Today we’ll try to realign our intuition with the standard textbook proof of “ is a Euclidean ring”.
denotes the set of all complex numbers of the form where and are integers. .
Let us take two complex numbers . Let and , where and . Then if there exists a complex number such that , we’ll be done. This is because if and , then , which in turn implies . This is what we wanted.
According to the conditions stated, we have and . From these two equations, we need to determine the values of and , and prove . The equations simplify to and . Are and integers, provided and are? A quick verification should tell you this is not true (try substituting and ). Then is there no such element such that this condition is satisfied? The fact that we’ve failed to find such an element does not imply one cannot exist. Note that our approach had several restrictions like , etc.
Let us now go to the book’s proof. The book first proves it for integers. A noteworthy step is ensuring the absolute value of the remainder is . Then it takes , which is a positive integer, and uses it in a particular fashion to prove that where . Don’t worry if you don’t get what I’m talking about. This proof is present in its entirety on pg.150 in “Topics in Algebra” by Herstein.
How different is this from our earlier method? Earlier we were concerned with quotients and remainders of and . The element and were, for some reason, thought to be of a pre-set nature; was thought to be and was thought to be . Here, we have no such pre-sets. There are two main techniques here, which are analyzed below:
1. and . What does this do? It ensures that . Had we not ensured these bounds, it is possible that would have exceeded . Note that the bounds of and cannot be made much sharper; in other words, we can’t ensure that , etc.
2. The observation that . This is only possible under the current definition of . If , then this wouldn’t be true.
Let and have the same descriptions as given at the beginning of the article. What ultimately is the nature of ? If where , then . Similarly, if where , then . This is different from the nature of that we’d initially thought of. No surprise then that our approach didn’t work.