Every Euclidean ring has a unique smallest Euclidean norm
Statement
Every Euclidean ring (and in particular, every Euclidean domain) has a unique smallest Euclidean norm . More specifically, given a commutative unital ring
that possesses a Euclidean norm, there is a Euclidean norm
with the following properties:
- For any Euclidean norm
on
, and any
,
.
- The elements of norm zero are precisely the units and the elements of norm one are precisely the universal side divisors.
-
is a multiplicatively monotone Euclidean norm, i.e., it is multiplicatively monotone: if
,
.
-
is an automorphism-invariant Euclidean norm.
Related facts
- Element of minimum norm in Euclidean ring is a unit
- Element of minimum norm among non-units in Euclidean ring is a universal side divisor
- Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm
Examples
- Over the ring of rational integers, the smallest Euclidean norm is the greatest integer of the logarithm to base two of the absolute value. Further information: Ring of rational integers is Euclidean with norm equal to binary logarithm of absolute value.
- Over the polynomial ring over a field, the smallest Euclidean norm is the usual degree function. Further information: Polynomial ring over a field is Euclidean with norm equal to degree
Facts used
- Element of minimum norm in Euclidean ring is a unit
- Minimum over principal ideal of Euclidean norm is a smaller multiplicatively monotone Euclidean norm
Proof
Given: A commutative unital ring with a Euclidean norm
.
Construction of the sloewst growing Euclidean norm
We define the norm inductively as follows.
For , define
as follows:
-
is the set of units.
- For
,
is defined as the set of those
such that
,
, and for every
, there exists
such that
.
For any ,
is defined as the unique
such that
.
Note that is well-defined for all
,
is Euclidean by definition: for every
with
, there exists
such that
with
or
, which is precisely saying that
or
. Thus, we can write:
with or
.
Proof that this Euclidean norm exists for a Euclidean ring and is smaller than any other Euclidean norm
We prove that if for the Euclidean norm
, then
. This will prove that
is well-defined and
for all nonzero
.
Given: with
.
To prove: .
Proof: We prove this claim by induction on .
Base case: When , we have
. By fact (1),
is a unit, so
by definition.
Induction step: Suppose , and the statement is true for
. We have
.
- If
, we are done.
- Suppose
. For any
, the division algorithm yields:
where or
. By the inductive assumption, Failed to parse (syntax error): r \in \{ 0 \} \cup \bigcup_{m=0}^n-1} R_m
. Thus, for any
,
for some
.
Thus, .
Proof that the norm is multiplicatively monotone
Given: such that
.
To prove: .
Proof: One way of proving this is to invoke fact (2), which states that given any Euclidean norm, we can obtain a smaller multiplicatively monotone Euclidean norm. Since the norm is the smallest Euclidean norm, it must be equal to the multiplicatively monotone Euclidean norm obtained from it using fact (2).
We can also check the condition directly. If , then either
for some
, or
satisfies all the conditions for being in
. In either case,
. Similarly,
.
Proof that the norm is automorphism-invariant
The way the norm is defined is clearly automorphism-invariant, since it depends on no choices.