 Monday the 22nd is I'll distribute, just like I did for exam one, a sort of list of things that you should know, a practice exam which will be the exam that I gave last year when I taught this course and solutions to it and some sort of, you know, idea of how to go about preparing for it. So that's what you'll see next Monday. Also next Monday folks, the homework assignment that I gave two days ago will be due next Monday. So it's a little bit quicker turnaround schedule on that particular assignment than has been in the past. So try to be aware of that. What that means of course is that you won't have the standard Tuesday, Wednesday SI sessions to ask questions about, but feel free to either, you know, make an appointment to see me on Friday or Monday sometime or if you want to email me with a question that's perfectly acceptable to do, so please feel free to do that. Also I have a little bit of graded stuff that some of you haven't picked up. It's nothing new. Most of you got your stuff back, but if you are behind on some stuff, I've got all that with me today and you can just grab it after class. All right, questions? Just administrative stuff. All right, what we're going to do today is finish up section three, just some quick observations about isomorphisms and then we're going to, in effect, start something, I won't say completely different, but we're going to start in on a whole new topic and that new topic, this second topic, will basically take us to the end of the semester. And so let me at least do this sort of cleanup work on isomorphisms and then we will move on. So last time, right at the end, I wrote down the definition of isomorphism and isomorphic. So the definition was this, definition, if g and g primes are groups, a homomorphism, let's call it phi as usual, phi from g to g prime is called an isomorphism, that's the word we're defining. In case two things are true, phi is one to one and phi is on two. And then the other word, definition, we say that two groups, g and g prime are isomorphic, g and g prime, g and g prime are isomorphic in case there's an isomorphism. There is isomorphism from g to g prime. So the two words are obviously related, but isomorphism is the function. And isomorphic is an adjective, I guess, meaning that there is an isomorphism from this to this. When you have an isomorphism from one group to another, intuitively what you're thinking is because you've written down a function that preserves the group structure, that's what homomorphism means, and that by requiring the function to be both one to one and onto, you've in effect simply paired up each of the elements in g with an element in g prime. So now you've got sort of a set pairing that preserves the group operations. The idea is that in a sense, all that means is that you've relabeled the elements of g to become the elements of g prime, so that the structure of g and g prime are the same. So the definition that we've been using all semester of isomorphic groups, same structure, in effect means that there's some labeling of the elements that gets you from one group viewed as a table, if you want, to the other group, and that relabeling is simply sort of formally written down as this isomorphism. So think, this means that they have the same structure, they have the same structure because there's a function from one to the other that preserves the structure, and that's isomorphic versus isomorphism. All right. What we did at the end of last time, I said, well, if you have two groups that are isomorphic, then they have to share certain properties, like if one of the two groups is a billion and they're isomorphic, then the other one has to be a billion, and if one is cyclic, the other has to be cyclic, and if one is not a billion, the other is not a billion. If one has the property that the number of solutions to the equation x star, x star, x star, x equals e, if there's, I don't know, seven solutions to that in the group g, then there has to be seven solutions to that in the group g prime. So if the question is, here are two groups, are the two groups isomorphic, if you can find one property of one group that's not shared by the other group, then the answer is no. You think, well, that's good because in order to show two groups are isomorphic, you have to write down an isomorphism from one to the other, which isn't typically an easy thing to do, but at least there's a goal here to show that two groups are isomorphic. Just on the surface, by the definition, you have to show that there isn't an isomorphism from one to the other, which means presumably you have to convince me that there's no possible function that's a homomorphism of one to one and onto, which on the surface looks pretty hard. But the way we typically show two groups are not isomorphic is to show that there's some structural property of one that's not shared by the other. On the other hand to show that two groups are isomorphic, the task is to somehow write down a function from one to the other and show that the function's a homomorphism and then show the function is one to one and onto. So, let's instead of sort of giving the negative point of view, here's how you show two groups aren't isomorphic. Let's write down some pairs of groups that turn out to be isomorphic and maybe it's a little bit surprising. So here's an example and this follows almost directly from a homework problem that you did two weeks ago. Example, here's a group, the group of real numbers with addition, so I'll call this group g. Here's another group, the positive reals under multiplication and the claim is, the claim, I'll call this g prime, and the claim is that g is isomorphic to g prime. Here's how that's written, g is isomorphic to and I've given you this notation before g prime. So rather than writing out the word isomorphic to each time, I just put that sort of equal little squiggly on top. Alright, so let's see, so to convince you that this is true, what do we have to do? Well, we have to write down a one to one and onto function from g to g prime that happens to be a homomorphism. And here it is, proof, define the following function, phi from g to g prime, here's the definition of phi, phi of x equals 2 to the x power. Alright, well first we have to make sure that the function really does go from that group to that group. See, regardless of what real number you hand me, it makes sense to take 2 to that power, because I know what 2 to the x looks like. So at least this thing is a real number, is it a positive real number? Yeah, if you take 2 to any power, you get a positive real number. So at least this is a legit function from here to here. Is it one to one? Yeah, if 2 to the x equals 2 to the y, then what? Log base 2 of 2 to the x equals log base 2 of 2 to the y and you get x equals y. So it's a one to one function. Is it an onto function? Can every positive real number be written as 2 to the x? Yeah, just let x equal log, you know, log base 2 of whatever real number you want, log base 2 of y and you get, this function is onto as well. So it turns out it's not too bad. I'll say easy, just use logs. Phi is one to one and onto, but phi is also homomorphism, but phi is a homomorphism. This was shown for homework. Can anybody pull out a reference? That's the one that came back on Monday, right? Let's see if I've got a homework problem number. Right? Problem 8 or something like that? I forget. Number 6? Thank you. Section 13, number 6. You're right down a function from G. Okay, so maybe that's a little bit counter-intuitive. If there's a group consisting of all real numbers, it's the group of real numbers under addition, then I don't know if your intuition says, and here's half the group, just the positive real numbers with a completely different operation multiplication, but really the structure of these two things is the same. Reals with addition and positive reals with multiplication. You get intuitively the same sort of structure. So maybe that's a little bit counter-intuitive at first glance because you're thinking, well, how can addition be multiplication? The answer is we know how to turn addition into multiplication. You just raise something to a power. So what's going on is once you've got the addition happening in the exponent, it translates down to multiplication. That's the point. There's another example. This one maybe is counter-intuitive as well, but it turns out to work just fine. Here is a situation where even using the same binary operation, I'm going to give an example of a group that's isomorphic to a subgroup of itself. So here's the group G, Z with plus. Of course, I don't really have to put the plus sign in here because we know what the only operation is that makes Z into a group. Here's G prime. I'll give you a specific one and then make the comment at the end that there's nothing special about the number four. Here's G prime. And the claim is that G is isomorphic to G prime. That the group of integers is isomorphic to or has the same structure as the group consisting of multiples of four. Reason? Proof? What do you need to do? You need to write down a one-to-one and on-toon function from G to G prime and convince me that it's a homomorphism. Well, here is such a function. Define phi from Z to 4 Z. Guess what it is? It's just phi of little Z equals 4 times Z. Take any integer and simply multiply it by four. Well, if you multiply an integer by four, you certainly get a multiple of four. So the fact that this function lands inside this group is an on-issue. Is it one-to-one? Let's see. If 4 times Z equals 4 times Z prime to Z equals Z prime, sure, just multiply both sides by a fourth. That's no big deal. Is it the case that this function is on to? In other words, is it the case that every element of this group can be realized as phi of something? Sure. If you hand me an element of this group, it's a multiple of four. What multiple of four? It's four times Z. I'm just plugging in. So the fact that this function is one-to-one and on to is, again, an easy observation. Easy phi is one-to-one and on to. But then it's the same observation as before, but phi is a homomorphism. We prove this in class and done in class. Heck, let's just write out a proof real quick. Here it is again. What's phi of one thing plus another? Z1 plus Z2. By definition, whatever goes in gets multiplied by four plus Z2, which is 4 Z1 plus 4 Z2 by the easy of arithmetic as you could ever want, which is phi of Z1 plus phi of Z2. So check. So these two groups are isomorphic. So don't, just because the two groups don't on the surface look the same, it doesn't necessarily mean that they're not the same structure that they're not isomorphic. In fact, here, admittedly, the first time I remember seeing this, I thought this was totally counterintuitive that you could have a group that's isomorphic to a subgroup of itself. But once you get past finite groups, once the groups that you're talking about are infinite, lots of things can happen that maybe you don't expect. Now, folks, if I handed you a finite group, a group with only finite many elements in it, then it's impossible for that group to be isomorphic to a proper subgroup of itself. Because if I hand you a group with so many elements in it and you take a proper subgroup, then there's necessarily fewer elements in it so that just from a function point of view, you can't have a one-to-one and on-to function from a set to a proper subset of itself if the set happens to be finite. But in this case where the set happens to be infinite, then it's possible to write down a one-to-one and on-to function from it to a subset of itself. And in fact, what we've shown here is that it's possible to write down a function that actually preserves the group structure. And if you, I don't know, I'm hoping to help you develop some intuition about this stuff. What is 4z? What is the collection of multiples of 4 look like? Well, it's 4, 8, 12, 16, blah, blah, blah, blah, and 0 and negative 4 and negative 8, et cetera. So the thing that we happen to call this has the property that if you add it to itself a bunch of times, you sort of get everything on this side and then you throw in 0 and then you throw in all the negatives of that. So at least on the surface, the fact that I've called the thing that generates everything else that versus if I had called it that, I mean, they do the same things. At least from the point of view of what's happening with respect to the operation plus. You start with this and you keep going and you get everything on that side and then you throw in 0 and you get everything on the other side but putting in the negatives. That's exactly what happens to get this group and that's exactly what happens to get this group. So they sort of walk and talk the same way. You just got to get over the fact that one was inside the other. Well, that's a little bit deeper. All right. Other questions or comments, remarks? Okay. What you'll see for one of the homework problems is a situation where you're going to show that the group of nonzero complex numbers, so this thing I called C star, where the operation is multiplication, is isomorphic to a subgroup of the group of 2 by 2 matrices with entries in the real numbers. So on surface looks totally bizarre because somehow you're talking about complex numbers, you know, a plus b i's and then you're talking about just matrices with entries in the real so you're thinking I don't see i anymore and you're thinking, wait, I got entries, I mean matrices, so typically I'm thinking those aren't commutative but this, but in the end it turns out you get a representation that's not just sort of cute, oh, let's see how this works out but actually gives you a way of thinking about the complex numbers as matrices rather than simply as a plus b i's and it turns out that there is some merit to doing that in certain settings and so that's what that homework assignment, that particular problem, I think it was 33 or something like that in section three is about, all right, two more comments about isomorphisms and then we'll move on to a new topic. Yeah. Well, there's, yeah, this is good, I'll do this now. It turns out it's possible, possible to write down isomorphisms where the groups G and G prime happen to be the same group. In other words, it's possible to write down a one-to-one and on to homomorphism from a group to itself. You're thinking, well, of course it is, just write down the identity function from the group to itself. Yeah. So any group is certainly isomorphic to itself, it's sort of a no-brainer, of course it is, any group has the same structure as itself. Well, I can write down an isomorphism, the identity map, but it turns out in certain situations there might be more than just the identity isomorphism from a group to itself and we've seen some examples of those in previous homework assignments. So example, let G be an element of any group G, notice I'm not specifying it's finite or infinite or billion or not a billion, I don't really care, define, let's see this was a homework assignment, phi sub G and the definition of phi sub G was phi sub G of X is what G, XG. Let's see, you showed for homework that this is a homomorphism. In fact, it's also pretty easy to show that this function phi sub G is both one-to-one and on-to, it's a little bit more doing and I'm not going to spend the time here to do it, but it turns out you can show phi sub G is an isomorphism. It's a one-to-one function that happens to be on-to and happens to be a homomorphism. Okay, I'm leaving out the details, but the details are totally straightforward. The word that we usually use, folks, in a situation where you have an isomorphism, but the groups that are involved happen to be the same, both as the input group and the output group, which we haven't done an example of yet, but here is a first one. The word that's usually used is, we call such an isomorphism an automorphism. So it's an isomorphism where the auto means the same, right? So auto or inherent or internal or something like that. So there's an example of an automorphism of a group. I think an automorphism basically is just a shuffling of the group. It's a permutation of the group that also happens to be a homomorphism. Another example of an automorphism of a group, if G is abelian. So if you start with any abelian group, then define this function, let's call it psi, from G to G, psi of x is x inverse. Let's see. First, homework, you showed that this was a homomorphism. Of course, it was only a homomorphism if you assumed that the group was abelian, but as long as the group was abelian, this thing turned out to be a homomorphism where you take an element and you assign it to its inverse, it turns out it's pretty easy to show that psi is one to one and on to. That's easy. And you showed for homework that because G is abelian, that psi is a homomorphism. And so what we've just shown is we've got an isomorphism from a group to itself, so psi here is an automorphism of G. Let me finish up this comment about automorphisms. A couple of times along the past two months or so that we've been together, I've mentioned the task is or the game is or the goal is to take existing groups and somehow cook up or come up with new groups. We've done that a lot. You look at subgroups, you look at factor groups, you look at product groups, you look at this, that and the other thing. Here's a new way of coming up with a group. You start with a group. And what you do is you write down all the automorphisms of that group. Well, it turns out, folks, if you take two automorphisms, in other words, if you take two isomorphisms that go from the group to itself from G to G, if you take two of them, well, this function goes from G to G and this function goes from G to G. If you compose them, you get a function from G to G and guess what? If you compose two automorphisms, you get another automorphism of the group. So it turns out function composition is a binary operation on the set of automorphisms of the group. So I'm looking at a set where the elements in the set are functions. The operation composition gives a binary operation on that set and it turns out that this set with this binary operation is a group in its own right. It's the group of automorphisms of G. So here's another way of building a group from something. Well, the something here is you take the original group and you look at all of its automorphisms. And it turned out this type of construction was really key or central or at least played a large part in the stuff that I talked about a little bit last time, this classification of the finite simple groups question. If you have some group, can you say something intelligent about, well, various properties of that group? And one question is, can you tell me what the automorphism group of that group is? And by analyzing properties of it, you're able to say something about groups being isomorphic or not isomorphic, et cetera. So here's another way of building groups. Final comment about isomorphisms. There is a result that we won't need in here, but is at least worth mentioning. Back in your linear algebra course, maybe you didn't prove all the details, but back in Math 313, you talked about the, it's usually called the rank plus nullity theorem or something like that, or the dimension theorem for matrices or something like that. It says that if you have an n by n matrix that if you look at the dimension of the kernel and you look at the dimension of the image or rephrased, if you look at the dimension of the row space and you look at the dimension of the null space and you add those two numbers together, you get n. It turns out that result is one very specific consequence of the result that we're about to write down. The result that we're about to write down is this. Look, if I start with, so final comment about isomorphisms is this. Take any homomorphism from, let's say, phi from g to some other group, I don't care. Well, associated with any homomorphism, starting with any group you want, there's always, well, there's a lot of different groups, subgroups, factor groups that we can talk about in this context. Here's one. Let's see. Kernel of phi, that's a subgroup of g. In fact, we proved in class that the kernel is always a normal subgroup of g. We can also talk about, and I mentioned this only briefly in the past, but we'll at least pull it up here. Just in terms of being a function, look, there's a function from that set to that set. You can ask what things actually get spit out in g prime. What's the image inside g prime under the function phi? Rephrased, what things are actually in the range. That's typically the word that's used, the range of phi. I am stands for the image. That's just the fancy way that we typically write this particular subset. Well, it turns out that the image is always a subgroup of g prime. Not necessarily a normal subgroup, but at least a subgroup of g prime. And it turns out that the three groups, g, kernel of phi and image of phi are related in the following way. This is usually referred to as the fundamental isomorphism theorem. Isotherm says regardless of what homomorphism you start with, here's what happens. You take g, you've got this normal subgroup that naturally arises, kernel of phi. So it makes sense to form the factor group that's at least doable because this subgroup is a normal subgroup. Here's another group that's floating around and it turns out that the fundamental isomorphism theorem says that regardless of what homomorphism you start with, if you look at the factor group of g by the kernel, you get a group that looks like its image. And I'm simply going to throw that out here. It turns out that we've done a lot of examples of situations where if you take g mod at kernel that you get something that looks like a group that we already know, the best example, at least to mention here is, yeah, I'll just mention it here. When g was Sn, permutations of the letters one through n or numbers one through n, if this was Sn and this was the group Z2, then we had a homomorphism from Sn to Z2. That was the homomorphism that said if you're an even permutation, spit out zero and if you're an odd permutation, spit out one. It turned out to be a homomorphism. The kernel, well, the things that go to zero are precisely the even permutation. So in that case, the kernel is just An and what this fundamental isomorphism theorem says is that if you take Sn and you look at the factor group of it by An, you get, well, what things actually come out in Z2? There's only two things in there and they both come out. One gets spit out if you run in an odd permutation, zero if you run an even permutation and so the image is all Z2. All right, that is the final comment. Nice. Okay, so what we've just done is, at least for now, finished up our study of groups as groups, what we're going to do for the rest of the semester is study a second type of algebraic structure. It turns out that the notion of a group plays a big role in this second type of algebraic structure. Let me tell you the sort of thing that you maybe have had in the back of your mind but haven't really sort of asked concretely, at least nobody in here has asked concretely yet in class. You know, the examples of groups that we've given are maybe the integers with addition or the reels with addition or Z sub n with addition and there've been some other examples of groups involving multiplication but we've always had to sort of throw zero out whenever there's been a zero around in order to get groups under multiplication just because the thing that we've called zero never had any sort of inverse to it. All right, that's fine. But in all the examples that I just mentioned, you know, the integers with addition or the reels with addition or, you know, take the complex numbers and throw out zero under multiplication, etc. The sets that I've talked about, even the sets, you know, Z sub n, zero through m minus one, all of those sets have always had two separate binary operations on them. Typically, they've been called plus in times and if I hand you the integers, you can make sense to add two integers. It also makes sense to multiply two integers. Adding two integers gave a group structure on the integers. Multiplying two integers at least was a binary operation even though it didn't give a group structure. Similarly, if I hand you the two by two matrices over the reels, something like that. Well, hey, if I hand you a couple of two by two matrices, it makes sense not only to add them to get another two by two matrix but also to multiply them to get another two by two matrix. The additive operation made the two by two matrices into a group. The multiplicative operation did not because there's issues of inverses when you're multiplying, but if you threw out the things that don't have inverses, if you threw out the things that have determined zero, then what's left over at least gave you a group. So sort of the intuition of the philosophy is, hey, you know, all these sets that we've looked at so far, except as it turns out, the permutation groups, the SN groups. But all the other examples of groups that we've had have actually come equipped with two different binary operations, one of which made the collection into a group and the other of which the multiplication sometimes came close to being a group, like if you threw out zero, maybe things became a group. Or if you threw out things that you didn't want to have around for the multiplication, like the non-invertible things, you want to put a group. What we're going to study for the rest of the semester are systems that have two binary operations. One of which will make the given set into a group, just like addition always did. And the other of which won't make the set into a group. Sometimes it'll make the set into a group if you throw out a lot more stuff. And so the general idea or the structures that we're now going to look at are the following. It turns out we will now focus, again for the rest of the semester, our attention on structures or sets having two binary operations. And again, I've just given you a lot of examples of such sets and they're certainly not esoteric, they're totally concrete, the integers, the rationals, the reels, the complexes, the two by two matrices, the ZN sets, whatever you want, all of those always came equipped with two binary operations. And here are the things that we're going to talk about. They are called rings. So what I'm going to do is simply start by giving you the definition of what a ring is, then we'll write down a whole bunch of examples of these things and then start looking at some of their properties. So here's the definition. I'm going to ask you to start with a set. And I'm going to assume that that set has not just one but two binary operations on it. And what I should technically do, folks, is call one of the binary operations stars of one and call the other one stars of two. But in all the examples that we're going to do in here, the first binary operation is typically going to be simply addition or will be denoted by plus sign. And the second operation will typically be denoted by multiplication. In the integers, plus and times. In the reels, plus and times. In the two by two matrices, plus and times. In the Z sub N sets, plus and times. Where the plus is done mod N and the times is done mod N. So rather than do it esoterically a star one and star two, similar to what we may have done on day one of the semester, we're simply going to call the two operations plus and times or plus and dot. Is called a ring in case the following. Well, folks, I only need to list out three things in order to define what a ring is. But I only need to list out three things because one of the things I'm going to list out contains a whole lot of information packed in already. In case first, if you look at the set R, and you only look at the binary operation plus, that this is an abelian group. That's the first property of what it means to say that this set together with these two binary operations is a ring. Look, I can write out in one line, because we're algebraically mature in here now, because we spent two months talking about groups. I can write down in one line what at the beginning of the semester would have taken me about five lines to write down. What does this mean? It means that this is a binary operation on that. In other words, if you add any two things in R, you get something back in R, and it's associative, and there's an identity element, and each element has an inverse, and it's the case that if you take two elements and you add them in either order, you get the same output. I mean, I could list that out in five lines, but all that information is now packed into this one line. That's good. Here's the second hypothesis that when you look at the same set and you look at this other thing that all you get is an associative binary operation. So this says a lot. This says there's not much action in general happening in the multiplication. All you require is that the multiplication is associative, and that if you take two things and you multiply them together, you get something back in the set. I'm not requiring an identity element. I'm not requiring inverses. I'm not requiring this as the other thing. I'm not requiring that they commute. And the third line hopefully won't be too surprising. I've got this set. I've got two binary operations. At this point in time, though, folks, the two binary operations are completely unrelated to each other, and what this third property says is that there has to be a reasonable relationship between them, but it's a relationship you're totally familiar with. The only relationship or the only sort of connection between addition and multiplication that you know is the distributive law. And so all we're going to do is assume that the distributive law holds. In other words, if you consider for every triple of elements A, B, C in R, if you do A multiplied B plus C, you get A dot B plus A dot C. And technically, I have to tell you that this is true on either side. And if you do B plus C times A, that you get B dot A plus C dot A. In other words, that you have the distributive laws, laws of multiplication over addition. So there's the equations that somehow connect the two binary operations. And again, because I'm not assuming that the multiplication operation is commutative, that's not part of the hypothesis here, I technically have to tell you that if you do the multiplication in either order, either you multiply times the sum, or you take first the sum and then multiply, that you get exactly what you'd expect you're supposed to get. So that's what it means for a set together with two binary operations to be a ring. And what I'm about to do here, folks, is list out at least a half a dozen examples that will follow almost immediately from all of the work that we've done all semester. Examples, and there are a lot of them. How about the set of integers together with the usual addition of multiplication? Is that an abelian group? Yes, it is. Is the product of any two integers an integer? Sure. In other words, I mean, all I need is closure of the multiplication operation. I don't require much of it. Well, I require that it's associative, but that's not too big a deal. Multiplication is associative. And I require distributive law, which we all, well, at least were told was true back in third grade. And I'll continue the tradition of simply saying there is a distributive law. You want some more? How about the rationales with addition and multiplication? How about the reels with addition and multiplication? How about the complex numbers with addition and multiplication? How about the, oh, the collection of n by n matrices over the reels with addition. And when I write down addition here, I mean addition of matrices and multiplication. And when I write down Now multiplication here, I mean multiplication of matrices. Remember, is that in the Boolean group? Yeah, addition of matrices is fine. Multiplication of matrices is not commutative. So what? The multiplication operation isn't assumed to be in sense anything. All you need to make sure is that if you take two things in the set and you multiply them, you get something back in the set. Hmm. Let's see. Another example. Oh, yeah. If I look at Zn. So 0 through n minus 1. We're here, the operation is addition, but I'll remind you it's addition mod n. Remember, this was the notation we used for. And the multiplication, where the multiplication is mod n, there's another example. These are easy. Technically, you've got to worry that the multiplication distributes over the addition, but that's a non-issue for matrices. It's a non-issue for multiplication in addition of mod n and Z sub n. So there are lots of rings around. I'll give you one more example. It's an example that's technically something that satisfies these three conditions, but it's one that we won't work with too much in here. The example is, if I look at the set of multiples of 4 with usual addition and multiplication, that's a perfectly good ring. Let's see, is 4z with plus and b in group? It certainly is. Is it the case that if I take any two multiples of 4 and I multiply them together, I get another multiple of 4? Yeah, I do. In fact, I get a multiple of 16, but that's sort of irrelevant. I get another multiple of 4. That's fine. And does multiplication distribute over addition? Sure, it does, because it's just an operation inside the integers. So there's another example of a ring. Yeah. Oh, yeah. Let me give you another example that's going to play a huge role. I'll show it to you now, and then we will definitely come back to it in about two weeks and look at some significantly more interesting properties. Example 8. So this is on the list. Example 8. What I want you to do is take the reals, change this in two weeks, but for now, take the reals and what I want you to do is look at all possible polynomials where the coefficients on the polynomials come from the reals. So this is the notation for the set of polynomials with coefficients or having coefficients in the real numbers. So here's an example of something in there, 1 plus x squared. There's a polynomial with real coefficients. Here's another example of something in there. Pi minus 3x squared plus 4x to the seventh minus 9.2x to the ninth or something. Just polynomials. Remember, polynomials, by definition, have positive or at least non-negative integer coefficients, non-negative integer exponents on the x term. So x to the 1 half is not in there. x to the negative 2 is not in there. Dividing polynomials doesn't necessarily give you polynomial. All it is is look at things that are of the form polynomials. Well, it makes sense to take any two polynomials with coefficients of the real numbers. Certainly makes sense to add them together. Just addition of polynomials, so it's no big deal. And it certainly makes sense to multiply two polynomials together and get another polynomial. And even though you've never written it down formally, your math 104 students, your algebra 1 students, know that if you take polynomials, they behave just like real numbers do, at least vis-a-vis distributive laws and multiplication laws and commutative laws and all that sort of stuff. Turns out this is a ring. So what do we have to check? We have to check that at least under the addition that this thing becomes a B in group, that's no big deal. If you add two polynomials, it doesn't matter what order you do it in. There's certainly a special polynomial that behaves as the identity. It's just the zero polynomial. If you have a polynomial, it doesn't have an inverse. Sure, just take whatever the polynomial isn't, but negative sign in front of all the coefficients. That'll allow you to add to zero. So here's a really nice example of a structure that will be of great importance later on. And I'll mention it now, but we'll look at it more deeply in about two weeks. There's really nothing special about putting R here. If I ask you to look at polynomials where the coefficients come from the rationals, you get another ring. Or if I ask you to look at polynomials where the coefficients come from the integers, I mean it makes sense to add whole number coefficients polynomials together. You get another whole number of coefficient polynomials. Or to multiply them, you get another. So in fact, this is just sort of looking down the road a little bit, it turns out any time I start with, well actually I don't even care if it's reals or rationals or, any time I start with a ring, I can talk about the collection of polynomials where the coefficients come from that ring and you get another ring. The usual suspects to use as coefficients are the reals or the rationals or the complexes, but it doesn't preclude me in that structure to use matrices or Z sub four or something like that is the coefficient. All right, and we get another, what will for us be a very important example of a ring. Okay, question so far? This is sort of an interesting one. You know, what is that? Well, it's that, okay. But I mean, why is that word used to describe these structures? And the history on this isn't really clear, but the German school, as I mentioned with groups, the German school in Gürigan in the mid 1800s was sort of the first to really start describing these objects as a single manifestation of a whole lot of examples. You know, here are the integers and addition, here are the rationals, and here are matrices, and here are polynomials, and here are all these structures. Let's take a step back and see if we can sort of collect all those things under one umbrella. Well, we've just done that. Those three things together collect a whole lot of structures under one umbrella. Now, why use this word? Someone suggested, and this is as good as, at least makes sense to me, that this word was significantly more popular a hundred to 150 years ago, and it turns out it's the same word in German. And the usage of it today has become a little bit co-opted, because if I use this word in the phrase drug ring, you know what that means, right? It's a bunch of things working together to, okay, well, a hundred years ago, this word was used in situations to mean that a collective of things working together, but it was also used in sort of non-co-opted situations or non-illegal situations, I guess. But through the years, for some reason, the only phrase I can think of now that uses the word ring as some collective thing working together is drug ring, so maybe you can think of another one. But apparently that's historically why this word is used here. It's meant to denote a collection of things working together. And well, that's what these things are. I mean, the integers work together, they have to work together two different ways. In addition to multiplication, the rationals do the same thing, matrices do the same thing, et cetera. All right, what we will do, at least initially when we're looking at properties of rings is, now that we've written down some examples, the question will be how, what sorts of properties are similar when you look at rings and what sorts of properties are different when you look at various rings? So for instance, how is it that the ring of integers differs from the ring of rational numbers, which in turn differs from the ring of real numbers, et cetera? And the first sort of descriptor or the first property that most rings of interest have, but not necessarily all of them, is the following. You see, by requiring that under the addition that the structure gives an obedient group means that by default, because that's one of the properties of group, that the plus operation has an identity to it. And we typically denote it by zero. But you'll notice there's no such requirement on the multiplication operation. It's not required that the multiplication have an identity element to it. And here's an example of a ring for which the multiplication doesn't have an identity element. There's no element in this ring with the property that you take something else in that ring or anything else in that ring and you combine it via this binary operation, in other words, you do multiplication, that you get the other thing back. And the intuition is one's not in there and one should be the multiplicative identity. So this ring, it turns out, is different than all the other rings I've written down because this ring has the property that the multiplication does not have an identity element. This one does. It's called one. That's called one. It's called one. It's called one. It's called i sub n, the identity matrix sub n. The identity element here is called one. The identity element over there in the polynomials is just the polynomial. It looks like the constant one with zeros everywhere else. So the first sort of filter is this. We say definition that the ring, the ring, and I'm already gonna be sloppy, folks, because you've seen me do this before and we're at least comfortable with it already. Technically, I should say the ring r comma plus comma dot because a ring is a set together with two binary operations, but I'm simply gonna say r plus dot. It just gets too cumbersome to keep writing down plus dot, understood, is has unity in case when you look at r with dot, this has an identity element. And if it does, then the identity element is called unity. This notation is terrible and it's pretty non-standard. I mean, most algebra books would call a ring for which the multiplication has an identity. They'll simply call it a ring with identity because there's no mistaking which of the two binary operations you're talking about, the addition binary operation default always has an identity because you're assuming r with pluses and b in group. So if you're gonna tell me the ring has an identity, the only statement of interest is it's an identity for a multiplication operation, but this author likes to use this word. So all of the rings that we've written down, except for example seven, are rings with unity or we'll say the ring has unity. And it typically looks like one, but it could look like an n by n identity matrix. Or in some situations, and I haven't written down this example, but I will probably a little bit more next Monday. You know, and back in section one to back in section four, we looked at these collections of functions, from the reels to the reels. It made sense to add functions from the reels to the reels, made sense to multiply functions from the reels to the reels, it turns out that gives a ring. And in that ring, that ring has unity and the unity happens to be the constant function one. So it could take on many different forms. Well, probably what that means is not talk much about the ring consisting of the multiples of four, just because it doesn't have a unity and having a unity turns out to be really helpful. It's not that it's not a ring, it's just not a ring that's as common or as useful as the other rings that will come up. So we'll slide it down the rug a little bit. Let me give you another example. In this example, we'll be one that we'll yeah, refer back to for at least the next two weeks. Well, once you've written down a bunch of rings, then you can ask, are there other rings floating around? Well, a good place to look for more rings is inside things that you already know are rings. So what I'm about to do is write down a bunch of real numbers, not all real numbers, but a bunch of them and convince you, hopefully, that the collection of real numbers that I've written down is in fact a ring. So here's the example. I'm gonna call this set S and the set S consists of the following real numbers. Real numbers of the form, real numbers of the form a plus b times the square root of five, where the numbers a and b, where a and b are rationals, i.e. a, b are rationals. Well, look, if a is a rational number, rational, you know, just to remind me, means the quotient of two integers and b is a rational number and you form the number a plus b times the square root of five. Well, folks, the square root of five, it turns out it's not a rational number. You can't write it as the quotient of two. But it's a perfectly good real number, that's fine. So I'm multiplying a rational number times a real number. So that's a real number, that's a rational number. If you add a rational number to a real number, you get a real number. So this set sits inside the reals. It turns out it'll be really easy to show that it's not all of the real numbers. For example, square root of two is not in that set. Pi is not in that set. It's the only things that you can form by taking a quotient of integers, a fraction of integers, and then adding that possibly to some other fraction of integers times this special irrational number, square root of five. Okay, turns out, claim s with the usual addition and multiplication of real numbers so I don't really need to write that down is a ring with unity. Well, showing something as a ring, well, is easier hard, I guess, depending on whether or not you view showing something as an obedient group is easier hard. Well, let's make sure at least that if I look at s with addition, that this set with addition, that I actually get an obedient group. Well, first thing you gotta convince me of is that it's actually a binary operation. So we gotta go back to sort of ground zero here. Is it the case that, so let's look at one, is s with plus and a obedient group? A obedient group. Well, is it even a binary operation? In other words, if I take two things of the correct form, here's one of them, and I add it to something else of the correct form, so I've handed you two things in the set. Here's what one looks like, here's what another one looks like, where the hypothesis is that A, B, A prime and B prime are rationals. Well, then let's see, well, I know how to add, these are just real numbers, folks, so this is just plus in the real numbers. So there's nothing mysterious going on here, but the point is I can just by rewriting things, I can certainly rewrite it that way. That's a complete non-issue. But the point is, by writing down this expression, what I've hopefully convinced you of is that if I take something in s, and I take something else in s, and I combine them using addition, that the result, well, looks like this, but if both A and A prime are rational, if you add two rational numbers, you get a rational number back. If you add two rational numbers, you get a rational number back, so the point is that the sum of two things of the correct form is of the correct form. In other words, I get something back in s. Correct form means you have to convince me that you can write the thing as a rational number plus some other rational number times a square to five, and I've just done that here. So at least it's a binary operation. Is it associative? Yeah, no problem because it's just, all we're doing here is adding real numbers, not all real numbers, some of them. And since addition of real numbers is associative, addition is associative, not a big deal. So associative, I'm just gonna say check, associative, check, how about identity element? Well, what is the identity element gonna look like if the operation is addition? Typically it looks like zero. So the question is, is zero in this set? Sure, zero, the real number zero can certainly be written in the correct form, correct form means you have to put a rational number here. And technically, as to be formally correct, I have to convince you that zero is a rational number. I mean, it's an integer, so it can certainly be viewed as a rational number. So I've just written zero in the correct form. Okay, I have to make sure that zero is in the set. Is it? Well, yeah, it is because I've written it in the correct form. Final, and not final, I got two more steps here. Let's see, how about inverses? If you hand me something in the set, if, well, let's call it A plus B squared to five, that's what a generic element in S looks like, then what is its inverse? In other words, inverse with respect to the addition, well, it's just it's negative, then I'll call it, let's call it little s prime, let's call this little s, now let's call it something else because my s is in my five. Similar, how about t? Then what's t inverse or t prime is minus A plus B root five? That's how you get the inverse of something under addition, but I can rewrite this as minus B root five, and the point is this is rational, it's rational because A is rational, this is rational because B is rational, and so the point is you've written the inverse of this element in the correct form. Finally, is it the case that plus is commutative? Sure, addition of real numbers is commutative, no big deal. So we get a group, in fact, we get an ambient group, so at least this set satisfies the first property of being a ring, that when you just look at the addition, that you get a group. Of course, technically what we've shown is that we've got a subgroup because this thing happens to be living inside the reels with addition, which we already know is a group. So to show that S with plus is an ambient group really just boils down to using a subgroup theorem. Closure, identity, inverses, that's what we're really worried about. Okay, second, now the second one is gonna be maybe a little bit screwier than you'd think. What do we have to now do to show that this set S forms a ring? We have to show that the set together with the multiplication operation is an associative binary operation. Well, the multiplication is just multiplication of real numbers, so associativity comes for free. What we're not yet sure of, though, is whether or not multiplication is a binary operation on this set. So we have to show that multiplication, or I'll say S with dot, is an associative binary operation, but again, the associativity comes for free because it's all happening inside the reels. So we have to show it's closed. In other words, we have to show that if we take, I don't know, let's call this one A plus B root five, and we multiply it by something else in the set, and what does it mean to say each of these is in the set where A, A prime, B, and B prime are rational, that that, let me just not say yes, show that that is in S. Show that the product of two things in the set gives me something back in the set. All right, well, folks, that's not clear because it's not clear that this expression can be written in the correct form. The correct form means you have to be able to write it as a rational number plus some other rational number times the square root of five. Well, let's multiply it out. Equals, what do I call this? Foil it out, A, A prime plus B, A prime B root five plus, what do I do here? So it's A times that plus that times that plus A, B prime root five plus B, B prime root five, root five. So I clearly did a little bit more arithmetic than I needed to, but I wanna show you what the real issue is. Look, this is A, A prime, this incredibly luckily is five B, B prime. And I'm gonna group those together, and then I'm gonna group these two together, A prime B plus A, B prime, all times the square root of five. So all this is is arithmetic, but the point is this piece is rational, and the reason it's rational is A, A prime, B and B prime are all rational. If you multiply two rational numbers together and then you multiply by five, you still get a rational number. So this is rational, so the whole thing is rational. The beauty of what just happened here was because I'm multiplying the square root of five times itself, it somehow gets kicked back down into the rational. And then this other piece, well, it's fine. I sort of factored out the square root of five piece, but all that stuff is in Q. And so the point is this whole thing is in capital S because it's of the correct form. And the correct form is that's a rational number plus some rational number times the square root of five. And a quick remark for contrast, if I had put something like the cube root of five here, which would have been perfectly legit to do, here's a collection of real numbers, those that can be written in this form. It turns out this set is not a ring because it's not closed under multiplication. If I had had a cube root of five times a cube root of five here, the best I could have done is written it as the cube root of 25 and that's not of the correct form. So it's here that things turn out to be sort of very closely tied with the fact that I've used a symbol like the square root of five. I've used a symbol where when you multiply it times itself you get something back in the rational. So at least this thing is a binary operation. I'm gonna ask that was the key part and then property three, the distributive law is fine. Folks, if you ever have a subset of something that you already presumably know as a ring, like the reels, then the distributive law is a complete non-issue. If you know the distributive law holds in the larger set and we certainly know that it holds in the reels, then it by default has to hold in the smaller set. So I mean, I'm not gonna assign you a homework assignment today. I'll probably assign one Monday but it won't be due until after the exam. The point is though on a homework question about showing something is or isn't a ring, if you know that what you're looking at already sits inside a ring, property three is completely trivial or irrelevant. It holds by default, since simply asks as a subset of something that we already know to be a ring, namely the reels. Okay, yeah, let me, let's prove a couple of propositions quickly and then we'll get out of here and I'll give you some words next Monday. So, what we've done so far is we've given, in fact, what we've done so far, folks, is exactly the same procedure that we did when we talked about groups. We wrote down the definition of a group, then we wrote down a bunch of examples. And then what we did for groups is we said, hey, groups have certain properties. Like, you know, if you take the product of two things, A, B, and you want to find its inverse, that turns out to be B inverse A inverse. That's a property of a group. Or another property of a group that's not given explicitly in the definition is that the identity element is unique or something like that. All right, so you start proving some propositions about groups. Let's prove two quick propositions about rings. And the two propositions I want to prove, and they'll each take me only about two minutes each, is well, look, we have these two binary operations. They're connected by the distributive law. You know, this is sort of interesting. In all the systems that you can think of, think about this, the thing called zero. Well, for us, zero is the identity element in the group R with plus. It's the additive identity element, right? It's the thing that has a property that when you add it to anything else, it doesn't change the other thing. But in all the systems you know, you actually know something about how that special symbol behaves with respect to the other operation, with the multiplication operation. Zero times anything always turned out to be zero. So even though this thing comes up technically as the additive identity, it's sort of the multiplicative killer in all the systems that you know. And that property turns out to be true in any ring. If you identify the additive identity, and typically we call it this, if you take that and you multiply it times anything in the ring, you'll always get it back. In other words, you always get zero. Conversely, okay? Proposition, proposition in any ring R, ring R, if you identify the additive identity, and you call it that, and you multiply it by any other thing in the ring, that you always get it. And for every R and R. Multiplying by zero gives zero. This is not news to anybody. Certainly in all the examples that we've written down, it's the case that you already know that. It's also the case, folks, that the thing called zero might take on various forms, depending on what ring you're looking in. If you're looking in the ring of two-by-two matrices, for example, the thing called zero really is the matrix 0, 0, 0, 0. But that's okay. I mean, we're familiar with identifying it. And what this result says is I don't care what ring you're in. If you're multiplying by zero, you're gonna get zero back. Here's the proof, and the proof is sort of nice. Remember, the addition operation gives me an abelian group, which necessarily then gives me all of the structural properties that I've been able to prove about groups in here. Let's go ahead and prove this one. The other one is just the same. You just flip things around. Look, zero is zero plus zero. Why? Because in any group, see how I did my identity, if you take the identity element and you combine it with itself, you get itself back. The identity element of a group is the unique idempotent element in there. So this is just a statement about groups. Take the identity, combine it with itself, get itself back. Now what I want you to do is combine both sides of this equation by multiplying on the right side by this thing called r. If these two things are equal, then necessarily multiplying both of them by any element r that you'd like. Gives you any, oh, but wait a minute. But this thing, ooh, this is where I get to use the distributive law. The distributive law says if you take the sum of two things and you multiply it by something, that you can group them as, so there's the distributive law. And hopefully, folks, you're not surprised to see the distributive law here because we're trying to say something about how the additive identity behaves with respect to the other operation, with respect to the multiplication. And the only piece of information that relates the two operations together is in the distributive law. Okay, but wait a minute. The thing called zero is the additive identity of the group. In other words, if you add zero to anything, you get the thing back. That's the definition of what it means to be the identity in this group. So if I do this, this is this. Why? Because I'm simply adding zero to something in this group. And adding the identity doesn't change it. And now I'm just gonna haul out a property that we proved of groups, in particular the group r with plus. And we call it left cancellation law. No, right cancellation law. If you have something, let's call it a, and you have something called c, and you have something called b, and you have the same thing called c, there's c. And if you have a plus c equals b plus c, then what do you get to do? You get to cancel the c's, cancellation in groups, a equals b. In other words, zero equals zero dot r, cancellation in groups, in r with plus. It's really powerful to be assuming that the plus operation gives a group. And that's what we've tried to prove, that zero is zero times r. So if you multiply anything by zero, you get zero back. And the other side works the same way. You simply use the other distributive law. All right, so here's the connection between the additive identity and the multiplication operation. It turns out there's a relationship in a ring with unity between the unity, the multiplicative identity, and the additive operation, and I'll just go ahead and state it for you. I won't go ahead and prove it. The proposition is, if r has unity, if r has unity, and we usually denote the unity by that symbol, even though just like with the zero thing, it might not take on exactly that form. It might look like one zero zero one or an identity matrix or an identity function or something like that. Then the following is true. If you take any element in the ring, call it r, then I can perform two different operations on r. One operation is tell me what its inverse element is with respect to the addition. In other words, tell me what its additive inverse is. Necessarily it exists because r with plus is an imbuing group. Tell me what its negative is. Here's another thing that I can ask you to compute. Tell me what the additive inverse of the unity is and then multiply that by the original element r. And in the end, those are always the same. Negative one times r is negative r. This is two separate statements, folks. This is a statement about multiplication and this is a statement about additive inverses. But it turns out these two are always the same for all r in the ring. So at least the distributive laws, and I'm not gonna prove that for you, but it turns out you just used the distributive law in much the same way that I just did here. It turns out there then is enough relationship between the addition and the multiplication that all of the stuff that you sort of know that you're bringing to the table here typically is true. Zero times anything is zero. Negative one times anything is the negative of that thing. Okay, that's a good place to quit. So no new homework tonight because it's Wednesday, but if you've got the homework that's due tonight, I'll let you leave it on the table here. Next Monday I'll be handing out exam two review materials. I'll also be posting them on the web next Monday.