 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 Missildine. In this lecture, lecture 15, I want to talk about the field of fractions, sometimes called the fraction field. It turns out that if you have an integral domain, every integral domain can be embedded inside of a field, and I want to explain how that happens. To begin with, I want to remind everyone, the important difference between a fraction and a rational number. So thinking of like integers, ratios of integers, simply put, a fraction is just an ordered pair of integers. So when I say something like two-fifths as a fraction, this really just means two comes first and five comes second. And so with that in mind, I can list lots of fractions. Two-fifths, I can take five-ninths, I can take seven-halves. It could be an improper fraction for all I care. Three-fourths, seven-sevenths, something like that. I can also talk about zero-thirds. I can talk about four-first. I can even talk about bump, bump, bump, 13 over zero, right? As a fraction, this is perfectly fine. Heck, I could even go zero over zero. This is a fraction that shows up in calculus all the time, but that's all that a fraction is. It's just an ordered pair of integers. Now, with that in mind, we don't usually write them that way. We write them like this, but that's not what one means by a rational number. A fraction is just this combination. And now we often are introduced to fractions as in the setting of rational numbers. So we're like things like, oh, things like this don't make any sense. Or we should write this as one, or we should write this as an improper fraction or whatever. I'm not gonna go through all those, but fractions themselves are just the first number. You have a second number and those numbers are integers. Now, of course, when it comes to the rational numbers, we're thinking of it as a ratio. We're thinking of it as division. And so we have to worry about some things. Like we don't wanna divide by zero. There's reasons to do that. We also have equivalency. Things like one half is equal to two fourths. Now, it's not the same fraction, but it is the same rational number. We consider those the same. And how do we know that? Well, one way to think of it is like, oh, you end up, you know, you go with two on top, two on bottom. And so two times one is two, two times two is four. So it's the same in that regard, but there's a lot embedded into that. It's like, well, why do we multiply the fractions top across top and bottom across bottom? That's not how you add fractions. Like it might take, for example, one third plus one fifth, this is not equal to two eighths as much as many early fraction users like to think. You don't just add across the top and bottom, but that's how we multiply. Why the difference, right? And why is two over two equal to one after all? So the real idea here is actually cross multiplication. These two fractions are equal to each other, because if you look at the cross product, one times four, that's the same thing as two times two. And so whenever that happens, we say the two fractions are equal to each other. Sorry. Two rational numbers are considered equal when their cross product is the same thing. So A over B equals C over D if AD is equal to BC, okay? And so really the rational numbers form an equivalence relation on the set of fractions where of course the denominator is never allowed to be zero. So you have to throw out some of the possible fractions, but then those which are considered acceptable fractions, including improper ones, right? Being improper can be acceptable sometimes. In that set of acceptable fractions, we define an equivalence relation where we say these are the same. And so really when we write a fraction, what we're meaning is this is the equivalence relation for the ordered pair AB, but there are other ordered pairs that can be considered the same thing and we write them as fractions because the fractions is to suggest division. So there's a lot loaded into that. And the reason I shared it is because I want to, I want to then generalize this to the setting of an integral domain. So note, if you're in a field F, then there's really in a field, since it's, every field is an integral domain, there's no proper divisors of zero because if there were, you could times it by a unit and that would imply that one is a proper divisor of zero. No, no, no, no. So there's no zero divisors inside of a field. So in particular, every subset of a field likewise has no, has any zero divisors, right? So in particular, if you take a subring of a field, it doesn't matter that subring has to be an integral domain. I mean, it could be a field, it could be something else, but every subset, every, excuse me, every subring of a field is necessarily an integral domain because if there were zero divisors here, then there would be zero divisors here, but fields don't have that. Okay, so that's an important observation. Every subring of a field is an integral domain. It might not be a field because the subring might not contain units. Like for example, Z sits inside of the field of rational numbers, it's an integral domain. That sits inside of that as two Z, sitting inside of that as six Z. All of these subrings are integral domains because they live inside of field, okay? What we wanna show is that the reverse direction is also possible, that every integral domain can be extended into a field. But how do you do that? How do you extend it? And it's like, there's nothing there. We have, where do I go? It's like you're in the darkness. What we have to do is we actually have to construct the field and we're gonna do that using fractions. So if you have an integral domain, say D, then, and again, this is the same D that's inside of a field. Take two elements, R and S, such that the second number isn't zero. Then we can define a fraction. What does this symbol mean, R over S? This means, oh, multiply by the reciprocal, right? R divided by S is the same thing as R times S inverse. So division can be introduced to a field because you multiply by the multiplicative inverse. Just like subtraction, when we define a field, we don't ever talk about subtraction because subtraction just means, if you take A minus B, this means A plus the additive inverse of B. Subtraction division are defined using inverses. Now, zero doesn't have an inverse, multiplicatively speaking, in a field so that we don't divide by zero, okay? So that's what a fraction is. So if we take the collection of all the fractions where the top's called the numerator and the bottom's called the denominator, so we take all the fractions where the denominator can't be zero, then the collection of fractions forms a subset of F and this subset is going to be a field, right? And we'll explain why that is a little bit later, but this collection of fractions, fractions coming from the integral domain, it forms a field and in fact, it's gonna be the smallest field that contains the integral domain, all right? So we can talk about fractions if we already have a field, but it turns out we can talk about fractions even if we don't have a field, right? Can we reverse this process? If we have an integral domain, is there necessarily a field that contains it? How do we construct such a thing? We're gonna construct it using fractions because a fraction is just, after all, an ordered pair of elements from the domain here, R and S, and we just apply an equivalence relation to that ordered pair and then those can make fractions and we can make a well-defined operation of addition and multiplication that forms a field. And so this is what we mean by the field of fractions. So this set, Fract D is gonna be this collection of fractions. After all, if you take the fraction field of the integers, you get back the rational numbers because after all the rational numbers are all the possible fractions of the integers excluding division by zero, of course, all right? So there's a couple of limits we have to prove before we can establish that this field of fractions is in fact a field. So what's the setting we're gonna live in? The setting we have to consider is an integral domain. So the first thing we have to do is you have to basically define what is Fract of D. So we're gonna first construct an equivalence relation on the set D cross D, but we remove zero, zero, okay? We wanna remove zero, zero. So we don't wanna divide by zero. Honestly, this zero over zero, if you remember from Calculus, this is an indeterminate form, it could be anything, right? It's sort of the same problem that we're gonna throw it out here. Now I want you to be aware that this set does contain things like, oh, I could have r comma zero in there. So in some respect, we are allowing division by zero temporarily. We'll throw it out later on. But we're gonna define an equivalence relation on this set D cross Z where we throw out zero, zero. And the equivalence relations what we saw before where AB is considered equivalent to the pair CD if the cross product AD is equal to BC. So we take A times B, excuse me, A times D and then B times C, that should be the same thing. If that's the same thing we claim this is an equivalence relation. So this is what we're gonna prove right now. Now, a little bit of notation that's not part of the proof, but this is really a definition. The equivalence class associated to the ordered pair A comma B, we denote that as A fraction B, right? For which we can write that using sort of like this notation or this notation, same basic thing, okay? So we're gonna denote the equivalence class of AB as the fraction. So a fraction is just an equivalence class of ordered pairs. We're also gonna show that if AB is equivalent to C zero then actually B was zero. So the only things that are equivalent to zero in the denominator were other fractions who have zero in the denominator. So with that then proven we will define the set of fractions of D to be all of these equivalence classes for which the denominator cannot equal zero. So we throw out when the denominator zero and since you can only be equivalent to zero in the denominator if your denominator was also zero that means we get the genuine fractions that we think about when we think of rational numbers. And so this generalizes the construction of the rational numbers. We can construct the rational numbers from the integers and it's this exact same construction. We build fractions as an equivalence relation on ordered pairs, all right? So why is this an equivalence relation? Well, we have to first prove it's reflexive. So why is the pair AB, why is it related to itself? Well, that's because AB equals BA. We're in a commutative ring. It's an integral domain after all. So since AB equals BA that means the ordered pair AB is related to AB. So it's reflexive. That was pretty simple, okay? Why is it symmetric? Suppose that AB is related to CD. That means that AD equals BC. But we're in a commutative ring. So AD is the same thing as DA and BC is the same thing as CB. And also it's an equation. So it's symmetric in that regard as well. Equality itself is an equivalence relation. So if AD equals BC, that means CB equals DA and that equation implies that CD is related to AB. So this relationship is symmetric. Why is it transitive? Well, suppose that AB is related to CD and suppose that CD is related to EF. Well, the first one implies that AD equals BC and then the second one implies that CF equals DE. This one I have to actually separate into a couple cases. So let's suppose that C is equal to zero. If C is equal to zero, that means B times C is equal to zero. And so AD is equal to zero. But also if C is equal to zero, that means CF is equal to zero. So DE is equal to zero. In that situation, we then get AD is equal to DE. And so since, you know, since D is not zero, notice here, if C is equal to zero, then D can't be zero. We threw out that possibility. So since C is zero, D is not zero, we can cancel out the D. And so you get that A equals zero. And so since A equals zero, that means the product AF is equal to zero. The product BE is also equal to zero. So we get that AB is related to EF. That's sort of a trivial case, but we have to consider it. Now let's consider the case where C is not zero. Why do I care about that? Well, if it's not zero and I'm in a domain, I can cancel it, all right? So let me actually scooch back up a little because I want to see the original equations all on the same screen so that you can follow along here. So consider the product AD times E, all right? So the original assumption is AD is the same thing as BC. So ADE, it becomes BCE, right? Redu parentheses on the left-hand side. This is the same thing as A times DE. And I'm just gonna remove the parentheses over here, of course, it's associated, it doesn't matter. Well, notice DE shows up as well. DE is the same thing as CF. So ADE is the same thing as ACF, which is the same thing as BCE. So moving the C around, we have C times AF, we have C times BE. Since C is not zero, aha, we can cancel it. And so we end up with AF is equal to, AF is equal to BE, which is what we're trying to check. If AF equals BE, that then means that AB is related to EF. And so we then have that this relationship is transitive. Since it's reflexive, symmetric, and transitive, this is in fact an equivalence relation like we were trying to show earlier, all right? Then the next part that's important to point out here, the other thing we gotta prove is that what if AB is related to C zero? So the denominator is zero. Well, in that situation, you have BC is equal to A zero, which A zero is equal to zero. So BC is equal to zero, okay? Now, because D is zero, C can't be zero because zero zero is not inside the set. So C has to be non-zero, so we can cancel the C, right? Because zero is the same thing as zero times C, we can cancel the C, and that's gonna get that B is equal to zero, which then proves this lima. This lima wasn't hard, it just was lengthy. It's some things we have to establish. So we know that we have this equivalence relationship on ordered pairs, and we then denote that equivalence relation using fractions. And so the set of fractions will be all of those equivalence classes where the denominator is not able to be zero, the denominator can only be zero. I should say it's only related to a zero denominator if you already have a zero denominator. So we've set up a situation which will never quote unquote divide by zero, even though we've never introduced division yet. All right, so we now have this, we have this set. So fraction D is in fact a set of equivalence relationships, equivalence classes, I should say, we then wanna put operations onto this. So we're gonna define the operations of addition and multiplication. And in this lima, we're gonna prove that these things are well-defined. And then we're gonna prove some other common properties here. So for example, if you, so okay, let's first define the operations here. We're gonna define addition of fractions in the following way. A plus B, excuse me, A over B plus C over D, that'll be defined to be AD plus BC over BD, right? So imagine when you add integers together if you don't have a common denominator, what do you end up doing? So you have one half plus like one third or something like that. You would rewrite this like, oh, I need a common denominator. This is going to be three sixth plus you're then gonna get two thirds. So this becomes five sixth, where six of course is two times three, the product of the original denominators. Where did the five come from? Well, the five is three times one plus two times one. So that's where five came from. So that's how we add these things together. So this is considered addition of fractions. Is this the usual rule we have when we add fractions together? Now I wanna point out that if the fractions already have a common denominator, you can just add the numerators together. We'll prove that as well. So this is the definition of the addition of fractions and then we'll prove that adding a common denominator is a consequence of that definition. But when it comes to multiplying fractions, we actually just multiply together the numerators and we multiply together the denominators. This of course will have the consequence that if you multiply a fraction A over B by C over C, right? Well, by definition you get A C over BC. I claim that this is then equal to A over B. So these two order pairs are equivalent to each other and these order pairs are equivalent to each other. Now, since the set of fractions is defined, well, fractions themselves are equivalent classes, there's always the concern that, well, maybe these operations aren't well defined. That's what we're gonna prove here in this lemma. So let's take two fractions, AB and CD. And of course, as they're fractions, we can now assume that the denominators are not equal to zero. So it should be clear that since B and D are non-zeros and we're in interval domain, the product of two non-zero elements is not zero as well. And so when you look at these fractions, AD plus BC over BD and AC over BD, their denominators are not zeros. So these are fractions that belong to the set. So as a function, it makes sense what it's saying. But of course we're concerned about, is it well defined? So suppose that AB equals A prime B prime. So these are two fractions that represent the same quantity. So like one half equals two fourths. We have that setting, that's what we're thinking of right now. So in that setting, if A over B equals A prime over B, then that means that AB prime equals A prime B. That's a statement in the domain D. Likewise, let's see that, where is the other one? What, C over D. C over D is equal to C prime over D prime. That would imply that C D prime equals C prime D prime. Okay, so imagine we have two representations for our two different fractions here. We have to prove that these different representations don't make a difference in terms of how we compute things. All right, so we have to prove, with regard to the addition here, we have to prove that A over B plus C over D is equal to A prime over B prime plus C prime over D prime. That's what we're trying to prove here, which of course this is defined to be AD plus BC over BD. This is defined to be A prime D prime plus B prime C prime over BD, B prime D prime like so. And so we have to argue that these two things are equal to each other. So what we need to show is that AD plus BC times B prime D prime. We have to show that's equal to BD times A prime D prime plus B prime C prime. So start with the left-hand side here. We can distribute the B prime D prime onto each of these and I'm gonna rearrange things. Take some liberty there using associative distributive commutative properties here. So we won't be distributed, we're gonna have A times B prime times D D prime added to B B prime C times D prime. Well, notice here, A B prime is the same thing as A prime B. So the prime basically moves to the other side. Same thing here, C D prime is the same thing as C prime D like so. And so then once we have these we're gonna move factors around. So put the A prime D prime together, put the B prime C prime together. So we have A prime D prime times BD. Then we have B prime C prime times BD. We can factor out the B prime and so then we get this. So that then proves the claim. These things are actually equal to each other, all right? So therefore that proves addition is well-defined. What about multiplication? Well, the same basic thing here. We have to prove that A B times C over D is equal to A prime B prime times C prime D prime. But what does the first one mean? It means A C over BD. And what's the second one mean? It means A prime C prime over B prime D prime. So we have to show that these things are the same fraction. So by cross multiplication, you should have that A C times B prime D prime is equal to BD times A prime C prime. So we're trying to prove this equation right here. So start with the left-hand side. A C times B prime D prime. If we reorder things, we're gonna get A B prime times C D prime. Because of the original assumptions, we can move the primes. So A B prime becomes A prime B and C D prime becomes C prime D. Then if we reorder things one more time, we're gonna get BD and A prime C prime. That then establishes what we're trying to prove there. So multiplication is likewise well-defined. Okay? Finally, I should mention that because we're associative or commutative, A times B C is the same thing as B times A C. And so notice what we have here. If you have A B times C C, by definition that's A C over B C, right? But then we cross multiply, you're gonna get A B C. And if you look at the other one, you get A B C. That's the same thing. So yeah, you can reproportion fractions in the usual way. And then next, what about adding a common denominator? Well, by definition A over B plus D over B, that becomes A B plus B D over B B squared there. You can factor out the B. So this becomes B times A plus D over B. And because of the previous property, we can cancel across the fraction bar. And this becomes A plus D over B. And so this proves the usual properties of fraction addition and multiplication. The main topic for this video lecture here, the idea of the field of fractions of an integral domain. Let D be any integral domain. Then this set of fractions that we talked about before for which we've now put upon it a well-defined operation of addition and multiplication. This set of fractions is in fact a field called the fraction field of the domain or the field of fractions. And in particular, we can visualize D as a subring of the fractions. And when we identify D with the set of fractions of the form A over one where A is some element of that. And so we can embed every single integral domain inside of a field of some kind. A similar proof can also be done for domains can be embedded inside of a skew field. But we're not gonna worry about that. Having the commutivity makes things a little bit easier. And that's where we're going to land in this video right here. So because of the way we defined addition and multiplication for fractions, it essentially comes from the associativity and particularly the commutivity of addition and multiplication in the integral domain. So we're gonna have that multiplication and addition are commutative operations, okay? Second, these operations are gonna have additive and multiplicative identities. The element zero over one, in fact, you could take zero over any non-zero element of the domain, that's gonna serve as the additive identity, right? Because if we take anything, any element A over B, if you add to it zero over one, this is gonna become one times A plus zero times B over B times one. That simplifies just to be A over B. It's the additive identity. Same thing we get for multiplication. If you take the fraction one over one or in fact A over A for any non-zero element, you'll get the same thing. A over B times one over one, by definition of the multiplication, you get one times A over B times one, you get A over B. So we have identities, multiplicative and additive. Like we said, we have commutivity. We're going through all the axioms of a field here, all right? Then the next one to consider is going to be the idea of inverses, right? So we have additive and multiplicative inverses. In particular, the additive inverse of the fraction AB is going to be negative A over B. Notice if I take A over B and you add to it negative A over B, by definition, this will be AB plus negative A over B over B squared. AB and negative AB, these are additive inverses in the domain D, so you get zero over B squared. And like we saw before, this is the same thing as zero, the additive inverse. What about multiplicative inverses? Well, this is just the common rule of times by the reciprocal, right? If you take A over B, you're going to times that by B over A. Now, since this is not the zero element when we talk about multiplicative inverses, we know that A is not zero, so its reciprocal actually does make sense. You end up with A over B, excuse me, AB over BA. This simplifies just to be one with how multiplication works. So we have identities, we have inverses, we have associative, excuse me, we have commutivity. Commutivity was pretty straightforward. Associativity does require a little bit of an argument here. Multiplication is fairly straightforward because really multiplication is essentially just component-wise multiplication like we would with a direct product. Addition's a little bit more complicated, so let's look at it. If you have A over B plus C over D, and you add to it E over F, we'll do the first two first. This becomes AD plus BC over BD. Then when we add these two together, we end up with AD plus BC times F plus BD times E over BDF. I'm using associativity throughout here and commutivity as well. So distribute the F in this situation. We get ADF, BCF and ADE as terms in the numerator. So then we're going to regroup these things. ADF can be written as A times DF. The second two terms have a common factor, B factor it out. So we get B times CF plus DE. We still have BDF in the denominator. So notice that this right here is exactly the sum of A over B plus CF plus DE over DF, which of course this right here is then the sum of CD plus E over F. So it's fairly straightforward, just work through it. So addition is associative, like I said, multiplication is associative as well. So the last thing that we need to do to prove that it is a field is to prove the distributive laws. Since we've already established that multiplication is commutative, we only have to prove one of the distributive laws. So we're going to have here A over B plus C over D times that by EF. And we want to argue that this distributes throughout here. Let's first add the two fractions together. This becomes AD plus BC over BD. Then when we times it by EF, we'll times the numerator by E, the denominator by F, distribute that E throughout the numerator. We get ADE plus BCE over BDF. As we learned earlier with addition, we can break this fraction into two fractions. So we have ADE over BDF and then BCE over BDF, like so. And each of those are some cancellation we can do. There's the D that cancels here. There's a B that cancels here. So we have AE over BF plus CE over DF for which each of those we can factor for which you're going to get AB times EF and you're going to get CD times EF. And voila, we then have the distributive property. So all of the axioms of a field are satisfied. So this set frack of D is in fact a field and it will contain an isomorphic copy of D. So up to isomorphism, this is going to be a field that contains D. And we can also argue it's the smallest field that contains D. So I want to say a little bit more about that. So let F be any field and suppose that D, an integral domain necessarily, is a subring. Then we can identify up to isomorphism, the fraction field of D as a subfield of F as well. And the identification simple, every fraction A over B is identified to the product A times B inverse. So if you take a fraction like this, be aware that A and B are elements of the domain and necessarily B is not zero. So it has an inverse in the field. So A times B inverse is an F. And so this is going to show you that you're a subset right there. So we can identify the fraction field with a subset of the larger field like so. And so what about our operations? Does addition and multiplication of fractions coincide with addition and multiplication in the field? The answer is yes. If we add together these fractions, we make the identification A over B becomes AB inverse, C over D becomes CD inverse like so. I can insert in between the A and the B inverse, A, D, D inverse. I can insert between the CD inverse, a BB inverse. Really what you want to be thinking here is I'm multiplying by B over B and I'm multiplying by D over D, all right? And so removing things around by associativity and commutativity we are in a field. This becomes AD times B inverse, D inverse. And the second one becomes BC times B inverse, D inverse for which we have a common factor of B inverse, D inverse. We can factor it out by the Shusot principle. The inverse of BD is just B inverse, D inverse we have so. And so then if you have a product times an inverse, this is, I should say a factor times an inverse, that's the same thing here. So multiplication, excuse me, addition of fractions is the same thing as addition inside of this field. Likewise, when it comes to multiplication, if we multiply two fractions, well A over B becomes AB inverse, C over D becomes CD inverse. This is commutative, we can move things around. So this becomes AC times BD of B inverse, D inverse for which of course that's the same thing as BD inverse. I mentioned the Shusot principle earlier, but as we're multiplicative, the multiplications commutative, the order doesn't matter whatsoever. So we have AC over BD inverse, that's the fraction AC over BD, all right? So this fraction, this set of fractions behaves like, as to say they have the same operations, addition and multiplication are the same with the field, whether it's a field of fractions, it doesn't make any difference. So in fact, the fraction field of D is not just a subset, it's a sub ring of F, same multiplication, same addition, and associative to commutative distribution, this all is gonna work just the same, no problems with it whatsoever. And so one last thing I wanna mention is this idea of a prime field, okay? Because what the previous theorem told us, I guess the previous corollaries, what we call it, told us is that if any field contains an integral domain, then it contains its fraction field as well. So the fraction field is the smallest field that contains it up to isomorphism, of course. And so we can actually categorize the smallest field based upon a characteristic. Suppose F is a field of characteristic zero, that field will contain an isomorphic copy of the integers, okay? Because we just take the unity of the field, and so we take the smallest ring generated by one, the unity, that'll give you something that's isomorphic to the integers. Therefore, the field F will then contain an isomorphic copy of the field of fractions of the integers, which of course is the rational numbers. So every field of character six zero contains an isomorphic copy of the rational numbers. And so this will be referred to as the prime field of character six zero. It is the smallest field of character six zero, because any other, well, as we saw here, any field of character six zero will contain Q, it's the smallest one. We can do this for a field of characteristic P as well. Because if you have a field of characteristic P, we're gonna do the same thing. We're gonna look at the subring generated by the unity of the ring. That's gonna be isomorphic to the field, the finite field ZP, where P is some prime, of course. Now, the fraction field for ZP is itself ZP. It's already, ZP is already a field, so it's its own fraction field, it's important to mention. And every finite domain is in fact a field. So if I take the unity in F, it'll generate a finite integral domain. That's gonna be a field, it's gonna be ZP. And so we then refer to ZP as the prime field of fields of characteristic P. Every field of characteristic P will contain a copy of ZP. And so Q, the rational field, and the finite field ZP, where P could be any prime number, these are referred to as the prime fields because they are the smallest fields of each characteristic, a consequence of this field of fractions.