 Here are some comments first is this in this problem number 12 to show that this Proposed ring is a field Well the the instructions were really the only difficult part is to show that Multiplicative inverses are appropriate, but some of you took that is that that's the only thing you have to show Well, there's some more you have to show in order to show this thing as a field you first have to show It's actually a ring a lot of the work that you have to do to show that it's a ring is Check because these are simply real numbers And we know that the following things are true for real numbers like commutativity of both addition and multiplication like Distributive law of multiplication over addition like Associativity like center center center so a lot of things you just click off you technically haven't shown But there were a few things that you had to show in addition to the multiplicative inverse issue One is for instance you have to show closure under addition of multiplication in other words you had to show that the two operations addition of multiplication are actually binary operations The addition part is I mean it's one step But it's something you got to show that if you take two things of the given form you add them together Not just that you get a real number that that you actually get something of the given form and the interesting one Which is similar to the one that we did in class is you have to convince me that if you take two things of that form and you multiply Me get something of that form That's why I did that one example for you in class where we had a plus b root five or something like that You have to show that the identity is in there That's I mean it's a no-brainer, but it's something you have to know the identity both the additive identity and the multiplicative Identity zero and one are of the correct form zero is zero plus zero times root two one is one plus zero times root two Therefore both zero and one are the correct form the The task to show that multiplicative inverses exist and Are of the appropriate form? Well at least all of you attempted to do that most of you got to the end But a couple of you missed what admittedly doesn't hit you over the head like a sledgehammer But is something you have to take care of in the end if you start with something like this You're going to try to convince me that Inside the set s there is something so that when you multiply it times this you get the multiplicative identity Which is this because one plus zero times square root of two and so what did you find you found that if you took? it You could write it in the correct form just by doing a rationalize the The denominator and you eventually got this and a couple of Comments to be made here and remind me if I get this right a over a squared minus 2b squared or something like that minus I'm going to write it this way plus negative b over a squared minus 2b squared times root 2 how do I do okay? all right Remark one Some of you wrote a minus sign here. Okay. That's technically alright But remember what you're trying to show the goal is to show that this thing Is of the right form and what do the things in s look like the things that are of the right form are a Rational number plus some rational number times the square root of 2 by writing it this way I've at least made it look like something Plus something times the square root of 2 Okay, so it looks like it could possibly be in the right form and then what some of you did was sort of quit I mean you put a minus sign here. I wasn't too irritated But if you're going to tell me this a rational number, I sort of believe you Except what? Got convinced me the denominator is not zero for carnal out If nothing else you haven't started with an arbitrary element in the set You started with an arbitrary Non-zero element in the set so boy, that's useful here at least to convince me the denominators not zero There's another thing that you need to convince me of you need to convince me that if you take Whatever a is and you square it that you don't get the same as if you took B and squared it and multiplied by two which is Sort of non-trivial, but it's certainly something that you've got to you know deal with and the way you deal with it is Can the denominator equals zero answer if the denominator was zero note? a squared minus 2b squared is not equal to zero because if it was If it equals zero then what would we get we'd get a squared is 2b squared Which would mean that? a squared over b squared is 2 Which would mean that the square root of 2 is? Rational square to 2 is not rational So it's actually important that you're starting with something that has quotient not rational it turns out If you look at s equals, you know a plus b times root 5 Looking expressions like we did in class then this observation will work But if you looked at something like you know a plus b times root for Well, then this could be zero of course if you look at a plus b root for you're not really getting anything new So so there was more to be done there then some of you did I Yes, sort of cause I apologize for maybe leaving in the wrong direction thinking that that the only thing you had to do Was show that inverses existed, but there was certainly more to the Task here then simply showing that Multiplicated of inverses are in the set, but there was some stuff that you had to do there All right, let's see Next comment. Oh, yeah in number 37 The one about the units Most all of you did okay there a couple of you fell into the trap of telling me that the inverse of a b is a inverse b inverse B inverse a inverse remember you got to switch the order when you're doing inverses of products, but The thing that I was a little bit more concerned with is some of you said all right if I take a unit I Need to convince you that the inverse of this unit Well is in the set Well, all of you convinced me that the inverse of the unit exists because the original element is a unit But wait a minute If you is in you Then you have to show its inverse doesn't just exist, but actually lives in the set that you're interested in So why is you inverse a unit? You have to show me it has an inverse Does it? Yeah, what is it? You so what Tony said was well and therefore because you is a unit you inverses in the ring You know you haven't told me a lie, but you haven't told me the whole truth yet either so what I don't care that you Inverses in R. I had need to show that you inverses in the set I'm interested in and the reason is you inverses in you because It's inverse Exists and why does it exist because it equals you and how do you show that you take the thing you're interested in? You multiply it by the thing that's proposed to be the inverse do you get one yeah, you do So it's I mean in some sense It's no big deal, but telling me that the inverse exists isn't enough to convince me that you've You're showing that this thing is a group or at least satisfies the third property for being a group Okay next this next one is style point. Oh question. Sorry money Uh-huh. Oh, I'm sorry. Yes, because one because one is the identity element of you Yeah, the implication and so the implication is that you inverse is a unit and the reason it is is because I found something else Happens to be that so that when I combine those two and technically for completeness. I should have written it out this way, too If you're thinking well, wait a minute, that's the equation that you have to write down to convince me that You as a unit Right, but also turns out to be the same equation that you write down to show me you inverse is a unit. All right So let's see next in section 19 number 23 Style comments. I won't write out what some of you did but essentially it looked like this Take an item potent a so a squared is a so a squared minus a is zero so a times a minus one is zero So a equals zero or one Period and then at some other point in that problem you wrote R is an integral domain Maybe you wrote it before you listed out those equations or after you listed out those equations If you wrote down because ours an integral domain or we know ours an integral domain and then you gave me that Information and then you moved on You haven't told me what it is about R being an integral domain that allows you to conclude that a is zero or one So if you're going to write all this stuff out a squared is a that's the given information So a squared minus a is zero. That's just arithmetic in a ring So a times a minus one is zero. That's just the distributive law so a is Zero or well, you know what? I'd really like to see at this stage folks or a minus one is zero because R is an integral domain It's this implication Going from here to here Where you use the fact that the assumed ring is an integral domain and the rings assumed to be an integral domain Because here you've got the product of two things equaling zero and the only way that can happen in an integral domain is if a is zero or The other one a minus one is zero which implies arithmetic That either a is zero or a is one and you're done But if you wait to down here to put this statement in or if you put the statement because ours an integral domain up here You've just sort of thrown the hypothesis in somewhere where it just Doesn't play a role. You have to tell me getting from here to here is where you've used the fact that ours an integral domain right No, oh A quick remark for the three of you that didn't change the instructions to our being an integral domain the original Instruction is something about let R be a division ring or something like that Okay, a couple of you forgot to do that in the two of the three of you that forgot to do that Had a proof that looked like take this and then multiply both sides by the inverse of a That's technically what you want to do assuming a is not zero But what some of you used for the notation for the inverse of a was of that folks in general That's not good notation for the inverse of an element the inverse of an element a general ring We'll always denote by a little minus one on top of it even though in Most of the situations that you knew about coming into this course that was the same as one over a a inverse is just Better notation if nothing else. I mean we've seen a lot of fields where the elements Don't look like that like z7 Makes no sense to talk about one-third inside z7 But we can talk about three inverse because we interpret that as the multiplicative inverse of three in that ring All right final comment on the homework coming back The the one that I talked about in class a week ago Show that in zp you have this freshman's dream a to the p plus b to the p is a plus b to the p Some of you said well, okay Here's what you do you take a plus b to the p and you use because the zp is commutative You crank it out using the binomial theorem and the Pascal's triangle whatever it is you get this and then y'all sit Well, I'm not y'all some of you said well because I can factor a p out of each of those things in the middle Here it is then all these things look like zero and z sub p and so all you get is the insides and the outsides Folks there should have been some red lights flashing if you did that Because nowhere along the way did you use the fact that p is prime? It turns out the only place the only time that you can conclude that the integer n divides all the binomial Coefficients in the nth row of Pascal's triangle except for the ones on the two ends when p is a prime So that's where things sort of you know sort of the head bells go off or whistles go off But that's one reason why I wanted to give you that alternate method of getting there Which was used for mausoleum theorem which we know holds for p prime Okay questions All right. Here is what we're up to tonight What we looked at starting last Wednesday is sort of an in-depth analysis of polynomial rings and In particular what we're going to do is focus on polynomial rings where the coefficients are always assumed to come from a field so That's going to be the standing hypothesis between now and the rest of the or the end of the semester that the coefficients that we're looking at the The the polynomials coming from our field now in general It's not going to matter what field it is. It might be the reals. It might be the complexes It might be z2. It might be z7 We know a lot of fields now that aren't simply the complex numbers of the real numbers heck the field might be this Field s equals, you know the things that look like a plus b root 2 or a and b rational So there's lots of choices for what the Coefficients can look like What we saw last Wednesday was a couple of things that we get by assuming that the coefficients come from a field first of all Because any fields in integral domain We can at least conclude that f bracket x is always an integral domain Let me show that if the coefficients are an integral domain Okay, so at least f bracket x is an integral domain But it's also the case that even though we're starting with f as a field f bracket x is never a field There are always things in there that don't have multiplicative inverses the good example to keep in mind is just the thing called x There's nothing you multiply times x I'll rephrase that there's no polynomial There's nothing in f bracket x that you can multiply times x to get one because the degrees of products Become sums of degrees On the other hand though f bracket x because f is a field behaves a lot like the integers And I tried to play that Analogy up on Wednesday We'll continue the analogy maybe by the end of today, but definitely on Wednesday with some more sort of Similarities between these two structures, so integral domain So Z not a field So Z not a field Has a division algorithm You just wrote out for me in quiz four here There is some notion of what it means to take two polynomials divide one into the other in such a way that you write the first one as some Multiple of the second one plus some remainder where the thing that you've left over somehow smaller than the thing that you've divided by When you do that in the integer Smallness is simply meant to measure by the size of the integer when you do that for a field Smallness is measured by the degree of the polynomial that you're leaving as a remainder And what we're going to see is the existence of that sort of division algorithm inside f bracket x is going to be key later on okay Now we get to an idea that I mentioned briefly three lectures ago It's an idea that came up in the homework that you'll turn in on Wednesday It is Well, it's an idea that we saw at least in the context of groups Corresponding to what it means to talk about a function between two groups that preserves the structure That's what we call the homomorphism the same sort of thing Can be investigated in the context of rings and we talked about what you can if you want just call them homomorphisms But for emphasis sometimes when we call them ring homomorphisms The general definition was simply a function from a ring to another ring that preserves both the addition Equals phi of r plus phi of r prime and Preserves the multiplication phi of r times our prompt is phi of r times phi of r prime For all r and r prime that's what ring homomorphism means at this level folks when you talk about a Ring homomorphism if you have a ring homomorphism that preserves the additive structure here All this is saying is that the given function is a group homomorphism Because remember the plus operation in a ring gives you a group happens to be in a billion group This second piece though is simply a statement that whatever your function is preserves the multiplication or respects the multiplication Because you remember in general the multiplication in a ring doesn't form a group Heck even in the nicest situation where you have a field the multiplication still doesn't form a group Maybe the non-zero elements form a group on occasion So the second line really isn't the statement that phi is a group homomorphism with respect to the multiplication because the underlying But that's fine You can still talk about is it the case that phi of a product is a product of the two fees and the answer turns out to be That's what's required and we gave one example of a ring homomorphism a quick one the function that takes Z to z sub n and simply takes the mod n or the remainder on division by n the most important example of ring homomorphism so for us Yeah for us Are what we call it evaluation ring homomorphism so here is a sort of cruel example of Ring homomorphism and the example is this What I'm going to ask you to do is start with A field e let e be a field and Let f be another field be a field that contains e Which contains Now what I mean by the word contains I'll make more formal later on but the intuition is for example the complex numbers are a field the real numbers are a field and Should I call your intuition or what you've been led to believe up until this course? Is that the real sit inside the complex numbers? Well, that's Not That's sort of true. It's not exactly true though Why because I mean the complex numbers are at least the way you come to the table in this course is the complex numbers The things that look like a plus b times root i Where a and b are real right? That's what the complex numbers are so are the real numbers inside there? Well sort of if you give me a real number what you're doing technically is Associating it with the complex number a plus zero times root i So there's a little bit of a rub in actually viewing the reals as sitting inside the complexes But for now that's what we're gonna have you do okay, so think of that example or maybe think of the Rational numbers as sitting inside the real numbers. Okay, that's good But again, there's still a little bit of a rub because the rational numbers really look like equivalence classes of quotients of integers Real numbers look like decimals or something Okay, but I mean If somebody says the reals are inside the complexes you're not in your head and saying move on you know tell me what's next So we've got here And here's the situation We're gonna take Some element in I'll call the bigger field So I'm thinking of this thing that I'm gonna call a subfield subfield Sitting inside and I should put quotes on this sort of sitting inside a bigger field Reals inside the complexes rations inside the reals etc I'm gonna ask you to play the following game pick any element you want in the bigger field. Let's call it alpha If you want to keep a good example in mind, I'll offer two either one is good here You have the real sitting inside the complexes Pick something in the complexes. I don't care what you pick the interesting situation Will be pick something in the complexes that aren't in the reals. You don't have to in what we're about to do But that's like pick I That's not real think of that as alpha or another good example is think of e as the rationals think of f as the reals Pick something in the reals. It's not the rationals like the square of two something like that Then here's the definition we define a function Fee with an alpha in the lower right hand corner. It's a function that inputs Polynomials having coefficients in the smaller field and what it spits out Aren't polynomials, but actually elements of the larger field It's a little bit weird to get your sort of arms around first time you see this what I'm going to ask you to do is take Whatever polynomial you want, but make sure you've built those polynomials with coefficients coming from the subfield from the smaller field So if you're thinking of e as the reals and f is the complexes give me any real valued polynomial and What I'm going to ask you to do is take that polynomial and in effect everywhere you see an x in that polynomial plug in alpha Fee sub alpha of something in here. Let's call it a zero plus a one x plus Dit dit dit plus a and x to the end Equals just plug in alpha everywhere see an x a zero plus a one alpha plus dit dit dit plus a and alpha to the end It's the plug in alpha everywhere you see an x function Now the point is this folks At least in the setup that I've described here this thing Makes sense. Why because the things a zero a one through a n are Taken from e because I've asked you to plug in polynomials with coefficients coming from e When I do something like this alpha a one times alpha This is something in the field e Times something that's come from the field f But the point is that that product that multiplication makes sense because if you've given me something in e Presumably it's also an element of f so that this product is happening inside the field out The idea might be if this is a real number and that's complex I It makes sense to talk about a real number times I Get a complex number it makes us to talk about to I it's the complex number to I Even though that thing didn't come from the complex numbers it came from the reels this thing did and you can multiply me get something in the complex So this is guaranteed to spit out something enough and the punchline is I'm not gonna prove it for you It's just gutting it out proposition Proposition these things fees of alpha is ring homomorphism the proof is nothing more than I'll just say in words if I hand you two polynomials and you add the two polynomials together In other words, you just add the appropriate coefficients, right? And then you drop some number in for acts Is that the same as having taken the two polynomials individually drop that number in for acts and then added them together? I'm sure you get the same answer Whether you want to call the sum of the polynomials first and then drop in the element or whether you want to drop in the Element and add them together. No big deal. It's the same thing with multiplication if I take two polynomials and multiply them Just you know, all right. It's a nastier process is bigger. You foil the whole thing out and then drop something in for acts It's the same thing as having dropped that thing in for acts before you multiply and they just multiply everything and you drag the excellent So the proof is just multiplied out proof just tedious. I'll call it. Just crank it out you do a specific example of One of these ring homomorphisms. This is called the evaluation at alpha homomorphism At alpha homomorphism and let's do an example for example Phi sub i from r bracket x to the complex numbers So the point is I start with one field that sits inside another We understand how the real sit inside the complexes There's a little subtlety that if somebody hands you a real number like two that technically they're viewing it as a complex number By writing it as two plus zero i not a problem if I do for example phi sub i of I don't know three plus four x minus nine x squared The definition is you take this thing you just plug it in everywhere you see an x is three plus four i Minus nine i squared. That's an element of the complex numbers now this thing happens to Quote and quote simplify a little bit because this is three plus four i let's see Minus nine times i squared happens to be minus one So this happens to simplify as 12 plus four i But that's what you get if you take this polynomial and you plug it into this Homomorphism this particular homomorphism now there's nothing special about i could have picked any element in the complex numbers I wanted for each element in the big field You get a different homomorphism you get the phi sub i homomorphism the phi sub two i homomorphism the phi sub six i Plus seven homomorphism It technically folks you also get the phi sub I don't know phi sub two homomorphism the homomorphism that asks you to drop a two in everywhere you see an x But intuitively that's not going to be very interesting if you hand me something that's already in here the intuition is What's going to get spit out is nothing bigger than the field that you're starting with that's the intuition So those just aren't going to be that interesting again It's not that you can't talk about one of these evaluation homomorphisms if you start with something in the smaller field It's just that they just won't produce much fruit. They would just you know won't be interesting These are homomorphisms now what we're What we're going to do tonight is Look at some results that you're familiar with Because you're used to working with polynomials with coefficients either in the reels or in the complexes And what we're going to show you is that a lot of the things that you already know In those settings actually hold true For polynomials regardless of what the field is that you're using for the coefficients So again, most of you most of you all of you are familiar with familiar polynomials Omules are bracket X Maybe C bracket X Maybe Q bracket X what we're going to be interested in doing is we show that many but not all But not all polynomial rings rings of the form F bracket X share Many but not all Properties that you used to properties you Have learned over the years about these two special ones Our bracket X and C bracket X. Let me give you some examples for instance you proved You've used this thing called the I forget what it's called the zero Theorem or something I'll give any Here's what you learned about polynomials in our bracket X and you've taught this to your algebra two students That if you hand me a polynomial with real coefficients, and if you find somehow some number it's called alpha and If you plug alpha in and zero comes out Then necessarily X minus alpha is a factor of that Some name for that So that's something that you somehow know is true for C bracket X or our bracket X turns out That's going to be true for any F bracket X. That's going to be a nice one contrast oh similarly Yeah, similarly What you prove about Polynomials in R of X is that if you have me a polynomial of degree n It can have it most n zero Might not have any zeros in R But even if you allow complex zeros if you have me a polynomial of degree two there can only be two zero Okay, that's another result that you come to the table knowing about polynomials in R and C that turns out that'll be true for general fields Quick sidebar We showed that that wasn't necessarily true if the coefficients that you start with Don't come from a field because you showed for the homework of assignment that I just passed back that you can have polynomials of degree two Where the coefficients come from Z6 not a few and that those polynomials can have four different zeros So there's something that's not true for general systems, but it'll be true regardless of those Let me tell you something. It's not true in general if I hand you polynomial in the Set in the setting of our bracket X the coefficients are the reals And I tell you the derivative of the polynomial is zero and you can tell me what the polynomials constant Constants have derivative zero and constants are the only actually the only functions From the reals or reals that have derivative zero turns out folks There are situations Even if you start with a field at bracket X where app is a field like Z2 Where there are things that have derivative equal to zero that aren't just constant Let me give you a good example if I ask you to start with the field Z2 and I hand you the polynomial X squared and I ask you to take its derivative This derivative is 2x But the coefficients are coming from Z2 since I Z2 2 is Zero so the derivative X squared is zero if you're working in Z2 X So there's something that isn't shared by all fields and the reason that's a hint on the homework problem due Wednesday is In past semesters. I've seen students submit solutions to that problem where they don't use all the given hypotheses So make sure especially on that particular problem, of course on all problems But I'm not wondering in particular that you somehow use all of the given hypotheses on that question. Let's begin. Okay Here's what we show first What does he call it? Oh, yeah, the factor theorem So here's the first one factor this first one is one that Extends stuff that you already know about this and this to all polynomials F-bracket x regardless of what field f is factor theorem says this let Let little f of x in capital f of x have Yeah, yeah, yeah degree F of x at least one all I'm trying to do is make sure that you're not plugging in the zero polynomial I think then the punchline is this then I'm sorry let alpha also in f. So what I'm doing here folks is something slightly different than what we did in Describing these evaluation homomorphisms in the evaluation homomorphism setting I took a field and then I had some sort of subfield or extension field in a smaller field or larger field Here what I'm asking you to do is take a polynomial where the coefficients come from some field and then take some element in that Same field then here's the punchline then Hmm if you evaluate f of x at zero, I'm sorry at alpha and You get zero Let me remind you in words what that means if it's the case that you take this polynomial And you plug that thing in and zero comes out If you want to write this as alpha of alpha That's how you'd usually write it and now it's for two class if feet alpha of f of x is zero this happens if and only if X minus alpha is a factor of Little f of x in the polynomial ring capital F. Let me rephrase this in other words if you Plug in alpha everywhere. You see an x in the polynomial little f and zero came out that happens if and only if you can write the polynomial f as x minus alpha times some other polynomial where G of x is Some polynomial so this is something that you've been teaching your students since forever in Algebra 2 in effect it says if somehow you've started with a polynomial and you've By hook or crook or somehow Realize that by plugging some real number into that polynomial everywhere you see an x at zero came out Then necessarily if you take x minus alpha and you long divide it into f of x that there'll be zero remainder I see a lot of you nodding your heads Well turns out folks that's nothing special having to do with the real numbers the complex numbers That's true regardless of what field you work over Here's the proof This is Really the most important place where we're going to use the division algorithm for polynomials that you just wrote down for me Proof is this so I'm going to assume First that it's the case that f of x is a polynomial that has the property that when you drop alpha and you get zero And I'm going to show that x minus alpha is a factor Turns out the other direction is almost trivial Let's see. So let's do that. So assume that'll be this direction assume that Fees of alpha of f of x is zero What do we want to do? We want to show that x minus alpha is a factor in other words We want to show that f of x can be written as x minus alpha times g of x. Well, here's the point Since alpha is assumed to be in f that's given information You've started with something that lives Inside the field that you're using as the coefficients for your polynomials. The point is that this Is a perfectly good polynomial in f bracket x. This is a polynomial. I'll tell you what its degree is one I'll tell you what its leading coefficient is one It's a polynomial is an f bracket x y have to make sure that all of the coefficients come from this field Well that coefficient certainly does it's the identity element every field containing an identity element and minus alpha does also Why because like pot this is that alpha is an f So minus alpha is also an f additive inversions are in there. And so this thing is also an f So both of the coefficients of this polynomial happen to be enough So does that mean it means I've got a polynomial in f bracket x I've got another polynomial in f bracket x and because this one is degree at least one It makes sense to divide this one into the other one as in the division out. So now use the division algorithm for polynomials Algorithm she just written out for me theorem 23.1 divide this one into f of x and we get this that f of x can be written as q of x Times this thing. This is what I'm using in the role of g of x Plus some remainder and I'll tell you something about the remainder where what where either the degree of r of x is less than the degree of the thing that you've divided by there it is or Or R of x is the zero polynomial It's the division algorithm. So y'all nodding your head. That's good Because you remember that all I've done is use the division algorithm where I happen to have chosen that thing to be the thing that I'm dividing by question so far Okay, but look folks. I know what the degree of x minus alpha is one The highest power of x that appears as one so that was easy So what I've asked what I've guaranteed is that this thing I've written down as the remainder has Degree less than one I Don't lose one possibility. It means they have degrees zero or that the polynomials zero to begin with So because of this inequality We actually have either the degree of r of x is zero or Or r of x is zero itself and remember from last time we had to make this distinction This statement folks simply means that the polynomial r of x is a constant but not zero And this statement means that polynomial r of x is a constant but is zero We had this distinguish these two out because we didn't want to assign a degree to the zero polynomial because it mucked up some of the formulas But I can rephrase this this way I e r of x is some element in the field Because degree zero means it's a constant and r of x equaling zero means well, it's in the field because zero's in the field Phrased another way. Whatever this remainder is doesn't have any x's on it It's just some element from the field. I said wait you've written it as r of x well Yeah beforehand when I've stated the division algorithm when I hauled out the division algorithm I don't really know what that thing is it's some polynomial and f bracket x But given the constraints of this particular use of the division algorithm in the end what we've actually shown is that r of x Doesn't contain any x's so we can write write R of x as let's call it C where C is some constant like we're doing calculus here where C is something Which contains no x's Something in f which contains no x terms You're still not in your heads. Maybe you're slightly Just believing in this day I'm just say this one more way The thing that I've tacked on the end here this thing called r of x has no x's in it in other words I e if I do fees of alpha of Of r of x if I ask you to plug alpha in Everywhere you see an x In r of x. That's what this means. There are no x's in r of x. So you just get r of x You want to do it this way maybe it's easier r of x is just some element of the field if you plug alpha in everywhere You see an x and see an x so you just get the element of field back So whichever point of view you'd prefer to take doesn't matter they both said the same thing That's nice and here's why What are we trying to do? We're trying to eventually show that if I Take f of x that I can write it as x minus alpha times something So the goal is to really show that r of x is zero at this stage All I know is that something it's something in the field, but here's how we show that it's actually zero so we show That in fact r of x is zero and here's how we have Alpha of x is a q of x times x minus alpha plus r of x What we concluded from the division algorithm now here's what I want you to do now apply the Evaluation homomorphism to both sides in other words drop out then everywhere we see an x So I do fees of alpha of the left-hand side Is that necessarily fees of alpha the right-hand side q of x times x minus alpha plus r of x Folks the two things are equal then the same things get spit out if you run them through the same function That's all we've said here, but wait a minute. Let's see. This is fees of alpha of r of f of x That's no big deal, but fees of alpha is a homomorphism Okay, I didn't prove that for you that was given so fees of alpha of One thing plus another is fees of alpha of that plus fees of alpha of that and fees of alpha of one thing times another is fees of alpha of that plus so I'm going to use the fact that fee is a Homomorphism fees of alpha is a ring homomorphism Meaning that it respects both the addition and the multiplication that gives me this fees of alpha of q of x times fees of alpha of x minus alpha plus fees of alpha of r of x So I've used both that fees of alpha respects the addition and that it respects the multiplication That holler if you're uncomfortable here That's nice to know it's a ring homomorphism. All right, let's do one more step No, let's do two steps No, let's do three steps all in one line. Watch this. What's the hypothesis the hypothesis is that? Fees of alpha of f of x is zero so we're assuming the thing on the left is zero So here's the hypothesis Hypothesis given Not this thing. Well fees about the Q of x who that knows what that is. I know What does fees of alpha mean it means drop alpha in everywhere you see an x Okay Done I dropped an alpha in everywhere. I saw an x. I see one x but alpha Plus ooh Fees of alpha of r of x Is r of x the sub alpha of r that oops right is r of x previous observation Boy, those are convenient So I've dropped alpha in use the hypothesis that when you drop it into f you get zero I've simply dropped it in because that's what this instruction tells me to do and then a previous Observation says if you happen to drop alpha into this thing it doesn't change it because that thing r of x even though it's written like a polynomial We've established actually doesn't have any x's in it Now we're almost done. Yeah, Lonnie question That's what we're assuming That's the starting point. Yeah one more step. That's almost interesting alpha minus alpha better known as zero Zero times anything is better known as zero So this thing turns out to all boil down to zero And so I get zero equals r of x. That's exactly what we wanted to show So the punchline is in fact we have f of x is q of x times x minus alpha plus zero In other words just rewriting and technically i'm doing two things at once but we can do that plus zero means I don't have to worry about it And multiplication inside f bracket x is commutative so I can switch things around And I then have written f of x as x minus alpha times something Where this is in f bracket x We're done So this fact that you've been imparting to your algebra two students that if you have what's called a zero for the polynomial If you have a number that when you plug it in spits out zero from the polynomial Then necessarily if you try to factor out an x minus alpha necessarily x minus alpha is in fact a factor of the original polynomial questions comments technically we're not done because Stated this theorem as an if and only a statement. We've shown if this then this that's the hard part. Let's do it the other way Suppose x minus alpha is a factor of f of x which means in equation form just this What do we want to show we want to show that if we drop alpha in that zero comes out well folks If you drop alpha in everywhere you see an x into that polynomial Means you've dropped in an alpha there and you've dropped in an alpha there, but as soon as you've dropped in an alpha here Get alpha minus alpha which is zero so you get zero times who cares what and you get zero So knowing that x minus alpha is a factor Immediately gives that when you plug alpha in zero comes out because then you've got a factor that looks like alpha minus alpha In other words, you got a factor looks like zero So the converse is easy. Just go ahead and plug it in. So just for completeness. I'll say the other way Is just plug in alpha And all right, so there's the factor theorem what you knew about polynomials before At least polynomials over our bracket x In fact hold true regardless of what the coefficients look like Question is no, yeah, no, all right um Let me give you A Yeah, let me give you a second corollary To the division algorithm you can think of this one the factor theorem as the most important consequence of the division algorithm for f bracket x And we certainly used it. We wrote down this this equation um I won't spend the time it'd take about 15 minutes or so to prove it, but here's a second consequence So another consequence of the division algorithm of the division algorithm Is proposition If we're working in a field If The uh f of x is in Capital f bracket x and again the underlying understood hypothesis is that capital f always stands for field and the degree The degree of f of x is n Give me something. See do I need it? Yeah Bigger than or equal to one I guess it technically is true for degrees zero polynomials, but those are uninteresting So give me a polynomial that has some guts that has some x's in it. Then the punch line is then there are at most n zeros of f In in other words, there are At most n elements I'll write them out this way alpha one alpha two through alpha sub t Where t is necessarily less than or equal to n that have the property So that when you plug this particular polynomial Into one of these evaluation homeomorphisms that you get zero And I'll list as the proof simply use the division algorithm division algorithm the factor theorem The thing that we just proved And what it boils down to and the fact that Uh f bracket x is an integral domain Okay, so I won't go through the whole proof of why it's true that In f bracket x where f is a field there can be at most n zeros for a polynomial degree n I will though contrast this result to the result that you already know Which is true for polynomials with coefficients in the reels but At least the only proof that I was exposed to growing up before I got to an algebra course was the calculus proof And the calculus proof is let's see. What are you trying to do? You're trying to convince me that if you hand me a polynomial degree n That from a graphical point of view it can only cross the axis at most n times That's what this result says that there's at most n zeros for it There's at most n different things that you can plug in that spit out a y value of zero And the proof of that is sort of by by induction you show that when you take the derivative of this thing That the polynomial can only change direction n minus one times and if you've got a function that can only You know change direction n minus one times Then you've guaranteed that it can only cross the axis at most n times But here, you know, we got no concept of what the graph of a function might look like or changing directions or up and down But this result turns out to be true regardless of what the underlying fields look like So regardless of whether you can do any sort of analysis or geometry or draw pictures of it It's the case that you can only pull out at most n zeros For those of you that saw the number theory course with me last spring We actually gave a relatively rigorous proof of this result in the particular case where The underlying field happens to be z sub p where p is prime But for those of you that didn't it's I mean, it's just no big deal Just suffice it to say that we could get there if we chose to Okay, all right Don't want to do this now Let me in the last Yeah, let me do this and then the last five minutes So I've been trying to play up this analogy or the similarities between Rings of the form f bracket x where f is a field So these polynomial rings with coefficients coming from a field and the ring of integers They're both integral domains neither of them is a neither of them is a field They both have this notion of a division algorithm or a notion of size of elements size of elements inside the integers It's just how big is the integer size of an element inside the polynomial ring is what's its degree Well, there's a totally important concept Inside the integers at least inside the positive integers, but I'm going to ask you to sort of extend things out as well It's the notion of primeness prime numbers And that somehow the prime numbers at least for multiplication are sort of the building blocks of all of the Admittedly positive integers, but for now we're just going to talk about all of the non zero integers You can somehow get as products of primes, right? Maybe you have to slap the negative sign from there. No big deal The question is if this analogy is really a good one. Is there some sort of corresponding notion Inside polynomial ring To the notion of being a prime number inside the integers the answer is yes I mean intuitively what is a prime number? It's one that you just can't break down any further Rephrased it's one that you actually can break down. It's just when you do break it down It's only done in sort of a trivial way. I mean seven you can break down as seven times one All right What turns out and we now have the language to talk about prime numbers Inside the integers where we include positive or negative. I'm not going to talk about zero in this context But if you hand me a non zero integer, I don't care again if you're starting with positive or negative Here's what I'm going to intuitively think of as a prime integer It's one where You might be able to factor it as the product of two other integers But if you do at least one of the other integers that you factor it into Is a unit in the integers So seven is prime Because the reason is There's actually a lot of different ways to write seven It's one times seven. It's also negative one times negative seven So there's two ways of writing seven as a product of two integers But notice in both of those situations If I write it as one times seven one is a unit in the integers Obviously, it's the multiplicative identity But I can also write it as negative seven times negative one, but negative one is a unit inside the integers So there's a sort of extended version of what it means to be a prime number And it completely agrees with what you understand to be a prime number It's something you can't factor further down It's just we can talk about that idea in a slightly more general context to include the negatives Well, if we look at things in that slightly more general point of view An integer is prime If it's the case that Well, I don't want to include one as a prime. It's easy to do An integer is prime If it's first of all not zero, so forget zero. There's just a mess. It is not zero If it's not a unit So that throws out one in minus one And if it has the property that anytime you can write it as a product of two things That necessarily one or the other of the things is a unit So definition of prime integer So again, I'm playing up this sort of connection between z and f bracket f bracket x An integer p is prime In case First of all p is not zero Secondly p is not One or minus one. I'm going to rephrase that as i e p is not A unit in z and third Uh if You can write p as a times b Where a and b are in z Then either A is a unit Or b is a unit. So that's the definition of a prime integer Admittedly, I've extended the definition now of prime to include some negatives But I mean it's not going to cause you to agree This says that negative two is prime negative three is prime negative five is prime I mean the negatives of all the things you know to be prime or primes and nothing else is So that's what a prime integer means So the question is what's the corresponding idea in f bracket x and the answer is this we simply call them Well, things that you can't break down any further. We call them irreducible So a polynomial Little f of x in capital f of x is called Irreducible in case Well in case it does exactly the same things that the prime numbers do inside the integers In case it's not zero not a unit and the only Factorizations are uninteresting our units All right now the first question is this if you're sitting inside f bracket x What are the units? What are the polynomials that have multiplicative inverses? Well, it's pretty easy if you hand me a polynomial It's actually got some guts to it if it's got some x's or x squared or something like that It's impossible to find another polynomial to multiply times this one That gives an output of just one because you can't reduce degree at all So the only things that have a chance of being a unit inside f bracket x Are the things of degree zero and those are precisely units If you hand me the polynomial two in our bracket x, it's a unit. It's inverse is a half It's no big deal Just tell me what the multiplicative inverse was anyway. You're thinking well two is not a polynomial. The only way that is It's a polynomial of degree zero. It's two plus zero x plus zero x squared blah blah blah Can you find something so that when you multiply two times it that you get one sure the polynomial one half plus zero x plus zero Sort of silly, but it works. But those are the only ones that work So when we talk about things being a unit inside f bracket x what we're really looking at are the non-zero constant polynomials And that's exactly what irreducibility is going to correspond to in case first of all f of x is not equal to zero Secondly f of x is not a unit Is not a unit In f bracket x i'll rephrase what that means i e f of x Oh the degree of f of x is not zero So those are the elements the field or the degree zero things and third finally if You've factored f of x as let's call it g of x times k of x Where the two polynomials g of x And k of x are in the underlying Ring of polynomials f bracket x then either Then either that one is a unit or that one is a unit Which phrased in the context of f bracket x means then either g of x is in f or k of x is in f Think Irreducible means the only way you can factor this polynomial is sort of by cheating Cheating means You know if I hand you the polynomial, let's say x minus three Can you factor it? You know, what do you mean factor it the way I can factor it watch x minus three is Two x minus six times one third. I mean two x minus three times a half. Sorry six minus three It's cheating. I mean you can't just multiply by constant And what we do is say all right you can always factor any polynomial if you allow me the sort of silly factorizations Irreducible means though that those are the only factorizations that that can happen So these are sort of like saying yeah, you know if you give me the number seven I can certainly factor it. It's one times seven or it's negative one times negative something but None of the factorizations have any guts okay The intuition is that the polynomials that somehow don't break down anymore Play the role of the prime numbers inside the integers What we're eventually going to do is take every polynomial and write them as products of Irreducible polynomials and a huge question for us In fact, it's a question that literally will take the rest of the Semester and we'll never get a complete answer to it But we'll at least get some good partial answers to it is if you hand me a field f And you hand me a polynomial In that field is the polynomial Irreducible and it's somewhat similar to the question inside z which turns out to be a hard question You know here's a number. Is it prime or not? Well, you can sort of do some you know just beat on the thing But it's hard to tell or can you build another prime number? You know this question of where's the next prime number? It's a hard question So there's going to be some Similar degree of difficulty in answering questions about irreducibility, but I have to be a good place to to make one comment and quit It turns out that the notion of irreducible and I I probably should have put it here Let me go ahead and and go back and put it in and then show you why it's important It's called irreducible in f bracket x I mean, I asked you to take a polynomial in f bracket x But it's important to talk about irreducibility in the context of a specific field and here's what example It turns out We'll show Next time look at this polynomial f of x is the polynomial x squared plus one in q bracket x So here's a polynomial Question can you factor this polynomial in q bracket x? In other words, can you write this thing As something other than Yeah, I can write it as two x squared plus two times a half But forget that can you write it as some sort of factorization Where each piece has some guts to it turns out the answer will be no So we'll show next time It turns out that f of x is irreducible in q bracket x but this time I'll show you but take the same polynomial If I give you f of x same polynomial But I now ask you to view it as a polynomial Inside the polynomials where you're allowed to use complex coefficients Then it's not irreducible then f of x is not Irreducible In c bracket x in other words, there's actually a way to take this thing and factor it down Where each of the factors has some guts to it. I'll show you how to do it x squared plus one is x minus i times x plus i And of course the idea here is that by allowing me to use Somehow a larger collection of things as the coefficients of the polynomials That then allows me to write down this thing as a polynomial in the system and allows me to write down this thing as a polynomial system Of course, if you multiply this out you'll get exactly that This isn't of course a factorization in q bracket x because these coefficients done this So the notion of whether or not a polynomial is irreducible depends highly on what the underlying field might be And that turns out well that means that we're going to have to be somewhat careful in talking about irreducibility in general, but it turns out this question of Well, if you're working over a field And you've got a polynomial that's irreducible Is it possible to somehow make the field bigger? And find a field where the polynomial isn't irreducible anymore where the polynomial eventually takes down question honey Either Not necessarily exclusive, I mean this Both well, but if both were from the field then f of x would be in the field Because then it would be a product of two things in the field And so then f of x would have been degree zero to begin with And that that would put you here, but we've explicitly excluded things from the field Okay It's monday. So here's a homework assignment, but folks next wednesday We have no class because school is closed on wednesday for Thanksgiving holiday So this will be due on a short turnaround. So due next monday, but it'll be a shorter assignment Just one section here. So that is the 19th And the homework is this in section 23 problems one through four nine through 16 27 through 30 and 34 and I want you to turn in these five two twelve 16 17 and 28 that'll be it I know we have a homework assignment due wednesday and some of you're still working on that You know enough to start at least the first two problems on this assignment and by the First out in a 20 minutes of wednesday's lecture, you'll be able to at least attack the remaining three problems