 Let's see, first here's some comments on the homework coming back. So the homework being returned, this is a good assignment, mostly you did really well on it. It's one of the more difficult assignments because you're actually getting your hands dirty computing with these coset or factor groups and that's in general not easy. Let's see, on the extra sheet, problem number three, you're asked whether or not is A L equal to L A for all, well you're asked to determine whether or not the subgroup is normal, I mean that's the question just rephrased and what some of you said was A L is not equal to L A for all A and G. This is how some of you answered the question and technically this is not a correct answer. You're asked to determine whether or not the left cosets are always equal to the right cosets. It turns out folks sometimes they are, it's just not always. When you make this statement what you've told me is that they never are equal and that's not a correct statement because for instance E times L does equal L times E. So the correct statement is not that one. It's A L is not equal to L A for some A and G and therefore L is not normal because the negation of are the two equal always is, are they not? Alright let's see, yeah second comment coming back, I've started to abbreviate this phrase because I'm using it relatively often maybe more often than I'd like to see. If you see this folks A slash O means apples and oranges. It means what you've done is you've somehow told me something like, well something like things that we talked about when I returned the assignment last time a week ago. Maybe you said something like kernel of phi equals one so that you've somehow told me that a set equals a number, something like that. Or that you've told me something like a collection of numbers equals a group or just something that can't be true simply because you're trying to compare two things that, well one which looks like apples and the other looks like oranges. And this happened similarly on this assignment in a couple of situations like the one I just mentioned you'll say something like kernel of phi equals one or the number of elements in the kernel of phi is the number of elements or anyway. Alright let's see, a couple of additional comments. In question one C on the sheet some of you correctly said D4 slash K is isomorphic to V. That's a true statement. Tell me why it is. In a previous problem you were asked to look at D4 mod some other subgroup and in that situation you got a group, the factor group had only two elements. Well folks once you get a group of two elements then there's no issues involved there is only one group having two elements it's Z2. But when you look at D4 mod K group with four elements and when you get a group with four elements then you know there's a choice. So just looking at the group table it's not clear right away whether or not you're looking at V or whether you're looking at Z4. So what you need to do is at least tell me enough but why? Specifically why is it not Z4 for example? That's a group of four elements. Heck it's an appealing group of four elements. All you need to do is tell me enough to distinguish the group that you've got from the group Z4. All you need to do is say look on the main diagonal every element has square equal to itself and therefore since Z4 doesn't have that property it's V. So there's really not much to say but something needs to be said before you can simply conclude that the answer is V or Z2 plus Z2 they're the same thing. Here's the style comment. It's probably Uber style it's more than just a style comment. Let me show you what most of your proofs look like for the statement if G is a billion then G slash H, G mod H is a billion. Most of your proofs look like A H star B H equals A B H equals B A H equals B H star A H. These three equations or four expressions connected by three equations contain all of the mathematical content that you need to have written down a correct proof. But here's what some of you did. Some of you wrote the phrase since G is a billion comma that. Folks you've made three different statements here. You've told me this equals this and this equals this and this equals this. Which one of those equations is the one that somehow follows from the fact that G is a billion is it this equals this? Is it this equals this? In fact it's the middle one. So in fact what some others of you did was you wrote that same phrase out here so that you wrote the proof A H star B H is that. Not good style why because the equality that you're verifying or justifying because G is a billion is this one. That's how you're getting from A B H to B A H. So the style statement is if you're going to tell me that you're getting from one step to the next by using one of the hypotheses point exactly to where you're using it. Don't just give me four statements and say well it's true because G is a billion. Which one is true because G is a billion. Now if you wanted to write that on the end and put the arrow here that's fine or if you wanted to as some of you did put it above the equal sign great idea. That's where it belongs. Or if you wanted to write it in front just draw an arrow to why or which of the statements is. But just putting it here and then listing out three or four equations just isn't good style because you haven't specified which of the statements or which of the equations you've justified by using the hypothesis that G is a billion. Alright that's the homework stuff. Let me give you the exam prep stuff here. For those of you that are watching the video or online I'll post all this stuff to the website by tomorrow. I just have to scan it and put it in electronic form to get it back to you. So let's see. Let me just do this. One, two, three, four. A minute or so here. Two, we'll do this a little bit backwards. Four, that's right about that. Six, let's see here I'm just going to have you pass these back. There you go. That's much more efficient. One, two, three. Pass those straight back. One, two, three. And then here is the... Everybody's got three documents now. Okay, exam two info. Well hey, it looks isomorphic to exam one info. It's the same structure here. Here's the date. Here's the sections it'll cover. Here's what you need to know and we'll go from there to give you the practice exam which is the exam that I gave last year. Will exam two this year look similar to practice exam two? Sure it will. Will it look as similar as exam one looked to practice exam one? Yeah, close. I mean if you've really got a good understanding of what's happening on practice exam two you'll probably do pretty well on real exam two but don't get tunnel vision here just because there's questions on the practice exam doesn't mean that the same or similar questions will necessarily show up on the exam and conversely there certainly might be questions on exam two that the real exam two that don't look like they have any direct corresponding questions on the practice exam but if you do all the sort of redo all the homework make sure you've got all the homework stuff under your belt go back through in your notes and make sure that you've picked out all of the key ideas that we studied in here so in case there's some true false questions or sort of big picture questions then you should be completely well prepared for exam two. Alright, questions about exam two? Let's see what I want to do here. There are SI sessions Tuesday and Wednesday. Let me email Jen and see whether or not we can run an SI session either before class on Monday or maybe Friday afternoon or something like that so I'll send an email out to you. Let me write myself a note here to see whether or not we can get an SI session together Friday or Monday before the exam. I mean I'll be in my office Monday before the exam but maybe we can get a little bit more work done on this stuff. Alright, remarks? So, here's what's up tonight. Well, we're going to continue on this second major theme, new topic that we introduced last Wednesday, this topic called rings and we'll continue in that, well definitely through the end of the semester so let me remind you what a ring is and then we'll continue looking at not only examples of things that are rings but various properties that certain rings possess. So, a reminder of the definition and I'll write it out in shorter hand or using more concise notation than I used on last Wednesday because, well, we've already seen the idea. In effect, a ring is a set with two binary operations, one typically denoted by plus, the other typically denoted by dot where the following three things are true about the set together with these two binary operations. First, that if you look at r together with the plus operation that you get in a billion group, that's a very strong statement. Here's a very weak statement that r together with the multiplication is an associative binary operation and third, I'll just write distributive laws, dot over plus as usual. So, a ring in general is simply a system where you've got two binary operations. The good intuition about rings is you know a lot of them. The integers, the rationals, the reals, the complexes, the n by n matrices, the zn's, the, I mean, et cetera, et cetera, et cetera, the functions from the reals to the reals. All of these are rings because in all of those settings, typically the set together with addition was in a billion group and typically we were able to combine things by some other operation called multiplication, but we never got a group in that sense. That's okay here. I don't really care if the underlying set together with the multiplication is a group or not. All I care about is that if you multiply two things in the underlying set, you get something back in the underlying set, that's a binary operation, okay, and that the multiplication is associative, but that'll be the case in all of the systems that we look at here. Okay, so big examples, lots of them. Examples, you know, z together with its two operations, the rationals, the reals, the complexes, the two by two matrices. In fact, I made this observation on Wednesday, but I sort of made it under my breath and I'll make it a little bit more prominent now. It turns out, folks, start with any ring you want. I don't care if it's the real numbers or the complex numbers or the rational numbers or the integers, it makes sense to talk about the two by two matrices where the entries in those matrices come from the underlying ring. Well, how do you add and multiply in the ring of two by two matrices over a ring? Answer exactly the same way you'd do if it was integers or rationals. Because matrix multiplication involves only two processes, addition and multiplication, and those are precisely the two processes that you get in any ring. So I don't care if here you put our bracket x, the ring of polynomials. I don't care if you put functions. I don't care if you put the complex numbers. I don't care if you put the three by three matrices over the two by two matrices or something like that. So it doesn't really matter what appears in the parenthesis. In fact, it doesn't even matter what size it is, the n by n matrices over any ring you want. The ring, oh, here's another one that we'll spend a lot of time looking at. The ring of polynomials, that's what our bracket x stands for, where the coefficients are taken from any ring you want. The collection of z n rings where the operations are... So look, you know, lots of systems where you can add and multiply. You can add and multiply real-valued functions. We know what that means. You can add and multiply matrices. I don't care what the underlying system is. You can add and multiply polynomials. I don't care what the underlying scalars are. So there's lots of different systems where this happens, and in each case it produces rings. So we have example upon example upon example of rings. Now, what makes rings interesting, it's fair to say that what makes rings interesting is trying to determine how rich or how much structure there is in the multiplication side of the ring. So the intuition, folks, is that the addition side of the ring is sort of uninteresting. It's as nice as you could want it to be. Not only is it a binary operation to do the addition, it's actually an embellion group, but what can differ between various rings is, for example, how close the multiplication comes to giving a group. Now, what we showed last time is if you look at the thing called zero, in other words, the thing that behaves as the additive identity, which is zero, zero, zero, zero, zero, zero, zero, all zeros, the zero polynomial, you know, it comes in slightly different notations but it all stands for the thing that you add to anything else that doesn't change the other thing. What we showed is that if you choose to take that thing and multiply it by anything else in the ring, you always get zero back. In particular, you never get, if there happens to be an identity, you never get one back. So if you've got a zero floating around, then this thing is never a group. So the question is, are with dot a group, is a complete non-starter? The answer is no, it's never a group. But what becomes interesting, especially in the examples that we already know, is if you take the underlying set and you throw the zero thing out, you might get a group. So let's just go through this list and at least intuitively ask this question. Suppose you take any one of these examples of a ring and you throw out zero. Then is the ring throw out zero together with dot a group? Well at least it's an associative, ooh, now I've got to be a little bit careful. If I throw zero out, is it the case that I even get a binary operation? I might not, let me give you an example. If I look inside the ring Z6, so 0, 1, 2, 3, 4, 5, there's Z6, it's easy to write down. I'm going to ask you to throw zero out and the question is, are the elements that remain, there's only five of them, do those form a group? The answer is no because the thing never even gets off the ground, it's not even a binary operation, here's why. If I take two and three, those are certainly non-zero elements and I multiply them together, it gets zero. I get something outside the set that I've just looked at. So boy in some situations it's a disaster. But in other situations, for example if I look at Q and I throw out zero, well let's see if I take any two non-zero rational numbers and multiply them together, alright, at least I get a non-zero rational number, just two non-zero rational numbers multiplied together and do non-zero, oh, so at least it's a binary operation on the set of non-zero elements and now we can ask, alright, is it the case that if you form, well let's see, is there an identity in there, yeah, is it the case that if I hand get any non-zero rational number that you can find, something that multiplies to give the identity, sure, just flip it over. So we've seen sort of, you know, one end of the spectrum to the next, there's situations where it doesn't even make sense to ask whether throwing zero out leaves a multiplicative structure that gives a group, because you don't even get a binary operation and on the other side of the coin, not only does it make sense to ask, do you get a group if you throw out the zero element, in fact you do, it's pretty easy to show, okay, so the big, let's call this philosophy, philosophy, it's this, how nice is R with dot, note that it's never a group is never a group and the reason it's never a group is because zero element, zero will never have a multiplicative inverse, that's what we showed last time, since we showed last time that zero dot R equals R dot zero equals zero for all R and R, in particular, regardless of what element of the ring you write down, if you combine it with zero, you never get one because you always get zero back, so here's the next reasonable question to ask, reasonable question to ask, how about, look at R, throw away zero, remember what that notation was, star with dot, is this a group, let's see, R, just to remind you just for emphasis, this is R, throw out zero, answer, sometimes yes, sometimes no, sometimes yes, give some yes examples, example, if I take Q, rationales, then the question is, look at the non-zero rationales together with multiplication, does this form a group, turns out folks we've seen this many times and we know the answers, yes, but the point is, I've now viewed Q as a ring, I'm now asking you to forget the addition operation for a minute, find inside Q the zero element, throw it out and ask what's left over, well in order to verify that this is a group, technically what do we need to do, we need to step zero, show that dot is a binary operation, dot binary operation on Q star, okay, why, because the product of two non-zero rationales is again non-zero, is again non-zero, what do you want to call that, the log zero products or something like that, property of integers, however you want to justify that, if you take two non-zero rational numbers and you multiply them, in other words, when you combine two things via that operation, if you take two things from this set, you get something back in this set, all right, associativity, yeah, associativity, not an issue, check, identity, sure, it's one divided by one or just better known as one, and then inverses, sure, sure, if I hand you A over B and it's in Q star, not in Q, but in Q star, that means by definition A is not zero, because the fraction isn't zero, and because A isn't zero, it means that A over B inverse exists, and I'll tell you what it is, it's B over A, and that's also in Q star, the reason it's in Q star, the reason this isn't zero is because by definition rational numbers, the denominator of the expression that we started with wasn't zero, all right, so we've just shown that if we start with this particular set together with two operations plus and dot, and we just focus on the dot operation and we throw zero out that we actually get a root, in fact we get an abelian group, because it turns out multiplication of rational numbers obviously is mutative, okay? Similarly, if I start with this ring and play exactly the same game, if you look at the collection of non-zero reals under multiplication, you get a group, if I look at the collection of non-zero complex numbers under multiplication, I get a group, in fact they're both abelian groups, that's fine, so the three examples that I listed out here turn out to be relatively special, let's look at a few more examples like how about Z, so these are groups, and I'll put in parentheses abelian groups, next level of example, example let's look at the integers, so here's the question, if I look at the non-zero integers together with multiplication, is it a group? The answer is no, we observed that way back when, but let's see what's preventing it from being a group, well in order to determine whether or not something is a group, step zero is determine whether or not the given operation is really a binary operation of the set, is it the case that if you take two elements of this set, in other words two non-zero integers and you multiply them together, do you get another non-zero integer? The answer to that question is yes, so is dot a binary operation on Z star? The answer is yes, so at least the question gets off the ground, it makes sense to decide whether or not Z star with dot is a group, associativity of course is fine, check, identity, is there some special integer with the property that of course it looks like one, how about inverses? Well no, that's where things fall apart, turns out not every element has an inverse, in fact most elements don't, has an inverse and the element for example two doesn't have an inverse, does not, not have an inverse in Z star, it obviously does in Q star and in R star and in Z star but there's no, one half is not, one half's not an integer that's the key, so not a group but I'm gonna put in parentheses but at least a binary operation, it turns out and I'm not going to go through all the details that there is another collection of elements on this list that behave like the integers do and what I'm going to do is look at a specific example because this is the one that I looked at last time, I'm going to look at the specific example of the collection of polynomials where the coefficient that I'm going to ask you to use come from the real numbers so it turns out if I look at this ring the ring of polynomials with coefficients in the real numbers, right this down in the same form that I had the other ones so here's the ring that I want you to consider and now the question is if I look at the collection of nonzero polynomials under multiplication what do we get? Well let's see, step zero is to make sure that multiplication of nonzero polynomials is a binary operation in other words if I take two nonzero polynomials where the coefficients are coming from the real and multiply them together well you always get a nonzero polynomial so the fact that this is a binary operation is okay again think about it if you take two nonzero polynomials and you multiply them together in fact I'll convince you that the result has to be nonzero in effect it's just the degree argument if you hand me two nonzero polynomials tell me the degree of the first one it's three tell me the degree of the second one maybe it's four hey if you multiply a degree three times a degree four you can't get the zero polynomial in fact what you get is a degree seven polynomial so it turns out there is it turns out and we'll use this terminology later there's what we'll call a law of zero products that acts both in the integers if you take two nonzero integers and you multiply them together you get a nonzero integer in the same way that in the same sort of way that we see happening inside the nonzero polynomials with real coefficients if you take two nonzero polynomials with real coefficients and you multiply them you get another nonzero polynomial and in fact for those of you that have taught an algebra course before when you use that a lot I'll show you where that happens later on associativity sure multiplication of polynomials is associated no big deal is there an identity for the multiplication in other words is there some special nonzero polynomial with the property that if you multiply it times any other polynomial it doesn't change the other polynomial sure it's typically called this it's the polynomial of degree zero that has one is its constant term and zero is its externals zero's it's x square two so I'm asking you to view the number one here's a polynomial how about inverses falls down for example here's a polynomial does it have an inverse in other words can you multiply this polynomial times something to give you this if your answer is one over x okay well that's correct but one over x isn't in here this means the collection of all polynomials nonzero that have real value coefficients polynomial by definition means a constant plus some real number times x plus real number times x squared plus da da da da you don't get to use negative exponents in the definition of polynomial so this thing has no inverse at least in our bracket x it's the same sort of issue as with the integers you at least get off the ground multiplication turns out to be a binary operation on the set it turns out to not give a group and in both cases it falls short because well various elements lack inverses all right now the sort of third piece of the puzzle of the third flavor of rings are rings in which when you throw out the zero element you don't even get a binary operation to give them set so here's sort of the third flavor of example example let's look at how about z6 it's the one that I just talked to you through let's write down the details so if I look at z6 star in other words throw out zero with multiplication then this not even a binary operation not even a binary operation so the question does it form a group is a non-starter why not because for example if I take two times three each of these individually are nonzero elements of z6 but I get zero which is then not in z6 star not a group similarly if I look at the let's say the two by two matrices over the reels if I throw out zero and look at the multiplicative operation not even a binary operation not a binary operation is it possible to take two nonzero matrices multiply them together and get zero of course I mean just for example there's a whole lot of different examples you could use folks just not in okay I'm gonna give you some words the words in effect correspond to these three flavors of rings in some cases you get when you throw out the zero element you get a group when you look at the nonzero elements under multiplication sometimes even though you don't get a group you at least get a binary operation on the nonzero elements sometimes you just I mean sometimes it sort of falls apart and you don't really get much of a structure it's just too much too much bad stuff there's just too much stuff that winds up being zero to allow multiplication to be a binary operation of nonzero elements so here's some words we sort of describe each of these three types of rings with some words let me give you the words the first is skew field skew field means this folks ring is a skew field means that a ring for which if you look at the nonzero elements ring R for which take the nonzero elements and look at the structure under the multiplication is a group now there's two other words that are germane here this is also called this is another phrase I won't say one is more standard than the other but they're certainly both used in literature this is also sometimes called a division ring intuitively it's a ring in which division always makes sense as long as you've thrown out zero so there's the first phrase if and this will be an important example if it's not only the case that when you throw out zero and you look at the structure on the multiplication that you get a group if you happen to get an abelian group in other words if the multiplication is commutative if dot is commutative in other words if r with dot is an abelian group ie if r star with dot is an abelian group then we call this and we call r a field so skew field means that the nonzero elements form a group under multiplication field means more field means not only is it a group in fact it's an abelian group so that in the three examples that we gave in that first collection the rationales the reels and the complexes those are all skew fields or division rings but in fact they're slightly more I won't say slightly more in fact they're somewhat more in fact our fields because the multiplication in each of those is in fact commutative so for example the collection of rational numbers that ring is a field this ring is a field this ring is a field our field and for one of the homework problems let's see that I'll assign tonight and it won't be due till after the exam so not to worry in one of the homework problems that I'll assign tonight example you'll do this for homework but you'll do a problem that's essentially identical to it for homework instead of square root of 5 it'll be square root of 11 or something like that remember we looked last time at the following ring the collection of elements inside the real numbers this was a subset of the reels I want you to just look at those real numbers that you can form by taking a rational number and adding it to a rational number times the square root of 5 we spent a little while last Wednesday showing that at least this thing is a ring turns out not only is it a ring this is actually a field show that for homework this one's sort of interesting alright well you have to make sure that if you throw zero out of this set that the resulting set together with multiplication is a group and as we've looked at these examples step zero is to show that at least it's a binary operation well folks I don't care what two real numbers you take whether they're in this set or not if you take two non-zero real numbers and you multiply together you get something non-zero so that's a non-issue so the fact that you get a binary operation on s star comes for free because things sit inside something that already have that property question that'll be alright is it associative yeah that's no problem because multiplication is happening inside reels already know that's associative does this particular set have an identity element in other words is well the question is is one in here well yeah because one is of the correct form it looks like one plus zero times the square root of five there's one so I found one in there now the real question is does every element in this set have an inverse under multiplication in the set and that becomes a little trickier you might say well wait a minute if you're looking at a non-zero element in the set it's just a non-zero real number so I'll tell you what its inverse is one over a plus b times the square root of five and what folks you haven't convinced me that that is in the set you have to write one over a plus b root five as something of this form can you do that I don't know maybe you can maybe you can't turns out since I've given you an answer the answer is yes but just stopping saying yeah every element has an inverse if I hand you a plus b root five in this set here's its inverse one over a plus b root five I'm completely non impressed by that because you haven't convinced me that it's in the set you've convinced me that the inverse is a real number but you haven't convinced me that's of the right form how do you do that quick hint what's it called rationalize the denominator and after a lot of algebraic work which turns out to have some merit to it you'll eventually be able to convince me that if a plus b root five is in s star then one plus over a plus b root five is also in s star arithmetic and it turns out when I hand you this similar question for homework that that will really be the only non-trivial issue that you have to deal with in deciding whether or not that set is a field or not we already shown it's a ring and then showing the other stuff all except for the last piece follows directly from the fact that you're just multiplying real numbers two non-zero numbers multiply the other is going to be non-zero so satiety comes for free that the identity element is in there comes for free and you just have to show that inverses are the correct form all right so that's the first collection of words skew field or division ring and then more importantly a field it will take us a little while folks and we may not even get there by the end of the semester but we'll get there in the middle of next semester if we can't get there this semester it'll take a little while to be able to write down an example of a skew field in other words a ring where when you throw out zero and you look at what's left over under multiplication that you get a group that's not an abelian group in other words it'll be relatively difficult to come up with an example of a this that's not actually that all the examples that we've written down so far are not only skew fields they actually are fields and the difficulty in doing so it is probably going to bubble to the surface by the end of tonight because it turns out that if there are some division rings out there in other words if there are some skew fields out there that aren't fields then they're gonna have to be of a very special form all right questions coming all right let's see do I want to do this next flavor yet no yeah let's do this proposition opposition let me phrase it as a question first question question all right question so I'm gonna list out a bunch of rings for you ZN where N is a natural number so here's a lot of rings Z2 Z3 Z4 Z1 technically the rings pretty uninteresting just zero by itself Z2 Z3 Z4 Z5 Z6 so I've all list of them we've given an example of a ring on this list Z6 that's not only not field in fact doesn't even get off the ground when you throw zero out you don't even get a binary operation that was this so if you ask the question are the rings on this list fields the answer is certainly some of them are not Z6 is a good example but it turns out there are rings on this list that are fields so the question is which if any of these any of these rings are fields quick remark Z6 certainly is not see this computation throughout zero then not only do you not get a group you don't even get a binary operation but example let's look at something small how about Z yeah Z3 so here's Z3 and the operation is addition of multiplication mod 3 so if I look at Z3 star together with multiplication well I know what that set looks like it's the set one and two there are the non-zero elements of Z3 question is this set closed under multiplication well see one times two is two and one times one is one two times one is two and two times two is one because the operation is mod three so two times two is four but mod three that's one so in fact here's what the multiplication table looks like one times one is one one times two is two two times one is two and two times two is four mod three which is one so we get a group that we're gonna be angry so field Z3 is a field so one is one isn't sort of cut to the chase here so the question is can we distinguish between the integers the values that produce fields versus those that don't produce fields and the answer turns out to be yes it turns out proposition that Z sub n is a field precisely when precisely when n is prime Z3 is a field through the prime Z4 is not a field four is not prime Z5 is a field five is prime Z6 is not Z6 seven is etc. I'm gonna prove this by the end of today I think but I'm not gonna prove it right now because we need to introduce one more term one more definition and then we need to look at properties of Z sub n where n is prime that somehow gets to that definition and then we'll want to be able to move from that definition back to the definition of a field so proof soon but if nothing else one piece looks like this piece is easy it turns out that if n is not prime then if you look at not binary operation so if I start with a number that's not prime well the question is is it possible for when you throw out zero corresponding set together with multiplication to be a group and the answer is not even chance doesn't even get off the ground not only is it not a group in fact you don't even get a binary operation and the reason is this well wait a minute if n isn't prime so if n is not prime and what does that mean it means you can write n as let's call it a times b where a is less than n bigger than one and b is less than n but bigger than one so it means to be not prime that you can actually factor it well what does it mean to say that the integers a and b are each less than n it means that they're not zero in zn but then a is not zero and b is not zero in z sub n because they're each less than n and bigger than one and the only way you can be zero in z sub n is if you're multiple of n rephrased a and b are each in z n star but if we multiply a times b we get n but that's zero in z n in z n so a times b is not in z n star so we don't even get a binary operation so one direction is clear if you start with a non prime in other words a composite that you're going to use as your base uses your modulus then you never get a field the slightly more difficult question is all right if you start with a prime why do you get a field and that's the slight little tangent that we have to go off on for 10 minutes or so so the converse will be soon so what have we done so far we've we've sort of given one end of the spectrum which is if you start with a ring you throw zero out it might be the case that you wind up with a group think intuitively that all the non zero elements of the underlying set have multiplicative inverses because that's typically what gets in the way here secondly we showed some examples that don't even get off the ground but there's a sort of middle ground the ground that includes for example the integers or the ground that includes the polynomials with real coefficients the situations where if you take two non zero things and you multiply them at least you get something non zero so you get a binary operation but that somehow maybe you don't have enough inverses and we're about to give those types of rings a name so the new words or verbiage is this what we're going to call an integral domain integral domain this is a type of ring is the situation is a ring R for which let's see well I mean the silly statement of the blessing that I have to put in is first of all that R has unity in other words that it has something that we identify as one secondly that R is commutative I said well wait a minute R has two binary operations on an addition of multiplication which of the operations are you telling me as commutative folks there's not an issue here because always the plus operation is assumed to be commutative because it's always assumed to be in a billion group under addition so when we say R is commutative we mean R with dot is commutative and the key is that when you look at the non-zero elements under dot that you get a binary operation and I'll rephrase what that means in other words ie the product of two non-zero elements in R is again non-zero so I've just rephrased what it means to say that the non-zero elements in your multiplication form of binary operation or as they set with the binary operation that means that if you take any two things in here you combine them using that to get something back in here well we've given we've actually given a lot of examples of these things called integral domains good example the integers are an integral domain the integers are an integral domain the polynomials with real coefficients are an integral domain yeah let's see the integers certainly has a multiplicative identity the multiplication in the integers is certainly commutative and we showed that whenever you take two non-zero integers and you multiply them you get a non-zero integer the same was true as well but you know what folks the same is also true of the rationals the rational certainly has an identity element multiplication certainly commutative and that could say more about the rational fact not only was the rationals with dot a binary operation it actually turned out to be a group and the same is true of the reals and the same is true of the complexes let's see proposition proposition if r is a field then r is also an integral domain if it's the case that when you look at the non-zero elements under multiplication that you get an abelian group that's what this means is it necessarily the case that when you look at our and it's non-zero elements under multiplication that the collection has unity is commutative and forms a binary operation well yeah has unity means that under multiplication there's an identity element that comes for free if the non-zero elements for a group form a group is it the case that if you look at the multiplication that it's commutative yeah that's the definition of being a field that when you throw out zero that the multiplication is abelian might say well what about zero folks zero commutes with everything zero times anything is anything times itself which is always zero so commutativity with zero is a non-issue it's not interesting turns out to be a non-issue third is it the case that if the non-zero elements with multiplication form an abelian group that the non-zero elements of the multiplication form a binary operation of course that's what gets it off the ground so the proof is that the ingredients that give our star with dot and abelian group group give enough information implies all three properties on this list so there are lots of examples of integral domains out there anything that's a field but it's important to keep in mind because these are going to be the most interesting rings that we'll be working with there are many examples of integral domains that are not fields and the integers is the best example of such a ring it's a situation where the non-zero elements at least form a binary operation take two non-zero integers multiply together you get something non-zero but that not all of the elements have inverses and the same is true with polynomials over the reels so every field is an integral domain but not conversely there are many integral domains that aren't so what we're we're going to ask the question all right which z sub n rings our field well what we've just shown here is in fact if you have a value of n that's composite in other words event is not prime and in fact we've shown not only that z n is not a field in fact it's not even an integral domain because you wind up getting the multiplication not being closed under the non-zero turns out proposition proposition if is an integral domain a b c r elements in our with a not equal to zero and you have the following equation if you have a times b equals a times c then the conclusion is you can cancel the equals this is going to be probably the key property of integral domain so whether or not you happen to be in q or in r or in z or in r bracket x if you're multiplying by something that's non-zero and you're multiplying two separate things called b and c and you wind up getting the same thing out then b and c had to be the same things going in reason oh look if a times b is a times c then subtract from both sides that gives a times b minus a times c is zero now factor this is the distributive law a times b minus c is zero and now stare for a minute at what we've got we've got inside an integral domain a situation where I've got a non-zero element it's non-zero by hypothesis times something equaling zero look folks in an integral domain if you take two non-zero elements you have to get something non-zero but if the product's going to be zero it means one or the other of the two things that you've multiplied had to be zero to begin with because you're an integral domain but we've assumed that things not zero so the conclusion is that b minus c is zero so that implies b minus c equals zero put parentheses since r is an integral domain domain and we've assumed a is not zero but if b minus c is zero that implies b equals c so the upshot or the nice property of being inside an integral domain is that if you're in a situation where a times c is b times c and you've multiplied by something non-zero at least you can cancel the a's I'd say well you can always cancel as long as you start with something non-zero and in fact that's not true so just for contrast I'll show you that if you're working inside something it's not an integral domain you might not be able to cancel even if you're multiplying by something non-zero in z six if I do three times one and I do three times three all right that's not zero folks it's not even zero in z six and certainly that's not zero well are those the same yeah three times one is three three times three is nine but in z six that's three but certainly one is not equal to three so you can't cancel in general even if the thing that you're multiplying by is non-zero you may not be able to cancel but if you're in a situation where you're an integral domain then you can here's the key observation then proposition in fact z sub n is an integral domain if and only if n is prime and say well that looks sort of like the proposition that we're in the middle of proving yeah in fact what we're gonna wind up doing is this folks and we won't be able to complete it tonight but we will next Wednesday or we will on Wednesday this will be relatively easy what we'll show next Wednesday is if you happen to have an integral domain and the integral domain only has finitely many elements in it then in fact the integral domain is a field so what we're gonna do is we're gonna show that this in situation where we're looking at z sub n in other words a ring with finally many elements turns out to imply that z sub n is field and that's what's gonna allow it to finish up the proposition that we're in the middle of proving proof well the first piece will look just like the piece that we've already written down in the context of showing that z sub n is a field if and only if n is prime so we'll first start with the contrapositive here if n is not prime and we've already seen then by the previous proof proof we have that z sub n is not an integral domain so that gives us this direction if z n is an integral domain necessarily is prime how about for the converse now suppose n is prime then what I'm gonna do is well for about half of you remind you of a fact from number theory for the other half of y'all just hand you this property of prime numbers then it turns out then we have this is usually referred to as Euclid's lemma Euclid's lemma here's what Euclid's lemma says if n is prime and that's the situation that we're in if n divides evenly into the product of two integers then either n is a divisor of a or n is a divisor of b or both this is a property of numbers that is specific to primes it's not true in general so for example if three is a divisor of a times b the only way that can happen is if three is either a divisor of a or a divisor of b to begin with it's not true for primes for instance six is divisor of three times four three times four is 12 but six is neither a divisor of three nor divisor of four all right so why is this helpful well look then I need to try to convince you that if n is prime that the z sub n is an integral domain it's not too bad to do suppose I have a times b suppose a is not zero and b is not zero in z n rephrase suppose I start with two elements in z n star to convince you that z n is an integral domain I have to convince you that if I multiply those two things together that I get something not zero we'll do it by contradiction suppose a b is zero in z n contradiction we'll see what the contradiction looks like wait a minute if a times b is zero inside z n what does that mean it means that a b is a multiple of n yeah that's what it means to be zero inside z sub n it means you're multiple of n in other words n divides a times b go but ends assuming to be prime so we have Euclid's lemma which says if n divides the product then by Euclid we get that n divides a or n divides b yeah but if n divides a that means that a is zero in z n or b is zero in z n that's what it means to divide which is a contradiction to hypothesis contradiction because we assumed that we started with non-zero elements in z n so what we've just shown is on this list of rings z 2 z 3 z 4 z 5 etc etc etc the ones that are integral domains are precisely the ones that correspond to primes in other words the ones that aren't integral domains are the ones that correspond to non primes and this is a good place to quit let me give you a homework assignment as I've mentioned at least in words it didn't write down what we will do next time is use this proposition of integral domains to show that if you start with an integral domain that happened to have only finally many elements like z 5 or z 11 or z 13 or something like that that necessarily get a field questions comments sorry about tonight being a lot of words and verbiage but what we have to do start building the machines that we're gonna use from now until the end of the semester okay here's home and this will again be a slightly odd do date this will be due two weeks from tonight so that's Monday November 5th because you've got an exam between now and then and I want to give you a little bit extra time to do that and here's the assignment in section 18 problem 7 through 22 27 to 33 37 41 42 44 55 and 56 I only want you to turn in three of those though 12 37 and 41 and not surprisingly I'm gonna ask you to change the instructions slightly on each of these on number 12 actually show the set as a field so this folks is the problem that as promised I'm gonna ask you to do for homework it's the one that we looked at in relatively gory detail it'll be a question let us be a set that looks like a plus b times the square to 5 blah blah blah show it says the field the remark is the only real work is showing that inverses work the only real issue is showing that inverses are of the correct form of the correct form just as I alluded to in class so don't start proving associativity don't prove distributive law or anything like that all that comes for free because the collection of numbers that you're looking at live inside the real numbers and you already know all those things are true by virtue of that fact on number 41 what I want you to do here is the result also true in z6 justify if it is tell me why and if it's not give me a counter example and then in section 19 here's some problems that of course I'm going to change the instructions on as well one through four eleven through seventeen nineteen twenty three and twenty nine the turn in problems are three and twenty three and on three I want you to use instead of the polynomial they give you use the polynomial x squared plus three x plus two that's much more interesting and on twenty three wherever they say division ring yeah scratch that and replace it by integral domain integral domain I need to write an algebra book someday at least at least I'll get all the I'll get all the statements of the problems the way they should be all right and again that'll be due to