As always, given a ring we denote by the multiplicative group of units of
Definition. Let be a commutative domain with identity. An element is called a universal side divisor if for every either or for some Equivalently, is a universal side divisor if
Remark. If is a universal side divisor of then is a maximal ideal of The converse is not true.
Proof. Let be an ideal of and Then for some Since we have and so Thus and hence
To see that the converse is false, consider the example Then is a maximal ideal of but and so is not a universal side divisor of
Proposition. If is a Euclidean domain, then has a universal side divisor.
Proof. Suppose first that is a field. Note that is a Euclidean domain with the constant Euclidean function Now, since every nonzero element of is a unit, is a universal side divisor of
Suppose now that is a Euclidean domain which is not a field and let be a Euclidean function of Let
Since is not a field, and hence has a smallest element, say Let So there exist elements such that where either or If then If then by minimality of and so Thus and because
Example 1. The ring of integers is a Euclidean domain with the Euclidean function For every integer we have either or amd so is a universal side divisor of Similarly, are also universal side divisors of and has no other universal side divisors (why?).
Example 2. The ring of Gaussian integers is a Euclidean domain with the Euclidean function It is easy to see that So the set in the proof of the above Proposition is So the smallest element of is and hence are universal side divisors of
Example 3. Let be a field. Then the polynomial ring is a Euclidean domain with the Euclidean function By this post, So the set in the proof of the above Proposition is So is the smallest element of and hence any polynomial of degree is a universal side divisor of
Example 4. The polynomial ring is not a Euclidean domain.
Proof. The standard proof is that since every Euclidean domain is a PID and is not a PID, is not a Euclidean domain. But here we want to prove that using the above Proposition. First note that, by the post linked in Example 3, Suppose now, to the contrary, that is a Euclidean domain and let be a universal side divisor of So for every we have either or or Choosing we get that or But then choosing we must have or or which are all impossible. So has no universal side divisor hence is not a Euclidean domain.
Note. The Proposition in this post is a slightly modified version of Proposition 3.63 in Joseph Rotman’s book Advanced Modern Algebra. The Remark and Examples in this post are mine. Proposition 3.63 in Rotman’s book says that every Euclidean domain which is not a field has a universal side divisor. I have no idea why Rotman decided to exclude fields!