \tilde{N}(a), \tilde{N}(b)\}$ for $ab \ne 0$: This follows from the fact that the set of nonzero multiples of $ab$ is contained in the set of nonzero multiples of $a$, as well as in the set of nonzero multiples of $b$.
3. Proof that $\tilde{N}$ is Euclidean: Suppose $a,b \in R$ with $b \ne 0$. Then, $\tilde{N}(b) = N(bx)$ for some $x \in R$. Euclidean division of $a$ by $bx$ with respect to the original norm $N$ yields $a = (bx)q + r$ where $r = 0$ or $N(r) < N(bx)$. In particular, $r = 0$ or $\tilde{N}(r) \le N(r) \le N(bx) = \tilde{N}(b)$. Thus, we have $a = b(xq) + r$ where $r = 0$ or $\tilde{N}(r) < \tilde{N}(b)$.