 You get this is what it says is, I don't care if you're written down to zero element, it turns out to typically be non-interesting, but if you happen to have written it down and you raise it to the p power, you're certainly going to get the original element back. If you take any non-zero element, you raise it to the p power. This says you get the original element back. So it says take any element of zp. I don't care what you call it. Call it abxypqcd. I don't really care. Call it suggestively a plus b. So if I take something and I raise it to the p power, well, folks, I don't care what that is, something in zp. And what we've just proved is that if you take anything in zp and you raise it to the p, you get itself back. That's sort of sweet. You're thinking, well, wait a minute. What I'm supposed to show is that a plus b to the p is a to the p plus b to the p. Yeah, but what's a? It's a to the p. What's b? b to the p. That's really nice. That is really nice. Now, the slight disadvantage of this proof, this is really nice. I like this slide. The slight disadvantage of this proof is that it's only legit in a situation where you have Formosal Theorem working for you. And the only ring that we prove that happens in is zp. So technically, this proof, while really nice and significantly more subtle or maybe cleaner than this one, only works in zp, whereas the sort of grind it out proof works not only in zp, but will work for any ring of characteristic p. And we noted last Wednesday that, for example, zpx or something like that is a ring of characteristic p as well, even though it's not just zp itself. So there's a, maybe, grossly exaggerated hint on, I don't know what, I don't know what that, what is that problem? Oh, 40, that was 41, right? OK. The other problem that I'll give you a quick hint on is in section 19, number three, you're asked to find all the zeros of this polynomial. And the polynomial they gave you, I asked you to change to a different polynomial, x squared plus 3x plus 2. And the book gives you a sort of a method to systematically find all the zeros in z6 of a certain polynomial, but, and we'll see this as well. What we're doing a lot of folks is sort of setting up for the situation that when we get to the specific rings that look like polynomials with coefficients taken from a field that we've got some experience with those, what we'll wind up doing here is just take this polynomial. If you're asked to decide, well, view it as coefficients in z6, how many zeros does it have in z6? We've got six elements in it, so just drop all six elements in the polynomial and see which ones kick out zero. And what this example is designed to do is to show you that if we're working in a situation where the coefficients are taken from something other than a field like z6 is not a field, because 6 is not prime, that some of the things that you know about polynomials when the coefficients are taken from the reels or the complexions or something like that, like a degree two polynomial can only have at most two zeros. Here we're going to get a degree two polynomial and I'll give away the answer that has four different zeros. So you have more zeros than the degree of the polynomial and the reason that can happen is because you happen to be working in a setting where the coefficients are being taken from something that's not a field. So that's the point of question number three there. All right, other questions or comments? All right, all right, so here's what we're going to do. We're going to finish up some, this is sort of a, I don't know, hodgepodge of topics tonight. So we're going to finish up some things that we've already seen. I'll introduce a couple of new ideas tonight and then hopefully by the end I'll give you sort of the overview of where we're headed with the study of rings. There's lots of different directions that we could take, all of them good, but the one that we're going to choose to take is the direction that'll allow us to lead up to the material that'll be in the math four slash five 15 course next semester. Fortunately though, the material that we will wind up focusing on in rings turns out to be of interest in its own right. So if you're not planning to take the follow-up semester to this course, that's fine. The material that we'll look at this semester, I think is inherently of interest in, is actually sort of useful in some applications as well. So, okay, so here are some sort of cleanup ideas. Remark on characteristic of a ring. Characteristic of ring R. We officially defined this last Wednesday just as a brief review or refresher. Here's what this means. If you take a ring with unity, call it one, you add one to itself a bunch of times. Sometimes you eventually kick out zero when you do that and sometimes you don't. The characteristic of the ring is either the smallest positive integer so that when you add one to itself so many times you get zero or if you never get zero when you add one to itself, like if the ring is the integers or the real numbers or something like that, then we say that the characteristic of the ring is zero. And so N is characteristic of R means is the smallest positive N and with one added to itself, N times equal to zero. And the point is if I tell you that I've got a ring of characteristic two or characteristic three or characteristic four, whatever it is, intuitively it means that whatever the characteristic is, whatever that number is, is treated as zero in that ring. So for instance, if I tell you the ring is characteristic three, all right, so I've told you that if you take whatever the multiplicative identity is one and you add it to itself three times that you get zero, what that means is that somehow any element in the ring when added to itself three times also gives you zero. Then the punchline is if characteristic of R is N and A is any element of R, then if you add A to itself, N times, you get zero. Well, yeah, let's see. Let me make sure I understand the question line. So if the question is could it become zero sooner? Is that the question? Is there any other element? It might become zero sooner for some other elements. In other words, if N is the smallest number that makes one equal to zero, it might be the case that if you hand me something else that you get there quicker, I'll give you a quick example. So the characteristic of z sub eight is eight because you add one to itself and it takes you eight wax to get back to zero. But I could hand you the number four in that ring and four plus four is already zero after two wax. So there might be elements that get you there quicker. But the point is if you do this sum to any element in the ring that many times you're guaranteed to always be looking at zero. So the only minimum value is, it's the minimum value that takes the element one and gets you back to zero. And so the reason is, well, what is A plus A plus A N times? It's easy. It's factoring A out. But this is zero, which is A times zero, which is zero. You've got to be a little bit careful. So what that means is if I tell you the characteristic is two, anytime you add an element to itself, you get zero. Which is sort of maybe a little bit strange the first time you look at it because you're thinking, well, just that means the ring doesn't have very much guts to it. And the answer is no, the ring could in fact have infinitely many elements in it. Even though each time you add an element to itself you get zero, that's fine. You're doing a lot of adding of elements to other elements, not necessarily to itself. So a good example is, example, here's the ring. R is the polynomials with coefficients taken from Z two. It's an infinite ring. There's lots of things and there's lots of polynomials. Now admittedly, I haven't given you much choice on what coefficients to use in those polynomials. You only get to use zeros and ones as coefficients because those are the only elements sitting inside this ring. That's all right. But x is in there, x squared is in there, x plus x squared is in there, x plus x squared plus x to the fifth is in there. But if I hand you something like x plus x, that's zero because it's two x and two is the same as zero. Or if I hand you x cubed plus x cubed, you get zero. Or if I hand you one plus x plus one plus x, you get zero. Because I think it's two plus two x, but the two's always sort of just melt down to zero. So the ring is infinite, infinitely many polynomials with coefficients taken from Z two. But every element, let's call it a, a in R has a plus a equals zero. Because the characteristic of this ring is two. All right. There's a remark on characteristic of a ring. That was a good one. Yeah. Okay, we did that. We did that. Let's see. We did, I don't think I did this example for you. So recall, and I've already looked at this result already, this thing that we called little for maus theorem in Zp. If p is prime, then a to the p minus one equals one in Zp for all a not equal to zero in Zp. So if I hand you something that's not the zero element in Zp, and I raise it to the p minus one, then I get one. Well, what we're going to do here is interpret elements in Zp as simply being remainders on division by p. So if you take something, and I'm working in Z7 or something like that, and I ask you to look at the integer 24, but view it as an element of Zp, well, you just divide seven into 24, and you get remainder three. And so 24 is viewed as the element three inside Zp. But if we sort of take that, I don't know, if we allow ourselves to go back and forth between what's happening inside Zp and what's happening inside Z for all integers, then we can use this result to ask questions about various divisibility properties inside the integers. So we can use this, for example, to ask example, find the remainder, the remainder when, let's see what example I want to use here. When 8 to the 62 is divided by 13, I think I didn't have time to do this follow-up example when we looked at a little Fermat's theorem. Fermat's little theorem, OK? All right, so here's the idea. When we're taking an integer and viewing it as an element in Zp, it's simply asking, well, what integer do you get when you divide by the given modulus here? Here the modulus is P, which in this case is going to be 13. And tell me what the remainder is. All right, well, let's see. Here's the idea. What we know is this, that 8 is not equal to 0 in Z13. And that's an important observation to start with. It's not because 8's not divisible by 13. That's fine. Oh, so what does that mean? So Fermat's little theorem, FLT or LFT, however you want to, Fermat's little theorem says that if you take 8 to the P minus 1 power, well, here, 13 is prime. 13 minus 1 is 12 equals 1 in Z13. So here's what that tells me. I'm interested in saying something intelligent about 8 to the 62nd. Well, I can at least get something close to 8 to the 62nd power into the act just by judiciously picking a power to raise both sides of this equation to. How about let's raise both sides to the 5th. So 8 to the 12 to the 5th is 1 to the 5th in Z13. Yeah, if you take two expressions that are equal inside some ring and you multiply them by themselves the same number of times, then the resulting outcomes will be the same. Oh, so what does that mean? Well, we have exponentiation here. 8 to the 12 to the 5th, 8 to the 60th is 1 in Z13. And now we're in business because I'm trying to say something about, well, what happens when you divide 13 into 8 to the 62nd power? What we've now observed is that if you divide 13 into 8 to the 60th, well, inside Z13, that's the same as 1. So the remainder when you divide 13 into 8 to the 60th is 1. But that's not the question. It's what happens when you divide 13 into 8 to the 62nd. Well, just cleaning things up. Hey, if I multiply both sides of this equation by 8 squared in Z13, so 8 to the 62nd is then the same as 8 squared in Z13. And that reduces the multiplicative load significantly because I can probably figure out what 8 squared is. In other words, it's 64 in Z13. 12? Why is that? If you divide 13 into 64, well, let's see. 13 times 4, for those of you that play cards, that's an important computation. Plus, let's see, that's 52 plus 12. And so the remainder, when you divide 13 into 64 is 12. So the punch line is if you want the remainder when you divide 13 into 8 to the 62nd power, so the remainder gives remainder 12 on division by 13. So Fermat's little theorem is pretty nice. It allows you to compute with potentially large exponents just by saying, well, hey, every time you see a certain integer relative to the modulus that you're dividing by, you can replace that. Bless you, you can replace that by the number 1. Question, what happens if the modulus that we're working in isn't prime? Well, we know a lot about Z sub n when n isn't prime. But the information we know about it is sort of negative information. It's not an integral domain. It's got zero divisors in there. So your first reaction is, well, we probably can't say much. And that's not a bad intuition to have. But it turns out we can actually say something about similar types of results, a sort of faux-little-fermase theorem, even in situations where the modulus, where the Z sub n value isn't a prime. So it turns out, even though it's not as nice, even though the ring Z sub n is not a field, heck, it's not even an integral domain. Not even an integral domain when n isn't prime is, I'll use the standard terminology, when n is composite, that means not prime, that means you can write n as the product of two integers, each bigger than one, but less than n. We still have some control units Zn. So here's the point, folks. If I hand you one of these Zn rings, if n is prime, then we've proved that every non-zero element is a unit. In other words, that Z sub n is a field when n is prime. And that's precisely the situations that we're able to prove for Mauss-Wald-Thierman or a little from Mauss-Thierman. The question is, what if we're working inside something like Z6 or Z12 or Z10 or Z20 or Z25, something where the modulus is not prime, do we still have the same sort of result? Is it the case that if you take an integer and raise it to the n minus 1 that you get 1 and the answer is definitely no. So note here in general, the direct generalization of Fermat-Wald-Thierman does not hold composite. Let's do an example here. Let's see, in Z6, if I do 2 to the 5th, let's see what I get. I don't want to do that one. Let's see, let's do this one in Z4, that's better. This would be an easier example. I'm going to take something not zero and I'm going to raise it to the 4 minus 1 power. The question is, do I get 1? Well, let's see, 2 raised to the 4 is 2 cubed, which is 8. But inside Z sub 4 was that zero. So it's certainly not equal to 1. So the question is, do we just give up and say, well, OK, when we have a non-prime modulus all is lost, and the answer turns out to be no, we can't make as powerful a statement as we did with vermosetal theorem. But there's still something to be salvaged here, and here's what it is. The proposition is the following, proposition. We're going to prove something about the Z sub n rings that we haven't seen yet. Let's see, in, this was two lectures ago, we looked at the two ideas, units and zero divisors. Units and zero divisors. And what I proved for you is that in some sense, those two things live on the opposite ends of the spectrum. If you have a unit, it's not a zero divisor. If you have a zero divisor, it's not a unit. And in, I'll say most rings, those two things live on opposite ends of the spectrum, and there's typically a lot of stuff in the middle. For instance, if you look at the ring of integers, the only units are 1 and minus 1. There's not very many units. There's not very many elements of multiplicative inverses. There's no zero divisors in there. There's nothing that is non-zero that you multiply or something that gets zero. All the other integers are sort of in between. What we're about to show, though, is that if you happen to look in the particular rings that look like Z sub n, in the Z sub n rings, you have to be in one or the other of the two camps. There's nothing in between. That either you're a unit or you're a zero divisor. So the proposition is, and I'm going to preface this just so that you don't get the wrong idea. Unlike most rings, this is unlike the integers, it's unlike the rationals, it's unlike polynomials, it's unlike most rings. It turns out if you're looking in a specific ring that looks like Z sub n, every element, every non-zero element is either a unit or a zero divisor. And I think I may have pointed this out when we looked at that proposition that said units and zero divisors somehow live on the opposite ends of the spectrum. But I don't think I've proved the details and this will be of interest here. Let's see, so what's the proof? Well, what you're able to do in the particular rings that look like Z sub n in these sorts of rings is, well, you can list out all the elements. They look like, well, I don't want you to look at the zero element, that's just for notational reasons. We don't call it a zero divisor. I know what all the non-zero elements look like. They're the elements 1, 2, 3 up 3, n minus 1. And it turns out there's two possibilities and each of those two possibilities leads to either being a unit or not a unit. So the proof is this, if, so pick any non-zero element in there, pick any, I'm tired of calling things a, how about any, let's call it b in Z sub n. Case one, the G, C, D of b and n is 1. And case two is the G, C, D of b and n is bigger than 1. Remember, G, C, D is the greatest common divisor. So if you hand me this integer, this integer b, what I'm going to try to do is convince you that, well, let's see, one of these two cases has to hold and what I'm going to convince you of is if you're in this case, that b is a unit. And if you're in this case, that b is a zero divisor and that's what will give us the proposition. Let's go ahead and prove case two first. No, in fact, I might do that in time here. No, I just want to give you enough info so that you can do one of the problems. So case one, this then b is a unit in Z sub n. Yeah, this will be a better way to approach this. Case two, then b is a zero divisor in Zn. And I'm going to leave the proofs of these out, use some number theory, which I think in the interest of time I'm going to just suffice to show you by means of a specific example how this works out. So for instance, if we look in R equals, how about Z12? Yeah, this would be a good example. All right, so 12 is composite, obviously. Then the claim is, if I go through all the non-zero elements, what this proposition says is if you pull an element out of there, tell me what its GCD with 12 is. If the GCD with 12 is 1, that you've written down a unit. And if the GCD with 12 is bigger than 1, then you've written down a zero divisor. So let's see how that works. So if I take 1, well, the GCD of 1 and 12 is 1. The GCD of 2 and 12 is 2. So what I'm going to do is separate these out into two camps. The GCD of 3 with 12 is 3. So that's bigger than 1. The GCD of 4 with 12 is 4, which is bigger than 1. The GCD of 5 with 12 is 1. The GCD of 6 with 12 is 6. The GCD of 7 with 12 is 1. The GCD of 8 with 12 is 4. The GCD of 9 with 12 is 3. The GCD of 10 with 12 is 2, and the GCD of 11 with 12 is 1. So what I've just done here, folks, is I've split the integers from 1 to 11 up into the two cases that I've mentioned here. The situation where the GCD of the integer with 12 is 1, and the case where the GCD of the integer with 12 is bigger than 1. And what I'm just going to do in this particular case, rather than going through the details of the proof, is show you that all the things on this list, GCD of blank with 12 equals 1, and the things on this list, GCD of blank with 12, bigger than 1, all the things on this list are units, and all the things on this list are zero divisors. And then we can use this result that we proved two times ago that says, hey, if you're on this list, you can't be on this list in vice versa. All right, I'm going to prove that all the things on this list are units by exhibiting their multiplicative inverses, by writing down another element in Z12 so that when you look at their product, you get 1, 112. How about 1 times 1? Boy, that was easy. That's 1, not 12. How about this one? If I take 5 and I multiply it by 5, I get 25, which in Z12 is 1, in Z12. If I look at 7, turns out if you multiply by 7, you get 49, which is 1 in Z12. So 7's a unit. In fact, it turns out here, let's see if I take 11, guess what? 11 times 11 is 1 in Z12 because 11 times 11 is 121. So in fact, what I've shown you is that, not only is it the case that each of the integers from 1 to 11 that are relatively prime to 12 is a unit inside Z12, but it actually happens that each element is its own inverse. All right, now here's a homework problem that you're doing to turn in this time. What you showed was if you look inside any ring, not just this one, and you look at the subset consisting of the units, well, presumably I've just done that, that the units form a group under multiplication. So using that homework problem, I've now got a group, because here presumably are the units of Z sub 12. It's a group under multiplication, it's got four elements. Each non-identity element has the property that when you combine it with itself, you get one. Anybody want to venture, guess what? The group of units of Z12 looks like, isomorphic to what group? Z2 cross Z2, perfect, or V. If you said V, then you said the correct answer, too. It's a group with four elements having the property that every non-identity element has order two, or rephrased to where every element in the group has square equal to the identity. So there's some group theory coming up in the context of rings. Now on the other hand, let me convince you that each of the elements on this other list, the integers whose GCD with 12 is bigger than one, are zero divisors in Z12. So what I'm gonna do over here is rather than trying to pair each of these numbers up with something whose product is one in Z12, over here I'm gonna try to pair each of these up with some non-zero element in Z12 so that the product of the two non-zero elements is zero in Z12, that's the definition of zero divisor. Two times six is zero in Z12. Three times four is zero in Z12. Four times three is zero in Z12. Six times two is zero in Z12. Eight times three is zero in Z12. Nine times four is zero in Z12. And 10 times, who knows, six is zero in Z12, all in Z12. Again, I have not proved the general case for you, but it turns out what I've done is shown you in detail, at least inside Z12, how it is that in the Z sub n ring, you're either a unit or a zero divisor. For those of you that are interested in how the general proof works, it's really not too bad. You need to know about a half hours worth of elementary number theory. This statement follows from what's called the linear combination theorem. If you have two relatively prime integers, in other words, GCD equaling one, then you can find integers U and V such that B times U plus n times V is one. That's linear combination theorem, and that allows you to show that B is necessarily a unit. And on the other hand, if the GCD is bigger than one, then it's not too bad to do. I mean, all you then have to do is sort of pair up the integer with what its sort of divisor is. So the GCD of two with 12 is two and 12 divided by two is six. And that's how you can sort of write down its corresponding pair that would give you a product equal to 12 where neither of the two terms in the product is zero. All right, well, with this in mind, we can then actually write down the generalization of Fermat's little theorem and here's how it goes. The definition is this. Inside Z12, here's the notation that I think was used in the problem. The units or so for homework, you show that, which problem was that? The units of the ring problem is a group. Is it 37? No. 37? In section 18, number 37, that the units of R for any ring under multiplication is a group. I think they just call it U, but here I'll be more emphatic. It's the units of the ring R. So we're gonna give this thing a name. When R is Zn, the number of elements in the units of Zn, that's sort of interesting. So what I'm asking you to do is look inside the Zn ring and tell me how many units are in there. Well, if n is prime, then Zn is a field. So everything except for zero is a unit. So if n is a prime, then this number is n minus one. But if n is not a prime, like if n is 12, well then the number happens to be four. So question is there's some algorithm that allows you to in general write down what this number is. Yes and no. I'll say at least for our purposes, the answer's no, but we're gonna give this number a name. We're gonna call it is defined to be to be the symbol, symbol phi, it's Greek letter phi of n. And this is called the Euler phi function. Example, if p is prime, then phi of p is, well, it's the number of units in Z sub p. Well, I know what that number is, it's p minus one. Let's adjust it, I'll show you why in a minute. On the other hand, we just showed if n equals 12, then the number of units in Z sub 12, well, it would just look, and it turns out to be four of them. That's it. So we're led to, yeah, good. Okay, one last reminder of group theory. It's nice, because we're working inside rings, but we keep harking back to results that we proved about groups. Remember this Lagrange consequence inside groups? If you hand me a group, and you did this one for homework, you tell me how many elements are in the group. It's a finite group. Let's call that number, and I wanna use n, even though that's the notation that was used in the problem. If you have Z elements in your group, so if your group is Z is S3, you've got Z equals six, something like that. If you have a group of six elements, and you take any element in the group and you raise it to the sixth power, you get the identity. Consequence of Lagrange is there. So we proved, consequence of Lagrange is there. If you tell me the order of the group, and you take any element in the group, and you combine that element with itself, that many times you always get the identity in the group. That's how we proved for Maslittle theorem. Because we looked inside when P was prime, we looked inside the group consisting of the non-zero elements of the field. That group had P minus one elements in it. So Lagrange's theorem says you take any element in that group, in other words any non-zero element of Cp, and you raise it to the P minus one power, and you get the identity element. What we're about to do is simply write down the same idea here. So corollary, consequence is this. If A is a unit in Zp, I'm sorry, in Zn, then if you take A and you raise it to the number of elements inside this group, well by definition that's how many elements are in this group, that you get one in Z sub n. Notice what just happened here. If I happen to be in a specific case where n is prime, then all I'm writing down here folks is if you take a unit in Z sub p, in other words if you take any non-zero element in Z sub p, and you raise it to the P minus one power, you get one in Zp. That's just for Maslittle theorem restated. But it turns out that this same result can be extended. It doesn't matter whether you're working in Z sub p where p is a prime or not, just look at the group of units inside Zn. How many elements are in there? Might not be easy to count. Here's the notation we use for that number. For instance, that number happens before when n is 12. But the punch line is take any element in this group, in the units of Zn, raise it to however many elements are in the group, p of n, you'll get one in the group. And this is usually given a name, this is called Euler's generalization of for Maslittle theorem. And both of them just boil down to using Lagrange's theorem on the appropriate group of units, whether the group of units happens to be all the non-zero elements of Z sub n or not. All right, so for example, let's see example. Oh, well let's just look at the situation that we've already seen. There are four units inside Z sub 12. So that says if you can't see them, any of the units inside Z sub 12, and you raise them to the fourth power that you'll necessarily get one. And while that, in fact, is the case, in fact it happens to be the case that we get there quicker on all these elements. There will be situations where you can't get there quicker. In other words, you'll need the full p of n in order to get some elements, but not all the time. All right, questions? All right, questions? So the quick example might be, let's see, find the remainder. The remainder, let's see, 13 to the hundredth is divided by, by what, 24? Manifestally not a prime number, but that's okay because the element that I'm starting with, 13 is relatively prime to 24. So since the GCD of 13 and 24 is 1, that means that 13 is a unit in Z sub 24. That's this proposition here. If you hand me something in Z sub n, if the GCD of the two numbers is 1, then you've got yourself a unit. Oh, so what does that mean? Euler's theorem, so 13 to the phi of 24 is 1 in Z24. That's what Euler's generalization says. So now all we have to do is figure out what phi of 24 is. It turns out, if you've seen the number theory course, there are some relatively subtle ways of doing it, but at least in the context of the homework problem that I'll give you tonight, just pound it out. That's phi of 24. 24 is what? Well, just list out 1, 2, 3, up through 24 and find the units. In other words, find all the things on this list that are relatively prime to, oops, I don't want to give you 25 out of 0, up to 23, up to M minus 1, find the units. I'll show you how to do that. Let's see where the units are precisely not the zero divisors or the units are precisely those things that have GCD with 24 equal to 1. Well, 1 certainly does. Let's just go through them quickly. 2, no good. 3, no good. 4, no good. 5, yeah, has GCD 5 and 24 is 1. 6, no good. 7, good. 8, no. 9, no. 10, no. 11, 12, no. 13, 14. I mean any even number's not going to work because it's the GCD of something with an even number. 15, no because the GCD of 15 and 24 is 3. 16, no. 17, yeah. 18, no, it's even 19, yeah. 20, no. 21, no. 22, no. 23, yeah. So 1, 2, 3, 4, 5, 6, 7, 8. So here they are. So phi of 24 then is 8. Simply count up the units. Bless you. So here's what that tells us. So necessarily 13 to the 8th is 1 in Z24. So what? Well, I'm interested in 13 to the 100th. Now we just play the same sort of game that we played before. So 13 to the 8th to the 12 equals 1 to the 12, which is 1 in Z24. But let's see. I know what 13 to the 8th to the 12th is. That's 13 to the 96th. And then I'll just let you play the same sort of game that we played before. So now you've got 13 to the 96th is 1. Now tell me what 13 to the 100th is. Just multiply both sides by 13 to the 4th. So I'll say etc. That's how you go about finishing up a problem of that type and you'll see one of those again for the homework that I'll give you tonight. All right. Questions there? Comments? So I mean it's nice. The group theory results that we proved actually give us some nice results about the ring structure of these things. OK. All right. Questions? Comments? Did that? Did that? Yeah. Do we do all these? Yep. What do I have here? 10 minutes? Perfect. OK. So I've got three choices. Let's do this. This sort of information overload on rings so far. We wrote down examples of rings. What we've been doing is stressing the multiplicative properties of the rings because the additive properties of rings are presumably nice. We, by definition, always have an abelian group when you look at the ring together with the additive structure. And what we've seen over the last three lectures is that the multiplicative structures of rings can differ wildly. Some are commutative, some are not commutative. Some have unity and some don't, but at least the ones that we're going to focus on are all going to be rings with unity. Some are finite, some are not. Some have zero divisors and some don't. Some have the property that every non-zero element has a multiplicative inverse, those are the fields. Some are integral domains. In other words, some have the property that there's no zero divisors or some have the property that if you take two non-zero elements, you multiply them, you always get something non-zero. There are a lot of those, and there are a lot of those that aren't just fields. Even though we prove fields have that property, there are a lot of examples of those types of rings. The most important ones are the integers. Certainly there's a lot of elements that don't have multiplicative inverses. But there's no zero divisors in there. And the other important example, which we're going to start in on Wednesday, will be polynomial rings. We've already looked those a little bit so far. What we're now going to do is play games that are similar to the games that we played with groups in the sense that we're going to be interested in homomorphisms from one ring to another, just like we were interested in homomorphisms from one group to another. We're going to be interested in forming factor rings in the same sort of manner that we formed factor groups. The constructions won't be identical, but the ideas will be similar enough that we won't have to reinvent the wheel on these things. So that when we start looking at, and this is what I'll introduce in the last ten minutes here, so that when we start looking at things like a homomorphism from one ring to another, when I write down the definition, it won't be as if it's sort of a fresh idea. It'll be, oh, okay. I know what a homomorphism was in the context of a group, so once you give me the definition of what a homomorphism is in the context of a ring, you say, alright, yeah, that makes sense because I sort of understood philosophically what the heck a homomorphism was in the context of a group. So what we're about to do is we're going to look at some similar constructions, similar constructions to those in groups. And let me just give you the definition here, definition. We start with two rings. Let R and S be rings. Communitive, I don't care. With unity, I really don't care, although all of the ones that we'll be looking at will have unity. With zero divisors, I don't care. Fields are not, I don't care. Then the definition is this. A function, I'll suggestively call it phi from R to S, is called a homomorphism. So we're already used that word. Well, yeah, but now I'm asking you to write down a function from one ring to another. I sometimes will emphasize that the two underlying structures are rings by telling you that I'm writing down what I'm going to call ring homomorphism in case, and my guess is that all of you could probably guess what the definition is. Here, because we've got two binary operations, we need to stipulate two different equations corresponding to the function phi. In case phi of R plus R prime is phi of R plus phi of R prime. And secondly, phi of R times R prime is phi of R times phi of R prime for all R and R prime and R. All R prime. If this was a month and a half ago, I would not have been so cavalier on the notation. Technically, this addition is happening inside the ring R. Technically, the addition over here is happening inside the ring S. They may be two different additions. For example, the addition here might be an addition inside the real numbers, and the addition over here might be an addition inside the ring of matrices or something like that. Similarly here, technically, this is multiplication inside R, this is multiplication inside S. But we sort of understand where all these operations are taking place, and there's no ambiguity because I've told you where the function goes from and goes to. So if you're looking at a symbol that looks like phi of R, you know that it's an element of S and phi of R prime is another element of S. So I mean it's just the notion of a group homomorphism sort of expanded to include both of the binary operations that we have in the setting of rings. So that's a ring homomorphism. Big example. I'll give you the homework assignment. That'll be the next one. So we'll get out of here. Example. Let N be an integer. Define the following function, phi from the ring of integers to the ring Z sub N, where you define phi of little z equals z mod N. A proof I won't write out in the details, so proof, just pound it out, pound it out. All we're using is used properties of modular arithmetic in Z sub N. All this says, folks, is the following. If you hand me two integers and you add them together, and then you ask what's the remainder when you divide by N, is that the same as taking the two remainders and adding them together and asking what's the remainder? Sure it is. Similarly with multiplication, if you have two integers and you multiply them together and ask what's the remainder is the same as having taken the remainders of each of the two integers individually, multiplying them and then asking what's the remainder and the answer turns out to be yes as well. In fact, we've used those properties of Z sub N already when we were looking at groups. So this will be one of the important examples of a ring homomorphism. The significantly more important example, which is what we're putting off tonight, will be in the context of polynomial rings, so we'll put that off for a little bit, but we will return to it because that will wind up being our focus for the remainder of the semester. Okay, questions, comments? Alright, here then is the next chunk of homework and I think we're close to being back on a more standard schedule. So homework assignment on Monday, this will be due next Wednesday, Wednesday, November 14th, and it'll just be a couple of problems from two sections. Let's see, in section 20, problems one through seven and ten, and the ones I want you to turn in are three, four and ten, and then in section 22, problems one through 15, and 21 through 28, and the ones I want you to turn in are five, 21 and 27. If you're interested, you can do all the problems in section 20 already, and by the time we're done with the material this Wednesday, you'll be able to do all the stuff in section 22 as well. This is on Fermat's little theorem and Euler's generalization, and this stuff...