 Welcome back to our lecture series Math 4230 abstract Algebra 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 18, we want to talk about the idea of a Euclidean domain. So let's summarize what we've done so far. As we've discussed in this unit about factorizations inside of integral domains, we've introduced things like unique factorization domains, principle ideal domains. We've shown that every principle ideal domain is a unique factorization domain, and we promise we'll provide a unique factorization domain that's not a principle ideal domain. We've also seen that, well, obviously every unique factorization domain is a domain, but not every integral domain has unique factorization. The Euclidean domains are actually going to sit on top of this, that we will see by the end of this lecture that every Euclidean domain is a principle ideal domain, and this is a consequence of the division algorithm. So before we can generalize the notion of the division algorithm like we see with integers, we have to have a notion of a norm, or sometimes called a valuation map, which is what I'm going to define in this video right here. So if we have some integral domain D, then we're going to define a map, typically it's called new, for which this will be a map for the non-zero elements of the domain to the natural numbers, all right? And like I said, we call it a norm, which is typically what I've denoted as new here, or it can be called a valuation map. And at the very minimum, there's really only one thing we require for these valuation maps, is that if you take the norm of an element, that'll be less than or equal to the norm of a product of that same element with any other element in the domain there. And of course, as these are natural numbers, the ordering we have here is the usual natural number ordering. Now, of course, this by itself might not be as extremely useful as if we had other conditions to it. So we can say that a norm is in fact additive, if when you take the norm of a product, this is equal to a sum of two different norms, well, of the two factors, they're the norm of X plus the norm of Y. Already in this lecture series, we've played around with the idea of a multiplicative norm. So we say a norm is multiplicative. If first of all, the norm is never equal to zero, because after all, as we have this map to the natural numbers, that does include the number zero. We could allow for such a thing. That could be very common, for example, with an additive norm, but for multiplicative norms, we don't require that the norm can never be zero. So this is actually a map into the positive integers in that situation. And whenever we take the norm of a product, X, Y, this will equal a product of norms, the norm of X and the norm of Ys. So I just kind of wanna pause here for a second and mention here that when we talk about, when we talk about an additive norm, we really think of this as a homomorphism from the multiplication structure, the multiplicative structure of the integral domain to the additive structure of the natural numbers there. So you have this additive homomorphism. It would be a semi-group homomorphism, because with regard to multiplication, we don't necessarily have inverses or anything like that. And so that gives us an additive homomorphism in that situation. Then clearly a multiplicative norm is gonna be a multiplicative semi-group homomorphism from the multiplication of the domain to the multiplication of natural numbers. Now, one thing that's important as we discuss norms here is the idea of equivalent norms. We say that two norms, new and new prime, are equivalent. If there exists some function fee from the natural numbers to the natural numbers, such that V of X is less than or equal to V of Y, whenever V prime of X is less than or equal to V prime of Y. So this is an if and only of statement. If this inequality implies this inequality and this inequality implies that inequality, then we say that the two norms are equivalent to each other. And for all intents and purposes, a two norms which are equivalent are considered the same. It'll develop essentially the exact same theory for us. Now, I do wanna mention that, like I said earlier, we have these additive and multiplicative norms, but really these two things are naturally equivalent to each other. So give it any additive norm. We can always turn this into a multiplicative norm, right? So the idea is we can define a new norm, new prime of X just to be two to the new power of X. So you take the old norm and raise that as a power of two, and that gives us a new norm. And so if this first norm was additive, then this new norm is gonna be multiplicative. And this is an example of, in fact, equivalent norms. So every additive norm can be turned into a multiplicative norm. And so typically as we talk about norms, we always will just turn an additive one to a multiplicative one without any loss of generality there, because they're equivalent. I also want to comment here that when you have this right here, this norm of X is less than the norm of X, Y, this itself is, but even this is somewhat of a superfluous condition, because if you had any function from the non-zero elements of a domain into the natural numbers, even if it didn't satisfy this condition, we could construct a norm that does. If we had some function F, which that is to say if we had some already norm, but it doesn't have this property here, we can construct a new function, which we'll call this one F again from the non-zero elements to the natural numbers. You define F of X in this situation to be the minimum of let's say new of X, Y, as then Y ranges over the elements of D like so. Notice in this situation, so basically you take the new norm of X to be the minimum of the old norm as you allow a factor to range over the entire ring there. And I guess I should suppose we should say this is why it's gonna be non-zero in this situation. So notice of course the product will never be zero because we are in an integral domain here. Now this minimum is gonna be less than or equal to the minimum where we now allow Z to range over D, again non-zero, where you take new of X, Y, Z like so. All right, so this doesn't, and this isn't really gonna be much, I mean it potentially takes away some of the options and so the minimum gets bigger, but by associativity we can redo the product here, the norm of X, Y and Z for which this is the norm as we've defined it here, F of X, Y like so. And so even if we have a function between the multiplicative structure of the domain into the natural numbers here, even if we don't have this absolute value, or excuse me, this less than or equal to condition, we can always create a norm that has exactly that property from it. So the axiomatics of this notion of a norm are very, very weak in general, but nonetheless it can be a very valuable tool in studying the factorization over integral domains. So just as some quick examples here, we should mention that the integral domains we've been playing around with our lecture series are very applicable here. You have the integers, you have the Gaussian integers, and then you also have the integers join the square root of negative three. These are all integral domains which have as their norm the absolute value function are the modulus of the complex number. And so it's very, very easy to have an integral domain that has attached to it a norm. And that norm can then say something about the factorization there. So for example, for any integral domain, a trivial option is you could just take the norm to always equal one. This would give you a multiplicative norm. It's not really useful, but we saw in the past how we used the absolute value as a way of predicting units in these various domains and we looked at irreducible elements using these things. So in particular, if you have a multiplicative norm, it's extremely useful in your study of factorizations. Let me actually prove for us a proposition about norms of integral domains. So let new be a norm. So it's a map from the non-zero elements of the domain to the natural numbers. D is an integral domain, like I said. Then at the very least, we can say that the norm of one is in fact the smallest element in the image of this norm. In particular, if we call the norm of one equal to A, that number A is gonna be the smallest number, because A is a natural number here, it's gonna be the smallest number inside the image there. This is essentially a immediate consequence of the well-ordering principle of the natural numbers. And in fact, if any other, well, if you have a number in the domain that's a unit, then it will also obtain this minimum value of the norm. So the norm is a good way to detect units inside of an integral domain. Because if you have a norm, and this doesn't have to be a multiplicative norm, it doesn't have to be an additive norm, it just has to be a norm. It satisfies that original inequality, new of X is less than or equal to, less than or equal to new of XY for all Y inside, all non-zero Y in the domain. If you just have that property, then the norm of the unity will be the smallest and any unit will obtain that exact same norm. So if you figure out who can possibly obtain this minimum norm, then you've enumerated the units inside your integral domain. It's a very powerful tool that we've used already and will continue to use as we talk about factorizations in this unit here. So okay, why is the norm of the unit the smallest possible one here? Well, the thing about unity is that one times X is always equal to X. And so when it comes to this inequality here, one is essentially, it's norm is gonna be smaller than everything. If we say that the norm of one is equal to A, well, one by the property is gonna be, well, the norm of one will be less than or equal to the norm of one times X, which of course is equal to the norm of X. And therefore this shows us the minimality of this number A. That was pretty simple, pretty quick. The unity has the minimal element or had the minimal norm inside the entire domain. I'm gonna leave it as an exercise to the viewer here to prove the other one. It's basically just as easy as what we just saw a moment ago. And so a norm can be a very useful tool to find and identify units inside of a integral domain. Think about the Gaussian integers you've considered before. When we had our norm, the norm of a Gaussian integer A plus BI, this turned out to be A squared plus B squared. Well, if you took the norm of one, that had to be one. And so we then can search what are elements in the integral domain whose norm can equal one. Because if it's a unit, its norm will have to be one. Now be aware that this proposition doesn't tell us that if a number obtains this minimum norm that it's necessarily a unit. But it does tell us that all of the units have this, obtained this minimum norm. So with Gaussian integers, you ask yourself, well, how many ways can you solve the equation A squared plus B squared equals one where A and B are integers? In which case the only options you have are, you know, A is plus or minus one, B is zero. Or you have that B equals plus or minus one and A is zero. Thus giving you the four units of the Gaussian integers plus or minus one and plus or minus I, that principle of the norm helped us find all of those norms. In general rings that don't have a norm, that can be a much more difficult conversation to have, a more difficult calculation. So norms are very, very useful. And it's the defining characteristic of a so-called Euclidean domain. Because the Euclidean domain will be an integral domain with a Euclidean norm, which we'll define in the next video.