 Last week, last Wednesday, I have copies of all the paper stuff that I handed out then, the practice exam, the solutions to it, and the exam three and final exam information that's also online, so some of you may have downloaded it from there, but feel free to pick up an extra copy of the exam if you'd like to do that. The homework from last week is due by Wednesday at five o'clock. If you'd like to turn it in today, that's fine. The homework that just got passed back, most of it was right on target. In fact, there was really only one sort of style issue that I would say came up more than once. Some of you are writing correct mathematics, but not sort of, I don't know, stylistically beautiful mathematics. That's fine, it's sort of the next step, but the idea would be read back through whatever you're offering me as the proof of some result and ask yourself, do I need this, do I need this, have I used all the hypotheses, have I put in too much, and this was simply a matter of using too much when you needed to show if i is an ideal of r, of r, and i happens to contain the unity element, show that i is actually all of r, and what essentially all of you realized after sort of playing around a little bit with the properties of being an ideal, specifically the absorption property is that, well if you hand me an element in r, you have to show that it's in i, but the point is r is r times 1, that's just the definition of unity, but this is in r, that's what was chosen, this by hypothesis is in i, so that means that this is in i, and in effect that's the guts of the proof. What some of you then did was that oh, but we also know by definition of an ideal that 1 times r is in i, and that certainly is also r, so because it's part of the definition of the absorption property for an ideal in a ring, you know that regardless of which order you do the product in, whether you multiply the ring times the ideal or the ideal times the ring that you get something back in the ideal, some of you then went ahead and told me that this is true, well it certainly is true, but it's a complete non-issue, once you've told me that is true, that's all you need, you've convinced me that r is in the ideal and you're done, so you're done here, it's not that this isn't true, it's simply not necessary or it's extraneous once you've convinced me that r is in the ideal, so again just a style comment on something that I saw a couple of times, and you know obviously there's, I'm not surprised to see this statement only because that's a part of the definition of being an ideal, but it turns out you don't need the full force of the definition of an ideal in order to get there. Alright, here's what we're going to do, we're going to in effect put the finishing touches on what you just told me was the basic goal and how it's achieved, I'm going to re-explain some of the details there and then spend the rest of tonight simply running through a bunch of examples so that you hopefully start getting a feel as to how to manipulate with these things, I know after class on Wednesday a couple of you came back in here start staring at the board as to how, you know how'd you get from here to here and here to here, well what we'll do is play with enough additional examples tonight, I think that you'll all hopefully get the right you know intuition about how that's working, so here's the big idea for a minute just for now, pretend I'm not even working with an irreducible polynomial that I've simply given you a polynomial, so let's start with some polynomial, I'm at least going to give it some guts, in other words make it have degree at least one so it's got x's in it, and let's see in the statement of this thing in E and f bracket x, okay with degree of f of x at least one, now here's what I want you to do, you know it might be the case that f is irreducible or it might be the case it's not irreducible, that actually plays a key role in determining whether or not when you form the factor ring f bracket x mod the ideal that you get by looking at all multiples of little f whether or not that's a field, and since our goal is to write down a field that has certain properties we'd like f to be irreducible, but there's no guarantee that the polynomial that you get handed up front is irreducible or not, the same goal still applies though, what we're interested in doing is we want to build a field capital E so that two things are true, let's see so that E is a subfield of f, I'm simply going to write that as contained in, but sharp containment that's the notation that we've used when we've got some structure to each of the pieces involved, and secondly there's some element, there's some alpha in E, I'm sorry in, oops, got these backwards, I must do that, f in E, I'm not sure why the author writes it that way, but that's alright, some alpha in the big field so that when you plug it into f that you get zero, again intuitively what you're trying to do is find some element, it might already be in the given field, but more than likely it's going to have to be in this larger field with the property that when you plug it in that zero comes out, and the first remark is this, look if the polynomial you happen to start with, f of x is not irreducible in f rack of x, and the question still makes sense folks, find some element in some larger field so that when you drop it in you get zero, the question on the surface doesn't look like it has anything to do with whether or not the original polynomial is irreducible, okay, but our only tool to build fields requires us to have irreducible polynomials, it turns out that piece of it is not an issue and this is what piece one of this four step outline of the proof that you just wrote down for me talks about, then here's what I want you to do, simply write f of x as p of x times q of x where p of x is irreducible, can we do that, yeah folks, and here's this analogy that I've been playing up for the last five or six weeks, if I hand you an integer, positive integer, if the integer is not prime, it means you can always find a prime factor of it, I don't care whether the other thing's prime or not, if I hand you the integer 12 it's not prime, so that means somewhere along the way you can factor some prime number out, factor two out if you want, factor three out if you want, I don't care, and what's left over is it prime, I don't care, just give me a prime factor, well we can always do the same thing for polynomials, as long as you have a polynomial that's not irreducible it means that, well it actually means that you can write it as a product of irreducible things, but all I'm worried about is that just pull out something irreducible, now here's the point, if I can find some field that has an element inside it so that when you plug that element into p that this thing becomes zero, in other words if I can make p of alpha zero, then hey if you plug that same thing into f, then f of alpha is p of alpha times q of alpha, I don't care what q of alpha is, I just don't care because it's going to be zero times q of alpha which will be zero, so the point is if we can find a zero, a field f e containing alpha so that p of alpha is zero, then f of alpha will be p of alpha times q of alpha which will be zero times q of alpha which will be zero, in other words as long as I can find a zero for an irreducible factor of f then I will have found a zero for f, no big deal, so what that means is all we really need to do, all we really need to do is to build is to find a field f, no e, yeah e and some element alpha in e so that p of alpha is zero as soon as I can make p of alpha equal to zero then necessarily f of alpha is also zero because the goal is to find a zero for f, alright, so in effect the way that the mathematician would phrase it is we reduce the problem to one of finding a zero for an irreducible factor of f and well so why bother and the reason that we want to bother doing that is if the goal is to find a zero for an irreducible polynomial then I have a chance because I know how to build fields corresponding to irreducible polynomial, here's how we do it, so here's step two, so now complete this task, this task, can we do it, yeah, here's how we do it, method, let's see if I can get the number right, okay, here's step two, this is really where all the hard work has gone into, the statement since p of x is irreducible, irreducible in f bracket x, this thing, f bracket x factoring with the ideal consisting of all the multiples of p of x is a field, of the four steps in achieving the basic goal folks, this one is the one we had to work hardest to get, if you take f bracket x and you take a polynomial that's irreducible in f bracket x and you look at the ideal consisting of all the multiples of it then the factoring that you form f bracket x mod those multiples of the fixed element irreducible element p of x is a field, okay, all right, so what's step three, step three is I need to convince you of two things, I need to convince you of the fact that I can view f as living inside e, we view f inside e, it's a little bit of a ruse but it's completely it's the same way that we view the integers living inside the rationals, I mean they sort of do but they sort of don't, the integers live inside the rationals, how you associate each integer with the symbol z divided by 1, that's how every integer is a rational, I need to teach you how I want you to view every element of the original field of the smaller field as living inside the larger field, take an element in the smaller field and I'm trying to be totally consistent with the notation here, take any element a in the smaller field and associate it with, I'll put sort of a wavy line here, view it as its coset a plus px, well this certainly is an element sitting inside this ring, why, because a is an element of the field, so an element of the field can simply be viewed as a polynomial of degree zero, it is a polynomial, it's not a very interesting one admittedly but that's okay, once you've got any element in here you can talk about its coset and so what we're about to do and of course we denote this by a with a bar on it, if you hand me an element of the field I'm going to ask you to associate it with the element inside this factoring that corresponds to simply taking the field element viewing it as a polynomial of degree zero and then looking at its coset and then finally here's the punch line, let alpha be the element x plus px in E and I'm going to give it a name i.e. the specific element x bar, the coset corresponding to the specific polynomial that you write down in here, the polynomial 1x plus zero if you want to view it that way or of course we simply just call it x then here's the point then we have p of alpha equals zero and here's why and this was the question that was asked at the end of Wednesday and let me reiterate why it's true, in the end folks when we're talking about this ring this factor ring what does it mean to say that two cosets are equal inside the factor ring it's not necessarily the case that their coset representatives are equal in order to deem two cosets to be equal all you need to do is ask if you take this coset representative and you subtract this coset representative do you get something inside the subgroup that's the definition of what it means for two cosets to be the same because remember when we form these cosets we form them corresponding to the additive structure this lives inside here is an ideal that means that this thing is a subgroup of that under the plus operation so if we're trying to equate cosets saying two cosets are equal is simply a matter of asking whether or not the two coset representatives have the property that i'm thinking in the back of my mind ab inverse is in the subgroup but when we're using additive notation ab inverse comes becomes the expression a minus b so here's why here's why p of alpha is zero inside e well what i have to do i have to show i have to show that inside this factor ring that this coset is the same as this coset that's what it means to say that two things are equal inside this factor rate which happens to be a field by all the hard work we've done so what do we have to do we have to show that if we plug alpha in everywhere we see an x in the polynomial p we get an expression so that when we subtract zero from it that the result lands inside this ideal in other words is a multiple of p of x in fact what i'm going to show you folks is the coset representative that we get not only lives inside the collection of multiples of p of x i'm going to show you that it actually is p of x so it's an even stronger statement than we need but it turns out we get a little bit more here so what do we have to do well what does it mean to plug alpha in well what's alpha it's this so this is going to start looking a little bit weird so let's see if i take p of x so let me not show here yet so right well what does p of x look like i don't know p of x as hmm it's something in f bracket x so let's write it as a n x to the n plus a n minus one x to the n minus one plus dip dip dip plus a one x plus a zero so there's p of x so let's see what's p of alpha well now you have to be a little bit careful and here's why alpha is some element that lives inside this bigger field e it makes sense to multiply elements from the smaller field f each a sub i in f the smaller field by something in a larger field e the result is going to give you something in e but how am i asking you to view the elements of f as living inside e i'm asking you to associate each element of f with its bar in other words with the element a plus p of x inside e so let's do this operation plug in alpha everywhere c and x so what i'm doing is two steps at once here i'm plugging an alpha in everywhere c and x but because this thing is in the big field e and when i do this product it's a product of things inside e i'm then asking you to multiply that by something in f okay that's legit as long as you're viewing f as something living inside e and the way that we've been asked to do that is to form the bar expression okay so this is just how do you evaluate something when you're then asked to multiply it by something in the smaller field answer you view the thing in the smaller field as something in the larger field and we've asked you to do that just by associating with bar so plus let's see a sub n minus one bar alpha to the n minus one plus et cetera et cetera et cetera plus a one bar alpha plus a zero bar that's what p of alpha is and i need to somehow convince you that this coset is the same as the zero coset inside e bar i'm sorry inside e all right well let's see what this is this is well i know what a n bar is it's that's right at this right this is a n bar what's alpha it's x bar x bar to the n plus a n minus one bar i'm just plugging in what alpha is it is x bar plus et cetera et cetera et cetera plus a one bar what's alpha it's x bar a zero bar alpha is x bar i mean that's the particular coset that i've happened to have chosen inside capital e it is one because it exits off as a polynomial oh but wait a minute everything is a bar here and the point is when we did coset multiplication and addition the coset multiplication addition was simply take whatever the corresponding cosets were and multiply them together in other words when we're doing x times x times x times x times x n times and then barring it it's the same as taking a n x to the n and putting a big bar on top so i'm going to write it this way a n minus one x n minus one plus plus a one x plus a zero bar definition of multiplication inside the factory that's not only that's a buyer on all this that's just definition of multiplication of the factory oh but wait a minute that equals by definition a n x n plus a n minus one x n minus one plus did did did plus a one x plus a zero what does bar mean bar means look at the coset of that thing inside the factor ring and here's what the coset i'm sorry here's what the subgroup inside the factor ring was it was ideal generated by px but this looks familiar it is in fact it is p of x by definition so this piece is just p of x so somehow we've run this thing through the original polynomial and what got kicked out well eventually wasn't the polynomial itself but at least was the coset for the polynomial inside the factor ring now here's the claim equals zero plus px and this is what we were sort of looking at at the end of wednesday why is it the case that i can trade this thing in for this well two cosets are equal precisely when if you take the one coset representative and you subtract the other that's easy to do you just get p of x because i'm just subtracting zero whether or not you get something that's inside the subgroup so the question is whether or not p of x minus zero lives inside the collection of multiples of p of x p of x minus zero is just p of x itself is p of x a multiple p of x of course it is it happens to be one times p of x so these are equal precisely because that minus that is inside the subgroup in other words is zero bar or is zero in e and so what this computation shows is that if you plug alpha in everywhere you see an ax in the polynomial p of x here's what you get which by definition just allows you to start playing around with the appropriate cosets which eventually kicks out this expression and once you get this expression the point is that's zero so we've found a polynomial it has coefficients in the bigger field and it has the property that there's some element inside the bigger field with the property that when you plug it in everywhere you see an x into the original polynomial you get zero so there's the goal achieved now if you're still not comfortable with what the heck these fields look like the fields that you get by starting with a polynomial picking out an irreducible factor of that polynomial that's typically not a piece of the game that we'll need to worry about because typically the polynomials that you'll be handed will already be irreducible so you typically don't have to mess around with step one the idea is to simply perform this field get comfortable with viewing the original field is living inside here and then picking out the special element that behaves as the zero of the original polynomial and that special element is always this one example example here's a polynomial x squared plus one in q now let's do it in our bracket x in our bracket x so here's an example that i've mentioned a couple times already are the reels here let's see statement one x squared plus one is irreducible in our bracket x how do i know that well at least it's a degree two polynomial and those are much easier to deal with than degree four or higher so if you have degree two or three to convince me that the polynomials irreducible you simply have to convince me that there's no zeros for this polynomial inside the given field of coefficients and clearly there are not okay and i'd like you to get in this habit since degree is less than or equal to three because degree is two we need only check for zeros check that x squared plus one has no zeros in the reels in script r and that's clear are there any elements in the reels with the property that when you square them and you add one you get zero no all right so now i have an irreducible polynomial in our bracket x so here's what i can do so here's step two if we form the field e our bracket x mod the ideal generated by x squared plus one is a field step two here's step three the reels live inside this new field e i can view the real numbers as living inside here how do i do that i simply take my favorite real number what should we call it how about little a and i associate it with the coset a plus x squared plus one so if somebody says where's the number two living inside here the answer is technically you're viewing it as two bar or you're viewing it as the coset corresponding to the number two where you're adding the subgroup ideal generated by x square plus one and now here's the punch line where is the zero alpha equals this thing x bar in other words x plus quantity x square plus one and the claim is that if we plug that value that expression into the original polynomial that zero should come out so let's check or verify that it works that if we look at let's call this thing half here half of x half of x if we look at half of alpha that we get zero let's see what it gives what's the polynomial the polynomial is one times x square plus one so let's see i first have to view this polynomial is having coefficients inside the larger field that's easy to do it's one bar x plus i'm sorry x square plus one bar viewed as an element of the larger field larger field so you take each of the coefficients they happen to be one on the x square term zero here and one on the constant term you view them as living inside the larger field by associating the real numbers with their corresponding cosets in other words a bar and now what are we supposed to do we're supposed to plug in alpha so i'm sorry i should have had an alpha here so that's what it means to plug in alpha so let's see what we get well we get one bar is one plus x square plus one that's what one bar means what's alpha well written it down there it is explicitly it's x squared plus here's one bar it's coset here's alpha squared well alpha is just this thing so when i square this i get x squared plus its corresponding coset plus one bar which is one plus and now folks we're just going to do coset arithmetic the way you add or multiply cosets is you simply add or multiply the corresponding coset representatives in other words that times that is by definition x squared because it's one times x squared x squared plus one that's what that coset is plus it's added to this coset one plus x squared plus one definition of coset addition add the coset representatives i know what's the final step this is just like it was in the general proof zero why are these two cosets the same that one and that one because if i take that minus that do i get a multiple of x squared plus one i certainly do have sort of preloaded it to work out that way so if it seems like you know it's sort of cheating you'd sort of rig the polynomial to work correctly you do that's exactly what you do equals zero bar so i've just plugged in alpha and what came out was zero so i've just built a field it looks like this and that field has the property that there is that field has the following two properties first of all it has the property that the reels live inside there and secondly it has the property that there is something in there so that when you plug it into this polynomial that zero comes out so you're thinking it's i because i know at least some field one i'm relatively familiar with the complex numbers that actually has an element inside it that when you plug it into this polynomial zero comes out complex i works so if you want to think of this thing as complex i that's not a bad intuition of course you could just as validly think of this as minus i because minus i is also a complex number that has the property that when you square it and add one you get zero and if the question is which one is it the answer is you can choose and in fact just this little bit of publicity for next semester one of the things that will focus on is sort of how many choices there are of what you might identify this element inside this extension field of the reals as being is it i or minus i at this stage it just is it's some symbol that lives inside some field with the property that when you square it and you add one get zero call it i or minus i makes no difference whether what what you call it what your friend calls it corresponds to some choice there's sort of a two choices to be made and it turns out corresponding to this sort of construction there is a group having two elements that will play a role and that will be a big part of the discussion next semester you start with a polynomial that polynomial assuming it's irreducible and that'll be the situation in all the ones that will be of interest assuming the polynomial is irreducible will allow you to form a field inside that field there will be an element that element will behave correctly in other words that element will become a zero of the original polynomial another question is well have you built other elements that work yeah folks if instead of using alpha i would have used minus alpha that's another perfectly good zero for this polynomial as well there's actually two zeros in here and what will happen is among all the zeros that you can build of the original polynomial i've just built not only this one but there's another one as well it's negative the question will be how do you somehow let's say go between the zeros can you get from one of the zeros to the other zero yeah yeah it's multiplied by minus one that's how you get from this one to the other one and if you remember way back one of the original examples of groups that we looked at in here was under multiplication the group consisting of the two elements one and minus one under multiplication with a group and that group is somehow going to play a role here in identifying the things that behave as zeros for this polynomial and that actually fairly can be described as the the main idea or a piece of the main idea that we will spend most of next semester studying how the groups actually get back involved in the identification of these zeros okay now for a homework problem and i apologize uh young lead came to my office before class and asked about this on the website folks i didn't give a full picture of what one of the homework problems was so when i assigned this uh this problem last monday i think i gave you a full description in class but on the website the full description didn't appear it turns out in this particular case turns out that this field if you look at rx mod x squared plus one that field is isomorphic to the complex numbers and what i asked you to do for homework is convince me that it is by actually writing down the isomorphism so show this this is one of the turn in problems and i'll give you a hint you somehow have to write down a function from well from one to the other it turns out it's easiest to write down a function from here to here and you have to convince me that the function that you write down has the appropriate two properties to be a ring homomorphism that phi of a plus b is phi of a plus b of b and that phi of a times b is phi of a times phi of b and then you have to convince me that the function is one to one and on to so let me tell you what the function is i'll give you a little bit more phi from the complex numbers to this thing mod x squared plus one by taking this phi of well i know what any complex number looks like where a and b are reals equal to this trade in i for x trade in i for x what the heck does that mean yeah folks that's exactly what we did here the thing that behaves like i is what you get when you look at the coset corresponding to x so trading i in for x is not a bad idea plus this in other words take phi of a plus b i where i is complex negative one or complex square root of negative one traded in for a plus b x notice now it's become a polynomial and look at the corresponding coset here is the function from the complex numbers to the reals convince me that this function is in fact a nice one there'll be really only one hard thing to do and what's that that'll be to show that phi of a complex number times another complex number so call your other complex number something like a prime plus b prime times i or something like that when you do the multiplication that this corresponding function actually works as far as being a ring homomorphism the comment about the homework is on the website all i've put here is f bracket x or f of x and so on the website what i should have put is f of x is x squared plus one and i made that change right before class but for any of you that maybe have printed off what's on the website go ahead and add that here is the polynomial that i want you to use x squared okay questions there comments this is what we would call the elegant way of constructing the complex numbers from the real numbers now the complex numbers and the real numbers are actually very well they're they're obviously closely related somehow you can get from the complex numbers or from the real numbers the complex numbers just by throwing in one additional symbol the thing called i and you deem that new symbol to have the property that its square is negative one and then you simply look at all the possible symbols that you can form that include the reels and this thing called i so you throw in all the reels you throw in all the i times reels you know bi so all the impure imaginaries and then you throw in all the sums of those so that allows you to move around in the complex plane and that's the entire complex plane in other words that's what everything in the complex numbers is to find a look like and it turns out that that's built in a in a stylistic way from the reels by doing this particular construction on the other hand look we don't have to start with the reels and we don't have to start with x squared plus one we can start with any field we want in any polynomial that's irreducible inside the polynomials over that field and play the same game so for example here's example two here's a polynomial the polynomial x squared minus two in q bracket x now for what it's worth if i had tried to play this game in our bracket x it would be uninteresting because i already know a zero for this polynomial in our bracket x the square root of two but because the square root of two is not rational this thing has no zeros since x squared minus two has no zeros in the scalar field inside a q y since the square root of two is not rational by a proof that you could give us about two thousand years ago zero x squared minus two is irreducible in q bracket x i'll put in parentheses the degree is less than or equal to three of course there's other proofs to show that this thing is irreducible in q bracket x it can haunt eisenstein's criterion because we're asking for irreducibility over q all you need to do is check that there's some prime that behaves correctly the prime two has a property it doesn't divide that coefficient it does divide zero it does divide two but the square of the prime doesn't divide two so eisenstein's criterion would get you there uh let's see it has no zeros in z either obviously not so okay so it's irreducible so what do we know uh the field e consisting of the cosets of that polynomial is a field is a field that's nice and it contains a zero for the original polynomial and if we look at alpha equal to x bar in other words x plus the polynomials that are multiples of x squared minus two is a zero for f of x inside e in other words when we plug this thing into this polynomial zero will come out let's do the computation again why check that it does i mean i want you to see it's going to be exactly the same analysis that has come up in the general case and it just came up in the specific case that we did over our bracket x plug in alpha everywhere you see an x in the original polynomial in other words do alpha squared minus two of course the rub is you know you have to view the coefficients in the original polynomial is living inside the bigger field so really you should call it two bar and if you want it technically there's a one bar outside here but one bar is the same as one it is just the the unity element is what it's well what's alpha it's this thing so that's x bar squared minus two bar because that's what alpha is x bar oh but wait a minute the bars do the same thing as the original operations properties of coset multiplication the way you add a multiply cosets is the just you do them by combining the coset representatives so you want to do this this could be technically x squared bar minus two bar but then the minus signs but that's zero bar reason what does bar mean bar means just for reminder this is x squared minus two plus it's the definition of bar it's the coset corresponding to this thing as its coset representative and the question is why is that equal to this x squared minus two and the answer is two cosets are the same precisely when you take their coset representatives and you subtract them take that minus that is it in the subgroup yeah it's x squared minus two minus zero which is just x squared minus two so is that minus that inside here of course it is so we plugged in alpha and we got zero wow we always get the same thing yeah you always get the same thing because you preload the thing to to be exactly the the the form of expression that's going to give you a zero inside the factoring the intuition folks is that when you've built this factoring that anything that's a multiple of p of x is zero at least in the factoring because anything that's a multiple of p of x minus zero is going to be a multiple p of x so if you can build something that kicks out a multiple of p of x you've in effect called that zero all right let's do another one example example let's try yeah this one here's a polynomial f of x equal to x squared plus x plus one in z two x question what is it please let's see the the a meaning what i don't keep at it oh i see i see to yeah to i i didn't explicitly remind you that the things inside q were going to be identified with the things inside q bar so i could have done that as step three view the rationals as living inside the factoring by associating each one with the bar so i sort of did that without explicitly telling you that i've done that let two be identified with two bar and that was the only place i had to do it because the other coefficient didn't appear yeah so yeah technically i've done that here and if you'd like to throw in another step we're identifying all the elements of q with elements in here in this factoring inside e by associating with their bars let's do this one let's see the book i think actually gives you the details in this but it'll be i think instructive to knock out some of the details all right so here's the polynomial i'm viewing it as a polynomial in z two x so let's see oh it's a polynomial in z two x it's a polynomial degree less than or equal to three so i can check whether or not this polynomial is irreducible in z two it's simply by checking for zeros if i plug in well there's only two things to check here so it's nice working over finite fields because checking for zeros you just do it by brute force just plug them in see if you get in if i plug zero in one comes out if i plug one in i get one plus one plus one one comes out so in neither of the two possibilities do i get zero so since the degree of f of x is less than or equal to three and f has no zeros no zeros in z two two just check it in z two f is irreducible in z two x now in these finite field examples the question of when you form the corresponding field in which this polynomial has a zero how many elements are in that field comes into play and here's how you answer that question so step two we can form this thing the field e equals z two x coset x squared plus x plus one is a field it's a field let's see it contains the original field the original field here was z two how does it do that i ask you to take any element inside z two we'll call it a and associate it with its coset in other words with a plus x squared plus x plus one so that's where z two lives that this thing out right now what we're going to do is analyze this new field so what is e how many elements in e how big is that field and the answer turns out to be this it will always be if the field that you start with let's call it if capital f is finite and little f of x in capital f of x has degree in and let's say oh then this field the field assuming f of x is irreducible fx mod ideal generated by little f has this many elements tell me how many elements were in the original field f raise that to the nth power if you start with a polynomial of degree n and it happens to be irreducible then the statement is well this is a ring in general and the ring will have this many elements but if f of x happens to be irreducible then this thing is actually field has this many elements and you can systematically write them out here's how you do it you list out all the elements of degree zero in f bracket x in other words you list out all of the elements of the underlying field corresponding coset then you list out all of the elements of degree one and their corresponding cosets and then all the elements of degree two and their corresponding cosets et cetera all the way up to all of the elements of degree n minus one and their corresponding cosets and once you get to here you quit so for this particular field when we list things out what we're going to do is list out the constant terms then we're going to list out the linear terms the polynomials of degree one and then once we get to degree two it turns out because the coset is generated by polynomial of degree two that we will have wound up getting repeats at that point so here in this particular case we have four elements because the number of elements in the underlying field is two number of elements in f is two and the original polynomial has degree two original polynomial has degree two so that just happens to be coincidental that those two numbers are the same we certainly could have a polynomial of degree four that's irreducible we could have a polynomial of degree four where the underlying field has five elements z five something like that but in the end there will always be this many elements okay so let's list them out we'll list them out systematically the first things you list out are the zero degree polynomials in their cosets so plus f of x then you list out oh i haven't done that yet so there they are there's the first two next two elements you list out the degree one polynomials well here is one x and then you list out the other degree one polynomial there's not that many polynomials folks when the underlying field when the coefficients are coming from a very small field when you're asked to build polynomials of degree let's say one here well hey the only coefficient if it's going to be degree one is one if it was zero then it wouldn't be a polynomial degree one it'd be a polynomial of lower degree so there's only one possibility for the coefficient on x here and then you're either going to add it to well what constant term do you want to use there's only two choices for the constant term either zero or one so here are the four elements of this field e now it'll always be the case regardless of what irreducible polynomial you look at that one of the expressions that you write down is exactly the thing that we call alpha in the general proof or the general basic goal it'll look like this specific polynomial plus the ideal generated by up and so when we ask you to crank out a table for this field well it'll always be this particular element somewhere let's call it zero bar and one bar so i've just listed those out zero bar one bar you want to call this x bar that's fine but this is the name that we typically give it we call it alpha so alpha and we can call this x plus one bar or if you want to call it well let's see that's the same as what that's the same as alpha plus one bar and there are the four elements now presumably you have presumably you have both in addition and a multiplication table the addition table i'm not really interested in i mean it is some table but that's the less compelling of the two because you've got a field here what you know is well of course there's going to be zeros everywhere here if you multiply by zero you get zero bar but everything else that appears here let's see what happens when you form a group the non-zero i'm sorry when you form a field the non-zero elements form a group so what happens here when i look at the non-zero elements is i should get the structure of some group under multiplication now let's see if we can do the multiplication well a lot of the multiplication is going to be easy because a lot of it's just multiplying by one so in fact that's one bar that's alpha you want to call this x plus one bar i guess so i mean we started doing that but if in your homework you want to call this alpha plus one bar that's fine it's the same thing that's what this is this element and let's see i'm multiplying by one multiplying by one so most of this table is pretty easy now things get a little bit interesting i have to multiply alpha times alpha so that's alpha squared well alpha times alpha is okay but the problem is i don't see alpha squared here it's got to be somewhere so the question is as cosets what is alpha times alpha well wait a minute alpha times alpha is x bar times x bar in other words it's x squared plus x squared plus x plus one because that's what the corresponding polynomial is that we're using the factor out here so what we're interested in doing is taking this coset which is alpha squared and finding it somewhere on the list of elements all right so what i need to do is figure out what this thing is so that we get that's equal to this in other words i need to write down something here so that when i take x squared minus that thing i get a multiple of that and there's no real good way to do it in general but when we've got elements from z two i can rig things this is x plus one and i'll tell you why if i take x squared minus x plus one well folks when you're working in z two minus is the same as plus so minus x is the same as plus x minus one is the same as plus one if i do that minus that i'm going to get x squared minus plus x minus plus one which is certainly in here so the point is this thing is this so when i do this product i get the same as x plus one how did i get this uh just c to the pants i mean just look around we know it has to be one of the cosets in fact if you want you can maybe do it by by process of elimination but the point is you'll be able to rig some coset representative so that that minus that is inside here and it happens in this case to be x plus one bar now if you want you can use structure of a field to complete the table without actually doing any additional products because if we have a field we know the nonzero elements form a group and we know in any group table each row and column contains each element of the group exactly one so i know that this has to be one bar why because the second row here has to contain each of the nonzero elements of the field i each of the elements of the group exactly once and then i can sort of work you know across this way and see what else so i need here and alpha here if you want to check that your location is on target go ahead and actually knock out that times that in fact maybe that's instructive here look if i do x plus one bar times x plus one bar let's see what i get well i get x squared plus let's see oh what is this going to be oh x square plus but wait i'm working in z2 so that's zero in z2 that's nice so let's see x squared oh but wait a minute what did we just find out about x squared x squared is let's see x squared is alpha right that turned out to be x plus one all right as x bar squared plus one bar and i have to figure out what x bar squared is that's what this is having to be calling it alpha and that's fine x bar squared turned out to be x plus one so this is x plus two bar but two is zero in z2 which is what we call out okay so now we have this way of somehow producing fields that's really nice remark we've just written down a ring that has four elements this ring is manifestly not z4 all right because z4 is certainly not a field it's got zero divisors in it we've just written down a ring with four elements that's not a ring that you've seen before but is an elegant way to build a field that has four elements so look with this in mind that the number of elements inside a factoring of this form is tell me how many elements were in the underlying field how many elements you're allowed to use as coefficients tell me what the degree of the polynomial is if you form the corresponding coset ring you get a ring that has this many elements in it number of elements raised to a power and so what that allows you to do is as long as you're able to find irreducible polynomials having specified degrees it allows you the opportunity to build many different fields having in some sense a prescribed number of elements in it so for example if i wanted you to build a field that had 125 elements well all i need to do then is find a field of this form where the number of elements in f in the coefficients is five and the degree of the polynomial that i'm using is three because then the corresponding field that i'd form would have five cubed elements and i could presumably write it out so here's a recipe a recipe to form a field having having p to the n elements where p is prime step one start with the field f equals zp well at least that's a field when p is prime that's good step two find some polynomial let's call it little f of x inside zp x in other words inside f bracket x which has two properties degree of f equals n degree of f of x equals n and secondly is irreducible in z bracket x in zp x find a polynomial that does that then step three simply form the field here's the field let's call it e form a field e e is the field z plus x mod this ideal generator by that polynomial let's see that remark says that there will be number of elements in the field that's p elements raised to whatever the degree of the polynomial is number of elements in it so if i can do this then i'll be able to build a field having a prescribed number of elements you want to field with eight elements eight is two cubed two is prime obviously so all i need to do is find a polynomial with coefficients in z two x having degree three and being irreducible you've actually produced some of those so if you want to field with eight elements just play the game that way you want to field with sixteen elements find a polynomial with coefficients in z two having degree four that's irreducible so here's the recipe of course you might be thinking well how do i know there is one for instance if i challenge you to find a field that has 32 elements in it you now sort of know where to look 32 is two to the fifth so i'd like to use coefficients from z two to his prime that's good and what i somehow need to do is find a polynomial degree five that's irreducible in z two x if i can do that then i'm home free but there's no guarantees that there's one out there if you list out all the possible degree five polynomials some of them are not going to be irreducible like x to the fifth is not irreducible it's x times x to the fourth and x to the fifth plus one is not irreducible in z two x because you plug in one you get one to the fifth plus one which is zero so that's not irreducible you got to keep trying there's this really nice theorem out there we might do it next semester that says if you hand me any field that looks like z p and you hand me any integer and bigger than or equal to two that you can actually find an irreducible polynomial of degree n with coefficients are coming from z p x so it turns out regardless of what degree you hand me you're always guaranteed that there is at least one irreducible polynomial of that degree over z p x it's totally not clear but totally works well it totally works i mean it it's it's this really intricate constructive proof i mean it seems like a straightforward question right here given an arbitrary degree can you find an irreducible polynomial of that degree over z something and the proof turns out to not be constructive which is interesting they don't give you a recipe for actually writing it down it's it's one of these existence proofs there has to be one because proof by contradiction if there were new no irreducible polynomials of a certain degree then something else would happen that can't happen uh question if i hand you a specified field like let's say z two and i ask you to build a field of let's say eight elements well you have a pretty good recipe of how to do that if only i could find an irreducible polynomial of degree three with coefficients in z two then i'd be able to play this game and i'd get a field of eight elements well here's one way to do it i'm gonna get a couple of minutes here yeah here's an irreducible polynomial of degree three in z two x so this is irreducible in z two x you prove for homework this is irreducible so you want a field with eight elements i'll tell you how to get one take z two x mod the ideal generated by this particular polynomial you've got a field with eight elements but here's another one this is also irreducible in z two x it's a degree less than or equal to three and it's got no zeros in z two one does it work so here's another way of forming a field having eight elements you take z two x mod the ideal generated by this irreducible polynomial on the surface they look different the thing you're going to call alpha in here has the property that alpha q plus alpha plus one is zero the thing you're going to call out in here has the property that alpha q plus alpha squared plus one is zero so on the surface they look different but it turns out that the two corresponding fields will be isomorphic so it turns out even on the surf even though on the surface they don't look the same the corresponding factor fields fields are isomorphic that one's also not clear but it turns out to be true in fact oh yeah this is good and then i'll leave you with the following two really nice results and this will hopefully wet your appetite for next semester it turns out folks if you and your friend have both written down fields that have the same number of elements then in fact you've written down fields that are isomorphic to each other if you give me a field of having four elements and your friend writes down another field of four elements the two fields after possibly relabeling will wind up actually being the same field so even though you and your friend on the surface can get to this field of eight elements two different ways it turns out you'll actually have arrived at the same place that's first second again to wet your appetite here it will turn out gotta be careful yeah it will turn out the following is true we've just given you a recipe for building fields having a prescribed number of elements if you hand me a prime number and you hand me some power of that prime number here's the recipe for building a field having that many elements the theorem is for any prime and for any integer there is a field that has this many elements in other words for every prime there is some irreducible polynomial of degree n having coefficients in z p x so that you can play this game and thereby give yourself a field having p to the n elements and moreover this is the only way to build fields having a finite number of elements if you hand me a field the same thing you have a field it's got finitely many elements it turns out it has to have exactly this many elements for some prime and some exponent and it turns out as soon as you tell me what the prime and the exponent is you've told me what the field is because there's only one field that looks that way even though there might be many different ways of getting to it so the theory finite fields is really interesting theory finite fields plays the central role in what's called the coding theory course which will happen a year from this spring and that I hope you'll all be around long enough to take and with that that's the end of the semester so um yeah so if you need some of the paperwork here that I handed out last wednesday please feel free to come up and get it if you have the homework that's due wednesday if you have it tonight you know feel free to turn it in and i'll try to kick it back as soon as possible uh if i don't see you before then folks i will