In this post we are going to show that if is a Euclidean domain with a Euclidean function that satisfies the inequality whenever then is either a field or a polynomial ring over a field. This result was posted as a problem on the Art of Problem Solving website recently; you can see the problem and my solution here. I’m now going to give a slightly improved version of that solution here.
Let’s begin with the definition of Euclidean domains.
Definition. A Euclidean domain is a commutative domain with identity such that there exists a function with the following properties.
i) If then for some with either or
ii) if then
The function is called a Euclidean function for
Not all ring theorists are happy with having the condition ii) in the definition of a Euclidean domain since it can be shown that i) implies the existence of a function that satisfies both i), ii). But in this post, we are going to stay away from that argument and keep ii).
There are many examples of Euclidean domains but, for a reason you’re going to see in a moment, we are only interested in two of them in this post.
Example 1. Any field is a Euclidean domain. Just choose any nonnegative integer and define for all Then the condition ii) clearly holds and if then so the condition i) is also satisfied.
Example 2. The ring of polynomials over a field is a Euclidean domain. Here we define the Euclidean function by for all Then the condition ii) clearly holds because
The condition i) is just the familiar process of dividing two polynomials.
Remark. Both Euclidean functions in Examples 1,2 satisfy the inequality whenever Interestingly, there are no other Euclidean domains whose Euclidean functions satisfy the said inequality. The goal in this post, as mentioned in the first paragraph of this post, is to prove that result.
Notation. From now on, is a Euclidean domain with a Euclidean function and is the multiplicative group of units of
Next is to find This is a well-known result about Euclidean domains but since it is important and also for the sake of completeness, I’m going to prove it here.
Lemma. Let Then
1) i.e. is the minimum value of
2) if and only if is a unit. In particular,
Proof. 1)
2) If is a unit, then and so Suppose now that If then is a unit. Otherwise, there exists and so for some with either or If then which is not true. Therefore which is not possible because and so for some which then gives the contradiction
We are now ready to prove the result mentioned in the above Remark.
Theorem. Let and suppose that
whenever Then
1) for all with
2) is a subfield of
3) either or is a polynomial ring over
Proof. 1) That is clear if Suppose now that Then by the Lemma
Similarly,
2) The only thing that we need to prove is that if then Well, that’s clear if or If are nonzero, then, by 1), and so
3) Choose such that is a minimal element of Notice that is not algebraic over the field because then either or would be a unit, contradicting So we are done if we show that Let We prove by induction over that By the first part of the Lemma, is the minimum value of If then and we are done. Suppose now that We have for some with either or If then by the minimality condition on So either way, So if then and we are done. Thus we may assume that Then, by the Lemma and the first part of the Theorem, and so, by induction, Therefore and that completes the proof of the Theorem. Merry Christmas!