 Hold this notation out on Monday, but I'll remind you this is the apples and oranges notation. I like this a lot I don't know why I didn't think about it 20 years ago Here's how it Here's how played out in this particular assignment I saw a couple of instances of things like this in one of the problems You're asked to find what the order of the factor group or something like that The order of the element the factor group. So some of you said well if I take the coset 0 2 plus h With that notation that you told me that that was in H This is a coset Yeah, so it makes sense to talk about this being in G Slash H because it's a coset It would make sense Although it wouldn't be stylistically good to say that this is contained in H As a coset but to say that it's in H is no good because now you're comparing Cosets to Elements of a subgroup and that makes no sense. Okay, so the apples and oranges. That's not good similarly Some of you said something like the element 0 2 Is H? That's no good either. This is a set and that's a single element So this is another apples and oranges issue the way you want to phrase that is If you're going to convince me what the order of an element is if you wind up with an element that happens to be in the subgroup therefore 0 2 bar Is the identity? 00 bar If you have an element in the subgroup then necessarily that element Generates the same coset as any element in the subgroup in particular generates the same coset as the identity element in Particularly, it's the identity element of the coset group All right, so that's the apples and oranges comment for this homework assignment In that modified second part of the question that I had you do from section 3 showing that you get an isomorphism from the collection of non-zero complex numbers to the collection of non-zero elements inside H this thing that we called H star Most all of you Well had written down the correct function. So that was good but you wrote down the correct function in Part I forget the problem over 33 years 30 years in part a you wrote down well Here's what the function should do it should take a complex number and there's a generic form of a complex number where A and B are Real numbers and should spit out this matrix Okay, good So there was your function and then what you said in part B is well, let's just look at the same function in part B Okay, good idea the issue though in part B is you then technically have to make sure That if you start with something not zero here because that's the new input set that you actually get something not zero Here because that's the new output set so you have to make sure that in the original sort of Connection between input complex number zero and output matrix zero That when you throw out the zero and one of them and you throw out the zero and the other that you've actually thrown out The two things that were compared to begin with okay So technically and some of you didn't do this and I didn't nick you for it But it's a good thing to observe note that If a plus bi is actually in C star Then this thing a minus b b a is actually in h star The point being if this is in C star it means neither a it means both of a or b aren't zero in other words You can't have both a and b equaling zero That's how you get the zero element in the complex numbers Which means if either one or both of these are not zero then this matrix is also not zero and Conversely it turns out when you show that this thing is on to you have to convince me that if you take a non-zero matrix It actually comes from a non-zero complex number, and that's pretty easy to do so. It's not that the task was hard It's just technically you have to note that that's what's really going on here And so what do you eventually do you eventually show that this group is isomorphic to that group? Because you write down a homomorphism That's one-to-one and on to and remember the philosophy of what it means for two groups to be isomorphic It means in a sense they walk and talk the same way they behave the same So that means whatever sort of properties That you can write down about the complex numbers under multiplication I'm sorry the non-zero complex numbers under multiplication also holds true for this collection of matrices Which means in effect if all you're interested in is multiplication of non-zero complex numbers You can actually do it without worrying about the symbol I you can actually do it just by looking at certain matrices So it turns out you can describe multiplication in the complex numbers without cooking up a special symbol called I that has this magic property that when you Square it you get negative one or rephrase that when you take it to its fourth power that you get one But no lower power gets you to one I mean that's a property of this thing and you're a little bit uncomfortable with that the first time you see it because You know I can't I can't touch it. I can't hold it. Where is it? The point is well if you don't like I then just think of things as matrices because that's what this isomorphism lets you do So if you want a thing whose square is negative one or whose fourth power is one, but no lower power is one I'll give you one instead of thinking of it as I well. What does the specific? Non-zero complex number I correspond to under the size of morphism it corresponds to in particular in particular If you do phi of I well put in the right form, it's phi of Phi of zero plus one I Right, that's what I is and what gets kicked out if you plug in a equals zero and b equals one you get zero minus one One zero So the point is whatever properties this has vis-a-vis multiplication This one has the same properties Because the non-zero complex numbers under multiplication is isomorphic to this collection of matrices under matrix multiplication Again, there's no eyes here folks zeros and ones and minus ones Okay, it's matrix multiplication now But if you're more comfortable with that and then this number I then at least you're good chicken look if I take this thing Let's call this I don't know Z or something like that then what the Z squared Well intuitively what you're figuring is it's whatever I squared is Well, I squared is minus one so what you're expecting is Z squared to be minus one But minus one meaning sort of minus one in the context of matrices. Let's see what we get multiplied out which is minus one Where one means multiplicative identity at least in the system that you're working in there it is So what's Z to the fourth? There was it's e or I or one or whatever you want to call it It's the thing that behaves as the identity element in whatever system you're working in Which in the complex numbers we choose to call this but well, it's really one plus zero I In the complex numbers. Hey, if you want to treat it instead as matrices just view them as two by two matrices What this notion of isomorphism allows you to do sometimes is just sort of change your point of view You don't like cooking up a new symbol to work with that's fine Just review things is happening inside a different sort of system that maybe you understand a little bit better that you're a little bit more comfortable with Okay question yeah Yeah, good question you can talk about isomorphisms between Sets having binary operations you can do that But I'm not so interested in that I'm interested in having Binary operations that form groups and then isomorphisms of groups. That's why yeah, I mean as far as the order goes folks I would have loved to have seen chapter 3 as chapter 16 But he But he chooses to talk about the notion of isomorphism At a level sort of one step before we even get to groups He talks about isomorphisms between sets and it's perfectly mathematically legit to do It's just at the time that we're dealing with just sets with binary operations You don't have a good enough feel I don't think for what the structure of those things are and if we're going to talk about things being isomorphic having the same Structure to me it makes more sense to talk about things that actually have a richer structure to begin with Groups and then talk about isomorphism in that country. So so you're right I mean Richard's question is would you get an isomorphism if you remove the stars here and the answer is yes It's just the underlying sets that aren't groups anymore. They're just I Don't know it our binary operations on set. Yeah, that's a great question So as I said if I if I were to write I would not ever write an alphabet book, but It's just I have some colleagues that have written textbooks and it's just it's an incredible energy drain And you know in the end if you feel like you've got some big contribution to make then maybe if there's some merit to doing it but I think the only contribution I'd be able to make is If I could take if you know, Professor Fraley would allow me to take his book and just move things around a little bit Then I'd be happy doing that Not sitting down crying one from scratch wanting Was Okay, so Lonnie's question is sort of where did the I go and how did it sort of get how did it manifest itself over here? Matrix Well, I mean that the issue of the group table is I mean there's infinitely many things in here that you got to worry about so So trying to pound out a group table isn't gonna Get you there here, but but I think the better question is so all right, you know I have this thing called I and I'm trying to make it look like a matrix And that's what's going on you're taking a complex number and you're associating with a matrix And how do you do that in such a way that somehow? Preserves the structure of multiplication in the complex numbers Well in part a it was addition and then part. Oh, I'm sorry part a oh, I'm sorry I'm sorry. I totally missed the question then so in part a yeah when you're asked to set up this Correspondence under addition, so I'm sorry ask the question again then one so Yeah Uh-huh Let me see if I can answer it this way as far as the addition goes when you're adding complex numbers You don't care what the heck that symbol is it could be an axe Because if you're adding complex numbers you never use the multiplicative property of this thing Which is really what makes it different from the real numbers All right So this can be any symbol you want it to be and what we're saying is if you have this sort of extra symbol You can sort of lay it into the matrices any way you want all you need is something that distinguishes the thing that sits out here by Itself from the thing that's acting as the coefficient on this other term So if you wanted to call this a plus b times x or a plus b times y I don't really care what that would be all I'm really asking you to do is pick off a and b Now if the question is well, couldn't you just correspond those to a pair a comma b? Yeah, that would have been a good idea, too But it turns out because we're eventually interested in also asking about the multiplication structure That we write down the function on the additive level that looks like this sort of very odd cross piece so If I understand your question correctly if this was only to be a question with part 33 and no 33 b Then we just say well here's a function from the complexes to maybe r cross r to two copies of the reals send Fee of a plus bi just a comma b. Is that a homomorphism? Yeah, it turns out to be I said in fact, you know That's a great question. So at the at the additive stage when you're doing problem 33 a What we've shown is that the group of complex numbers under addition is isomorphic to this funny-looking group But it's also isomorphic to just send fee of a plus bi to a comma b Which is just r cross r in other words r2 and you have You have lots of ways of viewing r2 typically viewed as the plane So that's the way you usually think of the complex numbers is you draw it like this, right? And there's r cross r It's just here. There's another way to look at R cross r with twist to it. All right Let's see. Am I done with the comments the horn comes I think so Yeah, all right All right. Oh if you didn't get the email I sent out an email yesterday There will be an SI session on Monday before the exam General be able to be in Monday from 9 to 10 30 in the usual room engineering 1 77 She'll trade Wednesday for Monday, so there won't be an SI session on Wednesday But there will be one on Monday and then I'll be in before class if you've got sort of last-minute questions before the exam All right So here is what we're up to we are looking at these things called rings and where two lectures into rings already in a lot of senses were going back to square one and we're thinking of these as new Systems, but in other senses the things that we're looking at we've been some To some degree already studied for example even in the definition of a ring Ring is well. I could list it out as something that has like nine properties to it But a lot of those properties I can smash into one piece It's a set with two binary operations addition and multiplication where when you view this set with the binary operation called plus You get an abelian group, so it's huge and stuff that we've done, you know up until now In this course, so I don't have to worry about writing out e what each and every you know detail of a viewing group means and Then our with Dodd is simply an associate of binary operation and the addition And multiplication are related by some sort of distributive law And we have up until now written down a whole slew of different rings Let me just remind you of some of those the ring of integers the ring of Racial numbers the ring of real numbers the ring of complex numbers the ring of n by n matrices And what I'm putting inside the brackets here folks isn't just the real numbers But take any ring you want you can talk about the collection of n by n matrices where the entries come from that ring And then it makes sense to do addition and matrix multiplication as usual and you wind up with another ring We can talk about start with any ring and look at the collection of polynomials with coefficients in that ring We can talk about these mod an arithmetic rings And there are a slew of others that we Might mention between now and the end of the semester, but these will be sort of the major players These will be the rings that we're going to look at in most detail Looking ahead a little bit What we'll wind up doing with rings is quite similar to what we did with groups We developed a whole set of examples of groups And then we built new groups From the ones that we had already Discovered or constructed by various operations one was To look inside the given group to find subgroups well similarly we'll do that here We've already seen situations where we've done that for instance We've looked inside the ring of real numbers to write down this ring that we called script s Things that look like rational plus a rational times the square root of five or something. There's a sub ring We'll talk later on about Direct products of rings in fact later on will be in about 10 minutes We'll eventually talk about Factor rings just like we talked about factor groups where you take an appropriate Sub ring that has some sort of additional property That allows you to somehow get a well-defined operation on the cosets And you wind up getting a factor ring So a lot of the things that we'll study here won't be identical to what we've done in the past But we'll have enough similarity that we won't have to reinvent the wheel when we look at those ideas again Okay, and that'll be good Here is what we did On monday. Let me just remind you some of the words This is a slightly unfortunate portion of the course because I have to throw a lot of words at you So let me at least keep repeating them so that they start to sink in a little bit here were the words skew field Which was the same as division ring They're used Sort of equally in the literature. These are the same thing. This is a ring where If you look at the non zero elements Of the ring under multiplication is a group. That's what it means for the ring to be a skew field For the ring to be a field without the word skew in front of it Means that when you look at the ring Throw out the zero element and ask what sort of behavior is there under multiplication Is an a billion group That's what field means skew field and then this last term that We talked about at the end of monday is what's called integral domain An integral domain means the ring is with unity In other words has a multiplicative identity is commutative And when you look at the non zero elements And Look at that as a set It turns out that the Binary operation really is one in other words a multiplication is a binary operation on the set Is closed Ie get a binary operation another way to phrase this is An integral domain is a ring that has a unity element has a one Yeah, all these rings that we're going to look at this semester have a one So this will typically be a non issue our commutative Most of the rings that will look at this semester will be commutative But certainly not all if I look at matrices Matrix multiplication is not commutative, but For the most part the rings that will wind up focusing on have this property And what this says is if you're in an integral domain that if you throw out the zero element And then you multiply any of those two non zero elements that you again get a non zero element That's what integral domain means We can point to some integral domains. That's an integral domain Sometimes that's an integral domain in fact we proved a result last time that says exactly when it is This is an integral domain that is two and that is two. So let's see. This is integral domain domain Yeah, if you take any two non zero integers and you multiply them together you get something non zero That's no big deal. This is also and this is also and this is also Well, we proved last time that any field is an integral domain So just a reminder Integral domain That's easy to see if the non zero elements under multiplication form an abelian group Well, if it's an abelian group it has an identity element If it's abelian the multiplication is commutative. That's just another word that we use And if it forms in a billion group then it certainly has to be a binary operation has to be closed So that's sort of a no-brainer for free. We get that fields imply integral domains But the good thing to keep in mind is that there are many integral domains Which are not fields Like the integers Another good example of an integral domain that's not a field is this if I start with Now I have to be a little bit careful here. I'm going to ask you to look at the collection of polynomials But now I'm going to specify What coefficients I want you to use I'm going to ask you to use real number coefficients So these are the usual polynomials that you confront in a calculus one course, right? So polynomials where the coefficients are real numbers. This thing is an integral domain And we gave a sort of informal proof On monday, we're going to revisit these polynomial rings later on in more detail But for now the proof that r bracket x is an integral domain We're again r standing for the real numbers here was if you take two non-zero things Two non-zero polynomials and you multiply them together. Can you get zero? Well, no And the reason essentially is because tell me what the degree of the first polynomial is You know the highest power x term and tell me what the highest power x term of the other polynomial is Then when you do the product of the polynomials you get a term that looks like x to the You know m plus n where the degrees are m and n respectively So it turns out you can't get zero because you always get the degree going up There's a little bit more to it. But at least for now, this will be a good example of an integral domain That's not a field And z is an integral domain and neither of these are fields These will be good examples to keep in mind Okay, but not fields There's a couple of goals for this evening a couple of goals two larger goals bigger goals The rest of the goals for tonight are just to introduce some additional notation and constructions that we'll need later on The first is to show that every finite integral domain Is a field That'll be interesting. So what that'll do is in particular What we'll show is the result will be proposition z sub n is a field if and only if n is prime So that's eventually where we're going to get tonight. And then the second goal I want to phrase it here. Oh, yes to look at zero divisors in here Yeah Is to look at Look more carefully r bracket x where r here now doesn't have the extra line where I'm allowing A general or any ring that you'd like to appear as the collection of coefficients Okay, and along the way we'll introduce some additional terminology Which I'm trying to parse out in little chunks so that I don't just snow you with a bunch of new words at the same time Okay Let's go ahead and do this first result What's interesting folks is what I've written down as the two major examples of integral domains that are not fields It's important for you to note that well, there's obviously infinitely many integers There's certainly infinitely many polynomials. In fact, that's true regardless of what ring you start with I can always generate infinitely many different polynomials. Let me list out a few for you x. That's a polynomial So it's x squared and x cubed is a polynomial next to the fourth. So here's infinitely So this is always And it turns out that in order to be able to write down examples of integral domains that are not fields I have to be writing down Infinite rings because what we're about to show is if you hand me a ring that only has finitely many elements Here's a good example to keep in mind That if that ring happens to be an integral domain Then it necessarily is going to be a field and that's what we're about to prove here. So it turns out proposition Suppose capital r is an integral domain with the added property and suppose Suppose that the number of elements in r is finite And I'll put in parentheses example think Zn is a good example example and is prime because remember on Monday what we proved is If n is prime then z sub n is an integral domain and we proved also at the end of money that if Zn is an integral domain then and is prime in other words that these two statements are if and only if Then the conclusion is then necessarily r is a field now the intuition I want you to somehow start developing as to What the difference is between an integral domain and a field is this Think in an integral domain You have lots of pieces you at least can get off the ground in asking whether or not The non-zero elements of the ring under multiplication form a group It makes sense to ask that question if you're an integral domain because at least you've got a binary operation What else do you have well you're assuming in an integral domain that the multiplication has an identity element That's what it means to be with unity. So you've got a multiplication By definition of a ring the multiplication is associative You've got an identity element. You've got a billionness So in effect folks you get everything you need to show that the non-zero elements form an abelian group except what? inverses And the point is here there are some elements that don't have multiplicative inverses And the point is that here some elements don't have multiplicative inverses So the goal really in Any situation where you're trying to convince me that an integral domain is a field Really the only thing you have to prove is that each non-zero element has a multiplicative inverse So the proof is We Because r is an integral domain by hypothesis is an integral domain Don't mean in order to show it's a field To show r is a field All we need to do Is to show That for each non-zero element of r for each let's call it little a In the non-zero elements of r I mean that's what the notation is r star There exists an inverse and let's write out what that means there exists a b in r star With the property that When you multiply the two together So what the heck is that thing well that thing is The supposed unity that comes with the definition of an integral domain you want to call it e that's all right But usually we call it one in the context of rings because that's what form it usually takes on As we mentioned on Monday, it might look like the identity matrix. That's all right Or it might look like the identity polynomial just one, but That's a reasonable notation to use for this thing Let's see. Wait a minute. Uh in order to show that Inverses exist remember in the definition of a group Inverse means that when you multiply the two or combine the two in the one order you get the identity But then you're also required to show that if you Combine them in the other order that you get the identity as well But that comes for free too because we're assuming that the ring we're starting with is commutative In other words, we're already assuming that a b equals b a Regardless of which elements that you start with so if you're going to get a b is one Then i'll put in parentheses. This implies b a equals one automatically by commutativity So we don't have to worry about that second piece of what it means to be an inverse element Okay Somewhere along the line we need to use the fact that the ring is finite and we do that right off the bat Here's how so let's list the elements of our Well in fact, here's what I'd prefer to do throw zero out And just show me what our star looks like. Well, there's some elements in there if the original ring is finite If it's got m elements in it, then I'll tell you how many elements Our star has m minus one you just throw one of the elements out Okay, the point is though that our star then is also a finite set. You're just throwing an element away from r So list them out. How about? Hmm a one a two up through Let's close maybe ace of m There it is Where m is the number of elements in the ring minus one, but I don't care what m is just that there's a finite number of things in here If you want a specific example to drag along to keep in mind as we're doing this think Maybe r is I don't know z seven or something like that because remember we showed on monday that if you look at z sub a prime number That you necessarily get an integral domain in which case what i'm doing is i'm Riding out the elements of r star And i'm not claiming that every finite integral domain looks like a z sub n But at least we have some examples of finite integral domains and let's see if we can figure out what's going on All right, so now here's what I need you to do Pick any non-zero element A Okay, it's one of the things on this list, but it turns out. I don't really care It's one of the a sub i's, but that's turn going to turn out to not be an issue What's the task? We've picked something in there. We need to somehow cook up or show that there exists an inverse for it So here's what I want you to do form this list This list What I want you to do is take this element a and simply multiply it in order times everything in r star a a one a a two up through a a m That was easy to do Now here's two important properties of this list. First of all I'm in an integral domain That means by definition if I take two non-zero elements and I multiply them I get something not zero Well look each of these things is in r star the thing that I've picked is in r star So each of these products is a non-zero element times a non-zero element So by definition of an integral domain necessarily each of these things is non-zero So the point is that this set Lives inside not just the ring that's a no-brainer, but lives inside the non-zero elements of the ring since r is an integral domain We're using the fact that the product of any two non-zero elements is necessarily non-zero to claim that everything Here winds up back in the Collection of non-zero elements. That's the first observation. Here's the second observation We're going to use a result that we proved on Monday that says this but Let's see on Monday. We proved that if you're in an integral domain if a a i equals a a j If we have three elements in an integral domain and they're all non-zero That's exactly what we've got here a is non-zero and each of the a sub i's on the list is non-zero Then You can cancel the a So just recall proposition in an integral domain as long as you've taken three non-zero As long as you've taken three non-zero elements a a i and a j I think I called them a b and c on Monday that the notation is a non-issue If a times one is the same a times another then the two things are equal to begin with So here's what that means folks. It means if you're looking through the elements of this list There's no repeats. The reason there's no repeats is because you've originally listed out the set r star in such a way That there's no repeats multiplying by this particular element a means there's no repeats So conclude that there's no repeats on the list There are no repeated elements elements In the list Again, if there were repeated elements repeated elements would mean that you've got one of these things equal to another of them But that can't happen because if the two things were equal then they actually started off as the same element in r star to begin with So here now is the setup. I've got a subset of r star How many different elements are on this list? Answered the same number as there were different elements on the original list So it means that the number of elements in this set Is m and it was so what so you've written down a subset of a set But now I've convinced you that the subset that I'm interested in Actually has exactly the same number of elements as the original set And folks if you have a finite set and you have a subset something inside it But the subset contains the same number of elements as the whole set then the subset is the whole set So the conclusion is that the subset Is r star It's usually called the pigeonhole principle or something if I hand you a set with 10 elements And I tell you I have a subset with 10 elements. There's only one choice. The subset is the set. That's all we're using here quick remark It's exactly at this point that the proof would fail if we didn't assume the set was fine Because at this point all we'd be able to conclude is something like now I have an infinite subset of an infinite set Can I conclude that the subset equals the set in a way? I mean think if I had Listed inside the non-zero integers and simply asked you to pick the number four And asked you to multiply each of the integers by four I'd get a perfectly good subset of the non-zero integers I'd get the non-zero multiples of four which certainly doesn't equal all the possible integers But when we've got a finite set a finite subset No, let me rephrase that a subset of a finite set That contains the same number of elements as the set is the set and that's what we're about to use here. So what? So in particular What in particular? Well, let's see The identity element is in our star Yeah, because by hypothesis i'm in an integral domain. So there is a unity element So one of the elements on this list is the identity I don't really care where it is. Heck, you could have listed it out first if you want it So let's see One of the a sub i's let's call it of the a sub and of k's is one Just without loss of generality. Let's say a sub one is one Again, I kind of rigged the list that way to begin with but it doesn't really matter where it is So what does that mean? So Hmm One is in here One is in here But if it's in here and this set equals this set that means one is in here Okay So what are the things in that subset look like so one is in our star Which is a a one a a two up through a a m So the conclusion is so one equals a a t for some T In other words, if one is in this set, all right, so where is it? I don't care. Maybe it's there. Maybe it's there Wherever it is. It means that one has to look like that. So we're done. So done A sub t equals b works I've convinced you that there is some element inside the ring with the property that when you multiply it by you get one Okay, we have to do it in the other order too. No because we're assuming that the ring is commutative Have I actually produced The element that works for you. No So this is what we usually call an existence proof I've convinced you by properties of finite sets that there necessarily has to be an element that works But I haven't given you any sort of algorithm or formula as to how to produce it For example, what would happen over here if I wanted to find something like the inverse of four in z seven Then what I would ask you to do if a is four Then I'd ask you to multiply each of these elements through By the number four and let's see what we get we get four times one four times two Four times three that's 12, but in z seven that's five four times four is 16 Which in z seven is two four times five is 20 which in z seven is six Four times six is 24 which in z seven is three So in the particular case where I happen to start with z seven if I start with a equals four here is the list Does one appear in there? Yeah, there it is Do I happen to know where it appears? No, if I start with a different one, maybe I start with a equals I don't know three or something like that Here's what the list looks like if you start with a equals three three times one Three times two three times three is nine, but in z seven, that's two three times four is 12 Which is five three times five is 15 Which in z seven is one three times six is 18 which is so there's the list Where does the one appear I didn't know could appear But the point is folks when you list things out if you're starting inside an integral domain if you list things out you're necessarily going to see distinct or different entries in this list when you do the products And since you're looking at different entries all of which are non-zero necessarily this list is the original list So somewhere along the line you had to find the number one It's not too bad. So if somebody says give me an example of an integral domain that's not a field These are the two good places to look But it's key that you understand that you can't look at a finite ring You can't look at for example any of these z n rings to try to find such an example Because we've just shown that any finite integral domain is a field. So here is a corollary to a result that we proved on monday therefore Uh these statements are equivalent first n is prime Secondly z sub n is an integral domain third Z sub n Is a field last year I No, it was two years ago What I've seen happen to students is this You look at this List of equivalent statements about z sub n rings and you try to Generalize it to all rings. You're now looking at a result that says the ring is an integral domain precisely when it's a field And what students somehow do is they take that as some sort of ticket to a place that you don't want to be And the ticket is oh then any your integral domain is a field Uh-huh It's only these kinds of integral domains that happen to be fields. There's many that aren't Yeah, it's exhaustive of whatever whatever integral domain. I've asked you to start with That's by definition of the number of elements in the set assuming to be assumed to be finite It means that if there's a finite number of things in the ring Maybe there's six things in the ring or seven things in the ring like in this particular situation. There's seven things in the ring Then what I ask you to just throw out zero That means that there's going to be six things left over now just list those six things out So yeah, there's there maybe I should have played up this equal sign more here by definition Can be assumed to be all of the non-zero elements of r list amount So the equal sign here is sort of emphatic here. They are and here's all of them And the point is I can exhaust them all because I've assumed that the set is finite okay Well, there's a number of you in this class that were in the number theory course last spring or in dr Son's number theory course two springs ago what I want to do is spend I don't know five minutes on a little detour And show you that What turns out to be a nice and maybe somewhat surprising result from the number theory course pops out almost immediately as A corollary to the stuff that we just proved and Lagrange's theorem So here is an aside Turns out here's a result from The number theory course math 311 This is usually referred to as fermat's little theorem Little theorem It's a pretty nice result says the following let p be a prime. Let p be prime Let a be any integer Uh And suppose p doesn't divide a in other words take any integer That's not a multiple of the prime number that you started with so if you start with the number five take anything It's not a multiple of five Then here's the punch line then If you take that integer then a Raise to the p minus one power Some of you remember this result is congruent to what? One mod p For those of you that aren't familiar with this number theory notation This means if you take this thing you raise to the p minus one power And then you divide it by p using long division that the remainder will always be one So for example if you take 783 Which is not divisible by five and you raise it to the five minus one you raise it to the fourth power The remainder is going to be one mod five Fermat spent a lot of time proving this thing In fact, he gave a number of different proofs And it turns out we'll be able to prove it Extremely quickly proof Well, what we're saying is the result rephrased is result rephrased Is that That if you take this element And this element That those two things are equal in the ring zp Because that's what it means to say two things are congruent mod p It means that if you consider them as elements in the set Zero through p minus one where the arithmetic is done mod p that you've got the same two things All right, so I need to somehow convince you that if I take a the p minus one That I get one mod p but wait a minute, but here's what we've just shown p is prime So what does that mean? It means that z sub p is a field A field previous result. That's what we just proved Which means by definition if you look at the non zero elements Under multiplication that this is an abelian group in particular. It's a group I won't actually need that. It's a billion but turns out we get that for free That's what it means for a ring to be a field it means if you throw out the zero element that you get a group Well, look folks. I've asked you to take an integer That p doesn't divide in other words that isn't a multiple of p rephrased in terms of z sub p language This simply means that I've asked you to take an integer that's not zero in z sub p. That's not a multiple of p So the hypothesis is p doesn't divide a simply means That a when you view it mod p is non zero Think mod five if I hand you the number 10. Well, that's zero mod five or if I hand you the number 100 That's zero mod five and I'm asking you not to do that take something else Take 98. That's not zero mod five Me not zero means it's in the star. That's just the notation Okay, now we did all work assignment And that homework problem was Use Lagrange's theorem to show that if you're in a finite group And you tell me how many elements are in the group The notation that was using the problem was and there's n elements in the group and you take any element of the group And you raise it to the nth power that you get the identity of the group Sorry, I can't pull out the exact place where that was I think was section nine or section eight or something like that Show that in a finite group g if the number of elements in the group is n And you take any element in the group and you raise it to the nth power that you get the identity So now use an old homework problem old homework problem And I'll put in parentheses, which was done based on Lagrange's theorem On Lagrange's theorem And what you get to conclude is that if you have an element in a group I do I have a which is in this group And you raise it to the number of elements in the group. Well folks if I take z p And I've thrown out zero Then I've got p minus one elements left That I get the identity element in this group. That's exactly what fermository theorem says I like this so there is fermository theorem done from the point of view of Taking z p realizing that when you throw zero out Not only do you get A binary operation under multiplication you actually get a group And then hauling out Lagrange's theorem Lagrange's theorem simply says if you take well one Manifestation of Lagrange's theorem is that if you're in a finite group and you raise any element in the group to the Order of the group power that you get the identity element in the group And this is just recasting that more general group result in the specific context of the z p rings Questions comments There is What's called Euler's generalization of fermository theorem And it turns out we'll be able to get that result as well, but we've got to work a little bit harder first So we'll just sort of be You know satisfied that we're able to re prove fermository theorem Using some groups here. Okay uh Yeah Questions Lonnie Careful under multiplication and so the identity is yeah exactly Exactly So yeah, I gotta throw some words at you here and here they are So some words Words there are these First is a unit in A ring with unity. This is why I hate this notation Unity folks is what every other algebraic on the planet calls an identity element Or a one in a ring So a ring with one And we all know what one means A ring that has an element that behaves like the multiplicative identity is an element Call it little a for which There exists There exists An element b having ab Equals one and b a So a unit Well, the word unit is actually relatively standard notation Other people call these invertible elements an invertible element in the ring is An element that has a multiplicative inverse. It's a an element that you can somehow Find an inverse to where inverse means Under the multiplication And you're thinking well, how would I know that when you talk about the inverse of an element That you're talking about multiplication the answer is folks because remember the addition Operation inside a ring is always viewed as being as nice as you'd want it to be There's always inverse elements under the addition So asking whether or not there's an inverse element for an element ring always is a question about the multiplicative structure A quick side note If you happen to be in a commutative ring with unity a commutative ring of identity If you have this Property then obviously that one follows for free There are situations where you have that property And you actually don't have this property In those situations we might call b something like a right inverse for a Or if you have this property But not that one you might call b a left inverse for a So sometimes we talk about two-sided inverses or simply one-sided inverses if that happens or if that happens But our focus will be typically on commutative rings And so inside a commutative ring being a two-sided unit In other words, both of these equations holding is the same as being a one-sided unit You simply have to find an element that works here. That's a unit And the second word I put these two in the same list But in some sense they represent opposite ends of the spectrum as far as multiplication goes The second phrase is I mean it'll sound sort of odd because you've been trained over the many years to Think that you can't do this A zero divisor In a ring r Is an element Is a non-zero element non-zero element With the property that Element, let's call it a element A in the ring for which There exists There's pins running out for which there exists A non-zero element b in the ring having A times b equal to zero So a zero divisor means you've got an element in a ring That's not zero and you can pair it up with something else. That's not zero But when you multiply them together you get zero Think I don't know anything like that. Well, you do know some things like that. You've seen some things like that There aren't any in the integers Or in the rations or the reels or the complexes, but there's lots of things like that and we looked at some on Monday Inside the two by two matrices. It's easy to find things neither of which are zero individually that gives zero It's also easy to find examples of zero divisors In z sub n rings where n isn't prime for instance in z sub six two isn't zero and three isn't zero But two times three is six which is zero in z sub six So there's lots of different situations where we have these It turns out you can Define what it means for a ring to be an integral domain integral domain means It's a ring with unity and is commutative And instead of saying that the collection of non-zero elements is closed Under multiplication all you need to do is say that there are no zero divisors In other words that it's impossible to find two non-zero elements that multiply to zero So it's simply another formulation or another way to look at what the notion of an integral domain is and the proposition is That somehow these two ideas live at the end of the multiplicative spectrum in general Although in many situations there's lots of stuff in between So it turns out if you're a zero divisor Then you can't be a unit So the zero divisors is sort of here plus the units are here In some situations That's everything But in some situations there's lots of stuff that neither falls into this camp nor into this camp But it's always the case that if you're here, you can't be here and if you're here, you can't be here Other stuff maybe Sometimes here's an example if I give you a field for instance if I give you The Let's start with the rational numbers if I give you the rational numbers. That's a field And I ask you to find the units Well, let's see the units are those things that have multiplicative inverses which is everything except for zero So in that particular case the units are everything and there's actually no zero divisors It's impossible inside the Rationals to have two non-zero things that multiply to zero So in that particular case the units take up everything in the there there's no zero divisors zero itself has always viewed as exceptional In other situations there's lots of zero divisors like in z sub six There are many zero divisors like two is a zero divisor because two times three is zero And three is a zero divisor because three times two is zero and four is a zero divisor because four times three is zero in z sub six So you know ring like z six and the zero divisors actually play a big role. How about inside the integers? Well, it's the inside the integers. There's no zero divisors There's no non-zero elements that you can multiply together to give zero. So there's none of those the units There's not very many units inside the integers either units or those things that have multiplicative inverses Well, one has a multiplicative inverse itself Minus one has a multiplicative inverse itself and that's it nothing else has a multiplicative inverse So in the integers these two sets that one's empty and this one's really small So the way things break out element wise at least these are at the two ends of the spectrum And sometimes there's stuff in between let's prove that these can't overlap So the proposition is The collection of zero divisors zero divisors Intersect the set of units is always empty in any ring That means empty set in any ring and the proof is Let's see So I first have to show you that if I take a unit It's not a zero divisor and then if I take a zero divisor It's not a unit So first let let's call it you let you be a unit in the set of units So what does that mean? So there exists there is Something called b in r so that When you do the product u times b is one and that's the same as b times u. That's the hypothesis I want to show it's not a zero divisor show that u is not A zero divisor We'll do it by contradiction. Suppose u is a zero divisor suppose It is u is A zero divisor So what does that mean? So it means that if I So there exists there is Let's call it z in r Where with u z Equals zero And z not equal to zero. That's what it means to be a zero divisor Definition of a zero divisor is you're not zero And you can find something else. It's not zero so that when you multiply the two things together you get zero And what we're about to contradict is Z not equaling zero. Well, let's see if I have u times z is zero What i'm going to do is choose to multiply both sides of that equation on the left by b So b times u times z is b times zero If I have two things that are equal in a ring then I can multiply both sides as long as I keep the sides straight So i'm going to choose to multiply on the left by b Oh, but wait a minute We proved on the first day that we started discussing rings that in any ring if you multiply zero you get zero So here's what we get B times u. Oh, but wait a minute the multiplication in a ring is associative. I'm not commuting anything here folks. I'm just regrouping So b times u times z is zero But wait a minute b times u is one But one times anything is that thing So I've started with something not zero and I've led to the contradiction that it is zero. So there's the contradiction So the conclusion is that u is not a zero divisor Because by assuming it was a zero divisor I was led to the contradiction that z is both zero and not zero So what we've shown is that if you're in here you can't be in here And the converse will be just as quick and then we'll call it a night here conversely converse is suppose I take something that's a zero divisor what's called z Is a zero divisor Show that z Is not a unit So what does it mean by hypothesis to say that z is a zero divisor? So by hypothesis z is not zero And there exists There is Well, let's call it w not equal to zero so that z w is zero That's what it means to be a zero divisor That you're not zero and you can combine with something else not zero to give zero And what I want to do is show that z is not a unit And we'll do it as well by contradiction by contradiction By contradiction assume z Is a unit we'll see what sort of contradiction we get to in other words Ie assume That we can find there is a b So that when you combine it with z z times b Is one in either order Because as I mentioned we're going to assume the unit means two-sided unit that you can get it both ways And we'll see what sort of contradiction we can get to but let's see so I have z times w is zero But I've assumed that w is not zero so I have z w is zero That's the hypothesis But I'm going to play the same game that I just played now multiply both sides on the left by b But now I can regroup these b times zero zero But b times z was one One times w is w and so we get w is zero which contradicts W being not zero contradiction So the conclusion is that z is in fact not a unit And the conclusion then is that the proposition holds namely if you start with a unit It's not a zero divisor and if you start with the zero divisor it's not a unit All right questions there comments All right, this is a good place to quit then of course, there's no new homework assignment and the assignment that I gave you Last monday was already due the previous monday so that you could have it back in hand for the exam If anybody wandered in late and needs graded homework return. I've got that here. I've also got exam