 What we'll do tonight is The guts of section 27. We'll talk more about ideals. We'll talk more about factor rings And we'll get some sort of intuition going as to what these things look like Wednesday will do more game Wednesday will also be faculty course questionnaires and Wednesday will also be the day that I distribute exam 3 and final exam information So you'll definitely want to be here Wednesday to get all that stuff and to fill out the course questionnaires The following Monday We'll have a quiz. It'll be quiz five. I'll tell you about the quiz on Wednesday So Wednesday again is a really crucial day to be here then the Wednesday of the following week That's December 5th. There won't be any class. I'll just be in my office So it'll give you some time maybe to prepare review for the final in exam 3 and then exam 3 slash final Happens Wednesday the 12th from 430 to 7 so for those of you that maybe have to run out after class To go to work or something. Please make sure that you've made arrangements to be able to stick around till 7 All right questions on the overview here I mean I'll give you a whole lot of details and the stuff that I hand out on Wednesday But that's roughly where we'll be for the next Two and a half weeks till the end of the semester. Okay, so since it's been a week Let me remind you where we were where we wound up at the end of a week ago Monday and then Well, and then in effect Show you the basic construction that's going to allow us to get to this thing that I talked about last Monday this basic goal This place where we're headed So just to give you a brief recap of the goal recap is this I'm going to hand you field. We'll call the field e I'm going to hand you polynomial The coefficients of that polynomial are going to be in this field e And what we're interested in doing well, at least give me a polynomial that has some guts to it And the goal is this Try to find a field In which the polynomial has a zero in other words find a field in which there's an element So that when you drop the element in everywhere you see an x that zero comes out Find a field f So that two things occur The field that you start with lives inside the new field. Maybe it's the rational's inside the reels Maybe it's the reels inside the complexes. Maybe it's z2 sitting inside Something that we haven't sort of encountered yet, but we will over the next couple of lectures and secondly There is some element We'll call it alpha in the bigger field For which When you plug alpha into the polynomial zero comes out Feast of alpha of f of x is zero So the The sort of verbiage the The statement in English of the basic goal is take a polynomial find a zero for it somewhere That may already be that the field that you're working in contains a zero for it If I hand you the polynomial x square minus four and ask you to find some number In the rationals that you can plug in and have zero come out easy plug in two or plug in minus two makes no difference But if I hand you something like the field is the reels and I hand you the polynomial x square plus one Find a field for which That polynomial has zero where you got you got to look bigger You got to look maybe to the complexes Can you look at something smaller between the reels and the complexes? Turns out the answer will be no If I hand you the rationals and I hand you the polynomial x squared minus two and ask you to find a zero for it Find some number so that when you plug it in for x at zero comes out. You're thinking well square to two Square to two is not rational So you got to build a bigger field you got to find a larger field that contains the square to two Maybe it's the reels Yeah, maybe it's something smaller like this thing that we called capital s seeing it look like how do we build that new field? All right, so that's the goal goal basic goal All right, here's how method So the reminder is The following definition and then Again, this is something that we did a week ago Monday. I'm just trying to sort of Knock the Thanksgiving cobwebs out here start with the ring r We look at a subset of r Is called an ideal In case two things are true the first thing I can state easily if you look at the subset together with addition that you get a subgroup of r under addition Maybe I should have written this as a Proper inclusion like this so a subset is my deal and secondly this thing that we called the absorption property for every every element n in n And every element r in the ring When you multiply r n you get something back in n and when you multiply n r you get something back in n That's what it means for a subset of a ring to be an ideal So the first one requires that the subset is a subgroup And remember since addition in a ring is assumed to be commutative when we talk about a subgroup of a commutative or an Abelian group. It's automatically a normal subgroup So I could have put This it would have meant the same thing remember that was our notation for a normal subgroup when you Close the thing up here And what we looked at last monday was some examples of rings and corresponding ideals in those rings example Uh any n z inside z is an ideal example If you start with any polynomial you want What do you want to call it g of x or half of x or q of x? I don't really care. Let's call it g of x g of x in f bracket x where f is any field then look at this collection So what you're thinking is take any polynomial and multiply by this fixed one g Just like we're doing here take any integer and multiply That integer times all the other integers and tell me what comes out Take this polynomial multiply by all the other polynomials and tell me what comes out Intuitively folks what i'm asking you to do is look at all of the polynomials that you can factor this particular one out of And i don't care what polynomial you start with a good example to keep in mind is x squared minus two In the situation where f is the rational another good example to keep it minus x squared plus one Where the underlying field is the reals look at all the multiples of okay So it's the same idea is up here and as you'll remember over the last couple weeks i've been trying to You know keep this analogy going between what's going on inside the ring of integers and what's going inside in On inside these rings f bracket x We gave this subset of notation Is an ideal f bracket x just as a side remark because this verbiage came up in one of the homework problems Folks if i hang any ring No back up i hang any group Always inside any group There were two what we called trivial subgroups group consisting of Just the identity element and the subgroup consisting of the entire group Same thing happens inside rings if i hand you any ring commutative or not. I don't really care With unity or not. I don't really care. There are always inside any ring two ideals One ideal is the ideal consisting just of the subset zero And the other is the ideal of the subset consisting of the entire ring r so example in any ring For any r These two subsets the subset consisting of zero and the subset consisting of the entire ring are always ideals And i think he calls This author happens to call both of these trivial ideals In other parts of the literature you'll just call this one the trivial ideal Just okay, so they're always a Proof this one is a non issue. It just is it's a subgroup of itself under addition Of course because it's the entire group does the ring have the absorption property Well, yeah, if you take anything in the ring And you take anything in the ring and you multiply them together you get something in the ring So i mean the absorption property is a non issue for r the absorption property You have to work, you know a tiny bit for zero This certainly is a subgroup of that under addition. That's a non issue The question is whether or not this thing has the absorption property Well, is it the case that you take anything in this set? And anything in the ring and you do the multiplication that you get something back in the set Well, yeah, because remember we proved that anything ring times zero is again zero So zero sort of this this sort of uber absorber, you know just anything you multiply by it's going to suck it back in now Here is a really nice analogy between what's going on in the integers and what's going on in f bracket x It's something that doesn't go on in general rings I'll mention just sort of as a side remark an example of where it doesn't happen But at least for now to play up or to continue playing up this analogy. It turns out that In z Every ideal ideal Is of the form nz for some integer n So if somebody says i'm thinking of an ideal of the integers Well, you may not know exactly what it is, but you know what its form is Its form is all the multiples of some fixed integer Maybe it's all the multiples of three. Maybe it's all the multiples of ten Maybe it's all the multiples of one Which gives you all the integers of course, maybe it's all the multiples of zero Which gives you this one So this statement, you know includes the sort of trivial cases the cases on both end similarly or analogously Or you could probably anticipate what i'm about to write down inside f bracket x where it was a field Every ideal is of the form something that looks like this Pick some polynomial and multiply that particular polynomial times all the other polynomials you get an ideal And that turns out to be all the ideals in f bracket x Every Ideal Is of the form g of x times capital f of x I prefer this notation folks because it's analogous to this notation this notation with the brackets Yeah, I might use it, but I just prefer this because it sort of more directly tells you what's going on You're starting with whatever polynomial you want. You're simply multiplying it times all the other polynomials in f bracket x Is of this form for some g of x In f bracket x so if somebody says i'm thinking of an ideal inside f bracket x Then necessarily there's some special polynomial floating around G of x for which that ideal is precisely all the multiples of that particular polynomial g of x G might change for instance It might be the case that you're looking at q bracket x and you decide to look at all the multiples of the particular polynomial x square minus 2 Or you're looking inside r bracket x where r is the reals and you decide to look at all polynomials of the form Multiples of x square plus one or you're looking in let me throw our third one in the pile here You're looking in z 2 x and you've decided to choose the polynomial x cube plus x plus 1 As g of x and you're looking at all multiples of that particular polynomial. That is an ideal And it turns out those will be special now it is not Coincidental that the three examples that I just sort of spotted out there Happen to be situations where you have an irreducible polynomial Inside the given f bracket x and you decide to look at all multiples of it So x squared minus 2 inside q bracket x look at all multiples of it x square plus 1 inside our bracket x look at all multiples of it x cube plus x plus 1 inside z 2 x look at all multiples of it Those will eventually play special roles Those will allow us to get to this thing that we're calling the basic goal All right questions comments remarks So the hope is by sort of continuing to To spout these many examples you're starting to get the the feel for what we're going to eventually do Well now we have these two pieces No, let me say a little bit more and then we'll go then we'll go from there. So turns out proposition No Okay construction the construction is this folks we're gonna start with A ring so start with r This is a brief recap, but this is going to be sort of the springboard for the guts of tonight's presentation Start with any ring r And let n be No, let me play it up this way Let let's call it Mm-hmm How about h b uh subgroup Of r with plus So what I want you to do is just focus on the added instruction now forget the multiplication for a minute Just look r with plus is an abuting group Take your favorite subgroup you want to take the multiples of two inside z. That's fine all right, so here's the point then Necessarily h is normal In r Reason because addition is assumed to be commutative inside a ring Plus is commutative. So as soon as you have a subgroup it's necessarily normal. So we can form The factor group All right, that was somewhat stomach turning for a while, but then you started got used to it and All right, I know what these factor groups look like these coset groups. You just sort of Write down all the cosets and then Add them as if you're just you know adding all the cosets together and get another coset or you can just add coset representatives. That's okay So here's what they look like cosets look like Look like I don't know r plus h maybe r1 plus h r2 plus h r3 plus h Et cetera How many are there? I don't know it depends if you start with the integers and you let the subgroup be 4 z Then there's four of them if you start with the integers and you let the subgroup be zero then there's infinitely many of them If there's many cosets it turns out if you start with q bracket x And you look at the subgroup consisting of all the multiples of x squared minus two under addition There's infinitely many different cosets. It's okay But here's the point big question question can We define a multiplication multiplication on these cosets on the collection Of cosets and notice folks the cosets are i'm going to refer to them as the additive cosets It's the cosets that we're forming By looking at the cosets under addition And that's legit because the addition is commutative and we know how to form cosets for any normal subgroup So that This set r slash h Actually becomes a ring becomes A ring that's the goal Hmm And the answer turns out to be this if all you've told me is that you've picked out a subgroup Of the ring under addition And you form the factor group under addition And then you simply try to multiply cosets together The result in general is going to be a disaster and it's going to be the same sort of disaster as we encountered when we were trying to decide when Uh the operation inside the factor groups was well defined remember what did we need we needed something extra Well when we tried to to make the Operation on cosets well defined we needed that the subgroup was normal All right, it turns out If you want to try to define a multiplication on these additive cosets in general, it's not going to be well defined So it turns out in general Well things crash Just like they did in groups unless you have an extra condition in general if we try To define a multiplication A multiplication on cosets well You know if there's any justice here's what you do r1 plus h there's a coset r2 plus h I mean what do you suspect a good Choice for a definition of multiplication would be just multiply r1 times r2 which makes sense inside the ring and look at that coset r1 r2 plus h Good idea. I mean this makes sense because we're living inside a ring Okay, the issue is that might not be well defined Uh this might Not be well defined. I'm not going to show you the examples, but they are easier to write down But and here's the theorem theorem if the subgroup that you start with Happens to have this extra property this absorption property So we put it together in one word if the subgroup that you're starting with inside the ring Is actually an ideal then the punchline is this multiplication on the additive cosets is actually well defined And in fact now that you've got an addition coset addition and a multiplication that's well defined It turns out that makes the additive cosets into a ring if n is an ideal of r In other words if we have this extra hypothesis on n that it has this absorption property then this Collection of cosets together with coset addition and the multiplication defined above Given above just multiply as you'd hope take the two cosets multiply them together. It's well defined is a ring We want to call it the factor ring instead of the factor group Now i'm going to leave out the proof the proof is i mean it's Omitted or left to the reader or whatever appropriate phrases the point though is We can now build new rings from existing rings and ideals just form the factor The thing well that looks mysterious it turns out folks. It's not mysterious at all Because you're totally familiar with these constructions at least in the case of the integers example Well take z There's my ring take an ideal I don't know how about 4z 4z All right now. What are we going to do? We're going to form the factor ring Well 4z is an ideal so that makes sense. So we're going to look at z slash 4z The point is folks. We've already investigated what this thing looks like as A group under addition and we wrote down the table of that thing many times That's just z sub four of the cyclic group having four elements So is well known It's just z four it's Just z four cyclic group of order four Remember the operation here's plus But now the point is because 4z is an ideal it turns out we can actually do multiplication here and Well, it shouldn't be any surprise and the multiplication that turns out to be multiplication Given by this definition just take the coset representatives and multiply them together Is nothing more than multiplication mod four. So here is the table For multiplication in the factor ring Looks like this Well, let's see. What are the elements in the factoring look like remember they could be written as These cosets. This is zero plus four z one plus four z two plus four z three plus four z and What's the multiplication the multiplication is just multiply the two representatives together And write down its coset So zero. Well, so that's easy. Look folks zero times anything is just zero So the first row is just zero That's easy And one time zero zero and two times zero zero and three times zero zero The fact that this is also seven bar. This is four bar. It doesn't make any difference You can use whatever representative you want multiply together and just tell me what that coset is All right one times one is one So look, this is a this is a perfect lecture to Run this video by because i'm about to do multiplication here. Why one Two times one times one is one two times one is two three times one is three. That's good One times two is two new to anybody yet Let's keep going until we mess things up one times three is three. All right now. I get to mess things up two times two is zero Because two times two is four four is the same coset as zero two times three is six, but six bars the same as two bar because Two is congruent to six mod four three times two is six So i'd write down six bar, but six bars the same as two bar And three times three is nine Which is the same as One How do i do there? Three times three is nine, but we're doing arithmetic mod four And nine is one mod four So there's the multiplication table questions Now all we're doing is multiplication mod four. So you may have seen this maybe computer science courses don't like that star mod four multiplication mod four But putting the bar on there is really just coset representatives questions there big example Is this r is q bracket x So polynomials with coefficients and the rationals the ideal i'm interested in is this one All multiples of the particular polynomial x squared minus two Well, it turns out folks that this ring well will turn out to be a Extremely important piece of this big puzzle that we're trying to solve this basic goal The slightly unfortunate thing about that ring is i can't just write out what its table is because it happens to contain infinitely many elements Okay, but here's a point i want to make about it note Arm on n at least has infinitely many elements And before the Thanksgiving break i wrote out some of those you know one plus n or one bar two bar three bar Now remember what it means to say that two expressions are equal in The coset group in the factor group What does it mean to say two cosets are the same it means if you compare the two coset representatives If you subtract the two that you get something that's in the subgroup That's what it means for two cosets to be equal so that for example when we said six bar was the same as two bar The reason that's the case is because six minus two This thing we called ab inverse right, but we're using additive notation here So a minus b is in the subgroup while six minus two is four which is in the subgroup Nine minus one is eight which is in the subgroup So here's the point note That if we take Any constant term constant means any rational or Any linear term that these The cosets for these terms such terms Must be different The statement is essentially this If I look at one plus n Should we call it one bar? Yeah, I'd prefer to do that. It's a whole lot easier to manage here That's a particular element of this coset group because I've taken something in the ring One it's a polynomial of degree zero. That's fine. And I've looked at the coset that it generates So I've added it to everything in here Is not equal to this which is not equal to Oops bad notation, sorry Apples and oranges Three plus n That's a lot silly. How could they be the same? Well, I mean think about over here when n was four z You know one bar was the same as five bar, which was the same as nine bar The point is that can't happen here. The reason it can't happen is How do you determine when two cosets are the same? You take the two representatives You subtract them and you ask whether or not you get something inside the subgroup You ask whether or not in this case you get some multiple of x squared minus two Folks if you've got something inside this subgroup There's only two choices. One is you're either looking at zero That's a perfectly good multiple of x squared minus two is x squared minus two times zero But if you look at something that's not zero inside here, it has to have degree at least two Because you're multiplying by some polynomial of degree at least two So here's the question. Can I hand you two constant terms and subtract them? And get a polynomial that has degree at least two, of course not. I get polynomials of degree zero Or hey, you know, I mean one bar is the same as one bar But that's only because you're looking at the same element twice if I take two minus one I get a polynomial of degree zero. It's not in there because nonzero elements of this thing are degree at least two if I take x plus n That's a perfectly good element inside r slash ny because well, I've taken some element of the ring I've looked at the coset that it generates technically what I've asked you to do is add x to everything in here And the claim is that this coset and for instance that coset can't be the same Why because when you subtract x minus one You can't get a multiple of x squared minus two because any multiple of x squared minus two has to degree at least two And x minus one is a polynomial of degree one So as long as you're mucking around inside either constant terms or linear terms There's no way that any two such cosets can be equal to each other That's going to be an important property that we will use later on In effect, what it means is this you know when we're working in z mod 4z the number one sort of behaves like the number one Inside the integers. It's just If you're working inside z mod 4z and you add one to itself four times you get zero Well, that certainly doesn't happen in the integers the point over here is going to be this if I hand you a rational number It's going to behave inside the rational numbers the same way that it behaves inside this factoring You know three plus three is six Plus three is nine plus three is 12 and none of those things are going to be the same as zero inside this factor Similarly x plus x is two x plus three x x is going to behave like x always behaves It's not going to somehow magically become zero Because constant terms and linear terms somehow can't ever trip over each other or somehow become the same as each other Because you can't ever take constant or linear terms and subtract them and get a multiple of x square minus two because x square minus two has to be two All right But and here is the key idea and I showed you this Before the break and I'll show it to you again because If you can sort of grasp what's going on here, then in effect you'll have seen the key To achieving this basic goal, which is this note if we look at this in Qx mod So inside this ring That's the same ring that we've been looking at here. I just wrote it out In in well in this notation just to start getting used to it look at I'm going to call this element alpha. It's the element that you get by looking at x bar I'll remind you what that means in other words take The specific element x Take the specific element x. Well, that's certainly an element in here And tell me what coset it generates inside this factor ring inside this coset ring in other words take x and add it to everything in there That's a perfectly good element in here. Now. Here's what I want you to do Take alpha and inside this ring multiply it times itself Okay, well, hey, we know what the definition of multiplication is inside these factor rings You just multiply the coset representative together. Well, I've asked you to take the same element twice In other words use the same coset representative twice. So this is by definition x times x plus Non-issue just multiply the coset representatives together. But here is the key This is the same coset as this coset. Why how do you decide when two cosets are the same you take This coset representative you subtract that one from it and you ask whether or not you get something inside the subgroup If I take x squared minus two do I get a multiple of x squared minus two? Yeah, I get x squared minus two times one So that's no big deal So the point is this folks what I've found inside this factor ring Is some element called alpha I wrote it down. There it is. It's not some mysterious thing. It's that thing With the property that when you multiply it times itself You get the same thing as two In other words you get an element whose square is two You might say well, and it's not really two. It's like, you know, it's two plus that it's the coset of two Yeah, but you know two walks and talks Inside the coset ring the same way it walks and talks inside a cube It doesn't become zero if you add it to itself or anything like that. That was these comments over here So somehow we've built a coset ring With the property that there is an element in there So that when you square it and you subtract two you get zero That's the same thing as saying that when you square it you get two So it would not be unreasonable to call this a square root of two Which I've built without this symbol at all I've just built by computing some sort of ring Inside which there's an element that behaves like the square root of two So here eventually is what our basic goal will lead us to I've just taken a specific polynomial x squared minus two That polynomial is irreducible inside cube bracket x y well, we proved that in class last monday You could use eisenstein's criterion if you want or hey, it's a polynomial of degree two It's got no zeros and q so we can use that other result however where you want to get that And what we've done is we've then built this ring And inside this ring there is something that walks and talks like a zero of this polynomial There is something with the property that when you square it and subtract two you get zero that thing happens to look like this So we've built a ring that has a zero for x squared minus two Does this ring contain the rationals? sort of I mean, you know, it's sort of The cosets of the rationals, but the cosets of the rationals behave like the rationals do Write down the details Wednesday But then the punchline is but i've produced a thing I've concretely written down a thing that behaves correctly whose square is two So the only piece that's missing is if we're going to try to use this as this System that contains the square to two We have to try to show that this thing is a field And this is what's going to act as our what we call field extension of the rational This is going to be the larger field inside which the rational sit and inside which we've found a zero for the polynomial x square minus two And that I think is the most daunting thing for For students to sort of grab their heads around the first time you see these factor rings. They seem like a mass And you're thinking well, how the heck am I viewing the rationals is sitting inside here? We'll play with that more But what's more important for you to see at this moment is that there is something Inside this ring that behaves like the square to whose square is two All right, so it will take I don't know and we'll take 15 or 20 minutes more tonight to start down that road and then we'll Basically be able to finish it up by wednesday this ring This factor ring turns out to be a field be a field Which and I'll use the words in quotes for now contains the rationals and Contains zero For the polynomial x squared minus two It contains some element so that when you plug it in for x Here's the element it's called alpha and you square it and you subtract two you get zero zero bar It's zero so goal Show that this thing that qx mod x squared minus two qx is actually A field we can do that We won't write down all the details, but we'll write down most of them What i'm going to be more interested in doing folks is playing up this Analogy between what's going on in the integers and what's going on in f bracket x and when we saw recall That let's see the ring z sub n Is a field if and only if n is what prime All right, so we've analyzed what these z sub p rings look like We proved this what two weeks ago. Yeah two weeks ago. Let's see it turns out Z sub n is the same as I could use the word isomorphic, but I'd prefer not to here is the same as the ring z Slash nz When I for example wrote down the multiplication inside z slash four z and I used The coset notation zero bar one bar two bar three bar, etc The multiplication folks was nothing more than multiplication mod well mod four there, but in general mod n So if you'd prefer to think of the ring z sub n as the numbers zero through n minus one With addition of multiplication mod n, that's fine Or if you'd like to view the ring z sub n as the cosets zero bar one bar two bar up through m minus one bar Just put bars on everything and have the addition and the multiplication being inside z mod n z You're getting the same sets with the same operations So I should be using the word isomorphic just because in one copy We're using cosets and the other copy. We're just using integers from zero to n minus one But you can view these as the same so the point is this When is the factor ring of the integers by some ideal a field answer Precisely when the number that you're using to generate the ideal is prime Now what did we do in our analogy between the integers and the polynomial rings the things that played the role of primes Are precisely the irreducible polynomials and so eventually what we're going to get to and we'll do it in the next I don't know 10 or 15 minutes and so turns out So restated restated We get z slash n z is a field Precisely when if and only if n is prime So the primes correspond to those integers that you use to generate ideals That when you look at the factor rings the corresponding factor ring is a field so I mean it's sort of interesting when you Hand me the integers It turns out there's three possibilities that can happen Hand me the integers then hand me an ideal Well situation one is that you've handed me an ideal That's generated by some prime number in other words that consist of all multiples of some prime two or three or five I don't care which one if you then form the factor ring what you've written down is the field z sub p If you give me a knot prime But don't give me zero yet. Give me a knot prime like six if you form z six Or in our new notation z slash six z If you form this factor ring Then what you get is something that isn't A field heck it's not even an integral domain. It's got zero divisors Inside z sub six Or if you want to think of it as z slash six z Take two bar times three bar You get six bar Which in z slash six z is zero So you've taken two non-zero things two bar and three bar and so you get zero divisors So things sort of crap out there the third possibility And it's one that we should take into account is The ideal that you've handed me is the zero ideal If you look at z slash zero z you look at z slash zero it turns out that's just z again Yeah, okay z bar, but it turns out you get something that's isomorphic to z Which is sort of in the middle. It's not a field, but it's also, you know, it doesn't have any zero divisors So that's sort of the third rail here You get either a field or you get not not an integral domain Or if you happen to mod out zero you get an integral to man It's not a field, but you just get the original ring back The punchline is the exact same thing happens in these rings f bracket x. So it turns out similarly The following is true in f bracket x Where f is a field Uh If you look at the factor ring ring f bracket x mod Let's call it g of x times f bracket x Is a field If and only if Well, if there's an analogous statement to what happens in the integers If and only if the thing that you're using For all the multiples of that fixed polynomial should be a type of thing that plays the role of prime integer And that was the notion of an irreducible polynomial g of x Is irreducible In f and we'll spend some time proving this We will prove this over the next 10 or 15 minutes of tonight and into wednesday prove this But this sort of again plays up this analogy between the integers and f bracket x We can go about producing or finding fields In the following way instead of finding a prime number like we did in the integers and go ahead and mod out We instead hunt around inside f bracket x find an irreducible polynomial Look at the ideal consisting of all the multiples of that irreducible polynomial and then forming the factor ring That will turn out to be a field So oh this factor ring Will turn out to be a field Precisely from This result why because in the particular case we looked at in q bracket x the polynomial that I handed you x squared minus two Is irreducible in q bracket x? And therefore when we form this factor ring in fact that field so how do we get here how to do this? How to do that And the answer is we need to talk about certain properties of fields. I mean we're right here To We need to yeah, we need to describe some ideals We describe certain properties that some I'll think of them as the nice or the special or the prime ideals special Ideals have And here they are definition Some ideals will have this property and some won't We're going to assume for the remainder of the semester that r is a commutative ring so r commutative It turns out you can talk about these ideas in the setting of non commutative rings as well, but We won't be interested in that and the technical details to start exponentially increasing So I don't want to go there with this then an ideal an ideal n of r An ideal n inside r is called prime Notice i'm not talking about prime numbers per se. I'm talking about An ideal being prime or not obviously in the setting of the integers it will be related to that idea In case the following is true for every pair a and b in r If a b is in n then either At least one of or b is in n So here's what it means for an ideal to be prime Remember by definition it's an ideal so it already has the absorption property So it means that if you take anything in the ring And anything in the ideal and you multiply them You're guaranteed to get something in the ideal that's already given piece of information by definition But what we're now requiring is something more than that the requirement is If you happen to multiply two things together and the result is in the ideal That the only way that that can happen is if one or the other of the things that you started with was in the ideal to begin with That's what it means to be a prime ideal And the second definition is this we call the ideal maximal in case The following is true First of all the ideal is not the entire ring And the ideal has the following property If j is any ideal of r With this containment being true n contained in j Which in turn is contained in r then the only way that can happen is if n is j Or n is r Here's what in english it means for an ideal to be a maximal ideal It means in effect that the ideal is as big as possible In the sense that well you have the ideal By definition, we're going to assume the ideal isn't the trivial ideal consisting of the whole ring So take an ideal that actually has some guts to it meaning that it's not all of r Maximal means that you can't slide any more ideals in between the given one in the entire ring I phrased it in the contra positive language Maximal means if you happen to have an ideal that sandwiched between it and the entire ring Then the ideal you're looking at is either it or the entire ring Okay, you can't sandwich anywhere in between question This should be Then oh j equals r. Yeah, thank you or j equals r Thank you. Yeah, so if you have an ideal We're calling it in then this ideal is maximal if it's as big as possible in the sense that there's no We'll call them proper ideals that you can sandwich between it and the entire ring Let me well, we'll get out here a little bit early today Let me at least Give you a couple of examples of ideals that aren't prime or maximal Just for contrast And then we'll go ahead and write down a couple of examples of ideals that are prime or maximal inside the integers And then we'll call it a night. So example. Let's look inside z in z Oh Tuesday, let's look at uh six z is not prime Now notice i'm using this adjective prime Corresponding to an ideal not to an integer Why What does it mean for an ideal to be prime prime means every time you have a product that lands in the ideal The only way that can happen is if one or the other of the two things that you've used was in the ideal to begin with It's not prime because I can find two things Both of which are outside the ideal folks neither of these are multiples of six That when i'm multiplying them together give something that's in the ideal So there's a not prime ideal in fact in general in general If n is composite meaning it's the product of two integers less than itself then And z is not prime Now if you're thinking well the next statement is if you look at all multiples of a fixed integer and the corresponding ideal is prime Then the integer is prime. That's good intuition, but it's not 100 true. It's like 99 true on the other hand example This one's sort of interesting zero is a prime ideal Of z Well, it's an ideal. That's not an issue question. Does it have this property? Does it have the property that if you take any two things inside the integers And you multiply them together And the product is in the ideal Is it the case that one or the other of the things that you started with had to be in the ideal? Well, yeah If you take inside the integers two integers and you multiply them together and you get something inside this ideal It means you've gotten zero That's the product of the two integers folks The only way you can get zero is the product two integers Is if one or the other of the two integers was zero to begin with in other words if one or the other of the two integers Was already inside this ideal So the proof is reason there are no zero divisors inside z Or the fancy word is z is an integral domain If you have the product two things that give zero then one or the other has to be zero All right now let's finish up what's going on inside the integers turns out And i'm going to haul out a number theory result here. It'll be a straightforward one In fact, it's one that we used like in week three or something like that turns out if p is prime A prime number prime integer Then The ideal pz Is a prime ideal What does it mean to be a prime ideal? So it means if you take two things in the ring and you multiply them together and the result lands inside the ideal The only way that could happen is if one or the other of the two things Was inside the ideal to begin with so suppose you take two integers a and b And suppose that when you compute their product That you get something inside this ideal Then what i want to be able to conclude is one or the other of the two things had to be in the ideal to begin with In other words, what we're supposing is that when you look at the product ab You get some multiple of p I don't know where z is some integer So what does this mean in other words what we're saying is that the product ab is a multiple of the prime p ab is a multiple Of the prime p, but then the point is but since p is prime p is prime If you have the product of two integers is a multiple for prime then Either a Or b Is a multiple of p And that results given a name. This is called euclids one Have a multiple of two integers that turns out to be a multiple of some prime number than one or the other had to Be a multiple of that prime number to begin with and that is the result i e Either a is in pz Or b is in pz. So in fact what we've just done folks if we've identified all of the prime ideals inside z They're almost exactly the ideals generated by prime numbers Where we understand prime numbers to include the negatives of the things that we're used to calling prime numbers But we also have to throw zero into the mix too zero is the prime ideal inside the integers If for no other reason than the integers it turns out to be an integral domain questions comments All right, what I want to do then is I'm going to give you the last homework assignment of the semester. It'll be due It'll be due a week from wednesday. So it'll be the standard homework Scheduled, but I'll have them graded and available For return by the following friday so that you can use it to study for the exam and here it is at home Again last assignment of the semester is this uh in section 27 Problems five through nine 14 18 and 19. I want you to turn in 18 and 19 And then in section 29 Problem 18 a and b I want you to turn that in and two more Give the isomorphism We'll give that word on wednesday from the complex numbers to this factor ring mod x squared plus one times r bracket x So it turns out we will eventually show that that's a field We're on our way to doing that in fact that field turns out to be a field isomorphic to the complex numbers. That's first and secondly mimic Example 29.19 To find a field Field having eight elements And let's see. What else do I want you to do here? Uh, give The multiplication table for the field table x bar b denoted by Alpha as we did here All right, we already at least tonight sort of opened up this idea that inside a factor ring Well, if you look at a factor ring that looks something like this There's the coset generated by x That's this thing and it might actually have certain properties that we can identify like its square might be two It turns out you'll be able to play the same sort of game by looking at an appropriate factor ring of z2x And they show you how to do that in Example 29.9. I want you to turn both of those in as well. Okay, and again, that'll be the last homework assignment Let's see right, so So this will be yeah, this will be due Wednesday December 5th, but since there's no class Just get it to me email Or fax or however you want to get this assignment to me will be fine And then I'll have them great