 Yeah, they are really lucky, like by about a day I think, it's supposed to get really bad. No? I don't know. It's just tough. I don't know. It's tough. Somebody, I'm sure. I'm sure they're nice people. Their mothers love them. I'm sure. Let me see. Let's see. Let's see. I don't know. I don't know where she is. Jessica, here you go. Thanks, sir. Okay, let's see. Christmas. Okay, not too bad. A couple of quick comments on the assignment coming back. Most of them were pretty computational, so it didn't cause too much grief. The only issue that came up on the more computational stuff is what that one question asks you to do is find a generator for, not for Z17, well I know what a generator is for Z17 as an additive group. Just one works, because one plus one is two. So that's not an issue. What is at issue is this. If you look at the non-zero elements, look, Z17 is a field, so the non-zero elements form a group under multiplication. That group has 16 elements in it. Is that a cyclic group? Well, if somebody just hands you a group with 16 elements and tells you it's a billion, there's absolutely no guarantee that it's cyclic. But what will be shown, we won't do it this semester, but we will do it next semester. If you come back for another dose, then you'll find that if you take any prime and you take the field Z sub that prime and you throw out zero and you look at the corresponding abelian group, that that group is always cyclic. So what you're trying to do is find a generator for this under multiplication. That's the question. The other way to phrase it is, in other words, take the units of Z17 and as soon as you're told that you're supposed to be looking at the units of Z17, it means you're throwing out zero. And so necessarily you're looking at times as the operation. A couple of you just convinced me that Z17 under addition is cyclic, and that certainly could have been done about two months ago. You just write down the number one and show that if you keep going, you get all the elements in Z17. So a couple of you misinterpreted that question. The other one that caused some of you some grief was the last one. In effect, what you're asked to do is this. Take the following function from a polynomial ring to another polynomial ring, the function that what we call formally differentiates polynomials. Well, you come into the table knowing something about differentiation, but only in a situation where the coefficients come from the real numbers. And in that particular situation, differentiation of polynomials is just one of many differentiation ideas that you can talk about. You can talk about differentiation in the context of exponential functions or trigonometric functions or whatever it is. And just as one sort of byproduct of differentiation happening to mean limit as h goes to zero, blah, blah, blah, blah, you happen to get this particular formula for polynomials. But the point is that formula can be thought of as a formula that applies to any polynomial over any field of coefficients. Take a polynomial and simply associate it with this other polynomial. Take the polynomial a0 plus a1x plus a2x squared plus da, da, da, da, da, and associate it with, I'm just going to make this thing up, a1 plus 2 times a2x plus 3 times a3x squared plus 4 times, it makes sense. The fact that it happens to be apparently interesting because it happens to be this byproduct of a formula that comes from a limit expression, okay, it just happens to be. But folks, this formula has nothing to do with limits or anything. It's just do it. And with that in mind, the question then becomes part a, is this a, well, is it a homomorphism where the operation is addition? You're thinking, well, if that was working over the reals, in the particular case where the field is r, then yeah, because the derivative of the sum of two functions is the sum of the derivatives. There's no addition rule. Question one, part b was, is the derivative function, somehow a multiplicative function. In other words, it's the derivative of f times g, the derivative of f times the derivative of g. You've been thinking right away, well, no, because derivatives and multiplication don't work out well. There is the product rule that doesn't say just derivative of f times derivative of g. And most of you recognize that. That's good. Some of you wrote this sort of flowing prose about how it shouldn't be and it's not and there's not a prod, blah, blah, blah, blah, just folks, give me a counter example. Just write one down and you're done. That's all you need. Okay? Now parts b and c were interesting. I guess I should have known that I'd get this, you know, I mentioned in class, well, you need to use the characteristic of the field as zero in parts b and c and some of you just started off parts b and c saying because the characteristic of the field is zero, it just went on and basically didn't, it all used the characteristic of the field as zero. It's all right. Or some of you sort of put it in but it sort of appeared at a place that indicated that maybe you weren't really using it in a way that was, that compelled me to believe that you understood what was going on. Here's the idea. What things get, well, what things are in the kernel of the derivative function? That's the idea. And if you come to the table thinking, well, I know what things the derivative function kills. The derivative kills constants. Perfect. Okay. It's just, we're taking the derivative in contexts that are larger than the one that maybe you're used to working with. You're taking derivatives in the context that isn't just r bracket x, which has a certain set of properties associated with it. You're taking derivatives presumably where the field could be anything of characteristic zero, maybe the field's the complex numbers, maybe the field's the rational numbers, maybe the field, there's some other fields that we've seen. For instance, this thing that we called s, the a plus b root two field. So how do you know that the only thing that gets killed in that context are the constants? It's not anything to do with a limit process anymore. And the answer is this. Just write down what it means for something to be in the kernel and convince me that it has to be a constant. So what does it mean for a polynomial to be in the kernel of the derivative function? It means if you write f of x as a zero plus a one x plus a two x squared plus a and x to the n, that's what it looks like. Then the derivative is, well, it's a one plus two times a two x plus three a three x squared plus et cetera, et cetera, et cetera, plus n a n x to the n minus one. There's the derivative. No problem. Question is, what does it mean to say that the polynomial is in the kernel? It means that the derivative is zero equals zero. That's what it means by definition of it being in the kernel. So look, I have this polynomial equal to this polynomial. So all the coefficients have to be the same. That's how two polynomials are equal. Let's see the coefficient on the constant term over here is a one. The coefficient on the constant term over there is zero. So we get a one is zero. The coefficient on the x term over here is two times a two, and the coefficient on the x term over there is zero. Coefficient on the x squared term over here is, and the coefficient on the x term over there is, et cetera, et cetera, et cetera, the coefficient on the x to the n minus one term is that, the coefficient on the x to the n minus one term on the right side is zero. Obviously they're all zero on the right hand side, so I'll say equate coefficients because that's what it means for polynomials to be equal. So here's what this says. If f is in the kernel, then necessarily the constant term is zero. Okay. Ah, I said that wrong. If the derivative is zero, then necessarily the linear term is zero. In other words, the term that's being multiplied times the x term is zero. If f is in the kernel, it means that 2 times a2 is zero. Question, does that imply that a2 equals zero? I'll give you a hint. And I'll give you a hint. Why? Look, if you have the product of two things equaling zero, any fields in the integral domain, so one or the other has to be zero, is this equal to zero? No, why? Because the characteristic, the field is zero. Two is not zero in our field because we're told that inside this field, if you take one and you keep adding it to itself, you never get zero. In particular, two isn't zero. So it implies a2 is zero since the characteristic of the field is zero. How about three times a3? So that's zero. Does that imply a3 is zero? Sure does. Why? Because the characteristic of the field is zero. I've n times an is zero. That implies that an is zero. Why? Because the characteristic of the field is zero. So here's where you're using the hypothesis. So what we've just shown is that if you take something that's in the kernel that necessarily a1 is zero, a2 is zero, a3 is zero, etc., up through an is zero. In other words, if you have something in the kernel, then all this stuff has to be zero because we've just proved that each of the coefficients is zero. So if you have something in the kernel, it has to look like just that. In other words, it has to be a constant. So essentially, all of you got to the right place, but you didn't use the place where the characteristic of the field is zero appropriately. Okay, now for part c, I'll go through this one a little bit more quickly. Most of you recognize that the image, in other words, what actually gets spit out of the derivative is every possible polynomial. In other words, it told me the image is all of f bracket x. Well, that's true. Let's see what happens in the reals. If you hand me a polynomial, you have to convince me there's some other polynomial whose derivative is the one you started with. That's what it means to convince me that the range is all of f bracket x. Just as an example, if I hand you x squared, can you build me a polynomial whose derivative is x squared? Yeah, great. One-third x cubed works. I mean, one-third x cubed plus seven works. There's lots of different integrals. But just write down an integral. One-third x cubed. Derivative of one-third x cubed is x squared. So x squared is in the image. In fact, x to the n is in the image because I can write down something whose derivative is x to the n. One over n plus one, x to the n plus one, I'm just integrating. So you take the derivative of the integral and you get right back where you started. So that all seems well and good, except for what? Except for how do you know that one over n plus one times x to the n plus one makes sense? In particular, how do you know that one over n plus one is a legit thing to write down? I'll ask one more way. How do you know that one over n plus one isn't one over zero? And the answer is, because the characteristic field is zero, that n plus one is not zero. So the punch line is, because we have a field of characteristic zero, we can quote-unquote integrate any polynomial. So the point is that for part c, to show that every f of x is in the image of d, image of the derivative, if f of x is blah, blah, blah, blah, blah, blah, blah, then, so as above, then write down this, g of x as, what is it, a zero x plus a one x squared over two plus a two x cubed over three plus, plus a n, just in some sense artificially pulled out of thin air some polynomial. This is in f bracket x. The reason it's in f bracket x is because each of these symbols makes sense because none of them requires me to divide by zero because none of the denominator is zero and f because the characteristic of f is zero. Okay, since characteristic of f is zero. So none of the denominator is zero and you can show that the derivative of g is f and so every polynomial little f of x is in the image. If the question is well what if you couldn't do this, what if for instance you were working in the field z two x? Can you divide by two? No. So what that means is that there are polynomials in z two bracket x that you can't integrate. In other words that you can't find a polynomial for which when you take its derivative you get the original one. I'll give you a good example. Try to find for me a polynomial whose derivative is x where you're working over the field z two. Find it, have it x squared. That's no good because its derivative is zero. It's two times x, but two times x is zero. Say well divide by two. You can't divide by two because two is zero. So that's what's going on and that's what a couple of you picked up on and that was good but some of you just sort of threw down characteristic of f of zero and we're happy with that and went on. Alright well I mean you're listening so I can't fault you for that. That's good but there was a little bit more to that. Okay here's what we're up to tonight. There's two, oh let me just make sure I do some administrative stuff that I need to do here. Yeah the administrative stuff is after the break we're meeting in where Dwyer 121 so I don't know. Are you on next Monday? Yeah. Oh oops. Oh I'm sorry. Make sure they help. Okay thanks. Maybe that's where everybody is. Oh I take back all the bad things I said. Maybe they're up at Dwyer 121 instead of mile high stadium but that's okay. No I'm not. And right if you want to turn in the homework tomorrow you can do so that's fine. Just fax it to me by let's say five o'clock tomorrow or so or email it to me as an attachment that's probably the preferred mode. What we're up to tonight is we're going to finish up some sort of final tidbits about irreducibility of polynomials. We'll spend ten minutes or so talking about the main goal for the remainder of the semester it will give some good context to what we've been doing for the last two weeks or so in this discussion of polynomials and polynomials over various fields and then we'll start sort of heading up the path towards the final summit for the semester and start building the tools we need to actually get to this main goal. So here's some sort of final tidbits on factorization and irreducibility, final stuff on factors, irreducibility, etc. The first is last Wednesday's presentation was unlike one you've seen before probably in any math class because it was essentially just this sort of ad hoc list of here's the big question tell me if you hand me a polynomial with coefficients in the following field is it or is it not irreducible in f bracket x where f is the given field and the answer is just that there's just no good hard fast rule that always works you have to use some methods that are specific to various fields or specific to the degree of the polynomial or specific to the form of the polynomial and we just don't have that many general results a lot of times it's just a matter of taking whatever polynomial you're looking at and just trying to haul out whatever tools you might have at your disposal. So that's why the last Wednesday looked a little bit odd and why this homework assignment that's coming in today or tomorrow was just you know you sort of had to look at it and say well what field is it in what's the degree of the polynomial what are the coefficients look like etc etc etc to decide what test to use. Here's one last irreducibility criterion that I didn't mention last Wednesday and it's this these are called cyclotomic polynomials and even though they look like a very special form they come up relatively often in a specific context that those of you that are I'm going to take the math four slash five fifteen course next semester we'll see it turns out if you if you hand me a polynomial it looks like this f of x equals x to the make sure I've got his notation see the p minus one or p depending on where you want to start here yeah if you look at x to the p minus one plus x to the p minus two plus plus x squared plus x plus one where p is a prime p is prime so for example you might be looking at if p is five x to the fourth plus x q plus x squared plus x plus one that polynomial is irreducible in q bracket x just you know I didn't prove much for you last Wednesday and other than the results of that degree two and degree three polynomials but a lot of these just ad hoc so you can add that one to the list here's another one if you happen to see a polynomial of this form then this happens to be irreducible in q bracket x alright that's the first comment second is this you have been trying to play up the similarities between these rings f bracket x the polynomial rings and the integers and we've seen lots of similarities neither is a field both is an integral domain both have division algorithms associated with them both both have the property that inside them we can talk about some sort of notion of primeness or irreducibility we call it prime in the integers we call it irreducibility in the context of polynomials and those two ideas are essentially the same and the last comment is you know for the integers at least if you start with non-zero integers in fact let me be a little bit more precise if you start with an integer that's not minus one zero or one so if you start with not zero but not a unit then it turns out you can always take such a thing and either the thing is already a prime or if it's not prime you can always write it as a product of primes that's what we call the fundamental theorem arithmetic and moreover if you somehow line up the primes in increasing order there's really only one way to do that so somebody hands you and your friend the same integer bigger than or equal to two puts in separate rooms as write this as a product of prime numbers well you can do it and in fact you'll both do it the same way of course you know if somebody hands you six you could write as two times three and your friend could write as three times two but you know if you agree to write them in increasing order then in fact there's only one way to do it and the punch line is the sort of last similarity that I want to point out between the integers and the polynomials is there's a similar sort of result for polynomials vis-a-vis writing polynomials as products of irreducible so the remark is the analogy between f bracket x and z sort of continues has another piece to it it's this the analog to I'll write down what it's called at least for the integers fundamental theorem of arithmetic theorem of arithmetic says this essentially for f bracket x given any any little f of x in cap f of x whose degree is bigger than or equal to one with the degree of f of x at least one then there is I'll say essentially because I don't want to get into the details but I'll at least give you the analogy of why I have to use this word essentially a unique way to write f of x as g one of x times g two of x times times g sub t of x where each g sub i is irreducible in f bracket bracket x g sub i of x is irreducible in f bracket x for all i no words any polynomial as long as the polynomial some guts just don't don't hit my constant polynomial give me a polynomial you can always write it as a product of irreducible polynomials and moreover if you and your friend try to write it as a product of irreducible polynomials well you know it's like here if I ask you to write I don't know 14 as a product of prime so technically you could write it as seven times two or you can write as negative seven times negative two because remember you know inside the integers viewed as a ring we have to include the negatives and we have to include negative numbers as prime numbers there it's the same thing here you know you might take x squared minus one and write it as x plus one times x minus one okay that's good but your friend might write it as two x minus two times a half x minus a half or something like that I mean you know if you if you got enough fractions around you can sort of mess it yeah but alright the ideas you can just you know yank the constants out and what you're left with is essentially the same so that's what's going on I definitely don't want to spend the time or the energy to write up this statement precisely and I'm certainly not going to ask you to you know kick it back at me or have it on exam three but just know that there really is this very tight relationship between what's going on inside the integers and what's going on inside f bracket x the only the only real non similarity between what's going on inside the integers and what's going on in f bracket x has come up in this question of how do you find whether or not polynomials are irreducible in the given field and that happens to depend on the field and the polynomial et cetera and we don't really have that sort of behavior happening inside the integers okay now two final comments not about polynomials but rather about fields in general we have not folks and won't get to this semester maybe next semester but we won't get to this semester an example of the following form all of the remember what are called skew fields or division rings they are rings with the property that when you throw out zero that what's left over is a group it's all this required all of the examples that we've shown of such a structure has always had the property that when you throw out zero not only do you get a group but you get an ability group in other words the multiplication has been commutative there are examples of rings with the property that when you throw zero out you actually do get a group under multiplication but it isn't a commutative group it isn't an ability group those are called division rings or skew fields and it would just it would take us a half hour to 45 minutes to write down all the details of the example so I'm going to leave that out and suffice it to say they're out there but we just got to work a little bit harder to find those and the last comment about fields before we move on is this the only finite fields that we've looked at so far that is the only fields with the property that they contain only finitely many elements are the z sub p fields so there's a lot of them but it turns out there are many other fields having finitely many elements that aren't just z sub p and we'll actually get to those that'll be one of the final goals in this class we'll be able to write down fields for example that have four elements you think well z four no z four is not a field it has zero divisors it can't be a field two times two to zero but it turns out there's another ring out there that has four elements and it turns out it has the property that if you throw out zero that what's left over is a group having three elements it's not z four it's something else we haven't seen yet something else we're going to have to build but they're out there so the remark is there are finite fields finite fields which are not of the form the form z sub p p prime we haven't seen any of those yet but we'll see one by the end of the semester we'll see some by semester's end we well maybe surprisingly no I'm actually put it this way I could you know tonight write down the ring I could tell you know I write down the four elements I could tell you what the addition operation is just by giving you a table tell you what the multiplication operation is just by giving you a table but it would make no sense it would just sort of look like you know gobbledygook and what we'll be able to do by semester's end is actually quite naturally in fact sort of lead ourselves to that sort of structure and not only with four elements but with eight elements or sixteen elements or nine elements or twenty seven elements or a prime power number well okay so here's a big aside I'm going to call this this is what the author calls it the basic goal so here's a five minute whirlwind history of what has motivated much of mathematics for the last I don't know almost twenty five hundred years now you look at a certain system you get comfortable with an operation in the system and then you realize the system somehow isn't big enough to allow you to solve all the possible equations that might come up in that system here's a good example if I'm looking only at the positive integers which you know the sort of basic natural numbers the counting numbers if I can do an equation that has something to do with addition so if the system is the positive all numbers positive integers and the operation is addition and I ask you to solve an equation of this type you're golden the x is four so no big deal you can solve it in the system the problem is though if I ask you to solve this equation in the same system you obviously can't solve it in the system of positive integers that's sort of unfortunate so what do you have to do well if you want to somehow allow yourself to solve all the possible equations that might come up in the system that you started in with the operation that you're interested in you might have to invent or cook up new symbols or new numbers that behave in a way that would allow you to solve systems of this type or equations of this type so I'll say okay in z plus but this is not okay in z plus meaning there is no solution in z plus and you know we're so comfortable with it especially as math majors but even by you know sixth grade or something you're comfortable enough with the idea of a negative whole number it's simply a symbol that in effect allows you to solve an equation like this it allows you to go backwards but this is okay in an extended version version of z plus we call it z i.e. z and in fact you invent symbols the negatives which behave well which do what you want to do do what you want them to do namely solve equations of this type so we wind up with a nice system you know as far as it goes so whole numbers sort of become entire or become whole if you allow the negative numbers because then you can solve any equation it looks like this that's good of course then what do you want to do well if you turn your sites to a different operation multiplication you try playing the same game and if somebody asks you to solve this 3x equals 12 in z solve that in the system that we're now working in in the integers no problem x equals 4 but solve this in z and the point is folks all of the symbols that you're writing down are coming from your underlying system and the operation is one that's of interest in the underlying system multiplication it's just you can write down equations that you know in some sense look just like this it's just instead of the operation addition you've written down the operation multiplication and obviously not okay can't solve in an integer so what do you do you invent a system that contains solutions to all the equations that look like this and that system is called the rational numbers so you extend you invent symbols symbols that do what you want them to do rationales and now you've got a bigger system and inside that bigger system not only can you solve any equation that came from the original system looks like this you can actually solve any equation that comes from the bigger system so if you want me to solve 3 7s x equals 19 fourths you can do that in the bigger system the rational no way a bigger system but then we ask similar questions like all right now instead of taking just linear equations x plus 3 equals 7 so linear in addition or 3 x equals 12 linear multiplication take maybe quadratic equations like this x squared equals 9 so now I'm asking you if you want just view this as x times x I mean I'm still asking you to take the multiplication operation but now instead of having you look at a constant times x now I want you to look at a variable times itself can you solve an equation that sets that equal sure that's of course okay it might be more than one solution plus or minus 3 that's no big deal but then if I ask you to solve this inside this system no good and the Greeks already knew that that the rationales weren't big enough to allow them to write down a solution to this equation to the x squared equals 2 equation because the solution was square to 2 and Euclid wrestled with that and proved that you can't solve this equation inside the rational so that's 2,000 years ago so what did they do they invented some numbers that made the system whole again that included solutions to these things now that's a little bit of a larger leap it wasn't just write down all the fractions or write down all the negatives all right write down a system where you have a square to 2 where well one way to do that is to in effect write down that ring s that we talked about a plus b root 2 just cook up a symbol invent symbols like the square to 2 that's all it is and then you've got a solution to that thing of course if you also want a solution to x squared equals 3 or x squared equals 7 or x squared equals 11 in this same system you have to sort of keep bundling stuff on and so eventually you know fast forward to 1800 years or so eventually you invent the real numbers sort of rigorously invent the real numbers most people played with them and sort of knew what their properties were but it wasn't until Kroniker came along and or Dedekin came along and sort of did this you know Dedekin cut approach to the real numbers as limits of appropriate sequences of rational numbers all right so you invent a system where you can solve equations that look like that because even with all that work you still left some things out like solve that no good you still need to go bigger so you invent some symbols that symbol and it behaves the way you want it to so you sort of you know over 2200 years has been sort of piling on of different levels different levels different levels and eventually what we get to is a system called the complex numbers and it turns out well one of the fundamental properties of the complex numbers is that if you hand me not only linear equations or quadratic equations if you hand me any polynomial equation a n x the n plus a n minus 1 x the n below and set it equal to 0 you can always find a solution and port in the complex numbers so somehow the complex numbers is as big as you need it to be it's a phrase that we'll use next semester called algebraically closed any polynomial that you write down if the coefficients come from here you can always find a solution to it in the complex numbers that's not true of any of the previous systems that we've written down all right what I'm going to focus on though is this if I want a system that let's say contains at least the rational numbers so that I can at least do the basic operations addition subtraction multiplication and division on the whole numbers inside my system that's how I get the rational numbers if I want to solve or find a system in which I can find solutions to specific equations like this one or like this one well you don't have to sort of throw everything in if all you're focusing on is solving a specific type of equation a specific type of equation that can be viewed as setting a polynomial equal to 0 of course if I want you to find a solution to this it's the same as asking you to find a solution to the polynomial x squared minus 2 equals 0 similarly asking you to solve this is asking you to find a solution to the polynomial x squared plus 1 equals 0 if that's the task well we want to be able to do it in a more sort of surgical way in a more precise way without having to throw in all this other stuff that we might not be interested in we might also want to do it in contexts other than the rationales for example it might be the case that we're interested in starting with z2 or z3 or coefficients coming from some finite field the question might be can you solve a certain polynomial equation where the coefficients come from a finite field so maybe I'm going to ask you to solve this equation where the underlying field is z5 perfectly good equation to try to solve inside z5 can you do it or can't you if you can't how do you somehow enlarge the system so that you can and so what we're going to call the basic goal for the remainder of the semester is this we're going to hand you polynomial well that's exactly what we've been looking at and we're going to be interested in the question of whether or not the given polynomial has the property that something in the original system when you plug it in for x spits out 0 if you go back all of these equations that I looked at ask exactly that sort of question can you given a polynomial find something inside the system so that when you plug it in for x that 0 gets kicked out well for this one we could in z for this one we couldn't in z there was no 0 for this polynomial in I'm sorry there was a 0 for this polynomial in z namely x equals 4 there was no 0 for this polynomial in z so we invented a bigger system inside this system there is a 0 for this polynomial has to be called 11 thirds here's a polynomial x squared minus 2 I'm sorry start with this one the one that worked x squared minus 9 question inside the rationals is there a 0 for this polynomial sure x equals 3 also x equals minus 3 is there a 0 for this polynomial in q no so what do we do we invent a bigger system for which this polynomial contains a 0 is there a is there a 0 for this polynomial in the reels no but there is a 0 for that polynomial in here inside a bigger system so question was this just some sort of historical you know continuing to pile on the system that had been built all the way or is there maybe a more subtle or more precise or more you know I use the word surgical that's probably a good one here sort of you know direct approach to taking a polynomial and somehow building a bigger system in which the polynomial has a 0 so that's the basic goal basic goal start with start with a field I'm going to call the field let me sneak a book just because I want to be consistent with his notation and I forget whether he calls the big field e or the big field f let's see well done that's why I don't have it in my notes because he dances around it doesn't use any letters for the notation okay that's fine start with the field I'm going to call it e for now e f of x be a non-constant non-constant polynomial in e bracket x think e is the field q rationals maybe and f of x is the polynomial x squared minus 2 or x squared plus 1 something like that non-constant means it has degree at least one well folks it might be the case that the given polynomial has the property that you can find something in the given field so that when you plug it in 0 comes out and I'm going to phrase that in terms of these evaluation homomorphisms in a minute but for now just think it might be the case that inside the given system that you're working in you can find some value so that when you plug it in everywhere you see an x here the 0 gets kicked out good example x squared minus 9 in q bracket x of course you can find some element that when you drop it in for x you get 0 3 works so there's negative 3 but it might be the case that the polynomial that you started with doesn't have the property that you can find something in the original field and have a 0 get kicked out like x squared minus 2 where the system is the rationals here's the question can we always find a bigger system a system bigger than the one that we started with with the property that the bigger system well contains the system that we started with and moreover also contains something that you can drop into the original polynomial and have 0 come out so then here's the question can we find a field let's call it f so that two things are true first the field that we started with lives inside the bigger field and secondly there exists let's call it alpha inside the bigger field for which when you plug alpha into the polynomial that you started with that you get 0 in other words for which when you do the evaluation homomorphism or f of x that you get 0 I'll say systematically build a field with the property that starting with a given polynomial coefficients in perhaps a smaller field we may not have to build it anymore but with coefficients living inside some field that inside this new field inside this extension field that we find a 0 for the given polynomial everything well I already know how to do that in the Greeks told me how to do that well yes and no at least at least the people that built the complex numbers over the centuries were able to do that as long as you started with a polynomial with coefficients in the rationals or eventually in the reels or eventually in the complexes themselves but they didn't deal with the case that's actually going to be of great interest to us this semester next semester would start with a finite field and it's also the case again that they really were looking more broadly when they built the reels and they built the complexes it wasn't a case of well start with a specific polynomial and make it work it was well can I make all polynomials that look like maybe x squared equals a works or maybe have all polynomials look like x squared equal negative a so you get the real to the complex and you know we've seen situations where we're able to answer this question is significantly more precisely if I hand you the rationals and I hand you x squared minus two let's just look at that one for a minute can you find a field that contains the rationals with the property that it also contains some special element so that when you drop that element into x squared minus two that you get zero up if you're thinking the reels or the complexes that's not incorrect but let me do one better look at the field s that you looked at for a homework problem problem number 12 in section I forget 18 or 19 or something like that the thing that looks like symbols of the form a plus b times the square root of 2 that's a field certainly contains the rationals a plus 0 times the square root of 2 there is a rational and certainly contains the square root of 2 0 plus 1 times the square root of 2 so there's a field that works and boy that's a whole lot more efficient than just throwing all the reels in in order to answer the question find a zero for the specific polynomial x squared minus 2 so the main goal for the rest of the semester the basic goal is come up with a method by which if you hand me a polynomial with coefficients in some field if that polynomial maybe doesn't already have a zero from the given field extend the field somehow so that the new field actually does contain a zero for the given problem that's the idea and this little historical trip was just meant to show you that really that process has already been done in one particular context but it was done more from a global perspective find zeros for all possible polynomials of a given form rather than from a more local perspective here is a polynomial build a system where this particular polynomial has a zero how do we do that the answer is what we're going to do is build some structures if we were building these structures from scratch it would take us three or four weeks to do but fortunately these structures are ones that we've already seen before what we're going to do is take these polynomial rings f bracket x well they are a billion groups under the addition so it means we can talk about subgroups of that a billion group in the addition which means that we can form the factor groups which weren't your favorite topic but they are nice and they're important or we're going to do that again but eventually what we're going to wind up doing is trying to put not only a group structure on these factor groups well we know what that is it's going to be you know coset addition but we're going to also try to define a multiplication on these coset groups so that these coset groups become rings and those are the places that we're going to try to find these fields so what we're going to do both tonight and next Monday after the break is look at how we might go about constructing these things these new rings and asking when it is the case that these new rings actually become fields and armed with that we'll get enough information to try to figure out how we can go about giving a specific polynomial building a larger field in which that polynomial has a zero so I mean I'd write the word aside here because what we're doing is we're sort of taking a little detour but the detour will take the rest of today and most of next Monday in order to build the machinery in order to answer this question so it's more than an aside let's call it I know a 75 minute detour or something like that detour we need we need to look at specific construction of rings which will look familiar and here's what the construction looks like so eventually we're going to be interested in doing this construction the context where the ring we're interested in is f bracket x is this polynomial but we'll talk about the idea from a more general perspective so start with any ring R with any ring call it R it's probably the case that the ring has unity so I won't write that in explicitly but probably the case that all the rings that we'll wind up looking at are rings with unity what we're going to do is look at a certain subset of the ring and here's the definition of what that subset looks like definition is this a subset let's call it N of the ring R is called an ideal of R the Germans call these ideal subsets of R and eventually the word subset got just thrown out for efficiency I think and the word ideal simply remained in case N has two properties first of all well remember the ring is always an abelian group under addition and the first property that I'm going to require of this subset is that when you look at the subset under addition that you actually get a subgroup under addition of the ring because this is this is stuff that we did you know weeks one through eight in here here's a group group under addition here's a subset is the subset of subgroup well we know how to prove that you just use the subgroup there right you have to show it's closed under addition you have to show that the identity elements in there you have it's closed under additive inverses this condition on the subset is usually relatively easy to to demonstrate or to show or something like that this next one is much more interesting it's the condition that says this for every if you take any element of the ring and every element of the subset so I want you to take any element anywhere in the ring I mean this thing might be in the subset might not I don't know just grab your favorite one now grab your favorite element inside this subset then here's what I want you to do do this you know I'll put a dot in there to emphasize that now I want you to combine these things under the multiplication and I'm going to ask you to multiply them in the other order as well because remember in general they don't commute most of the ones that we look at will have this equal to this by properties of commutativity but in general they don't need to be and the requirement is that when you do that product will look folks because we're inside a ring if you look at the product of any two things necessarily get something back in the ring but what I'm going to require is that that product's always back in the subset and the word we use to describe this situation as we say and has the absorption property p absorption property meaning if you multiply anything in the ring times anything in the subset somehow that lands you back in the subset that's what it means to be an ideal of a ring let's start with the two main types of examples example let the ring be the integers and let n be the subset consisting of multiples of four and it turns out that n is an ideal of r then you love our lesson so what do we have to check we first have to check that if you look at the multiples of four under addition that you get a subgroup of the integers under addition well folks we've done this a number of times so check already done probably two months ago take two things in here I mean why does something subset the subgroup you just use the subgroup there take two things in here Adam do you get another multiple four sure is zero in there sure it's four times zero if you have something in there is it negative in there sure yeah four n then it's four times negative in this so that's right now how about property two you have to convince me that if you take any integer so let's call it z and take anything that's inside the subset let's call it four l if you do the product z times n you get z times four l which is four times zl because multiplication inside the integers is commutative right that's fine which is in the subset because it's four times some integer now I haven't written out all the window dressing with all the you know all the justification but that's what's going on if you multiply any integer by a multiple of four the point is you get another multiple of four I don't care what you started with as the first integer three times 12 is still multiple four I don't care that three wasn't as long as you so the idea is that somehow multiples of four have this absorption property multiply any integer by multiple four you get another multiple four and the other way n times z well of course I don't really have to check it here because n times z equals z times n inside the integers because the integers happen to be commutative but in general we may not have a ring with that property okay was there anything special about the number four absolutely not in fact if I hand you any positive integer I could probably hand you zero two and I look at the collection of multiples of that integer then I'll get an ideal so more generally generally let n be any integer then as usual let capital N denote all the multiples of whatever integer you've chosen in fact folks I'll even let you choose zero here look at all the multiples of zero there aren't very many just zero itself turns out that subset will work out just as well then n is an ideal of z it's given a fancy name it's called the principal ideal generated by little n the proof is essentially the same as above proof similar to the n equals four case so inside the integers it turns out there are a lot of ideals hand me any number and you'll get an ideal sometimes if you hand me two different numbers they might produce the same ideal for instance if you hand me the number three or if you hand me the number four and ask me to look at all the multiples of four your friend has me the number minus four you look at all the multiples of minus four you get the same collection of numbers the multiples for the multiples of minus four why because four and minus four well they only differ by okay negative sign but we're going to use ring theory language here they only differ by a unit they differ by multiplication by negative one and negative one is a unit in the integers has a multiplicative end all right but you know aside from that what that result says is hand me any integer look at all of its multiples you'll get an ideal okay now mumble under my breath turns out those are the only ideals somebody says I'm thinking of an ideal inside the integers it has to be one of those the proof is it turns out we didn't we didn't write out all the details two and a half months ago but we could have proved that inside the integers if you just ask what are all the subgroups forget rings and ideals we're all the subgroups of the integers the answer is they all look like n times z for some integer n so hey it turns out that just getting past step one only limits the sorts of things that you have to consider to the n times z subsets and it turns out they happen to have this absorption property as well in other words happen to be ideals okay here's the next big example example guess what if I've just given you an example that involves the integers you expect the next example to be one that involves f bracket x because we're trying to play up the relationship between those and that's exactly what's going to happen if the ring is f bracket x here's what I want you to do pick any polynomial I don't care pick zero if you want pick a constant polynomial if you want pick you know a cubic polynomial I don't care let let's call it little f of x in capital f of x be any polynomial so pick any element you want and now look at this set consider I'm going to call it this the set well if I wanted to notationally write this set down in analogy to the notation that I used for the set over the integers I simply write little f of x times capital f of x just like I wrote little n for the integer written next to the ring I'm interested in capital Z and what did the notation mean over there it simply meant look at all the things that were multiples of the thing that you started with in other words look at all the things that could be written as the thing you started with times something else in the system so the multiples of four are start with four and just multiply four times everything else in the system see what you get you don't get everything you only get some things that's what I'm going to ask you to do here start with your polynomial maybe it's x squared minus two and simply multiply it times every other polynomial in the system and see what you get so this is what we're going to call it it's also sometimes denoted by this little f of x with random bracket or triangle brackets around it and here's what it is it's the collection of polynomials let's call them g of x that you can write as with the property that g of x can be written as little f of x times h of x for some h of x in bracket x it's the collection of things that are multiples of that particular polynomial let's do an example example for instance if I want you to look at x squared minus two that's a polynomial in cube bracket x what I want you to do is look at this set folks there's no way I can write out everything in that set because what I'm asking you to do is simply take the polynomial x squared minus two and multiply it by whatever other polynomial you want and tell me what you get well I'll give you one example okay zero works why because it's x squared minus two times zero you get zero and x squared minus two works because it's x squared minus two times one and x cube minus two x works because it's x squared minus two times x and x not do this in my head because it's been a long day let's see but I mean hopefully you see what I'm doing here just take x squared multiply by anything you want tell me what you get well I happen to be at what plus x squared minus two x minus two that's in there x cube plus x squared minus two x minus two etc there's lots of things in there how do you determine whether something is in there oh it's in there in case when you divide x squared minus two into it that what you get is simply a polynomial with zero remainder so another way to view the collection of things I'm interested in is take the polynomial that you started with in this example happens to be x squared minus two divide that polynomial into the given polynomial the one you're interested in determining whether or not it's in this set and ask what the remainder is if the remainder is zero throw it in the set if the remainder is not zero then don't throw it in the set so let's just put a big containment sign here contains all those things but there's no way I can write them all out maybe for yeah hopefully this will be a little bit enlightening I can tell you something that's obviously not in that set folks if I hand you a polynomial of degree two there's only two possibilities I either multiply that polynomial by zero and I get that or I multiply this polynomial by something that's not zero and if I multiply by something it's not zero but has to get kicked out of something at least of degree two so if I hand you a polynomial of degree one like x plus three it's not in this set so there's a whole lot of things that aren't in the set any polynomial of degree one there's also polynomials of degree two that aren't in this set like the polynomial x squared minus three is not in the set can you write x squared minus three is x squared minus two times some polynomial well now I mean if you're going to try to rig a polynomial so that a polynomial times x squared minus two equals x squared minus three and the polynomial has to start with a one fact has to be one to make the leading coefficients match up if you're going to multiply by one you get x squared minus two you don't get x squared minus three so here are some things that aren't in there for example x plus four is not in there x squared minus three is not in there are not in this set and x squared minus two again the question of trying to determine those things that are or are not in there is going to boil down to some sort of division algorithm so note in this case let's see a polynomial let's call it g of x is in the subset of f bracket x that we've denoted by this if and only if or precisely when r of x is zero when you divide f of x into g of x and we've used it turns out the notation is slightly turned around from the original usage in the statement of the division algorithm that we learned but that's okay I'm going to ask you to divide the polynomial that you happen to be given which happens to be f of x here in the g of x if the remainder zero and it's in that subset if it's not zero that's not the subset guess what the proposition is proposition start with any such polynomial look at the corresponding subset consisting of multiples of that polynomial and then you get an ideal so for any little f of x in capital f of x this set the collection of polynomials which can be viewed as multiples of this particular one is an ideal just as the collection of multiples of four always gave an ideal of the integers or more generally just as the collection of all multiples of a particular fixed given integer given ideal inside z the collection of all multiples of any fixed given polynomial is an ideal inside f bracket x proof that's going to be the same idea same ideas before same idea as in the integers let's see I have to first show that this set is in fact a subgroup of this set under addition so what do you have to do you have to take two things in this set how about hmm g of x it's called the first one g1 of x and g2 of x in the set you have to show that the sum is in the set so what does it mean to say that the first ones in the set means I can write it as f of x times something h1 of x or h1 is a polynomial it's just the definition of what this set is it's the collection of things that can be written as f of x times something and g2 of x can be written as the same polynomial folks this is the one polynomial it's common to everything it's like factoring a four out we're just factoring out of x out times h2 of x where h2 is in f bracket x and so now Adam g1 plus g2 and look what's going to happen you just I mean I'll do two steps at once you do excuse me that plus that factor in f of x out h1 of x plus h2 of x I mean there's no big deal which is in the given set why because it looks like f of x times something in f bracket x that's what the first step of the proof of showing that this subset is actually subgroup looks like so that's closed under the addition is zero in here can you write the zero polynomial is f of x times some polynomial easy f of x times zero well that was totally easy third if you've got something in here you got something it looks like a multiple of f is it's negative also multiple of f sure if you've got something that's written as f of x times h of x then just look at the polynomial it looks like f of x times negative h of x hold the negative sign out you get it's negative so pieces two and three are easy so eventually we get that this set with addition is a subgroup of f bracket x with addition that part turns out to be essentially not an issue now for the second piece that we have to show in order to conclude that a subset of a ring is actually an ideal second piece is you have to show this absorption property you have to show that if you pick anything inside f bracket x what do you want to call it I don't care how about p of x and you pick anything inside the subset and you multiply them together you have to convince me that the result is back inside the subset so do p of x times g of x well wait a minute g of x is in here that tells me something the g of x can be written as f of x times h of x so this is then p of x times g of x is f of x times h of x and technically I'm using associativity now I'm about to use commutativity too because p of x times f I can switch those around because f bracket x is commutative oh that's f of x I don't have used associativity again p of x times h of x you know look folks that's then f of x times something that's all you need in f bracket x so check I mean absorption turns out to be in effect the identical proof to the absorption property when we showed that 4z had the absorption property inside z you multiply something by a multiple of 4 you get another multiple of 4 if you multiply a polynomial by something that's already a multiple of f of x you get a multiple of f of x I mean the arithmetic inside z and the arithmetic with these polynomials share lots of common properties that's just one of them now technically I have to show that if I multiply them in the other order that I get something in the subset but again we're working inside a commutative ring so once I've shown this one then this one in the context in which we're working comes for free these are ideals let me talk you through where we're headed and then I'll hand you a homework assignment which won't be due for a while I'll be doing a week and a half and then we'll get out of here look any time you have an ideal well ideal sits inside this group as a subgroup but the hypothesis on the addition inside a ring is that this is always commutative so by default as soon as you have a subgroup of a commutative group it's necessarily a normal subgroup because any subgroup of an obedient group is normal which means that we can automatically form this factor group where the operation is addition and the question is going to be is it possible to turn this group which will be an obedient group under addition just factor you know coset addition can we turn this into a ring you're thinking well R is a ring can we somehow use the multiplication inside R to turn the cosets into a ring by defining some sort of multiplication the answer is in general no but if the subgroup happens to have this second property this absorption property then the answers yes and that's why we're going to need this second property in order to eventually make the factor group the additive coset group into a ring and then the question is going to be all right once we form those rings some of which you're familiar with for instance when we start with R equals the integers these will just look like Z mod n or Z sub n so we already know what those look like in that one particular setting the question is going to be what do those look like if we start playing this game in the situation where R is f bracket X and n is one of these ideals generated by a polynomial what do the structure of those things look like and that in effect is what the basic goal and its answer will lead us to okay all right so here is home this will be the next to the last homework assignment that'll give you for the semester it'll be due well on normal homework schedule so Wednesday November 28th so the Wednesday after the break and here is the stuff I want you to look at in section 26 problems 9 and 11 through 15 I want you to turn into those 12 and 13 but then I'm going to hand you three extra problems and these are each related to the other so once you get one the others should follow relatively immediately so and these three additional problems first if I is an ideal of the ring R and it happens to be the case that the unity element of R is inside the ideal then show that the ideal is actually the whole ring show that I equals R that's one two if R is a field show that it's only ideals it's only ideals are zero and R and the hint is use problem one use the result of this and then third inside the integers if I hand you two integers A and B I conform the ideal consisting of all the multiples of A that's an ideal I conform the ideal consisting of all multiples of B that's an ideal and what I'm going to ask you to do is look at the intersection of those two ideals it turns out that's an ideal in fact it turns out that it's the ideal consisting of all the multiples of some fixed number and the question is what is C in terms of A and B terms of A and B and that's I guess 3A and part 3B is the same sort of question take two integers I'll call them C and D I guess I'll call them E and F and what I want you to do is add everything inside the multiples of E to everything inside the multiples of F you get a subset and in fact you get a subset that looks like all the multiples of some fixed integer G and the question is what is G in terms of E and F so take the intersection of two ideals of this form it turns out it is an ideal of the same form tell me what the and the way I want you to do this folks is just write out a number of examples write out two or three examples and you'll see the pattern after a while I'm not worried about the actual proof of this result in general just get to the answer for me and I'll have you turn all these in alright if you happen to have the homework that's due tonight that's great just leave it on the table here if you don't you want to turn it in by tomorrow at five that's fine as well just either fax it in or the preferred way would be to send it to me as an attachment to an email remember after the break so week from tonight we're meeting in Dwyer 121 and if I don't see you before then if I don't see you tomorrow please have a safe holiday and I'll see you a week from tonight