 Recall that a Euclidean domain is an integral domain, which has a norm attached to it such that that norm satisfies the division algorithm. I should say it has a division algorithm. There always exists some quotient and remainder up to some uniqueness conditions. What was a norm again for an integral domain? This is going to be a function of our domain is d here. A norm, typically we call it new, new for norm is a function from the non-zero elements of the domain to the natural numbers. It turns out that the Euclidean norm is a very powerful tool to use. Well, what happens if that norm is itself multiplicative? Now, in a previous video, I mentioned that if you take the norm of the one element inside of your integral domain, this is always going to give you the minimum element in the image of the norm. Without the loss of generality, you can assume this element is actually the number one itself. Because if it wasn't, what you can do is you can define a brand new norm. We'll call it norm prime, a new prime of x here. You're going to find this to be the norm of x minus the norm of one and then you add one to it. I do want you to be aware that this is in fact going to be a norm. This norm is actually equivalent to the original norm new and in fact, if you plug in x equals one here, you're going to end up with one. Without the loss of generality, you can always assume that the norm of the unity of a domain is always one because of this change. You can always swap out the norm in that manner. We also had previously mentioned that if you take the norm of a unit, you always have to get the same norm as the number one itself, which we can assume is one. We're going to strengthen that condition if you have a multiplicative Euclidean norm. In the Euclidean domain, if its norm is multiplicative, then we get that the norm of an element equals one if and only if it's a unit. We get both directions now. Furthermore, if all of the divisors of the norm of some elements, so if all the divisors, because this is a natural number after all, if all of the divisors of new of X are not contained within the image of new except for one and new of X here, which think about that. What are numbers who only have as its divisors one in itself? That's typically a prime number, right? Well, if all of the divisors of your norm for a specific element are missing except for one, which is always there and new of X of course is going to be there too, then it turns out that X is irreducible. In particular, if the norm of X is a rational prime, that is, it's a prime number of the usual sense of integers and natural numbers, then X has to be irreducible. So we can use the norm to locate units and we can use the norm to locate irreducible elements. This is the most effective setting is when you have this multiplicative Euclidean norm, such as the integers when you can use absolute value as your norm. That's what we're trying to base this upon. Now, we've already seen previously that if the norm of an element is equal to one, that is, if the norm of your element is equal to the norm of one, which we can assume as one, then you're going to have to be a unit. We've proven that previously. So it's the reverse direction that we're considering right now. So suppose that the norm of your element U, which we don't necessarily know as a unit yet, suppose it's equal to one. Well, what if you take then the norm of U squared? Since it's multiplicative, this will be the same thing as the norm of U squared, where now instead of squaring the element of the domain, we're squaring the natural number, new of U right there. So this would then be equal to one because it's going to be one times one. Well, as we showed previously in this lecture series, every non-zero ideal is in fact generated by an element of minimum norm. So what was this theorem here? Just because you probably don't memorize these numbers, right? In our lecture series, it's actually the very previous video in this lecture series. We prove that every Euclidean domain is in fact a principal ideal domain. And in the process of doing that, we had to argue giving a random arbitrary ideal that it was principal. We'd grab the element in the ideal of minimum norm and that's then a generator for it. If your norm is equal to one, that's necessarily the minimum norm. No one can get smaller than the unity, which of course is one in this situation. So the norm of U is equal to one. The norm of U squared is equal to one. And clearly, we have that U squared. U squared is inside of the principal ideal generated by U. But since U squared is also minimum norm, we get that these two principal ideals must be one and the same thing, all right? So U and U squared, if U wasn't a unit, they would have to be associates of each other. So in particular, there exists some element Y in the domain D such that Y U squared is equal to U, which of course, we're in a domain so we can cancel. If you cancel the U from the right-hand side of the equation, you get Y U equals one. So therefore, Y is in fact a unit. So studying units is very effective when you have this multiplicative Euclidean norm. So let's take a look at the second part of this proof here. So now we wanna consider the conditions that can guarantee when an element is irreducible, when we consider of course, what its norm is doing. So consider the natural number nu of X such that none of the divisors of nu of X appear in the image of nu with the obvious exceptions of one, because that's the norm of the unity. And then the norm of X of course is equal to the norm of X. So suppose there's no other factors of this norm inside of the image here. Now look at a factorization of X. So let's say that X equals AB is some factorization of X inside this integral domain. Since our norm is multiplicative, the norm of X would equal the norm of AB which is then equal to the norm of A times the norm of B. So we have a factorization of this natural number norm of X in particular, the norm of A divides the norm of X for which by our hypothesis right here that would imply that the norm of A is equal to one in which case that would imply that A is a unit. The other possibility is that the norm of A equals the norm of X and that situation you'd have to have the norm of B is equal to one which would then imply that it is a unit. So since A or B has to be a unit this would imply that X is an irreducible element inside the domain there. So in particular if the norm of a number from this domain, from this Euclidean domain is a rational prime then that guarantees that it's gonna be irreducible because primes don't have any factors whatsoever. So they can't be in the image because they don't exist. So that's a very effective tool to find irreducible elements, to find units inside of these Euclidean domains, look at their norms. Even if we don't have a Euclidean norm which again we're assuming multiplicative Euclidean norm here even if we don't have a Euclidean norm if we just have a multiplicative norm parts of this theorem still are applicable because the proofs are still applicable. And so these multiplicative norms are very, very useful as we study factorizations inside of integral domains. So that brings us to the end of lecture 18 in this lecture series Math 4230. Thanks for watching. If you learned anything about Euclidean domain in this lecture, please like these videos, subscribe to the channel to find out more stuff about mathematics. And if you have any questions, please post them in the comments below. Thank you.